1

Relations and Functions

Short Answer Type Questions
Q. 1 Let A = {a, b, c} and the relation R be defined on A as follows
R = {(a, a), (b, c ), (a, b)}
Then, write minimum number of ordered pairs to be added in R to make
R reflexive and transitive.
Sol. Given relation, R = {(a, a), (b, c ), (a, b )}.
To make R is reflexive we must add (b, b ) and (c, c ) to R. Also, to make R is transitive we must
add (a, c ) to R.
So, minimum number of ordered pair is to be added are (b, b ), (c, c ), (a, c ).

Q. 2 Let D be the domain of the real valued function f defined by

f (x ) = 25 - x 2 . Then, write D.
Sol. Given function is, f(x ) = 25 - x 2
For real valued of f(x ) 25 - x 2 ³ 0
x 2 £ 25
-5 £ x £ + 5
\ D = [-5, 5]

Q. 3 If f , g : R ® R be defined by f ( x ) = 2x + 1 and g( x ) = x 2 - 2, " x Î R,

respectively. Then, find gof .
K Thinking Process
If f , g : R ® R be two functions, then gof(x) = g { f(x)}, "x ÎR.
Sol. Given that, f(x ) = 2 x + 1 and g (x ) = x 2 - 2, "x Î R
\ gof = g {f(x )}
= g (2 x + 1) = (2 x + 1)2 - 2
= 4x 2 + 4x + 1 - 2
= 4x 2 + 4x - 1
Q. 4 Let f : R ® R be the function defined by f (x) = 2x - 3, " x Î R. Write
-1
f .
Sol. Given that, f(x ) = 2 x - 3, " x Î R
Now, let y = 2x - 3
2x = y + 3
y+ 3
x=
2
x+ 3
\ f -1(x ) =
2

Q. 5 If A = {a, b, c, d } and the function f = {(a, b), (b, d), (c, a), (d, c)}, write
-1
f .
Sol. Given that, A = {a, b, c, d }
and f = {(a, b ), (b, d ), (c, a), (d , c )}
f -1 = {(b, a), (d , b ), (a, c ), (c, d )}

Q. 6 If f : R ® R is defined by f (x) = x 2 - 3x + 2, write f { f (x)}.

K Thinking Process
To solve this problem use the formula i.e., (a + b + c)2 = (a2 + b2 + c2 + 2 ab + 2bc + 2 ca)
Sol. Given that, f(x ) = x 2 - 3x + 2
\ f{f(x )} = f(x 2 - 3x + 2 )
= (x 2 - 3x + 2 )2 - 3(x 2 - 3x + 2 ) + 2
= x 4 + 9x 2 + 4 - 6x 3 - 12 x + 4x 2 - 3x 2 + 9x - 6 + 2
= x 4 + 10x 2 - 6x 3 - 3x
f {f(x )} = x 4 - 6x 3 + 10x 2 - 3x

Q. 7 Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? If g is described by
g(x) = a x + b, then what value should be assigned to a and b?
Sol. Given that, g = {(1, 1), (2, 3), (3, 5), (4, 7)}.
Here, each element of domain has unique image. So, g is a function.
Now given that, g (x ) = ax + b
g(1) = a + b
a + b =1 …(i)
g(2 ) = 2 a + b
2a + b = 3 …(ii)
From Eqs. (i) and (ii),
2(1 - b ) + b = 3
Þ 2 - 2b + b = 3
Þ 2 -b = 3
b = -1
If b = - 1, then a = 2
a = 2, b = - 1
Q. 8 Are the following set of ordered pairs functions? If so examine whether
the mapping is injective or surjective.
(i) {(x, y) : x is a person, y is the mother of x}.
(ii) {(a, b) : a is a person, b is an ancestor of a}.
Sol. (i) Given set of ordered pair is {(x, y) : x is a person, y is the mother of x}.
It represent a function. Here, the image of distinct elements of x under f are not distinct,
so it is not a injective but it is a surjective.
(ii) Set of ordered pairs = {(a, b ) : a is a person, b is an ancestor of a}
Here, each element of domain does not have a unique image. So, it does not represent
function.

Q. 9 If the mappings f and g are given by f = {(1, 2), (3, 5), (4, 1)} and
g = {(2, 3), (5, 1), (1, 3)}, write fog.
Sol. Given that, f = {(1, 2), (3, 5), (4, 1)}
and g = {(2, 3), (5, 1), (1, 3)}
Now, fog (2 ) = f{g (2 )} = f(3) = 5
fog (5) = f{g (5)} = f(1) = 2
fog (1) = f{g (1)} = f(3) = 5
fog = {(2, 5), (5, 2 ), (1, 5)}

Q. 10 Let C be the set of complex numbers. Prove that the mapping

f : C ® R given by f (z) = | z |, " z Î C, is neither one-one nor onto.
Sol. The mapping f :C ® R
Given, f( z) = z , " z Î C
f(1) = |1| = 1
f(-1) = |- 1| = 1
f(1) = f(-1)
But 1¹ -1
So, f( z) is not one-one. Also, f( z) is not onto as there is no pre-image for any negative
element of R under the mapping f( z).

Q. 11 Let the function f : R ® R be defined by f (x) = cos x, " x Î R. Show

that f is neither one-one nor onto.
Sol. Given function, f(x ) = cos x , " x Î R
æpö p
Now, f ç ÷ = cos = 0
è2 ø 2
æ -p ö p
Þ fç ÷ = cos = 0
è 2 ø 2
æpö æ -p ö
Þ fç ÷=fç ÷
è ø
2 è 2 ø
p -p
But ¹
2 2
So, f(x ) is not one-one.
Now, f(x ) = cos x, " x Î R is not onto as there is no pre-image for any real number. Which
does not belonging to the intervals [-1, 1], the range of cos x.
Q. 12 Let X = {1, 2, 3} and Y = {4, 5}. Find whether the following subsets of
X ´ Y are functions from X to Y or not.
(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)} (ii) g = {(1, 4), (2, 4), (3, 4)}
(iii) h = {(1, 4), (2, 5), (3, 5)} (iv) k = {(1, 4), (2, 5)}
Sol. Given that, X = {1, 2, 3} and Y = {4, 5}
X ´ Y = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)}
f is not a function because f has not unique image.
(ii) g = {(1, 4), (2, 4), (3, 4)}
Since, g is a function as each element of the domain has unique image.
(iii) h = {(1, 4), (2, 5), (3, 5)}
It is clear that h is a function.
(iv) k = {(1, 4), (2, 5)}
k is not a function as 3 has not any image under the mapping.

Q. 13 If functions f : A ® B and g : B ® A satisfy gof = I A , then show that

f is one-one and g is onto.
Sol. Given that,
f : A ® B and g : B ® A satisfy gof = I A
Q gof = I A
Þ gof{f(x1 )} = gof{f(x 2 )}
Þ g ( x1 ) = g ( x 2 ) [Q gof = I A ]
\ x1 = x 2
Hence, f is one-one and g is onto.

1
Q. 14 Let f : R ® R be the function defined by f (x) = , " x Î R.
2 - cos x
Then, find the range of f .
K Thinking Process
Range of f = { y ÎY : y = f(x) : for some in x} and use range of cos x is [-1,1]
1
Sol. Given function, f( x ) = , " x ÎR
2 - cos x
1
Let y=
2 - cos x
Þ 2 y - ycos x = 1
Þ ycos x = 2 y - 1
2y - 1 1 1
Þ cos x = =2 - Þ cos x = 2 -
y y y
1
Þ -1 £ cos x £ 1 Þ -1 £ 2 - £ 1
y
1 1
Þ - 3 £ - £ -1 Þ 1£ £ 3
y y
1 1
Þ £ £1
3 y
é1 ù
So, y range is ê , 1ú.
ë3 û
Q. 15 Let n be a fixed positive integer. Define a relation R in Z as follows " a,
b Î Z , aRb if and only if a - b is divisible by n. Show that R is an
equivalence relation.
Sol. Given that, " a, b Î Z, aRb if and only if a - b is divisible by n.
Now,
I. Reflexive
aRa Þ (a - a) is divisible by n, which is true for any integer a as ‘O’ is divisible by n.
Hence, R is reflexive.
II. Symmetric
aRb
Þ a - b is divisible by n.
Þ - b + a is divisible by n.
Þ -(b - a) is divisible by n.
Þ (b - a) is divisible by n.
Þ bRa
Hence, R is symmetric.
III. Transitive
Let aRb and bRc
Þ (a - b ) is divisible by n and (b - c )is divisible by n
Þ (a - b ) + (b - c ) is divisibly by n
Þ (a - c ) is divisible by n
Þ aRc
Hence, R is transitive.
So, R is an equivalence relation.

Long Answer Type Questions

Q. 16 If A = {1, 2, 3, 4}, define relations on A which have properties of being
(i) reflexive, transitive but not symmetric.
(ii) symmetric but neither reflexive nor transitive.
(iii) reflexive, symmetric and transitive.
Sol. Given that, A = {1, 2, 3, 4}
(i) Let R1 = {(1, 1), (1, 2 ), (2, 3), (2, 2 ), (1, 3), (3, 3)}
R1 is reflexive, since, (1, 1) (2, 2) (3, 3) lie in R1.
Now, (1, 2 ) Î R1, (2, 3) Î R1 Þ (1, 3) Î R1
Hence, R1 is also transitive but (1, 2 ) Î R1 Þ (2, 1) Ï R1.
So, it is not symmetric.
(ii) Let R 2 = {(1, 2 ), (2, 1)}
Now, (1, 2 ) Î R 2 , (2, 1) Î R 2
So, it is symmetric.
(iii) Let R 3 = {(1, 2 ), (2, 1), (1, 1), (2, 2 ), (3, 3), (1, 3), (3, 1), (2, 3)}
Hence, R 3 is reflexive, symmetric and transitive.
Q. 17 Let R be relation defined on the set of natural number N as follows,
R = {(x, y) : x Î N , y Î N , 2x + y = 41}. Find the domain and range of
the relation R. Also verify whether R is reflexive, symmetric and
transitive.
Sol. Given that, R = {(x, y) : x Î N, y Î N, 2 x + y = 41}.
Domain = {1, 2, 3, K, 20}
Range = {1, 3, 5, 7, K, 39}
R = {(1, 39), (2, 37 ), (3, 35), K, (19, 3), (20, 1)}
R is not reflexive as (2, 2 ) Ï R
2 ´ 2 + 2 ¹ 41
So, R is not symmetric.
As (1, 39) Î R but (39, 1) Ï R
So, R is not transitive.
As (11, 19) Î R, (19, 3) Î R
But (11, 3) Ï R
Hence, R is neither reflexive, nor symmetric and nor transitive.

Q. 18 Given, A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of

the following
(i) an injective mapping from A to B.
(ii) a mapping from A to B which is not injective.
(iii) a mapping from B to A.
Sol. Given that, A = {2, 3, 4}, B = {2, 5, 6, 7}
(i) Let f : A ® B denote a mapping
f = {(x, y) : y = x + 3}
i.e., f = {(2, 5), (3, - 6), (4, 7 )}, which is an injective mapping.
(ii) Let g : A ® B denote a mapping such that g = {(2, 2 ), (3, 5), (4, 5)}, which is not an
injective mapping.
(iii) Let h : B ® A denote a mapping such that h = {(2, 2 ), (5, 3), (6, 4), (7, 4)}, which is a
mapping from B to A.

Q. 19 Give an example of a map

(i) which is one-one but not onto.
(ii) which is not one-one but onto.
(iii) which is neither one-one nor onto.
Sol. (i) Let f : N ® N, be a mapping defined by f(x ) = 2 x
which is one-one.
For f( x1 ) = f( x 2 )
Þ 2 x1 = 2 x 2
x1 = x 2
Further f is not onto, as for 1Î N, there does not exist any x in N such that f(x ) = 2 x + 1.
(ii) Let f : N ® N, given by f(1) = f(2 ) = 1 and f(x ) = x - 1 for every x >2 is onto but not
one-one. f is not one-one as f(1) = f(2 ) = 1. But f is onto.
(iii) The mapping f : R ® R defined as f(x ) = x 2 , is neither one-one nor onto.
 x -2
Q. 20 Let A = R - {3}, B = R - {1}. If f : A ® B be defined by f (x) = ,
x -3
" x Î A. Then, show that f is bijective.
K Thinking Process
A function f : x ® y is said to be bijective, if f is both one-one and onto.
Sol. Given that, A = R - {3}, B = R - {1}.
x -2
f : A ® B is defined by f(x ) = ,"xÎA
x-3
For injectivity
x1 - 2 x 2 - 2
Let f( x1 ) = f( x 2 ) Þ =
x1 - 3 x 2 - 3
Þ (x1 - 2 )(x 2 - 3) = (x 2 - 2 )(x1 - 3)
Þ x1x 2 - 3x1 - 2 x 2 + 6 = x1x 2 - 3x 2 - 2 x1 + 6
Þ -3x1 - 2 x 2 = - 3x 2 - 2 x1
Þ - x1 = - x 2 Þ x1 = x 2
So, f(x ) is an injective function.
For surjectivity
x -2
Let y= Þ x - 2 = xy - 3 y
x-3
2 - 3y
Þ x(1 - y) = 2 - 3 y Þ x =
1- y
3y - 2
Þ x= Î A, " y Î B [codomain]
y-1
So, f(x ) is surjective function.
Hence, f(x ) is a bijective function.

Q. 21 Let A = [ - 1, 1], then, discuss whether the following functions defined

on A are one-one onto or bijective.
x
(i) f (x) = (ii) g(x) = | x |
2
(iii) h(x) = x| x | (iv) k(x) = x 2
Sol. Given that, A = [-1, 1]
x
(i) f(x ) =
2
Let f( x1 ) = f( x 2 )
x1 x 2
Þ = Þ x1 = x 2
2 2
So, f(x ) is one-one.
x
Now, let y=
2
Þ x = 2 y Ï A, " y Î A
As for y = 1 Î A, x = 2 Ï A
So, f(x ) is not onto.
Also, f(x ) is not bijective as it is not onto.
 (ii) g (x ) = |x|
Let g ( x1 ) = g ( x 2 )
Þ |x1| = |x 2| Þ x1 = ± x 2
So, g (x ) is not one-one.
Now, y =|x| Þ x = ± y Ï A, " y Î A
So, g (x ) is not onto, also, g (x ) is not bijective.
(iii) h(x ) = x|x|
Let h(x1 ) = h(x 2 )
Þ x1|x1| = x 2|x 2| Þ x1 = x 2
So, h(x ) is one-one.
Now, let y = x|x|
Þ y = x 2 Î A, " x Î A
So, h(x ) is onto also, h(x ) is a bijective.
(iv) k(x ) = x 2
Let k(x1 ) = k(x 2 )
Þ x12 = x 22 Þ x1 = ± x 2
Thus, k(x ) is not one-one.
Now, let y = x2
Þ x = y Ï A, " y Î A
As for y = - 1, x = -1 Ï A
Hence, k(x ) is neither one-one nor onto.

Q. 22 Each of the following defines a relation of N

(i) x is greater than y, x, y Î N .
(ii) x + y = 10, x, y Î N .
(iii) xy is square of an integer x, y Î N .
(iv) x + 4 y = 10, x, y Î N
Determine which of the above relations are reflexive, symmetric and
transitive.
Sol. (i) x is greater than y, x, y Î N
(x, x ) Î R
For x Rx x > x is not true for any x Î N.
Therefore, R is not reflexive.
Let (x, y) Î R Þ x Ry
x >y
but y > x is not true for any x, y Î N
Thus, R is not symmetric.
Let xRy and yRz
x > y and y > z Þ x > z
Þ xRz
So, R is transitive.
 (ii) x + y = 10, x, y Î N
R = {(x, y); x + y = 10, x, y Î N}
R = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)} (1, 1) Ï R
So, R is not reflexive.
(x, y) Î R Þ ( y, x ) Î R
Therefore, R is symmetric.
(1, 9) Î R, (9, 1) Î R Þ (1, 1) Ï R
Hence, R is not transitive.
(iii) Given xy, is square of an integer x, y Î N.
Þ R = {(x, y) : xy is a square of an integer x, y Î N}
(x, x ) Î R, " x Î N
As x 2 is square of an integer for any x Î N.
Hence, R is reflexive.
If (x, y) Î R Þ ( y, x ) Î R
Therefore, R is symmetric.
If (x, y) Î R, ( y, z) Î R
So, xy is square of an integer and yz is square of an integer.
Let xy = m2 and yz = n2 for some m, n Î Z
m2 x2
x= and z =
y y
m2 n2
xz = , which is square of an integer.
y2
So, R is transitive.
(iv) x + 4 y = 10, x, y Î N
R = {(x, y) : x + 4 y = 10, x, y Î N}
R = {(2, 2 ), (6, 1)}
(1, 1), (3, 3), K, Ï R
Thus, R is not reflexive.
(6, 1) Î R but (1, 6) Ï R
Hence, R is not symmetric.
(x, y) Î R Þ x + 4 y = 10 but ( y, z) Î R
y + 4 z = 10 Þ (x, z) Î R
So, R is transitive.

Q. 23 Let A = {1, 2, 3, K , 9} and R be the relation in A ´ A defined by

(a, b) R (c , d ) if a + d = b + c for (a, b), (c , d ) in A ´ A. Prove that R is an
equivalence relation and also obtain the equivalent class [(2, 5)].
Sol. Given that, A = {1, 2, 3, K, 9} and (a, b ) R(c, d ) if a + d = b + c for (a, b ) Î A ´ A and
(c, d ) Î A ´ A.
Let (a, b ) R (a, b )
Þ a + b = b + a, " a, b Î A
which is true for any a, b Î A.
Hence, R is reflexive.
Let (a, b ) R (c, d ) a+d =b+c
c + b = d + a Þ (c, d ) R (a, b )
So, R is symmetric.
 Let (a, b ) R (c, d ) and (c, d ) R (e, f )
a + d = b + c and c + f = d + e
a + d = b + c and d + e = c + f
(a + d ) - (d + e ) = (b + c ) - (c + f )
(a - e ) = b - f
a+ f =b+e
(a, b ) R (e, f )
So, R is transitive.
Hence, R is an equivalence relation.
Now, equivalence class containing [(2, 5)] is {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.

Q. 24 Using the definition, prove that the function f : A ® B is invertible if

and only if f is both one-one and onto.
Sol. A function f : X ® Y is defined to be invertible, if there exist a function g = Y ® X such that
gof = I X and fog = I Y . The function is called the inverse of f and is denoted by f -1.
A function f = X ® Y is invertible iff f is a bijective function.

Q. 25 Functions f , g : R ® R are defined, respectively, by f (x) = x 2 + 3x + 1,

g(x) = 2x - 3, find
(i) fog (ii) gof (iii) fof (iv) gog
Sol. Given that, f(x ) = x 2 + 3x + 1, g (x ) = 2 x - 3
(i) fog = f{g (x )} = f(2 x - 3)
= (2 x - 3)2 + 3 (2 x - 3) + 1
= 4x 2 + 9 - 12 x + 6x - 9 + 1 = 4x 2 - 6x + 1
(ii) gof = g {f(x )} = g (x 2 + 3x + 1)
= 2(x 2 + 3x + 1) - 3
= 2 x 2 + 6x + 2 - 3 = 2 x 2 + 6x - 1
(iii) fof = f{f(x )} = f(x 2 + 3x + 1)
= (x 2 + 3x + 1)2 + 3 (x 2 + 3x + 1) + 1
= x 4 + 9x 2 + 1 + 6x 3 + 6x + 2 x 2 + 3x 2 + 9x + 3 + 1
= x 4 + 6x 3 + 14x 2 + 15x + 5
(iv) gog = g {g (x )} = g (2 x - 3)
= 2(2 x - 3) - 3
= 4x - 6 - 3 = 4x - 9

Q. 26 Let * be the binary operation defined on Q. Find which of the

following binary operations are commutative
(i) a * b = a - b, " a, b Î Q (ii) a * b = a2 + b2 , " a, b Î Q
(iii) a * b = a + ab, " a, b Î Q (iv) a * b = (a - b)2 , " a, b Î Q
Sol. Given that * be the binary operation defined on Q.
(i) a * b = a - b, " a, b Î Q and b * a = b - a
So, a* b ¹ b* a [Q b - a ¹ a - b]
Hence, * is not commutative.
 (ii) a * b = a2 + b 2
b * a = b 2 + a2
So, * is commutative. [since, ‘+’ is on rational is commutative]
(iii) a * b = a + ab
b * a = b + ab
Clearly, a + ab ¹ b + ab
So, * is not commutative.
(iv) a * b = (a - b )2 , " a, b Î Q
b * a = (b - a)2
Q (a - b )2 = (b - a)2
Hence, * is commutative.

Q. 27 If * be binary operation defined on R by a * b = 1 + ab, " a, b Î R.

Then, the operation * is
(i) commutative but not associative.
(ii) associative but not commutative.
(iii) neither commutative nor associative.
(iv) both commutative and associative.
Sol. (i) Given that, a * b = 1 + ab, " a, b Î R
a * b = ab + 1= b * a
So, * is a commutative binary operation.
Also, a * (b * c ) = a * (1 + bc ) = 1 + a(1 + bc )
a * (b * c ) = 1 + a + abc …(i)
(a * b ) * c = (1 + ab ) * c
= 1 + (1 + ab )c = 1 + c + abc …(ii)
From Eqs. (i) and (ii),
a * (b * c ) ¹ (a * b) * c
So, * is not associative
Hence, * is commutative but not associative.

Objective Type Questions

Q. 28 Let T be the set of all triangles in the Euclidean plane and let a
relation R on T be defined as aRb, if a is congruent to b, " a, b Î T .
Then, R is
(a) reflexive but not transitive (b) transitive but not symmetric
(c) equivalence (d) None of these
Sol. (c) Consider that aRb, if a is congruent to b, " a, b Î T.
Then, aRa Þ a @ a,
which is true for all a Î T
So, R is reflexive, …(i)
 Let aRb Þ a@b
Þ b@a Þ b@a
Þ bRa
So, R is symmetric. …(ii)
Let aRb and bRc
Þ a@b and b @ c
Þ a@c Þ aRc
So, R is transitive. …(iii)
Hence, R is equivalence relation.

Q. 29 Consider the non-empty set consisting of children in a family and a

relation R defined as aRb, if a is brother of b. Then, R is
(a) symmetric but not transitive
(b) transitive but not symmetric
(c) neither symmetric nor transitive
(d) both symmetric and transitive
Sol. (b) Given, aRb Þ a is brother of b
\ aRa Þ a is brother of a, which is not true.
So, R is not reflexive.
aRb Þ a is brother of b.
This does not mean b is also a brother of a and b can be a sister of a.
Hence, R is not symmetric.
aRb Þ a is brother of b
and bRc Þ b is a brother of c.
So, a is brother of c.
Hence, R is transitive.

Q. 30 The maximum number of equivalence relations on the set A = {1, 2, 3}

are
(a) 1 (b) 2 (c) 3 (d) 5
Sol. (d) Given that, A = {1, 2, 3}
Now, number of equivalence relations as follows
R1 = {(1, 1), (2, 2 ), (3, 3)}
R 2 = {(1, 1), (2, 2 ), (3, 3), (1, 2 ), (2,1)}
R 3 = {(1, 1), (2, 2 ), (3, 3), (1, 3), (3,1)}
R 4 = {(1, 1), (2, 2 ), (3, 3), (2, 3), (3, 2 )}
R 5 = {(1, 2, 3) Û A ´ A = A 2 }
\ Maximum number of equivalence relation on the set A = {1, 2, 3} = 5

Q. 31 If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is

(a) reflexive (b) transitive (c) symmetric (d) None of these
Sol. (b) R on the set {1, 2, 3} be defined by R = {(1, 2 )}
It is clear that R is transitive.
Q. 32 Let us define a relation R in R as aRb if a ³ b. Then, R is
(a) an equivalence relation
(b) reflexive, transitive but not symmetric
(c) symmetric, transitive but not reflexive
(d) neither transitive nor reflexive but symmetric
Sol. (b) Given that, aRb if a ³ b
Þ aRa Þ a ³ a which is true.
Let aRb, a ³ b, then b ³ a which is not true R is not symmetric.
But aRb and b R c
Þ a ³ b and b ³ c
Þ a ³c
Hence, R is transitive.

Q. 33 If A = {1, 2, 3} and consider the relation

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
Then, R is
(a) reflexive but not symmetric (b) reflexive but not transitive
(c) symmetric and transitive (d) neither symmetric nor transitive
Sol. (a) Given that, A = {1, 2, 3}
and R = {(1, 1), (2, 2 ), (3, 3), (1, 2 ), (2, 3), (1, 3)}
Q (1, 1), (2, 2 ), (3, 3) Î R
Hence, R is reflexive.
(1, 2 ) Î R but (2, 1) Ï R
Hence, R is not symmetric.
(1, 2 ) Î R and (2, 3) Î R
Þ (1, 3) Î R
Hence, R is transitive.

Q. 34 The identity element for the binary operation * defined on Q - {0} as

ab
a*b = , " a, b Î Q - {0} is
2
(a) 1 (b) 0 (c) 2 (d) None of these
K Thinking Process
For given binary operation * : A ´ A ® A, an element e Î A , if it exists, is called identity
for the operation *, if a * e = a = e * a, " a Î A.
Sol. (c) ab
Given that, a * b = , " a, b Î Q - {0}.
2
Let e be the identity element for *.
ae
\ a* e =
2
ae
Þ a= Þ e =2
2
Q. 35 If the set A contains 5 elements and the set B contains 6 elements,
then the number of one-one and onto mappings from A to B is
(a) 720 (b) 120 (c) 0 (d) None of these
Sol. (c) We know that, if A and B are two non-empty finite set containing m and n elements
respectively, then the number of one-one and onto mapping from A to B is
n!, if m = n
0, if m ¹ n
Given that, m = 5 and n = 6
\ m¹ n
Number of mapping = 0

Q. 36 If A = {1, 2, 3, K , n} and B = {a, b}. Then, the number of surjections

from A into B is
(a) n P2 (b) 2 n - 2 (c) 2 n - 1 (d) None of these
Sol. (d) Given that, A = {1, 2, 3, K, n} and B = {a, b}.
We know that, if A and B are two non-empty finite sets containing m and n elements
respectively, then the number of surjection from A into B is
n
C m ´ m!, if n ³ m
0, if n < m
Here, m = 2
n!
\ Number of surjection from A into B is n C 2 ´ 2 ! = ´ 2!
2 !(n - 2 )!
n(n - 1)(n - 2 )!
= ´ 2 ! = n2 - n
2 ´ 1 (n - 2 )

1
Q. 37 If f : R ® R be defined by f (x) = , " x Î R. Then, f is
x
(a) one-one (b) onto (c) bijective (d) f is not defined
K Thinking Process
In the given function at x = 0, f(x) = ¥. So, the function is not define.
1
Sol. (d) Given that, f( x ) = , " x ÎR
x
For x = 0,
f(x ) is not defined.
Hence, f(x ) is a not define function.

Q. 38 If f : R ® R be defined by f (x) = 3x 2 - 5 and g : R ® R by

x
g(x) = 2 . Then, gof is
x +1
3x 2 - 5 3x 2 - 5
(a) 4 2
(b)
9x - 30 x + 26 9x - 6x 2 + 26
4

3x 2 3x 2
(c) (d)
x + 2x 2 - 4
4
9x + 30 x 2 - 2
4
 x
Sol. (a) Given that, f(x ) = 3x 2 - 5 and g (x ) =
x2 + 1
gof = g {f(x )} = g (3x 2 - 5)
3x 2 - 5 3x 2 - 5
= 2 2
=
(3x - 5) + 1 9x - 30x 2 + 25 + 1
4

3x 2 - 5
=
9x - 30x 2 + 26
4

Q. 39 Which of the following functions from Z into Z are bijections?

(a) f ( x) = x3 (b) f ( x) = x + 2 (c) f ( x) = 2x + 1 (d) f ( x) = x 2 + 1
Sol. (b) Here, f( x ) = x + 2 Þ f( x1 ) = f( x 2 )
x1 + 2 = x 2 + 2 Þ x1 = x 2
Let y=x +2
x = y - 2 Î Z, " y Î x
Hence, f(x ) is one-one and onto.

Q. 40 If f : R ® R be the functions defined by f (x) = x 3 + 5, then f -1 (x) is

1 1 1
(a) ( x + 5)3 (b) ( x - 5)3 (c) (5 - x)3 (d) 5 - x
3
Sol. (b) Given that, f( x ) = x + 5
Let y = x3 + 5 Þ x3 = y - 5
1 1
x = (y - 5)3 Þ f(x )-1 = (x - 5)3

Q. 41 If f : A ® B and g : B ® C be the bijective functions, then (gof ) -1 is

(a) f -1og -1 (b) fog (c) g -1of -1 (d) gof
Sol. (a) Given that, f : A ® B and g : B ® C be the bijective functions.
(gof )-1 = f -1og -1

3x + 2
f : R - ìí üý ® R be defined by f (x) =
3
Q. 42 If , then
î5þ 5x - 3
1
(a) f -1( x) = f ( x) (b) f -1( x) = - f ( x) (c) ( fof ) x = - x (d) f -1( x) = f ( x)
19
Sol. (a) 3x +2
Given that, f( x ) =
5x -3
3x +2
Let y=
5x -3
3x + 2 = 5xy - 3 y Þ x(3 - 5 y) = - 3 y - 2
3y + 2 3x + 2
x= Þ f -1(x ) =
5y - 3 5x - 3
\ f -1(x ) = f(x )
 ì x, if x is rational
Q. 43 If f : [0, 1] ® [0, 1] be defined by f (x) = í
î1 - x, if x is irrational
then ( fof )x is
(a) constant (b) 1 + x (c) x (d) None of these
Sol. (c) Given that, f: [0, 1] ® [0, 1] be defined by
ì x, if x is rational
f (x ) = í
î 1 - x, if x is irrational
\ (fof )x = f(f(x )) = x

Q. 44 If f : [2, ¥ ) ® R be the function defined by f (x) = x 2 - 4 x + 5, then

the range of f is
(a) R (b) [1, ¥ ) (c) [ 4, ¥ ) (d) [5, ¥ )
K Thinking Process
Range of f = {y ÎY : y = f(x) for some in X}

Sol. (b) Given that, f(x ) = x 2 - 4x + 5

Let y = x 2 - 4x + 5
Þ y = x 2 - 4x + 4 + 1= (x - 2 )2 + 1
2
Þ (x - 2 ) = y - 1 Þ x - 2 = y-1
Þ x =2 + y-1
\ y - 1 ³ 0, y ³ 1
Range = [1, ¥ )

2x - 1
Q. 45 If f : N ® R be the function defined by f (x) = and g : Q ® R
2
3
be another function defined by g(x) = x + 2. Then, (gof ) is
2
7
(a) 1 (b) 1 (c) (d) None of these
2
2x - 1
Sol. (d) Given that, f( x ) = and g (x ) = x + 2
2
æ 3 ö
3 é æ 3 öù ç 2 ´ - 1÷
(gof ) = g ê f ç ÷ ú = g ç 2 ÷
2 ë è 2 øû çç 2 ÷÷
è ø
= g(1) = 1 + 2 = 3
 ì2x : x > 3
Q. 46 If f : R ® R be defined by f (x) = ïí x 2 : 1 < x £ 3
ï3x : x £ 1
î
Then, f (-1) + f (2) + f (4) is
(a) 9 (b) 14 (c) 5 (d) None of these
ì2 x : x > 3
ï
Sol. (a) Given that, f( x ) = í x 2 : 1 < x £ 3
ï3x : x £ 1
î
f(-1) + f(2 ) + f(4) = 3 (-1) + (2 )2 + 2 ´ 4
= - 3 + 4 + 8= 9

Q. 47 If f : R ® R be given by f (x) = tan x, then f -1 (1) is

p p
(a) (b) ìínp + : n Î Z üý
4 î 4 þ
(c) Does not exist (d) None of these
Sol. (a) Given that, f(x ) = tan x
Let y = tan x Þ x = tan-1 y
Þ f -1(x ) = tan-1 x Þ f -1(1) = tan-1 1
p p é p ù
Þ = tan-1tan = êëQ tan 4 = 1úû
4 4

Fillers
Q. 48 Let the relation R be defined in N by aRb, if 2a + 3b = 30. Then, R = …..
.
Sol. Given that, 2 a + 3b = 30
3b = 30 - 2 a
30 - 2 a
b=
3
For a = 3, b = 8
a = 6, b = 6
a = 9, b = 4
a = 12, b = 2
R = {(3, 8), (6, 6), (9, 4), (12, 2 )}

Q. 49 If the relation R be defined on the set A = {1, 2, 3, 4, 5} by

R = {(a, b) : | a 2 - b 2 | < 8}. Then, R is given by ……… .
Sol. Given, A = {1, 2, 3, 4, 5},
R = {(a, b ) :|a2 - b 2| < 8}
R = { (1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (4, 3), (3, 4), (4, 4), (5, 5)}
Q. 50 If f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, then
gof = ……… and fog = ……… .
Sol. Given that, f = {(1, 2 ), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}
gof(1) = g {f(1)} = g (2 ) = 3
gof(3) = g {f(3)} = g (5) = 1
gof(4) = g {f(4)} = g (1) = 3
gof = {(1, 3), (3, 1), (4, 3)}
Now, fog (2 ) = f{g (2 )} = f(3) = 5
fog (5) = f{g (5)} = f(1) = 2
fog (1) = f{g (1)} = f(3) = 5
fog = {(2, 5), (5, 2 ), (1, 5)}

x
Q. 51 If f : R ® R be defined by f (x) = , then ( fofof )(x) = ……… .
2
1+x
x
Sol. Given that, f( x ) =
1 + x2
(fofof )(x ) = f [ f{f(x )}]
æ x ö
ç ÷
é æ ö ù ç 1 + x 2 ÷
ç x ÷
= f êf ú= f ç ÷
ê ç 2 ÷ ú ç 2
÷
ë è 1 + x øû x
ç 1+ 2 ÷
è 1 + x ø
é 2 ù æ ö
x 1+ x ú= f ç x ÷
=fê
ê ú ç 2 ÷
è 1 + 2x
2 2
ë 1 + x ( 2 x + 1) û ø
x
1 + 2x 2 x 1 + 2x 2
= =
x2 1 + 2 x 2 1 + 3x 2
1+ 2
1 + 2x
x x
= =
1 + 3x 2 3x 2 + 1

Q. 52 If f (x) = [4 - (x - 7) 3 ], then f -1 (x) = ……… .

Sol. Given that, f(x ) = {4 - (x - 7 )3 }
Let y = [4 - (x - 7 )3 ]
(x - 7 )3 = 4 - y
(x - 7 ) = (4 - y)1/ 3
Þ x = 7 + (4 - y)1/ 3
f -1(x ) = 7 + (4 - x )1/ 3
True/False
Q. 53 Let R ={(3, 1), (1, 3), (3, 3)} be a relation defined on the set
A ={1, 2, 3}. Then, R is symmetric, transitive but not reflexive.
Sol. False
Given that, R = {(3, 1), (1, 3), (3, 3)} be defined on the set A = {1, 2, 3}
(1, 1) Ï R
So, R is not reflexive. (3, 1) Î R, (1, 3) Î R
Hence, R is symmetric.
Since, (3, 1) Î R, (1, 3) Î R
But (1, 1) Ï R
Hence, R is not transitive.

Q. 54 If f : R ® R be the function defined by f (x) = sin(3x + 2) " x Î R.

Then, f is invertible.
Sol. False
Given that, f(x ) = sin(3x + 2 ), " x Î R is not one-one function for all x Î R.
So, f is not invertible.

Q. 55 Every relation which is symmetric and transitive is also reflexive.

Sol. False
Let R be a relation defined by
R ={(1, 2), (2, 1), (1, 1), (2, 2)} on the set A = {1, 2, 3}
It is clear that (3, 3) Ï R. So, it is not reflexive.

Q. 56 An integer m is said to be related to another integer n, if m is a integral

multiple of n. This relation in Z is reflexive, symmetric and transitive.
Sol. False
The given relation is reflexive and transitive but not symmetric.

Q. 57 If A = {0, 1} and N be the set of natural numbers. Then, the mapping

f : N ® A defined by f (2n - 1) = 0, f (2n) = 1, " n Î N, is onto.
Sol. True
Given, A = {0, 1}
f(2 n - 1) = 0, f(2 n) = 1, " n Î N
So, the mapping f : N ® A is onto.

Q. 58 The relation R on the set A ={1, 2, 3} defined as R = {(1, 1), (1, 2), (2,
1), (3, 3)} is reflexive, symmetric and transitive.
Sol. False
Given that, R = {(1, 1), (1, 2), (2, 1), (3, 3)}
(2, 2 ) Ï R
So, R is not reflexive.
Q. 59 The composition of function is commutative.
Sol. False
Let f( x ) = x 2
and g(x ) = x + 1
fog (x ) = f {g (x )} = f (x + 1)
= (x + 1)2 = x 2 + 2 x + 1
gof (x ) = g { f(x )} = g (x 2 ) = x 2 + 1
\ fog (x ) ¹ gof(x )

Q. 60 The composition of function is associative.

Sol. True
Let f( x ) = x, g ( x ) = x + 1
and h(x ) = 2 x - 1
Then, fo{goh(x )} = f [g {h(x )}]
= f {g (2 x - 1)}
= f(2 x - 1 + 1)
= f(2 x ) = 2x
\ (fog ) oh(x ) = (fog ){h(x )}
= (fog )(2 x - 1)
= f{g (2 x - 1)}
= f(2 x - 1 + 1)
= f(2 x ) = 2x

Q. 61 Every function is invertible.

Sol. False
Only bijective functions are invertible.

Q. 62 A binary operation on a set has always the identity element.

Sol. False
‘+’ is a binary operation on the set N but it has no identity element.
 2
Inverse Trigonometric
Functions
Short Answer Type Questions
Q. 1 Find the value of tan -1 æç tan 5p ö÷ + cos -1 æç cos 13p ö÷.
è 6ø è 6 ø
K Thinking Process
æ p pö
Use the property, tan-1tan x = x, xÎ ç - , ÷ and cos -1(cos x) = x, xÎ[0 , p] to get the
è 2 2ø
answer.
æ p pö
Sol. We know that, tan-1 tan x = x; x Î ç - , ÷ and cos -1 cos x = x; x Î [0, p ]
è 2 2ø
-1 æ 5p ö -1 æ 13p ö
\ tan ç tan ÷ + cos ç cos ÷
è 6 ø è 6 ø
é æ p öù é æ 7 p öù
= tan-1 ê tan ç p - ÷ ú + cos -1 êcos ç p + ÷
ë è 6 øû ë è 6 ø ûú
æ pö æ 7p ö
= tan-1 ç - tan ÷ + cos -1 ç - cos ÷ [Q cos(p + q) = - cos q]
è 6 ø è 6 ø
æ pö é æ 7 p öù
= - tan-1 ç tan ÷ + p - êcos -1 cos ç ÷
è 6ø ë è 6 ø úû
{Q tan-1 (- x ) = - tan-1 x; x Î R and cos -1(- x ) = p - cos -1 x; x Î [- 1, 1]}
æ pö é æ p öù
= - tan-1 ç tan ÷ + p - cos -1 êcosç p + ÷ ú
è 6ø ë è 6 øû
-1 æ pö é -1 æ p öù
= - tan ç tan ÷ + p - êcos ç - cos ÷ ú [Q cos(p + q) = - cos q]
è 6ø ë è 6 øû
æ pö æ pö
= - tan-1 ç tan ÷ + p - p + cos -1 ç cos ÷ [Q cos -1 (- x ) = p - cos -1 x ]
è 6ø è 6ø
p p
=- + 0+ =0
6 6
æ 5pö 5p æ 13 p ö 13 p
Note Remember that, tan-1 çtan ÷ ¹ and cos -1 ç cos ÷¹
è 6 ø 6 è 6 ø 6
5p æ p p ö 13 p
Since, Ï ç - , ÷ and Ï[0 , p]
6 è 2 2ø 6
 é æ - 3 ö pù
Q. 2 Evaluate cos êcos -1 ç ÷+ .
ç 2 ÷ 6ú
êë è ø úû
é æ - 3 ö pù é 5p pù é 5p - 3 ù
Sol. We have, cos êcos -1 çç ÷ + ú = cos êcos -1 æç cos ö÷ + ú êQ cos = ú
÷ 6 è ø
ë è 2 ø û ë 6 6 û ë 6 2 û
æ 5p p ö
= cos ç + ÷ {Q cos -1 cos x = x; x Î [0, p ]}
è 6 6ø
æ 6p ö
= cosç ÷
è 6 ø
= cos(p ) = - 1

Q. 3 Prove that cot æç p ö

- 2 cot -1 3 ÷ = 7.
è4 ø
æp ö
Sol. We have to prove, cot ç - 2 cot -1 3 ÷ = 7
è4 ø
æp -1 ö -1
Þ ç - 2 cot 3 ÷ = cot 7
è4 ø
p
Þ ( 2 cot -1 3) = - cot -1 7
4
-1 1 p 1
Þ 2 tan = - tan-1
3 4 7
1 1 p
Þ 2 tan-1 + tan-1 =
3 7 4
2/3 1 p
Þ tan-1 + tan-1 =
1 - (1/ 3)2 7 4
2/3 1 p
Þ tan-1 + tan-1 =
8/ 9 7 4
3 1 p
Þ tan-1 + tan-1 =
4 7 4
3 1
+
Þ tan -1 4 7 = p
3 1 4
1- ×
4 7
(21 + 4)/28 p
Þ tan-1 =
(28 - 3)/28 4
25 p
Þ tan-1 =
25 4
p
Þ 1 = tan
4
Þ 1= 1
Þ LHS = RHS Hence proved.
 -1 é æ - p öù
Q. 4 Find the value of tan -1 æç - 1 ö -1 æ 1 ö
÷ + cot ç ÷ + tan êsin ç ÷ú .
è 3ø è 3ø ë è 2 øû
æ 1 ö -1 æ 1 ö -1 é æ - p öù
Sol. We have, tan-1 ç- ÷ + cot ç ÷ + tan êsin ç ÷ú
è 3ø è 3ø ë è 2 øû
æ 5p ö æ pö
= tan- 1 ç tan ÷ + cot -1 ç cot ÷ + tan-1 (- 1)
è 6 ø è 3ø
é æ p ö ù é æ p öù é æ p öù
= tan-1 ê tan ç p - ÷ ú + cot -1 êcot ç ÷ ú + tan-1 ê tan ç p - ÷ ú
ë è 6 ø û ë è 3 ø û ë è 4 øû
æ p ö æ p ö æ p ö
= tan-1 ç - tan ÷ + cot -1 ç cot ÷ + tan-1 ç - tan ÷
è 6ø è 3ø è 4ø
é -1 æ p pöù
êQ tan (tan x ) = x, x Î ç - ' ÷,ú
è 2 2ø
ê ú
ê cot -1(cot x ) = x, x Î (0, p) ú
ê and tan-1(- x ) = - tan-1 x ú
ê ú
ë û
p p p -2 p + 4p - 3p
=- + - =
6 3 4 12
- 5p + 4p p
= =-
12 12

æ 2p ö
Q. 5 Find the value of tan -1 ç tan ÷.
è 3 ø
æ 2p ö æ pö
Sol. We have, tan-1 ç tan -1
÷ = tan tan ç p - ÷
è 3 ø è 3ø
æ p ö
= tan-1 ç - tan ÷ [Q tan-1(- x ) = - tan-1 x]
è 3ø
p p é æ -p p öù
= - tan-1 tan = - -1
êQ tan (tan x ) = x, x Î çè 2 , 2 ÷ø ú
3 3 ë û
æ 2pö 2p
Note Remember that, tan-1 çtan ÷ ¹
è 3 ø 3
æ p pö 2p æ - p p ö
Since, tan-1(tan x) = x, if x Î ç - , ÷ and Ï ç , ÷
è 2 2ø 3 è 2 2ø

-p æ-4ö
Q. 6 Show that 2 tan -1 (- 3) = + tan -1 ç ÷.
2 è 3 ø
Sol. LHS = 2 tan-1(-3) = - 2 tan-1 3 [Q tan-1(- x ) = - tan-1 x, x Î R]
é 1 - 32 ù é -1 1 - x
2 ù
= - êcos -1 2ú
-1
êQ 2 tan x = cos 2
, x ³ 0ú
ë 1 + 3 û ë 1 + x û
é -1 æ - 8 ö ù é -1 æ - 4 ö ù
= - êcos ç ÷ = - êcos ç ÷
ë è 10 ø úû ë è 5 ø úû
é æ 4 öù
= - ê p - cos -1 ç ÷ ú {Q cos -1(- x ) = p - cos -1 x, x Î [- 1, 1]}
ë è 5 øû
æ 4ö é æ 4ö 4 3 3ù
= - p + cos -1 ç ÷ êlet cos -1 ç ÷ = q Þ cos q = Þ tan q = Þ q = tan-1 ú
è ø ë
5 è ø
5 5 4 4û
 æ 3ö ép æ 3 öù
= - p + tan-1 ç ÷ = - p + ê - cot -1 ç ÷ ú
è 4ø ë 2 è 4 øû
p -1 3 p -1 4
= - - cot = - - tan
2 4 2 3
p -1 æ - 4 ö
= - + tan ç ÷ [Q tan-1(- x ) = - tan-1 x]
2 è 3 ø
= RHS Hence proved.

Q. 7 Find the real solution of

p
tan -1 x (x + 1) + sin -1 x 2 + x + 1 = .
2
K Thinking Process
Convert the sin-1 x 2 + x + 1 into inverse of tangent function and then use the property
æ x+ yö
tan-1x + tan-1y = tan-1 çç ÷÷ .
è1 - xy ø
p
Sol. We have, tan-1 x (x + 1) + sin-1 x 2 + x + 1 = ...(i)
2
-1 1
Let sin 2
x + x + 1= q Ö– x 2 + x + 1

x2 + x + 1 q
Þ sin q =
1 Ö– x 2 – x
x2 + x + 1 é sin q ù
Þ tan q = êQ tan q = ú
- x2 - x ë cos q û

x2 + x + 1
Q q = tan-1
- x2 - x
= sin-1 x 2 + x + 1
On putting the value of q in Eq. (i), we get
x2 + x + 1 p
tan-1 x (x + 1) + tan-1 =
2
-x -x 2
æx + yö
We know that, tan-1 x + tan-1 y = tan-1 çç ÷, xy < 1
÷
è 1 - xy ø
é x2 + x + 1 ù
ê x (x + 1) + ú
ê - x2 - x ú p
\ tan-1 ê ú=2
2
ê 1 - x (x + 1) × x + x + 1 ú
ê - x 2 - x úû
ë
é x2 + x + 1 ù
ê x2 + x + ú
-1 ê - 1(x 2 + x ) ú p
Þ tan ê ú=2
2
ê 1 - (x 2 + x ) × (x + x + 1) ú
ê -1 (x 2 + x ) úû
ë
x2 + x + - (x 2 + x + 1) p 1
Þ = tan =
[1 - - (x + x + 1] (x 2 + x )
2 2 0
 Þ [1 - - (x 2 + x + 1)] (x 2 + x ) = 0
Þ - (x 2 + x + 1) = 1 or x2 + x = 0
2
Þ - x - x - 1 = 1 or x(x + 1) = 0
2
Þ x + x + 2 = 0 or x (x + 1) = 0
- 1± 1- 4 ´2
\ x=
2
Þ x = 0 or x = -1
For real solution, we have x = 0, - 1.

Q. 8 Find the value of sin æç 2 tan -1 1 ö÷ + cos (tan -1 2 2).

è 3ø
æ 1ö
Sol. We have, sin ç 2 tan-1 ÷ + cos (tan-1 2 2 )
è 3ø
é ì 1 üù
ê ï 2´ ïú é
ï 3 ïú + cos æ cos -1 1 ö 1 ù
= sin êsin í
- 1
ý ç ÷
-1
êQ tan x = cos
-1
ú
ê ï1 + æ ö ï1 2 ú è 3 ø ë 1+ x 2 û
ê ïî ç ÷ ú
ë è 3 ø ïþû
é -1 -1 2 x 1ù
êQ 2 tan x = sin 2
, - 1 £ x £ 1 and tan-1 (2 2 ) = cos -1 ú
ë 1 + x 3û
é æ 2 öù
ê -1 ç 3 ÷ ú 1
= sin êsin ç ÷ú + {Q cos (cos -1 x ) = x; x Î [- 1, 1] }
ê çç 1 + 1 ÷÷ ú 3
ë è 9 øû
é æ 2 ´ 9 öù 1 é 3 ù 1
= sin êsin-1 çç ÷ ú + = sin êsin-1 æç ö÷ ú + [Q sin (sin-1 x ) = x]
´ ÷ è 5 øû 3
ë è 3 10 ø û 3 ë
3 1 9 + 5 14
= + = =
5 3 15 15

Q. 9 If 2 tan -1 (cos q) = tan -1 (2 cosec q), then show that q = p , where n is any
4
integer.
K Thinking Process
æ 2x ö
Use the property, 2 tan-1x = tan-1 çç ÷ to prove the desired result.
2÷
è1 - x ø
Sol. We have, 2 tan-1(cos q) = tan-1(2 cosec q)
æ 2 cos q ö
Þ tan-1 çç 2 ÷
÷ = tan-1(2 cosec q)
è 1 - cos q ø
é -1 -1 æ 2 x öù
êQ 2 tan x = tan çç ÷
2 ÷ú
êë è 1 - x ø úû
æ 2 cos q ö
Þ ç ÷ = (2 cosec q)
è sin2 q ø
Þ (cot q × 2 cosec q) = (2 cosec q) Þ cot q = 1
p p
Þ cot q = cot Þ q =
4 4
 æ 1ö æ 1ö
Q. 10 Show that cos ç 2 tan -1 ÷ = sin ç 4 tan -1 ÷.
è 7ø è 3ø
K Thinking Process
1 - x2 2x
Use the property 2 tan-1 x = cos -1 and 2 tan-1x = tan-1 , to prove LHS = RHS.
1 + x2 1 - x2
æ 1ö æ 1ö
Sol. We have, cos ç 2 tan-1 ÷ = sin ç 4 tan-1 ÷
è 7ø è 3ø
é æ æ 1ööù
2
ê ç1- ç ÷ ÷ú
ç è7 ø÷ú é -1 1 ù
é 2 ù
-1 æ 1 - x ö
Þ cos êcos ç
- 1
÷ ú = sin êë2 × 2 tan 3 úû
-1
êQ 2 tan x = cos çç ÷
2 ÷ú
ê æ 1ö
2
êë è 1 + x ø úû
ê çç 1 + ç ÷ ÷÷ ú
ë è è7 ø øû
é é æ öù
æ 48 ö ù ê ç
2 ÷ú
ç ÷ú é öù
ê ç ÷ú -1 æ 2 x
Þ cos êcos ç -1 49
÷ = sin ê2 × tan - 1 3 -1
êQ 2 tan x = tan çç ÷ú
50 ÷ ú ê ç 2 ÷ú 2 ÷
ê ç
ç ÷ú çç æ
1- ç ÷
1ö
÷÷ ú êë è1 - x ø úû
ë è 49 ø û ê
ë è è 3ø øû
é æ 48 ´ 49 ö ù é 18 ù
Þ cos êcos -1 çç ÷ ú = sin ê2 tan-1 æç ö÷ ú
÷
ë è 50 ´ 49 ø û ë è 24 ø û
é æ 24 ö ù æ 3ö
Þ cos êcos -1 ç ÷ ú = sin ç 2 tan-1 ÷
ë è 25 ø û è 4ø
æ 3 ö
ç 2´ ÷
é -1 æ 24 ö ù -1 4 ÷ é -1 -1 2 x ù
Þ cos êcos ç ÷ ú = sin ç sin êëQ 2 tan x = sin 1 + x 2 úû
ë è 25 ø û çç 9 ÷
1+ ÷
è 16 ø
24 æ 3 /2 ö
Þ = sin çç sin-1 ÷
25 è 25/16 ÷ø
24 48 24 24
Þ = Þ =
25 50 25 25
\ LHS = RHS Hence proved.

æ 3ö
Q. 11 Solve the equation cos (tan -1 x) = sin ç cot -1 ÷.
è 4ø
æ 3ö
Sol. We have, cos (tan-1 x ) = sin ç cot -1 ÷
è 4ø
æ ö
Þ cos ç cos -1
1 ÷ = sin æ sin-1 4 ö
ç ÷ ç ÷
2
x + 1ø è 5ø
è
x
Let tan-1 x = q1 Þ tan q1 =
1
1 1
Þ cos q1 = Þ q1 = cos -1
2 2
x +1 x +1
3
-1 3 5 4
and cot = q2 Þ cot q2 =
4 4
4 4 q2
Þ sin q2 = Þ q2 = sin-1
5 5 3
 1 4
Þ =
2
x +1 5
{Q cos (cos -1 x ) = x, x Î[- 1, 1] and sin (sin-1 x ) = x, x Î [- 1, 1]}
On squaring both sides, we get
16 (x 2 + 1) = 25
Þ 16x 2 = 9
2
æ 3ö
Þ x2 = ç ÷
è 4ø
3 -3 3
\ x=± = ,
4 4 4

Long Answer Type Questions

æ 1 + x2 + 1 - x2 ö p 1
-1 ç ÷ = + cos -1 x 2 .
Q. 12 Prove that tan ç ÷ 4 2
2 2
è 1 + x - 1- x ø
Sol. We have,
æ 1 + x2 + 1 - x2 ö p 1
tan-1 ç ÷ = + cos -1 x 2
ç 2 2 ÷ 4 2
è 1+ x - 1- x ø
æ 1 + x2 + 1 - x2 ö
\ LHS = tan-1 ç ÷ . ...(i)
ç 2 2 ÷
è 1+ x - 1- x ø
[let x 2 = cos 2 q = (cos 2 q - sin2 q)= 1 - 2 sin2 q = 2 cos 2 q - 1]
1
Þ cos -1 x 2 = 2 q Þ q = cos -1 x 2
2
\ 1 + x 2 = 1 + cos 2 q
= 1 + 2 cos 2 q - 1 = 2 cos q
and 1 - x 2 = 1 - cos 2 q
= 1 - 1 + 2 sin2 q = 2 sin q
æ 2 cos q + 2 sin q ö
\ LHS = tan-1 çç ÷
÷
è 2 cos q - 2 sin q ø
æ cos q + sin q ö
= tan-1 çç ÷
÷
è cos q - sin q ø
æ p ö
ç tan + tan q ÷
-1 æ 1 + tan q ö -1 4
= tan çç ÷ = tan ç
÷ ÷
è 1 - tan q ø çç 1 - tan p × tan q ÷÷
è 4 ø
é æ p ö ù é tan x + tan y ù
= tan-1 ê tan ç + q ÷ ú êQ tan(x + y) = ú
ë è 4 ø û ë 1 - tan x × tan y û
p p 1
= + q = + cos -1 x 2
4 4 2
= RHS Hence proved.
Q. 13 Find the simplified form of
æ3 4 ö é -3p p ù
cos -1 ç cos x + sin x ÷, where x Î ê , .
è 5 5 ø ë 4 4 úû
é3 4 ù é -3p p ù
Sol. We have, cos -1 ê cos x + sin x ú, x Î ê ,
ë5 5 û ë 4 4 úû
3
Let cos y =
5
4
Þ sin y =
5 5 4
3 4 æ 4ö
Þ y = cos -1 = sin-1 = tan-1 ç ÷
5 5 è 3ø y 3
\ cos -1 [cos y × cos x + sin y × sin x ]
= cos -1 [cos ( y - x )] [Q cos ( A - B) = cos A × cos B + sin A × sin B]
4 é -1 4 ù
= y - x = tan-1 - x êëQ y = tan 3 úû
3

8 3 77
Q. 14 Prove that sin -1 + sin -1 = sin -1 .
17 5 85
8 3 77
Sol. We have, sin-1 + sin-1 = sin-1
17 5 85
8 3
\ LHS = sin-1 + sin-1
17 5
-1 8 -1 3
= tan + tan
15 4 17
8 8 8
Let sin-1 = q1 Þ sin q1 =
17 17 q1
8 8 15
Þ tan q1 = Þ q1 = tan-1
15 15
3 3
and sin-1 = q2 Þ sin q2 =
5 5
3 -1 3
Þ tan q2 = Þ q2 = tan
4 4
é 8 3 ù
+ é
ê ú -1 æ x + y ö ù
= tan-1 ê 15 4 ú -1 -1
êQ tan x + tan y = tan ç ÷ú
8 3 ë è 1 - xy ø û
ê1 - ´ ú
ë 15 4 û
é 32 + 45 ù
-1 ê ú æ 77 ö
= tan ê 60 ú = tan-1 ç ÷
60 - 24 è 36 ø 5
ê ú
ë 60 û 3
77 77
Let q3 = tan-1 Þ tan q3 = q2
36 36
77 77 4
Þ sin q3 = =
5929 + 1296 85
77
\ q3 = sin-1
85
77
= sin-1 = RHS Hence proved.
85
 Alternate Method
8 3 77
To prove, sin-1 + sin-1 = sin-1
17 5 85
8
Let sin-1 =x 17
17
8 8
Þ sinx =
17 x
2 15
æ 8 ö
Þ cos x = 1 - sin2 x = 1 - ç ÷
è 17 ø
289 - 64 225 15
= = =
289 289 17
3
Let sin-1 = y
5
3 9
Þ sin y = Þ sin2 y = 5
5 25
9 3
\ cos 2 y = 1 -
25 y
2
æ 4ö 4 4
Þ cos 2 y = ç ÷ Þ cos y =
è 5ø 5
Now, sin(x + y) = sin x × cos y + cos x × sin y
8 4 15 3
= × + ×
17 5 17 5
32 45 77
= + =
85 85 85
æ 77 ö
Þ (x + y) = sin-1 ç ÷
è 85 ø
8 3 77
Þ sin-1 + sin-1 = sin-1
17 5 85

5 3 63
Q. 15 Show that sin -1 + cos -1 = tan -1 .
13 5 16
5 3 63
Sol. We have, sin-1 + cos -1 = tan-1 ...(i)
13 5 16
5
Let sin-1 =x
13
5 13
Þ sin x =
13 5
and cos 2 x = 1 - sin2 x x
25 144 12
= 1- =
169 169
144 12
Þ cos x = =
169 13
sin x 5 / 13 5
\ tan x = = = ...(ii)
cos x 12 / 13 12
Þ tan x = 5 / 12 ...(iii)
 3 3
Again, let cos -1 = y Þ cos y =
5 5
\ sin y = 1 - cos 2 y 5
2
æ 3ö 9 4
= 1- ç ÷ = 1-
è 5ø 25 y
16 4 3
sin y = =
25 5
sin y 4 / 5 4
Þ tan y = = = ...(iii)
cos y 3 / 5 3
We know that,
tan x + tan y
tan (x + y) =
1 - tan x × tan y
5 4 15 + 48
+
12 3 36
Þ tan (x + y) = Þ tan (x + y) =
5 4 36 - 20
1- ×
12 3 36
63 / 36
Þ tan(x + y) =
16 / 36
63
Þ tan (x + y) =
16
63
Þ x + y = tan-1
16
5 4 63
Þ tan-1 + tan-1 = tan-1 Hence proved.
12 3 16

1 2 1
Q. 16 Prove that tan -1 + tan -1 = sin -1 .
4 9 5
1 2 1
Sol. We have, tan-1 + tan-1 = sin-1 ...(i)
4 9 5
-1 1
Let tan =x
4
1
Þ tan x =
4
1
Þ tan2 x =
16
1
Þ sec 2 x - 1 =
16
1 17
Þ sec 2 x = 1 + =
16 16
1 17
Þ =
cos 2 x 16
16
Þ cos 2 x =
17
4
Þ cos x =
17
16 1
Þ sin2 x = 1 - cos 2 x = 1 - =
17 17
1
Þ sin x = ...(ii)
17
 2
Again, let tan-1 = y
9
2 4
Þ tan y = Þ tan2 y =
9 81
4
Þ sec 2 y - 1 =
81
4 85
Þ sec 2 y = + 1=
81 81
81 9
Þ cos 2 y = Þ cos y =
85 85
2 2 81 4
Þ sin y = 1 - cos y = 1 - =
85 85
2
Þ sin y = ...(iii)
85
We know that, sin (x + y) = sin x × cos y + cos x × sin y
1 9 4 2
= × + ×
17 85 17 85
17 17 1
= = =
17 × 85 17 × 5 5
1
Þ (x + y) = sin-1
5
-1 1 -1 2 -1 1
Þ tan + tan = sin Hence proved.
4 9 5

Q. 17 Find the value of 4 tan -1 1 - tan -1 1

.
5 239
K Thinking Process
æ 2x ö æ x-y ö
Use the properties 2 tan-1x = tan-1 ç 2÷
and tan-1x + tan-1y = tan-1 ç ÷ to get
è 1 - x ø è1 + xy ø
the desired value.
1 1
Sol. We have, 4 tan-1 - tan-1
5 239
1 1
= 2 × 2 tan-1 - tan-1
5 239
é 2 ù
ê ú
5 ú - tan-1 1 é -1 æ 2 x öù
= 2 × ê tan- 1
2
-1
êQ 2 tan x = tan çè 1 - x 2 ÷ú
ê æ 1ö ú 239 ë øû
ê 1- ç ÷ ú
ë è 5ø û
é æ 2 öù
ê -1 ç 5 ÷ ú 1
= 2 × ê tan ç ÷ - tan-1
ç 1 ÷ú 239
ê ç 1- ÷ú
ë è 25 ø û
é æ 2 / 5 öù -1 1
= 2 × ê tan-1 ç ÷ ú - tan
ë è 24 / 25 ø û 239
-1 5 -1 1
= 2 tan - tan
12 239
 5
2×
-1 12 1 é -1 æ 2 x öù
= tan 2
- tan-1 -1
êQ 2 tan x = tan çè 1 - x 2 ÷ú
æ 5 ö 239 ë øû
1- ç ÷
è 12 ø
æ 5 ö
ç ÷ 1
= tan ç 6 ÷ - tan-1
-1
çç 1 - 25 ÷÷ 239
è 144 ø
æ 144 ´ 5 ö -1 1
= tan-1 ç ÷ - tan
è 119 ´ 6 ø 239
æ 120 ö -1 1
= tan-1 ç ÷ - tan
è 119 ø 239
æ 120 1 ö
ç - ÷
119 239 ÷ é -1 æ x - y ö ù
= tan-1 ç -1 -1
êQ tan x - tan y = tan ç ÷ú
çç 1 + 120 × 1 ÷÷ ë è 1 + xy ø û
è 119 239 ø
æ 120 ´ 239 - 119 ö
= tan-1 ç ÷
è 119 ´ 239 + 120 ø
é 28680 - 119 ù -1 28561
= tan-1 ê ú = tan
ë 28441 + 120 û 28561
æ pö p
= tan-1(1) = tan-1 ç tan ÷ =
è 4ø 4

æ1 3ö 4 - 7
Q. 18 Show that tan ç sin -1 ÷ = and justify why the other value
è2 4ø 3
4+ 7
is ignored?
3
æ1 3ö 4- 7
Sol. We have, tan ç sin-1 ÷ =
è 2 4ø 3
é1 æ 3 öù
\ LHS = tan ê sin-1 ç ÷ ú
ë2 è 4 øû
1 -1 3 3
Let sin = q Þ sin-1 = 2q
2 4 4
3 2 tan q 3
Þ sin 2 q = Þ =
4 1 + tan2 q 4
Þ 3 + 3 tan2 q = 8 tan q
Þ 3 tan2 q - 8 tan q + 3 = 0
Let tan q = y
\ 3 y2 - 8 y + 3 = 0
+8 ±
64 - 4 ´ 3 ´ 3 8 ± 28
Þ y= =
2 ´3 6
2 [4 ± 7 ]
=
2×3
4+ 7
Þ tan q =
3
 é4 + 7 ù
Þ q = tan-1 ê ú
ë 3 û
ì 4+ 7 1 p é æ1 3 öù ü
íbut > . , since max ê tan ç sin-1 ÷ ú = 1ý
î 3 2 2 ë è2 4 øû þ
æ 4 - 7 ö 4 - 7
\ LHS = tan tan-1 çç ÷=
÷ = RHS
è 3 ø 3
p 3
Note Since, - < sin-1 < p / 2
2 4
- p 1 -1 3
Þ < sin < p / 4
4 2 4
æ -p ö 1 æ -1 3 ö p
\ tan ç ÷ £ tan ç sin ÷ £ tan
è 4 ø 2è 4ø 4
æ 1 -1 3 ö
Þ -1 < tan ç sin ÷ <1
è2 4ø

Q. 19 If a 1 , a2 , a 3 , ..., a n is an arithmetic progression with common difference

d, then evaluate the following expression.
é æ d ö æ d ö æ d ö
tan êtan -1 çç ÷÷ + tan -1 çç ÷÷ + tan -1 çç ÷÷
ë è 1 + a 1a2 ø è 1 + a2a 3 ø è 1 + a 3a 4 ø
æ d öù
+ ... + tan -1 çç ÷÷ ú
è 1 + a n -1 a n ø û
Sol. We have, a1 = a, a2 = a + d , a3 = a + 2d
and d = a2 - a1 = a3 - a2 = a4 - a3 =... = an - an -1
é æ d ö æ d ö
Given that, tan ê tan-1 çç ÷ + tan-1 ç
÷ ç 1+ a a ÷
÷
êë è 1 + a1a2 ø è 2 3 ø

æ d ö æ d öù
+ tan-1 çç ÷ + K + tan-1
÷
ç
ç 1+ a × a
÷ú
÷
è 1 + a3 a4 ø è n -1 n ø úû
é a - a1 a - a2 a - an -1 ù
= tan ê tan-1 2 + tan-1 3 + K + tan-1 n ú
ë 1 + a2 × a1 1 + a3 × a2 1 + an × an -1 û
= tan [(tan-1 a2 - tan-1 a1 ) + (tan-1 a3 - tan-1 a2 ) + ... + (tan-1 an - tan-1 an -1 )]

= tan[tan-1 an - tan-1 a1 ]
é a - a1 ù é -1 æ x - y ö ù
= tan ê tan-1 n ú
-1 -1
êQ tan x - tan y = tan ç ÷ú
ë 1 + an × a1 û ë è 1 + xy ø û
a - a1
= n [Q tan (tan-1 x ) = x ]
1 + an × a1
Objective Type Questions
Q. 20 Which of the following is the principal value branch of cos -1 x?
p p p
(a) é - , ù (b) (0 , p) (c) [0 , p ] (d) (0 , p) - ìí üý
ëê 2 2 ûú î2 þ
Sol. (c) We know that, the principal value branch of cos -1 x is [0, p ].
Y
5p
2
2p
3p
2
p
p
2
1
X¢ X
–1 O p
–
2
–p
–3p
2
–2p
5p
–
2
Y¢
\ y = cos -1 x

Q. 21 Which of the following is the principal value branch of cosec -1 x ?

-p p ö p p p -p p ù
(a) æç , ÷ (b) [0 , p ] - ìí üý (c) é , ù (d) é , - [0 ]
è 2 2ø î2 þ êë 2 2 úû êë 2 2 úû
é -p p ù
Sol. (d) We know that, the principal value branch of cosec -1x is ê , ú - 0.
ë 2 2û
Y

2p
3p
2
p p
2 1 2
X¢ X
–2 –1 O – p
2
–p

Y¢
\ y = cosec -1 x
Q. 22 If 3 tan -1 x + cot -1 x = p, then x equals to
1
(a) 0 (b) 1 (c) -1 (d)
2
Sol. (b) Given that, 3 tan-1 x + cot -1 x = p ...(i)
-1 -1 -1
Þ 2 tan x + tan x + cot x=p
-1 p é -1 -1 pù
Þ 2 tan x = p - êëQ tan x + cot x = 2 úû
2
p
Þ 2 tan-1 x =
2
2x p é -1 2 x ù
Þ tan-1 = -1
êQ 2 tan x = tan , " x Î (-1, 1)ú
1 - x2 2 ë 1 - x 2
û
2x p
Þ = tan
1 - x2 2
2x 1
Þ 2
= Þ 1 - x2 = 0
1- x 0
Þ x2 = 1 Þ x = ± 1 Þ x = 1
Hence, only x = 1satisfies the given equation.
Note Here, putting x = -1 in the given equation, we get
3 tan-1(-1) + cot -1(-1) = p
é æ -p ö ù é æ -p ö ù
Þ 3 tan-1 êtan ç ÷ ú + cot -1ê cot ç ÷ ú = p
ë è øû 4 ë è 4 øû
æ pö æ pö
Þ 3 tan-1 ç - tan ÷ + cot -1 ç - cot ÷ = p
è 4ø è 4ø
æ pö æ pö
Þ -3 tan-1 çtan ÷ + p - cot -1 ç cot ÷ = p
è 4ø è 4ø
p p
Þ -3 × + p - = p
4 4
Þ -p + p = p Þ 0 ¹ p
Hence, x = -1 does not satisfy the given equation.

é æ 33p ö ù
Q. 23 The value of sin -1 êcosç ÷ ú is
ë è 5 øû
3p -7p p -p
(a) (b) (c) (d)
5 5 10 10
Sol. (d) We have,
æ 33p ö -1 é æ 3p ö ù -1 é æ 3p ö ù
sin-1 ç cos ÷ = sin êcos ç 6p + ÷ ú = sin êcos ç ÷ [Q cos(2np + q) = cos q]
è 5 ø ë è 5 øû ë è 5 ø úû
é æp p öù -1 æ pö
= sin-1 êcos ç + ÷ = sin ç - sin ÷
ë è 2 10 ø úû è 10 ø
æ pö
= - sin-1 ç sin ÷ [Q sin-1(- x ) = - sin-1 x ]
è 10 ø
p é -1 æ -p p öù
=- êQ sin (sin x ) = x, x Î çè 2 , 2 ÷ø ú
10 ë û
Q. 24 The domain of the function cos -1 (2x - 1) is
(a) [0, 1] (b) [-1, 1] (c) (-1, 1) (d) [0 , p ]
Sol. (a) We have, f(x ) = cos -1(2 x - 1)
\ -1 £ 2 x - 1 £ 1
Þ 0 £ 2x £ 2
Þ 0 £ x £1
\ x Î[0, 1]

Q. 25 The domain of the function defined by f (x) = sin -1 x - 1 is

(a) [1, 2] (b) [ -1, 1] (c) [0, 1] (d) None of these
Sol. (a) Q f(x ) = sin-1 x - 1
Þ 0 £ x - 1£1 [Q x - 1 ³ 0 and - 1 £ x - 1 £ 1]
Þ 1£ x £2
\ x Î[1, 2 ]

æ 2 ö
Q. 26 If cosç sin -1 + cos -1 x ÷ = 0, then x is equal to
è 5 ø
1 2
(a) (b) (c) 0 (d) 1
5 5
æ 2 ö
Sol. (b) We have, cos ç sin-1 + cos -1 x ÷ = 0
è 5 ø
2
Þ sin-1 + cos -1 x = cos -1 0
5
2 p
Þ sin-1 + cos -1 x = cos -1 cos
5 2
2 p
Þ sin-1 + cos -1 x =
5 2
p 2
Þ cos -1 x = - sin-1
2 5
-1 -1 2 é -1 -1 pù
Þ cos x = cos êëQ cos x + sin x = 2 úû
5
2
\ x=
5

Q. 27 The value of sin[2 tan -1 (0.75)] is

(a) 0.75 (b) 1. 5 (c) 0.96 (d) sin 1. 5
-1 æ 3ö é 75 3ù
Sol. (c) . )] = sin ç 2 tan-1 ÷
We have, sin [2 tan (075 . =
êëQ 075 =
è 4ø 100 4 úû
æ 3 ö
ç 2× ÷
= sin ç sin-1 4 ÷ = sin ésin-1 3 / 2 ù
çç 9 ÷÷ êë 25 / 16 úû
1+
è 16 ø
é æ 48 ö ù é æ 24 ö ù 24
= sin êsin-1 ç ÷ ú = sin êsin-1 ç ÷ ú = = 0.96
ë è 50 ø û ë è 25 ø û 25
Q. 28 The value of cos -1 æç cos 3p ö÷ is
è 2 ø
p 3p 5p 7p
(a) (b) (c) (d)
2 2 2 2
Sol. (a) æ 3p ö
We have, cos -1 ç cos ÷
è 2 ø
æ pö é æ pö pù
= cos -1 cosç 2 p - ÷ êQ cosçè 2 p - 2 ÷ø = cos 2 ú
è 2ø ë û
æpö p
= cos -1 cos ç ÷ = {Q cos -1(cos x ) = x, x Î [0, p ]}
è2 ø 2
æ 3p ö 3p
Note Remember that, cos -1 ç cos ÷ ¹
è 2 ø 2
3p
Q Ï(0 , p)
2

Q. 29 The value of 2 sec -1 2 + sin -1 æç 1 ö÷ is

è 2ø
p 5p 7p
(a) (b) (c) (d) 1
6 6 6
1 p p
Sol. (b) We have, 2 sec -1 2 + sin-1 = 2 sec -1 sec + sin-1 sin
2 3 6
p p
= 2× + [Q sec -1(sec x ) = x and sin-1 (sin x ) = x ]
3 6
4p + p 5p
= =
6 6

4p
Q. 30 If tan -1 x + tan -1 y = , then cot -1 x + cot -1 y equals to
5
p 2p 3p
(a) (b) (c) (d) p
5 5 5
4p
Sol. (a) We have, tan-1 x + tan-1 y =
5
p -1 p -1 4p
Þ - cot x + - cot y =
2 2 5
-1 -1 4p é -1 -1 pù
Þ - (cot x + cot y) = -p êëQ tan x + cot x = 2 úû
5
æ pö
Þ cot -1 x + cot -1 y = - ç - ÷
è 5ø
-1 -1 p
Þ cot x + cot y =
5
 æ 2a ö æ 2 ö æ ö
Q. 31 If sin -1 çç ÷ + cos -1 ç 1 - a ÷ = tan -1 ç 2x ÷, where a, x Î ] 0, 1[,
2 ÷ ç1 +a ÷2 ç1 - x ÷
2
è1 +a ø è ø è ø
then the value of x is
a 2a
(a) 0 (b) (c) a (d)
2 1 - a2
æ 2a ö æ 1 - a2 ö æ 2x ö
Sol. (d) We have, sin-1 çç 2
÷ + cos -1 ç
÷
÷
ç 1 + a2 ÷ = tan
-1 ç
ç1 - x2
÷
÷
è1 + a ø è ø è ø
Let a = tan q Þ q = tan-1 a
æ 2 tan q ö æ 1 - tan2 q ö -1 2 x
\ sin-1 çç 2 ÷
÷ + cos -1 ç ÷
ç 1 + tan2 q ÷ = tan 1 - x 2
è 1 + tan q ø è ø
2x
Þ sin sin2 q + cos cos 2 q = tan-1
-1 -1
1 - x2
2x
Þ 2 q + 2 q = tan-1
1 - x2
2x
Þ 4 tan-1 a = tan-1
1 - x2
2x
Þ 2 × 2 tan-1 a = tan-1
1 - x2
2a 2x é -1 2 x ù
Þ 2 × tan-1 = tan-1 -1
êQ 2 tan x = tan ú
1 - a2 1 - x2 ë 1 - x2 û
æ 2a ö
2 × çç ÷
-1 è 1 - a2 ÷ø æ 2x ö
Þ tan 2
= tan-1 çç 2
÷
÷
æ 2a ö è1 - x ø
1 - çç ÷
2 ÷
è1 - a ø
2a
\ x=
1 - a2

é æ 7 öù
Q. 32 The value of cot êcos -1 ç ÷ ú is
ë è 25 ø û
25 25 24 7
(a) (b) (c) (d)
24 7 25 24
é æ 7 öù
Sol. (d) We have, cot êcos -1 ç ÷ ú
ë è 25 ø û
7
Let cos -1 =x
25
7
Þ cos x = 25 24
25
2
æ7 ö
\ sin x = 1 - cos 2 x = 1 - ç ÷ x
è 25 ø
7
625 - 49 24
= =
625 25
 7
cos x 25 7
\ cot x = = = ...(i)
sin x 24 24
25
æ7 ö æ7 ö
Þ x = cot -1 ç ÷ = cos -1 ç ÷
è 24 ø è 25 ø
æ -1 7 ö æ -1 7 ö 7 é -1 7 -1 7 ù
\ cot ç cos ÷ = cot ç cot ÷= êëQ cot 24 = cos 25 úû
è 25 ø è 24 ø 24

æ1 2 ö
Q. 33 The value of tan ç cos -1 ÷ is
è2 5ø
5+2
(a) 2 + 5 (b) 5 - 2 (c) (d) 5 + 2
2
æ1 2 ö
Sol. (b) We have, tan ç cos -1 ÷
è2 5ø
1 2
Let cos -1 =q
2 5
2 2
Þ cos -1 = 2q Þ cos2 q =
5 5
2 2
\ (1 - 2 sin q) =
5
2 2
Þ 2 sin q = 1 -
5
2 1 1
Þ sin q = -
2 5
1 1
Þ sin q = -
2 5
\ cos 2 q = 1 - sin2 q
1 1 1 1
= 1- + = +
2 5 2 5
1 1
Þ cos q = +
2 5
1 1
-
2 5 = 5 -2 é sin q ù
\ tan q = êQ tan q = ú
1 1 5+2 ë cos qû
+
2 5
5 -2 1 2
Þ q = tan-1 = cos -1
5+2 2 5

æ1 2 ö 5 -2
\ tan ç cos -1 ÷ = tan tan
-1
è2 5ø 5+2
5 -2 5 -2
= ×
5+2 5 -2
( 5 - 2 )2
= = 5 -2
5-4
 æ 2x ö
Q. 34 If | x | £ 1, then 2 tan -1 x + sin -1 çç ÷ is equal to
2 ÷
è1 + x ø
p
(a) 4 tan -1 x (b) 0 (c) (d) p
2
Sol. (a) 2x
We have, 2 tan-1 x + sin-1
1 + x2
Let x = tan q
2 tan q
\ 2 tan tan q + sin-1
-1
[Q tan-1 (tan x ) = x]
1 + tan2 q
é 2 tan q ù
= 2 q + sin-1 sin2 q êQ sin 2 q = ú
ë 1 + tan2 q û
= 2q + 2q [Q sin-1 (sin x ) = x]
=4q [Q q = tan-1 x ]
= 4 tan-1 x

Q. 35 If cos -1 a + cos -1 b + cos -1 g = 3p, then a (b + g) + b(g + a ) + g(a + b)

equals to
(a) 0 (b) 1 (c) 6 (d) 12
Sol. (c) We have, cos -1 a + cos -1 b + cos -1 g = 3p
We know that, 0 £ cos -1 x £ p
Þ cos -1 a + cos -1 b + cos -1 g = 3,p
If and only if, cos -1 a = cos -1 b = cos -1 g = p
Þ cos p = a = b = g
Þ -1 = a = b = g
Þ a = b = g = -1
\ a ( b + g ) + b(g + a ) + g(a + b )
= - 1 (-1 - 1) - 1 (-1 - 1) - 1 (-1 - 1)
=2 + 2 + 2= 6

Q. 36 The number of real solutions of the equation

ép ù
1 + cos 2x = 2 cos -1 (cos x) in ê , p ú is
ë2 û
(a) 0 (b) 1 (c) 2 (d) ¥
ép ù
-1
Sol. (a) We have, 1 + cos 2 x = 2 cos (cos x ), ê , p ú
ë2 û
Þ 1 + 2 cos 2 x - 1 = 2 cos -1(cos x )
Þ 2 cos x = 2 cos -1 (cos x )
Þ cos x = cos -1 (cos x )
Þ cos x = x [Q cos -1 (cos x ) = x]
which is not true for any real value of x.
Hence, there is no solution possible for the given equation.
Q. 37 If cos -1 x > sin -1 x, then
1 1 1
(a) < x £1 (b) 0 £ x < (c) -1 £ x < (d) x > 0
2 2 2
Sol. (c) We have, cos -1 x > sin-1 x, where x Î [-1, 1]
Þ x < cos(sin-1 x )
é xù
Þ x < cos [cos -1 1 - x 2 ] -1
êëlet sin x = q Þ sin q = 1 úû
é Q cos q = 1 - sin2 q = 1 - x 2 Þ q = cos -1 1 - x 2 ù
ëê ûú
Þ x < 1 - x2
Þ x 2 < 1 - x 2 Þ 2x 2 < 1
1 æ 1 ö
Þ x2 < Þ x<± ç ÷ ...(i)
2 è 2ø
Also, -1 £ x £ 1 ...(ii)
1
\ -1 £ x £
2

–¥ –1 1/Ö2 1 +¥
Alternate Method
p
- sin-1 x > sin-1 x
2
p p
> 2 sin-1 x Þ > sin-1 x
2 4
1 1
>x Þ < x £1
2 2
é -p p ù
We know that, sin-1x Î ê , ú
ë 2 2û

Fillers
æ 1ö
Q. 38 The principal value of cos -1 ç - ÷ is ......... .
è 2ø
Sol. Q 0 £ cos -1 x £ p
æ 1ö æ 2p ö 2p
cos -1 ç - ÷ = cos -1 ç cos ÷ =
è 2ø è 3 ø 3

Q. 39 The value of sin -1 æç sin 3p ö÷ is ......... .

è 5 ø
p p
Sol. Q - £ sin-1 x £
2 2
æ 3p ö æ 2p ö -1 æ 2p ö 2p
\ sin-1 ç sin ÷ = sin-1 sin ç p - ÷ = sin ç sin ÷ =
è 5 ø è 5 ø è 5 ø 5
Q. 40 If cos(tan -1 x + cot -1 3) = 0, then the value of x is ......... .
Sol. We have, cos (tan x + cot -1 3 ) = 0
-1

Þ tan-1 x + cot -1 3 = cos -1 0

p
Þ tan-1 x + cot -1 3 = cos -1cos
2
p
Þ tan-1 x + cot -1 3 =
2
p
Þ tan-1 x = - cot -1 3
2
é pù
Þ tan-1 x = tan-1 3 Q tan-1 x + cot -1 x = ú
ëê 2û
\ x= 3

Q. 41 The set of values of sec -1 1 is ......... .

2
-1
Sol. Since, domain of sec x is R - (-1, 1.)
Þ (-¥, - 1] È [1, ¥ )
1
So, there is no set of values exist for sec -1 .
2
So, f is the answer.

Q. 42 The principal value of tan -1 3 is ......... .

æpö
Sol. Q tan-1 3 = tan-1 tan ç ÷
é -1 æ p p öù è 3ø
êQ tan (tan x ) = x, x Î çè - 2 , 2 ÷ø ú
ë û æpö
=ç ÷
è 3ø

æ 14p ö
Q. 43 The value of cos -1 ç cos ÷ is ......... .
è 3 ø
æ 14p ö æ 2p ö
Sol. We have, cos -1 ç cos -1
÷ = cos cosç 4p + ÷
è 3 ø è 3 ø
2p
= cos -1 cos [Q cos(2np + q) = cos q]
3
2p
= {Qcos -1 (cos x ) = x, x Î [0, p ] }
3
æ 14 p ö 14 p
Note Remember that, cos -1 ç cos ÷¹
è 3 ø 3
14 p
Since, Ï[0 , p]
3

Q. 44 The value of cos(sin -1 x + cos -1 x), where | x | £ 1, is ......... .

Sol. cos (sin-1 x + cos -1 x )
p é -1 -1 pù
= cos = 0 êëQ sin x + cos x = 2 úû
2
 æ sin -1 x + cos -1 x ö
Q. 45 The value of tan ç ÷, when x = 3 , is ......... .
ç 2 ÷ 2
è ø
æ sin-1 x + cos -1 x ö p/2 ö
Sol. Q tan çç ÷ = tan æç ÷
é -1 -1 pù 2 ÷ è 2 ø
è ø
êëQ sin x + cos x = 2 úû
p
= tan =1
4

æ 2x ö
Q. 46 If y = 2 tan -1 x + sin -1 ç ÷, then ......... < y < ......... .
ç 1 + x2 ÷
è ø
2x
Sol. We have, y = 2 tan-1 x + sin-1
1 + x2
2 tan q
\ y = 2 tan-1 tan q + sin-1 [ let x = tan q]
1 + tan2 q
é 2 tan q ù
Þ y = 2 q + sin-1 sin2 q êQ sin2 q = ú
ë 1 + tan2 q û
Þ y = 2 q + 2 q = 4q [Q q = tan-1 x]
Þ y = 4 tan-1 x
Q - p / 2 < tan-1 x < p / 2
4p
\ - < 4 tan-1 x < 4p / 2
2
Þ - 2 p < 4 tan-1 x < 2 p
Þ - 2p < y < 2p [Q y = 4 tan-1 x ]

æ x-yö
Q. 47 The result tan -1 x - tan -1 y = tan -1 çç ÷÷ is true when the value of
è 1 + xy ø
xy is ......... .
æx - yö
Sol. We know that, tan-1 x - tan-1 y = tan-1 çç ÷
÷
è 1 + xy ø
where, x y >-1

Q. 48 The value of cot -1 (-x) x Î R in terms of cot -1 x is ......... .

Sol. We know that,
cot -1(- x ) = p - cot -1 x, x Î R

True/False
Q. 49 All trigonometric functions have inverse over their respective
domains.
Sol. False
We know that, all trigonometric functions have inverse over their restricted domains.
Q. 50 The value of the expression (cos -1 x)2 is equal to sec 2 x.
Sol. False
2
é 1ù
Q [cos -1 x ]2 = êsec -1 ú ¹ sec 2 x
ë xû

Q. 51 The domain of trigonometric functions can be restricted to any one of

their branch (not necessarily principal value) in order to obtain their
inverse functions.
Sol. True
We know that, the domain of trigonometric functions are restricted in their domain to obtain
their inverse functions.

Q. 52 The least numerical value, either positive or negative of angle q is

called principal value of the inverse trigonometric function.
Sol. True
We know that, the smallest numerical value, either positive or negative of q is called the
principal value of the function.

Q. 53 The graph of inverse trigonometric function can be obtained from the

graph of their corresponding function by interchanging X and Y-axes.
Sol. True
We know that, the graph of an inverse function can be obtained from the corresponding
graph of original function as a mirror image (i.e., reflection) along the line y = x .

Q. 54 The minimum value of n for which tan -1 n > p , n Î N , is valid is 5.

p 4
Sol. False
n p n p
Q tan-1 > Þ > tan
p 4 p 4
n é p ù
Þ >1 êëQ tan 4 = 1úû
p
Þ n >p
So, the minimum value of n is 4. [Q n Î N and p = 314
. ...]

é æ 1 öù p
Q. 55 The principal value of sin -1 êcosç sin -1 ÷ ú is .
ë è 2 øû 3
Sol. True
é æ 1 öù é æ p öù
Given that, sin-1 êcosç sin-1 ÷ ú = sin-1 êcos sin-1 ç sin ÷ ú
ë è 2 ø û ë è 6 øû
-1 é pù
= sin êcos ú [Q sin-1 (sin x ) = x]
ë 6û
-1 3
= sin
2
-1 p p
= sin sin =
3 3
 3
Matrices
Short Answer Type Questions
Q. 1 If a matrix has 28 elements, what are the possible orders it can have?
What if it has 13 elements?
Sol. We know that, if a matrix is of order m ´ n, it has mn elements, where m and n are natural
numbers.
We have, m ´ n = 28
Þ (m, n)= {(1, 28), (2, 14), (4, 7), (7, 4), (14, 2), (28, 1)}
So, the possible orders are 1 ´ 28, 2 ´ 14, 4 ´ 7, 7 ´ 4, 14 ´ 2, 28 ´ 1.
Also, if it has 13 elements, then m ´ n = 13
Þ (m, n) = {(1,13), (13,1)}
Hence, the possible orders are 1 ´ 13, 13 ´ 1 .

é ù
êa 1 x ú
ê
Q. 2 In the matrix A = 2 3 x - 2
y ú , write
ê -2 ú
ê0 5 ú
ë 5 û
(i) the order of the matrix A.
(ii) the number of elements.
(iii) elements a23 , a31 and a12 .
éa 1 x ù
ê ú
2 3 x2 - yú
Sol. We have, A=ê
ê0 -2 ú
5
ê 5 ú
ë û
(i) the order of matrix A = 3 ´ 3
(ii) the number of elements = 3 ´ 3 = 9
[since, the number of elements in an m ´ n matrix will be equal to m ´ n = mn]
(iii) a23 = x 2 - y, a31 = 0, a12 = 1
[since, we know that a ij , is a representation of element lying in the
ith row and jth column]
Q. 3 Construct a 2 ´ 2 matrix, where
( i - 2 j )2
(i) aij = (ii) aij =|-2i + 3 j|
2
Sol. We know that, the notation, namely A = [a ij ] m ´ n indicates that A is a matrix of order m ´ n,
also 1 £ i £ m, 1 £ j £ n; i , j Î N.
(i) Here, A = [aij ]2 ´2
(i - 2 j )2
Þ A= ,1£ i £2 ;1£ j £2 ...(i)
2
(1 - 2 )2 1
\ a11 = =
2 2
2
(1 - 2 ´ 2 ) 9
a12 = =
2 2
(2 - 2 ´ 1)2
a21 = =0
2
(2 - 2 ´ 2 )2
a22 = =2
2
é 1 9ù
Thus, A = ê2 2 ú
ê ú
êë 0 2 úû 2 ´2
(ii) Here, A = [a ij ]2 ´2 =½-2 i + 3 j ½, 1 £ i £ 2; 1 £ j £ 2

\ a11 =½- 2 ´ 1 + 3 ´ 1½= 1

a12 =½-2 ´ 1 + 3 ´ 2½ = 4 [Q |-1| = 1]
a21 =½-2 ´ 2 + 3 ´ 1½ = 1
a22 =½-2 ´ 2 + 3 ´ 2½= 2
1 4
\ A=
1 2 2 ´2

Q. 4 Construct a 3 ´ 2 matrix whose elements are given by aij = e i ×x = sin jx.

Sol. Since, A = [a ij ] m ´ n 1 £ i £ m and 1 £ j £ n, i , j Î N
\ A = [e i ×xsin jx ]3 ´2 ; 1 £ i £ 3; 1 £ j £ 2
Þ a11 = e1×x × sin1× x = e x sin x
a12 = e1×x ×sin2 × x = e x sin2 x
a21 = e 2 ×x × sin 1× x = e 2 x sin x
a22 = e 2 ×x × sin 2 × x = e 2 x sin2 x
a31 = e 3 ×x × sin 1× x = e 3 x sin x
a32 = e 3 ×x × sin 2 × x = e 3 x sin 2 x
é e x sin x e x sin2 x ù
ê ú
\ A = êe 2 x sin x e 2 x sin2 x ú
êë e 3 x sin x 3x
e sin2 x úû
3 ´2
Q. 5 Find the values of a and b, if A = B, where
éa + 4 3b ù é2a + 2 b 2 + 2 ù
A=ê ú and B = ê ú
ë 8 -6û ë 8 b 2 - 5b û
K Thinking Process
By using equality of two matrices, we know that each element of A is equal to
corresponding element of B.
é a + 4 3b ù é2 a + 2 b 2 + 2 ù
Sol. We have, A=ê ú and B = ê ú
ë 8 - 6 û 2 ´2 ë 8 b 2 - 5b û 2 ´2
Also, A=B
By equality of matrices we know that each element of A is equal to the corresponding
element of B, that is aij = bij for all i and j.
\ a11 = b11 Þ a + 4 = 2 a + 2 Þ a = 2
a12 = b12 Þ 3b = b 2 + 2 Þ b 2 = 3b - 2
and a22 = b22 Þ - 6 = b 2 - 5b
Þ - 6 = 3b - 2 - 5b [Q b 2 = 3b - 2 ]
Þ 2b = 4 Þ b = 2
\ a = 2 and b = 2

Q. 6 If possible, find the sum of the matrices A and B, where

é 3 1ù éx y z ù
A=ê ú and B = ê ú.
ë 2 3û ë a b c û

K Thinking Process
We know that, two matrices are added, if they have same order.
é ù é x y zù
Sol. We have, A = ê 3 1ú and B = ê ú
ë 2 3û 2 ´ 2 êëa b 6úû 2 ´ 3
Here, A and B are of different orders. Also, we know that the addition of two matrices A and
B is possible only if order of both the matrices A and B should be same.
Hence, the sum of matrices A and B is not possible.

é3 1 -1ù é2 1 -1ù
Q. 7 If X =ê ú and Y = ê ú , then find
ë5 -2 -3û ë7 2 4 û
(i) X + Y.
(ii) 2X - 3Y.
(iii) a matrix Z such that X + Y + Z is a zero matrix.
é 3 1 -1ù é2 1 -1ù
Sol. We have, X = ê ú and Y = ê
ë 5 -2 -3û 2 ´ 3 ë7 2 4úû 2 ´ 3

é3 + 2 1 + 1 -1 - 1 ù é 5 2 -2 ù
(i) X + Y = ê =
ë 5 + 7 -2 + 2 -3 + 4úû ê12 0 1 ú
ë û
 é 3 1 -1ù é 6 2 -2 ù
(ii) Q 2 X = 2 ê ú=ê ú
ë 5 -2 -3û ë10 -4 -6 û
é2 1 -1ù é 6 3 -3ù
and 3Y = 3 ê ú=ê ú
ë7 2 4 û ë21 6 12û
é 6 - 6 2 - 3 -2 + 3 ù é 0 -1 1 ù
\ 2 X - 3Y = ê ú=ê ú
ë10 - 21 -4 - 6 -6 - 12 û ë -11 -10 -18û
é 3 + 2 1 + 1 -1 - 1 ù é 5 2 -2 ù
(iii) X + Y = ê ú= ê ú
ë 5 + 7 -2 + 2 -3 + 4û ë12 0 +1û
é 0 0 0ù
Also, X+Y+Z=ê ú
ë 0 0 0û
We see that Z is the additive inverse of (X+Y) or negative of ( X+Y ).
é -5 -2 2 ù
\ Z=ê ú [Q Z = - ( X + Y )]
ë -12 0 -1û

Q. 8 Find non-zero values of x satisfying the matrix equation

x é2 x 2 ù + 2 é8 5 xù = 2 é( x + 8) 24ù .
2
ê 3 xú ê 4 4xú ê ú
ë û ë û ë (10) 6 xû
Sol. Given that,
é2 x 2 ù é 8 5x ù é(x 2 + 8) 24 ù
xê
3 x ú + 2 ê 4 4x ú = 2 ê 10 ú
ë û ë û ë 6x û
é2 x 2 2 x ù é16 10x ù é2 x 2 + 16 48 ù
Þ ê 2ú
+ ê ú=ê ú
ë 3x x û ë 8 8x û ë 20 12 x û
é2 x 2 + 16 2 x + 10x ù é2 x 2 + 16 48 ù
Þ ê ú=ê ú
ë 3x + 8 x 2 + 8x û ë 20 12 x û
Þ 2 x + 10x = 48
Þ 12 x = 48
48
\ x= =4
12

é0 1ù é0 -1ù
Q. 9 If A = ê ú and B = ê ú , then show that
ë1 1û ë1 0 û
( A + B ) ( A - B ) ¹ A 2 - B 2.
é 0 1ù é0 -1ù
Sol. We have, A=ê ú and B = ê 1
ë 1 1û ë 0 úû
é 0 + 0 1 - 1ù é 0 0ù
\ ( A + B) = ê ú=ê
ë 1 + 1 1 + 0û ë2 1úû 2 ´ 2
é 0 - 0 1 + 1ù é 0 2 ù
and ( A - B) = ê ú=ê ú
ë 1 - 1 1 - 0û ë 0 1û 2 ´ 2
Since, ( A + B) × ( A - B) is defined, if the number of columns of (A + B) is equal to the number
of rows of ( A - B), so here multiplication of matrices ( A + B) × ( A - B) is possible.
é 0 + 0 0 + 0ù é 0 0ù
Now, ( A + B)2 ´ 2 ×( A - B)2 ´ 2 = ê ú=ê ú …(i)
ë 0 + 0 4 + 1û ë 0 5û
 Also, A2 = A × A
é 0 1ù é 0 1ù
=ê ú×ê ú
ë 1 1û ë 1 1û
é 0 + 1 0 + 1ù é1 1ù
=ê ú=ê ú
ë 0 + 1 1 + 1û ë1 2 û
é 0 - 1ù é 0 - 1ù
and B2 = B × B =ê ú. ê ú
ë1 0 û ë1 0 û
é 0 - 1 0 + 0 ù é -1 0 ù
=ê ú=ê ú
ë 0 + 0 -1 + 0û ë 0 -1û
é1 1ù é -1 0 ù é2 1ù
\ A 2 - B2 = ê ú-ê ú=ê ú ...(ii)
ë1 2 û ë 0 -1û ë 1 3û
Thus, we see that
(A + B) × (A - B) ¹ A 2 - B2 [using Eqs. (i) and (ii)]
é 0 0ù é2 1ù
Þ ê 0 5ú ¹ ê 1 3ú Hence proved.
ë û ë û

é 1 3 2ù é1 ù
Q. 10 Find the value of x, if [1 x 1] êê 2 5 1úú ê 2 ú = 0.
ê ú
êë15 3 2úû êë x úû
é 1 3 2ù é 1ù
Sol. We have, [1 x 1]1 ´ 3 ê 2 5 1ú ê2 ú =0
ê ú ê ú
êë15 3 2 úû 3 ´ 3 êë x úû 3 ´ 1
é ù
ê 1ú
Þ [1 + 2 x + 15 3 + 5x + 3 2 + x + 2 ]1 ´ 3 ê2 ú =0
êx ú
ë û3 ´1
é ù
ê 1ú
Þ [16 + 2 x 5x + 6 x + 4]1 ´ 3 ê2 ú =0
êx ú
ë û3 ´1
Þ [16 + 2 x + (5x + 6) × 2 + (x + 4) × x ]1 ´ 1 = 0
Þ [16 + 2 x + 10x + 12 + x 2 + 4x ] = 0
Þ [x 2 + 16x + 28] = 0
2
Þ [x + 2 x + 14x + 28] = 0
Þ (x + 2 ) (x + 14) = 0
\ x = -2,-14

é5 3ù 2
Q. 11 Show that A = ê ú satisfies the equation A - 3 A - 7I = O and
ë - 1 - 2û
hence find the value of A -1 .
é5 3ù
Sol. We have, A=ê ú
ë -1 -2 û
é5 3ù é5 3ù
\ A2 = A × A = ê ú×ê ú
ë -1 -2 û ë -1 -2 û
 é 25 - 3 15 - 6 ù é22 9ù
=ê ú=ê
ë -5 + 2 -3 + 4û ë -3 1úû
é 5 3 ù é15 9 ù
3A = 3 ê ú=ê ú
ë -1 -2 û ë -3 -6û
é 1 0ù é7 0ù
and 7I = 7 ê ú=ê ú
ë 0 1û ë 0 7 û
é22 9ù é15 9 ù é7 0ù
\ A2 - 3 A - 7 I = ê ú-ê ú-ê
ë -3 1û ë -3 -6û ë 0 7 úû
é22 - 15 - 7 9 - 9 - 0ù
=ê ú
ë -3 + 3 - 0 1 + 6 - 7 û
é 0 0ù
=ê ú
ë 0 0û
=0 Hence proved.
Since, A2 - 3 A - 7 I = 0
Þ A -1[( A 2 ) - 3 A - 7 I ] = A -1 0
Þ A -1 A × A - 3 A -1 A - 7 A -1I = 0 [Q A -1 0 = 0]
-1
Þ IA - 3I - 7 A =0 [Q A -1 A = I ]
Þ A - 3I - 7 A -1 = 0 [Q A -1 I = A -1 ]
Þ -7 A -1 = - A + 3I
é -5 -3ù é 3 0ù é -2 -3ù
=ê ú+ ê ú= ê
ë 1 2 û ë 0 3û ë 1 5 úû
-1 é -2 -3ù
\ A -1 =
7 êë 1 5 úû

Q. 12 Find the matrix A satisfying the matrix equation

é2 1ù é -3 2 ù é1 0 ù
ê3 2ú A ê 5 -3ú = ê0 1ú .
ë û ë û ë û
K Thinking Process
We know that, if two matrices A and B of order m ´ n and p ´ q respectively are
multiplied, then necessity condition to multiplication of A × B is n = p. So, by taking a
matrix of correct order we can get the desired elements of the required matrix.
é2 1ù é -3 2 ù é1 0ù
Sol. We have, ê3 2 ú A× ê ú =ê
ë û2 ´ 2 ë 5 - 3û2 ´ 2 ë0 1úû 2 ´ 2
éa bù
Let A=ê
ëc d úû 2 ´2
é2 1ù é a b ù é -3 2 ù é 1 0ù
\ ê3 =
ë 2 úû êëc d úû êë 5 -3úû êë 0 1úû
é 2a + c 2 b + d ù é -3 2 ù é 1 0ù
Þ ê 3a + 2c =
ë 3b + 2d úû êë 5 -3úû êë 0 1úû
é -6 a - 3c + 10 b + 5d 4a + 2c - 6 b - 3d ù é 1 0ù
Þ ê -9 a - 6c + 15 b + 10d =
ë 6a + 4c - 9b - 6d úû êë 0 1úû
Þ -6 a - 3c + 10 b + 5d = 1 …(i)
 Þ 4 a + 2 c - 6 b - 3d = 0 …(ii)
Þ -9 a - 6c + 15 b + 10d = 0 …(iii)
Þ 6 a + 4c - 9 b - 6d = 1 …(iv)
On adding Eqs. (i) and (iv), we get
c + b -d = 2 Þ d = c + b -2 …(v)
On adding Eqs. (ii) and (iii), we get
-5 a - 4c + 9 b + 7 d = 0 …(vi)
On adding Eqs. (vi) and (iv), we get
a + 0 + 0 + d = 1Þ d = 1- a …(vii)
From Eqs. (v) and (vii),
c + b - 2 = 1- a Þ a + b + c = 3 …(viii)
Þ a = 3- b -c
Now, using the values of a and d in Eq. (iii), we get
-9 (3 - b - c ) - 6c + 15 b + 10 (-2 + b + c ) = 0
Þ -27 + 9 b + 9c - 6c + 15 b -20 + 10 b + 10 c = 0
Þ 34 b + 13c = 47 …(ix)
Now, using the values of a and d in Eq. (ii), we get
4 (3 - b - c ) + 2c - 6b - 3 (b + c - 2 ) = 0
Þ 12 - 4 b - 4c + 2 c - 6 b - 3 b - 3c + 6 = 0
Þ -13 b - 5c = -18 ...(x)
On multiplying Eq. (ix ) by 5 and Eq. (x) by 13, then adding, we get
-169 b - 65c = -234
170 b + 65c = 235
b=1
Þ -13 ´ 1 - 5c = -18 [from Eq. (x)]
Þ -5c = -18 + 13 = -5 Þ c = 1
\ a = 3 - 1 - 1 = 1and d = 1 - 1 = 0
é1 1ù
\ A=ê ú
ë1 0û

é4 ù é -4 8 4 ù
Q. 13 Find A, if ê1ú A = êê -1 2 1úú.
ê ú
êë3úû êë -3 6 3úû
é 4ù é -4 8 4ù
ê 1ú A = ê -1 2 1ú
Sol. We have,
ê ú ê ú
êë 3úû 3 ´1 êë -3 6 3úû 3 ´3
Let A = [x y z]

é 4ù é -4 8 4ù
\ ê 1ú [x y z] = ê -1 2 1ú
ê ú 1´3 ê ú
êë 3úû 3 ´1 êë -3 6 3úû 3 ´3
é 4x 4 y 4 zù é -4 8 4ù
Þ êx y zú= ê -1 2 1ú
ê ú ê ú
êë 3x 3 y 3 zúû êë -3 6 3úû
 Þ 4x = -4 Þ x = -1, 4 y = 8
Þ y = 2 and 4 z = 4
Þ z=1
\ A = [-1 2 1]

é3 -4ù
é2 1 2ù
Q. 14 If A êê1 1 úú and B = ê 2 2 2
ú , then verify (BA) ¹ B A .
ë1 2 4û
êë2 0 úû
é 3 -4ù
é2 1 2 ù
Sol. We have, A = ê1 1 ú and B = ê ú
ê ú ë 1 2 4û 2 ´3
êë2 0 úû 3 ´2

é 3 -4ù
é2 1 2 ù ê1 1 ú
\ BA = ê ú
ë 1 2 4û 2 ´3 ê ú
êë2 0 úû 3 ´2
é 6 + 1 + 4 -8 + 1 + 0 ù é11 -7 ù
=ê ú=ê ú
ë 3 + 2 + 8 -4 + 2 + 0û ë13 -2 û
é11 -7 ù é11 -7 ù
and (BA) × (BA) = ê úê ú
ë13 -2 û ë13 -2 û
é121 - 91 -77 + 14ù é 30 -63 ù
Þ (BA)2 = ê ú=ê ú ...(i)
ë143 - 26 -91 + 4 û ë117 -87 û
é2 1 2 ù é2 1 2 ù
Also, B2 = B × B = ê ú ê ú
ë 1 2 4û 2 ´3 ë 1 2 4û 2 ´3
So, B2 is not possible, since the B is not a square matrix.
Hence, (BA) 2 ¹ B 2 A 2 .

Q. 15 If possible, find the value of BA and AB, where

é 4 1ù
é2 1 2ù
A=ê ú and B = ê2 3ú.
ë1 2 4û ê ú
êë1 2úû
é 4 1ù
é2 1 2 ù ê2 3ú
Sol. We have, A = ê ú and B =
ë 1 2 4û 2 ´3 ê ú
êë 1 2 úû 3 ´2
So, AB and BA both are possible.
[since, in both A × B and B × A, the number of columns of first is equal to the number of rows
of second.]
é 4 1ù
é2 1 2 ù ê ú
\ AB = ê ú . 2 3ú
ë 1 2 4û 2 ´ 3 ê
êë 1 2 úû 3 ´ 2
é 8 + 2 + 2 2 + 3 + 4ù é12 9 ù
=ê ú=ê ú
ë 4 + 4 + 4 1 + 6 + 8 û ë12 15û
 é 4 1ù
é2 1 2 ù
and BA = ê2 3ú ê 1 2 4ú
ê ú ë û2 ´3
êë 1 2 úû 3 ´2

é4 ´ 2 + 1 4+ 2 8+ 4 ù é 9 6 12 ù
= ê 4+ 3 2 + 6 4 + 12 ú = ê7 8 16 ú
ê ú ê ú
êë 2 + 2 1 + 4 2 + 8 úû êë 4 5 10 úû

Q. 16 Show by an example that for A ¹ 0 , B ¹ 0 and AB = 0.

é 0 -4ù é 3 5ù
Sol. Let A=ê ú ¹ 0 and B = ê 0 0ú ¹ 0
ë0 2 û ë û
é 0 0ù
\ AB = ê ú= 0 Hence proved.
ë 0 0û

é1 4 ù
é2 4 0 ù
Q. 17 Given, A = ê ú and B = ê2 8 ú . is ( AB)¢ = B¢ A¢ ?
ë3 9 6û ê ú
êë1 3úû
é 1 4ù
é2 4 0ù ê2 8ú
Sol. We have, A = ê and B =
ë3 9 6úû 2 ´3 ê ú
êë 1 3ûú 3 ´2
é2 + 8+ 0 8 + 32 + 0 ù é10 40 ù
\ AB = ê ú=ê ú
ë 3 + 18 + 6 12 + 72 + 18û ë27 102 û
é10 27 ù
and ( AB)¢ = ê ú …(i)
ë 40 102 û
é2 3ù
é 1 2 1ù ê4
Also, B¢ = ê ú and A ¢= 9ú
ë 4 8 3û 2 ´3 ê ú
êë 0 6ûú 3 ´2
é 2 + 8+ 0 3 + 18 + 6 ù é10 27 ù
\ B¢A ¢ = ê ú =ê ú ...(ii)
ë 8 + 32 + 0 12 + 72 + 18û ë 40 102 û
Thus, we see that, ( AB)¢ = B¢ A ¢ [using Eqs. (i) and (ii)]

é2ù é3ù é -8 ù
Q. 18 Solve for x and y, x ê ú + y ê ú +ê ú = 0.
ë1û ë5û ë -11û
é2 ù é 3ù é -8 ù
Sol. We have, xê ú + y ê ú + ê -11ú = 0
1
ë û ë 5û ë û
é2 x ù é 3 × yù é -8 ù
Þ ê x ú + ê 5 × yú + ê -11ú = 0
ë û ë û ë û
é2 x 3 y -8 ù é 0ù
Þ êx =
ë 5y -11úû êë 0úû
\ 2x + 3y - 8 = 0
Þ 4x + 6 y = 16 ...(i)
and x + 5 y - 11 = 0
Þ 4x + 20 y = 44 …(ii)
 On subtracting Eq. (i) from Eq. (ii), we get
14y = 28 Þ y = 2
\ 2x + 3 ´2 - 8 = 0
Þ 2x = 2 Þ x = 1
\ x = 1and y = 2

Q. 19 If X and Y are 2 ´ 2 matrices, then solve the following matrix equations

for X and Y
é2 3 ù é-2 2ù
2 X + 3Y = ê ú, 3 X + 2Y = ê ú.
êë4 0 úû êë 1 -5úû
Sol. We have,
é2 3ù
2 X + 3Y = ê ú …(i)
ë4 0û
é-2 2ù
and 3X + 2Y = ê ú …(ii)
ë 1 -5û
On subtracting Eq. (i) from Eq. (ii), we get
é -2 - 2 2 - 3ù
\ (3 X + 2 Y ) - (2 X + 3Y ) = ê ú
ë 1 - 4 -5 - 0û
é-4 -1ù
(X - Y) = ê ú ...(iii)
ë-3 -5û
On adding Eqs. (i) and (ii), we get
é0 5ù
(5 X + 5Y ) = ê ú
ë5 -5û
1é0 5ù é 0 1ù
Þ (X + Y) = ê ú=ê ú ...(iv)
5 ë 5 -5 û ë 1 -1 û
On adding Eqs. (iii) and (iv), we get
é- 4 0 ù
(X - Y)+ (X + Y) = ê ú
êë - 2 - 6 úû
é-2 0ù
Þ 2 X = 2ê ú
ë-1 -3û
é-2 0ù
\ X=ê ú
ë -1 -3û
From Eq. (iv),
é-2 0ù é0 1ù
ê ú+ Y = ê ú
ë- 1 - 3û ë1 -1û
é2 1ù é-2 0ù
\ Y =ê ú and X = ê ú
êë 2 2 úû ë-1 -3û
Q. 20 If A = [3 5] and B = [7 3], then find a non-zero matrix C such that
AC = BC.
Sol. We have, A = [3 5]1 ´ 2 and B = [7 3]1 ´ 2
éx ù
Let C = ê ú is a non-zero matrix of order 2 ´ 1.
y
ë û 2 ´1
éx ù
\ AC = [3 5] ê ú = [3x + 5 y]
y
ë û
éx ù
and BC = [7 3] ê ú = [7 x + 3 y]
ë yû
For AC = BC,
[3x + 5 y] = [7 x + 3 y]
On using equality of matrix, we get
3x + 5 y = 7 x + 3 y
Þ 4x = 2 y
1
Þ x= y
2
Þ y = 2x
éxù
\ C=ê ú
ë2 x û
We see that on taking C of order 2 ´ 1, 2 ´ 2 , 2 ´ 3, ..., we get
éxù éx xù éx x xù
C = ê ú, ê ú, ê ú ...
ë2 x û ë2 x 2 x û ë2 x 2 x 2 x û
In general,
ékù ék kù
C = ê ú, ê ú etc...
ë2 k û ë2 k 2 kû
where, k is any real number.

Q. 21 Give an example of matrices A, B and C, such that AB = AC, where A is

non-zero matrix but B ¹ C.
é1 0ù é2 3ù é2 3ù
Sol. Let A=ê ú, B = ê ú and C = ê ú [QB ¹ C ]
ë0 0û ë4 0û ë4 4û
é1 0ù é2 3ù é2 3ù
\ AB = ê úê ú= ê ú …(i)
ë0 0û ë4 0û ë0 0û
é1 0ù é2 3ù é2 3ù
and AC = ê . = …(ii)
ë0 0úû êë 4 4úû êë 0 0úû
Thus, we see that AB = AC [using Eqs. (i) and (ii)]
where, A is non-zero matrix but B ¹ C.
 é 1 2ù é2 3 ù é 1 0ù
Q. 22 If A = ê ú , B=ê ú and C = ê ú , verify
ë -2 1û ë3 -4 û ë -1 0 û
(i) (AB) C = A(BC).
(ii) A (B + C) = AB + AC.
é1 2ù é2 3 ù é 1 0ù
Sol. We have, A = ê -2 1úû
, B= ê 3 -4ú and C = ê -1 0ú
ë ë û ë û
é 1 2 ù é2 3 ù é 2 + 6 3- 8 ù é 8 -5 ù
(i) ( AB) = ê úê ú= ê -4 + 3 -6 - 4ú = ê -1 -10ú
ë -2 1û ë 3 -4û ë û ë û
é 8 -5 ù é 1 0ù
and ( AB)C = ê úê ú
ë -1 -10û ë -1 0û
é 8 + 5 0ù é13 0ù
=ê ú=ê ú ...(i)
ë -1 + 10 0û ë 9 0û
é2 3 ù é 1 0ù
Again, (BC ) = ê úê ú
ë 3 -4û ë -1 0û
é2 - 3 0ù é -1 0ù
=ê ú =ê ú
ë 3 + 4 0û ë 7 0û
é 1 2 ù é -1 0ù
and A (BC ) = ê ú ê ú
ë -2 1û ë 7 0û
é -1 + 14 0ù é13 0ù
=ê ú=ê ú ...(ii)
ë +2 + 7 0û ë 9 0û
\ ( AB) C = A (BC ) [using Eqs. (i) and (ii)]
é2 3 ù é 1 0ù é 3 3 ù
(ii) (B + C) = ê ú+ ê =
ë 3 -4û ë -1 0úû êë2 -4úû
é 1 2 ù é3 3 ù
and A × (B + C ) = ê úê ú
ë -2 1û ë2 -4û
é 3+ 4 3- 8 ù é7 -5 ù
=ê ú= ê -4 -10ú ...(iii)
ë -6 + 2 -6 - 4û ë û
é 1 2 ù é2 3 ù
Also, AB = ê ú. ê ú
ë -2 1û ë 3 -4û
é 2 + 6 3 - 8 ù é 8 -5 ù
=ê ú=ê ú
ë -4 + 3 -6 - 4û ë -1 -10û
é 1 2 ù é 1 0ù é 1 - 2 0ù é -1 0ù
and AC = ê úê ú= ê ú=ê ú
ë -2 1û ë -1 0û ë -2 - 1 0û ë -3 0û
é 8 -5 ù é -1 0ù
\ AB + AC = ê ú+ ê ú
ë -1 -10û ë -3 0û
é7 -5 ù
Þ AB + AC = ê ...(iv)
ë - 4 - 10úû
From Eqs. (iii) and (iv),
A (B + C ) = AB + AC
 é x 0 0ù éa 0 0ù
Q. 23 If P = ê0 y 0ú and Q = êê0 b 0úú, then prove that
ê ú
êë 0 0 z úû êë0 0 c úû
é xa 0 0 ù
PQ = ê 0 yb 0 ú = QP .
ê ú
êë 0 0 zc úû
é x 0 0ù é a 0 0ù é xa 0 0ù
Sol. PQ = ê 0 y 0ú ê 0 b 0ú = ê0 yb 0ú ...(i)
ê ú ê ú ê ú
êë 0 0 zúû êë 0 0 c úû êë 0 0 zc úû
éa 0 0ù é x 0 0ù é ax 0 0ù
and QP = ê 0 b 0ú ê 0 y 0ú = ê0 by 0ú ...(ii)
ê úê ú ê ú
êë 0 0 c úû êë 0 0 zúû êë 0 0 zc úû
Thus, we see that, PQ = QP [using Eqs. (i) and (ii)]
Hence proved.

é -1 0 -1ù é 1 ù
Q. 24 If [2 1 3] êê-1 1 0 úú êê 0 úú = A, then find the value of A.
êë 0 1 1 úû êë -1úû
é -1 0 -1ù é 1 ù
Sol. We have, [ 2 1 3 ]ê -1 1 0 ú ê 0 ú = A
ê úê ú
êë 0 1 1 úû êë -1úû
é -1 0 -1ù
\ [2 1 3 ] ê -1 1 0 ú = [-2 - 1 + 0 0 + 1 + 3 -2 + 0 + 3]
ê ú
êë 0 1 1 úû
= [-3 4 1]
é1ù
Now, [ - 3 4 1 ]ê 0 ú = A
ê ú
êë -1úû
é1ù
\ A = [ - 3 4 1 ]ê 0 ú
ê ú
êë -1úû
= [- 3 + 0 - 1] = [-4]

é5 3 4ù é -1 2 1ù
Q. 25 If A = [2 1], B = ê ú and C = ê , then verify that
ë8 7 6 û ë1 0 2úû
A(B + C) = (AB + AC).
Sol. We have to verify that, A (B + C) = A B + AC
é 5 3 4ù é -1 2 1ù
We have, A = [2 1], B = ê ú and C = ê ú
ë8 7 6û ë 1 0 2û
 é 5 - 1 3 + 2 4 + 1ù
\ A (B + C ) = [2 1] ê ú
ë8 + 1 7 + 0 6 + 2 û
é 4 5 5ù
= [2 1] ê ú
ë 9 7 8û
= [8 + 9 10 + 7 10 + 8]
= [17 17 18] ...(i)
é 5 3 4ù
Also, AB = [2 1] ê ú
ë 8 7 6û
= [10 + 8 6 + 7 8 + 6] = [18 13 14]
é -1 2 1ù
and AC = [2 1] ê ú
ë 1 0 2û
= [-2 + 1 4 + 0 2 + 2 ] = [-1 4 4]
\ AB + AC = [18 13 14] + [-1 4 4]
= [17 17 18] ...(ii)
\ A (B + C ) = ( AB + AC ) [using Eqs. (i) and (ii)]
Hence proved.

é1 0 -1ù
Q. 26 If A = êê2 1 3 úú , then verify that A 2 + A = ( A + I ), where I is 3 ´ 3
êë0 1 1 úû
unit matrix.
é 1 0 -1ù
Sol. We have, A = ê2 1 3 ú
ê ú
êë 0 1 1 úû
\ A2 = A × A
é 1 0 -1ù é 1 0 -1ù é 1 -1 -2 ù
= ê2 1 3 ú ê2 1 3 ú = ê 4 4 4ú
ê úê ú ê ú
êë 0 1 1 úû êë 0 1 1 úû êë2 2 4 úû
é 1 -1 -2 ù é 1 0 -1ù
\ A 2 + A = ê 4 4 4 ú + ê2 1 3 ú
ê ú ê ú
êë2 2 4 úû êë 0 1 1 úû
é2 -1 -3ù
= ê6 5 7 ú ...(i)
ê ú
êë2 3 5 úû
é 1 0 -1ù é 1 0 0ù é2 0 -1ù
Now, A + I = ê2 1 3 ú + ê 0 1 0ú = ê2 2 3ú
ê ú ê ú ê ú
êë 0 1 1 úû êë 0 0 1úû êë 0 1 2 úû
é1 0 -1ù é2 0 -1ù é2 -1 -3ù
and A ( A + I ) = ê2 1 3 ú × ê2 2 3 ú = ê 6 5 7 ú ...(ii)
ê ú ê ú ê ú
êë 0 1 1 úû êë 0 1 2 úû êë2 3 5 úû
Thus, we see that A 2 + A = A( A + I ) [using Eqs. (i) and (ii)]
 é 4 0ù
é0 - 1 2 ù ê ú
Q. 27 If A = ê ú and B = ê1 3ú, then verify that
ë 4 3 - 4û
êë2 6úû
(i) (A ¢)¢ = A
(ii) (AB)¢ = B ¢A ¢
(iii) (kA)¢ = (kA ¢).
é 4 0ù
é 0 -1 2 ù ê 1 3ú
Sol. We have, A = ê ú and B =
ë 4 3 -4û ê ú
êë2 6úû
(i) We have to verify that, A ¢ = A
é0 4ù
\ A¢= ê -1 3 ú
ê ú
êë 2 -4úû
é 0 -1 2 ù
and A¢ = ê ú= A Hence proved.
ë 4 3 -4û
(ii) We have to verify that, AB¢ = B¢ A¢
é3 9 ù
\ AB = ê
ë11 - 15úû
é 3 11 ù
Þ ( AB)¢ = ê ú
ë 9 -15û
é 0 4ù
é4 1 2ù ê é 3 11 ù
and B¢A ¢ = ê - 1 3ú = ê
ë0 3 6úû ê ú ë 9 -15úû
êë 2 -4úû
= ( AB)¢ Hence proved.
(iii) We have to verify that, (kA)¢ = (kA¢)
é 0 -k 2 k ù
Now, (kA) = ê ú
ë 4k 3k -4k û
é0 4k ù
and (kA)¢ = ê - k 3k ú
ê ú
êë2 k -4k úû
é0 4k ù
Also, kA ¢ = ê - k 3k ú
ê ú
êë2 k -4k úû
= (kA)¢ Hence proved.
 é1 2ù é 1 2ù
Q. 28 If A = ê4 1ú and B = êê6 4úú , then verify that
ê ú
êë5 6úû êë 7 3úû
(i) (2A + B)¢ = 2AA + B¢.
(ii) (A - B)¢ = A¢ - B¢.
é1 2 ù é1 2 ù
Sol. We have, A = ê 4 1ú and B = ê 6 4ú
ê ú ê ú
êë 5 6úû êë7 3úû
é2 4 ù é1 2 ù é 3 6 ù
(i) \ (2 A + B) = ê 8 2 ú + ê 6 4ú = ê14 6 ú
ê ú ê ú ê ú
êë10 12 úû êë7 3úû êë17 15úû
é 3 14 17 ù
and (2 A + B)¢ = ê ú
ë 6 6 15 û
é 1 4 5ù é 1 6 7 ù
Also, 2 A ¢ + B¢ = 2 ê ú+ ê ú
ë2 1 6û ë2 4 3û
é 3 14 17 ù
=ê ú = (2 A + B)¢ Hence proved.
ë 6 6 15 û
é1 2 ù é1 2 ù é 0 0ù
(ii) ( A - B) = ê 4 1ú - ê 6 4ú = ê -2 -3ú
ê ú ê ú ê ú
êë 5 6úû êë7 3úû êë -2 3 úû
é 0 -2 -2 ù
and ( A - B)¢ = ê ú
ë 0 -3 3 û
é 1 4 5ù é 1 6 7 ù
Also, A ¢ - B¢ = ê ú-ê ú
ë2 1 6û ë2 4 3û
é 0 -2 -2 ù
=ê
ë 0 - 3 3úû
= ( A - B)¢ Hence proved.

Q. 29 Show that A ¢ A and A A¢ are both symmetric matrices for any matrix
A.
K Thinking Process
We know that, for a matrix A to be symmetric matrix, A¢ = A . Also by using the result
(AB)¢= BA¢, we can prove that A¢A and AA¢ are both symmetric matrices for any matrix
A.
Sol. Let P = A ¢A
\ P ¢ = ( AA¢)¢
= A¢( A ¢)¢ [Q ( AB¢)¢= B¢A ¢]
= A ¢A = P
So, A ¢A is symmetric matrix for any matrix A.
Similarly, let Q = A A¢
\ Q ¢ = ( AA ¢)¢ = ( A ¢)¢( A)¢
= A ( A ¢)¢ = Q
So, AA ¢ is symmetric matrix for any matrix A.
Q. 30 Let A and B be square matrices of the order 3 ´ 3. Is ( AB)2 = A 2 B 2 ? Give
reasons.
Sol. Since, A and B are square matrices of order 3 ´ 3.
\ AB2 = AB × AB
= ABAB
= AABB [Q AB = BA]
= A 2 B2
2 2 2
So, AB = A B is true when AB = BA.

Q. 31 Show that, if A and B are square matrices such that AB = BA, then
( A + B)2 = A 2 + 2 AB + B 2.
Sol. Since, A and B are square matrices such that AB = BA.
\ ( A + B)2 = ( A + B) × ( A + B)
= A 2 + AB + BA + B2
= A 2 + AB + AB + B2 [Q AB = BA]
= A 2 + 2 AB + B2 Hence proved.

é 1 2ù é 4 0ù é2 0 ù
Q. 32 If A=ê ú ,B =ê ú ,C =ê ú , a = 4, and b = -2, then show
ë -1 3û ë1 5û ë1 -2û
that
(i) A + (B + C) = (A + B) + C
(ii) A (BC) = (AB) C
(iii) (a + b)B = aB + bB
(iv) a (C - A) = aC - aA
(v) (A T ) T = A
(vi) (bA) T = b A T
(vii) (AB) T = B T A T
(viii) (A - B)C = AC - BC
(ix) (A - B) T = A T - B T
é 1 2ù é 4 0ù
Sol. We have, A=ê ú, B = ê 1 5ú
ë -1 3û ë û
é2 0 ù
C=ê ú and a = 4, b = -2
ë 1 -2 û
é 1 2ù é 6 0ù é7 2ù
(i) A + (B + C ) = ê ú+ ê2 3ú = ê 1
ë -1 3û ë û ë 6úû
é5 2ù é2 0 ù
and ( A + B) + C = ê +
ë0 8úû ê 1 -2 ú
ë û
é7 2 ù
=ê ú = A + (B + C ) Hence proved.
ë 1 6û
 é 4 0ù é2 0 ù é 8 0 ù
(ii) (BC ) = ê úê ú= ê ú
ë 1 5û ë 1 -2 û ë7 -10û
é 1 2 ù é8 0 ù
and A (BC ) = ê úê ú
ë -1 3û ë7 -10û
é 8 + 14 0 - 20 ù é22 -20ù
=ê ú =ê ú
ë -8 + 21 0 - 30û ë13 -30û
é 1 2 ù é 4 0ù é 6 10ù
Also, ( AB) = ê ú× ê ú= ê ú
ë -1 3û ë 1 5û ë -1 15û
é6 10ù é2 0 ù
( AB) C = ê
ë -1 15úû êë 1 -2 úû
é22 -20ù
=ê = A (BC ) Hence proved.
ë13 -30úû
é 4 0ù
(iii) (a + b ) B = (4 - 2 ) ê ú [Q a = 4, b = -2]
ë 1 5û
é8 0 ù
=ê ú
ë2 10û
and aB+bB = 4B - 2B
é16 0 ù é 8 0 ù
=ê ú-ê ú
ë 4 20û ë2 10û
é8 0 ù
=ê ú
ë2 10û
= (a + b) B Hence proved.
é2 - 1 0 - 2 ù é -2 ù
1
(iv) (C - A) = ê ú = ê2 -5 ú
ë 1 + 1 - 2 - 3û ë û
é 4 -8 ù
and a(C - A) = ê ú [Q a = 4]
ë 8 -20û
é8 0 ù é 4 8 ù é 4 -8 ù
Also, aC - aA = ê ú-ê ú= ê 8 -20ú
ë 4 -8û ë -4 12 û ë û
= a (C - A) Hence proved.
T
é 1 2ù é 1 -1ù
(v) AT = ê ú =ê ú
ë -1 3û ë2 3 û
T
é 1 2ù
Now, ( AT )T = ê ú
ë -1 3û
=A Hence proved.
T
é -2 -4ù
(vi) (bA)T = ê ú [Q b = -2 ]
ë 2 -6û
é -2 2 ù
=ê ú
ë -4 -6û
é 1 -1ù
and AT = ê ú
ë2 3 û
é -2 2 ù
\ bAT = ê ú = (bA)
T
Hence proved.
ë -4 -6û
 é 1 2 ù é 4 0ù é 4 + 2 0 + 10ù é 6 10ù
(vii) AB = ê úê ú =ê ú=ê ú
ë -1 3û ë 1 5û ë -4 + 3 0 + 15û ë -1 15û
é 6 -1ù
\ ( AB) T = ê ú
ë10 15û
é 4 1ù é 1 -1ù é 6 -1ù
Now, BT AT = ê úê ú= ê ú
ë 0 5û ë2 3 û ë10 15û
T
= ( AB) Hence proved.
é 1 - 4 2 - 0ù é -3 2 ù
(viii) ( A - B) = ê ú= ê ú
ë -1 - 1 3 - 5û ë -2 -2 û
é -3 2 ù é2 0 ù é -4 -4ù
( A - B) C = ê úê ú= ê ú ...(i)
ë -2 -2 û ë 1 -2 û ë -6 4 û
é 1 2 ù é2 0 ù é 4 -4ù
Now, AC = ê ú ê ú = ê 1 -6ú ...(ii)
ë -1 3û ë 1 -2 û ë û
é 4 0ù é2 0 ù é 8 0 ù
and BC = ê ú ê ú= ê ú ...(iii)
ë 1 5û ë 1 -2 û ë7 -10û
é 4 - 8 -4 - 0 ù
\ AC - BC = ê ú [using Eqs. (ii) and (iii)]
ë 1 - 7 -6 + 10û
é -4 -4ù
=ê ú
ë -6 4 û
= ( A - B)C [using Eq. (i)] Hence proved.
T
é 1 - 4 2 - 0ù
(ix) ( A - B)T = ê ú
ë -1 - 1 3 - 5û
T
é -3 2 ù é -3 -2 ù
=ê ú =ê ú
ë - 2 - 2 û ë 2 -2 û
é 1 -1ù é 4 1ù
AT - BT = ê ú-ê ú
ë2 3 û ë 0 5û
é -3 -2 ù T
=ê ú = ( A - B) Hence proved.
ë 2 -2 û

é cos q sin q ù é cos 2q sin 2q ù

Q. 33 If A = ê ú , then show that A 2 = ê ú.
ë - sin q cos q û ë - sin 2q cos 2q û
é cos q sin q ù
Sol. We have, A = ê ú
ë - sin q cos q û
é cos q sin q ù é cos q sin q ù
\ A2 = A × A = ê ú× ê - sin q cos q ú
ë - sin q cos q û ë û
é 2 2
cos q - sin q cos q × sin q + sin q cos q ù
=ê ú
ë - sin q cos q - cos q sin q - sin2 q + cos 2 q û
é cos 2q 2 sin q cos q ù
=ê [Q cos q - sin2 q = cos2 q]
2

ë -2 sinq cos q cos 2q úû

é cos 2q sin2q ù
=ê ú [Q sin2 q = 2 sin q × cos q] Hence proved.
ë - sin2q cos 2q û
 é0 - x ù é0 1ù
Q. 34 If A=ê ú ,B =ê ú and x 2 = - 1, then show that
ëx 0 û ë1 0 û
( A + B)2 = A 2 + B 2.
é 0 -x ù é 0 1ù 2
Sol. We have, A=ê ú, B = ê 1 0ú and x = - 1
ë x 0 û ë û
é 0 - x + 1ù
\ ( A + B) = ê
ë x + 1 0 úû
é 0 - x + 1ù é 0 - x + 1ù
and ( A + B)2 = ê ú ê
ëx + 1 0 û ëx + 1 0 úû
é1 - x 2
0 ù
=ê ú …(i)
ë 0 1 - x2û
é0 -x ù é 0 -x ù é-x 2 0 ù
Also, A2 = A × A = ê ú ê ú =ê ú
ëx 0 û ë x 0 û ë 0 -x 2 û
é 0 1ù é 0 1ù é 1 0ù
and B2 = B × B = ê úê ú= ê ú
ë 1 0û ë 1 0û ë 0 1û
é-x + 1
2
0 ù é1 - x 2
0 ù
Now, A 2 + B2 = ê 2 ú= ê ú [using Eq. (i)]
ë 0 - x + 1û ë 0 1 - x2û
= ( A + B)2 Hence proved.

é0 1 -1ù
Q. 35 Verify that A = I, when A = êê4 -3 4 úú.
2

êë3 -3 4 úû
é 0 1 -1ù
Sol. We have, A = ê 4 -3 4 ú
ê ú
êë 3 -3 4 úû
é0 1 -1ù é 0 1 -1ù
\ A2 = ê4 -3 4 ú × ê 4 -3 4 ú [Q A 2 = A × A]
ê ú ê ú
êë 3 -3 4 úû êë 3 -3 4 úû
é1 0 0ù
= ê0 1 0ú = I Hence proved.
ê ú
êë 0 0 1úû

Q. 36 Prove by mathematical induction that ( A ¢) n = ( A n )¢ where n Î N for

any square matrix A.
Sol. Let P (n) : ( A ¢)n = ( A n )¢
\ P(1) : ( A)1 = ( A)¢
Þ A ¢ = A ¢ Þ P(1) is true.
Now, P(k ) : ( A ¢)k = ( A k )¢,
where k Î N
and P(k + 1) : ( A ¢)k + 1 = ( A k + 1 )¢
 where P (k+1) is true whenever P (k) is true.
\ P(k + 1) : ( A ¢)k . ( A ¢)1 = [ A k + 1 ]¢
( A k )¢. ( A)¢ = [ A k +1 ] ¢
( A × A k )¢ = [ A k +1 ]¢ [Q ( A ¢)k = ( A k )¢ and ( AB) = B¢A ¢]
( A k + 1 )¢ = [ A k + 1 ]¢ Hence proved.

Q. 37 Find inverse, by elementary row operations (if possible), of the

following matrices.
é 1 3ù é 1 - 3ù
(i) ê ú (ii) ê ú
ë-5 7û ë-2 6 û
K Thinking Process
To find the inverse of a matrix A, we know that A = IA is used for elementary row
operations. So, with the help of this method we can get the desired result.
é 1 3ù
Sol. (i) Let A = ê ú
ë -5 7 û
In order to use elementary row operations we may write A = IA.
é 1 3ù é 1 0ù
\ ê -5 7 ú = ê 0 1ú A
ë û ë û
é 1 3 ù é 1 0ù
Þ ê 0 22 ú = ê 5 1ú A [Q R 2 ® R 2 + 5R1 ]
ë û ë û
é1 3ù é 1 0 ù éQ R ® 1 R ù
Þ ê0 = A
ë 1úû êë 5 / 22 1 / 22 úû êë 2
22 2 úû
é1 0ù é7 / 22 -3 / 22 ù
Þ ê0 = A [Q R1 ® R1 - 3R 2 ]
ë 1úû êë 5 / 22 1 / 22 úû
0ù 1 é7 -3ù
é1
Þ = ê0 A
1úû 22 êë 5 1 úû
ë
Þ I = B A, where B is the inverse of A.
1 é7 -3ù
\ B=
22 êë 5 -1úû
é 1 -3ù
(ii) Let A = ê ú
ë -2 6 û
In order to use elementary row operations, we write A = IA
é 1 -3ù é 1 0ù
Þ ê -2 6 ú = ê 0 1ú A
ë û ë û
é 1 -3ù é 1 0ù
Þ ê 0 0 ú = ê2 1ú A [Q R 2 ® R 2 + 2 R1 ]
ë û ë û
Since, we obtain all zeroes in a row of the matrix A on LHS, so A -1 does not exist.
 é xy 4 ù é8 w ù
Q. 38 If ê ú =ê ú, then find the values of x, y, z and w.
ë z + 6 x + y û ë0 6 û
é xy 4 ù é 8 wù
Sol. We have, ê z + 6 x + yú = ê 0 6 ú
ë û ë û
By equality of matrix, x + y = 6 and xy = 8
Þ x = 6 - y and ( 6 - y) × y = 8
Þ y2 - 6 y + 8 = 0
Þ y2 - 4 y - 2 y + 8 = 0
Þ ( y - 2 ) ( y - 4) = 0
Þ y = 2 or y = 4
\ x = 6-2 = 4
or x = 6- 4 = 2 [Q x = 6 - y]
Also, z + 6 =0
Þ z = - 6 and w = 4
\ x = 2, y = 4 or x = 4, y = 2, z = -6 and w = 4

é1 5 ù é9 1ù
Q. 39 If A = ê ú and B = ê ú, then find a matrix C such that
ë7 12û ë7 8 û
3A + 5B + 2C is a null matrix.
é1 5 ù é 9 1ù
Sol. We have, A = ê ú and B = ê7 8ú
ë7 12 û ë û
é a bù
Let C=ê ú
ëC d û
\ 3A + 5B + 2C = 0
é 3 15 ù é 45 5 ù é2a 2bù é 0 0ù
Þ ê21 36ú + ê 35 40ú + ê2c 2dú = ê 0
ë û ë û ë û ë 0úû
é 48 + 2 a 20 + 2 b ù é 0 0ù
Þ ê 56 + 2c 76 + 2d ú = ê 0
ë û ë 0úû
Þ 2 a + 48 = 0 Þ a = - 24
Also, 20 + 2 b = 0 Þ b = - 10
56 + 2c = 0 Þ c = - 28
and 76 + 2d = 0 Þ d = - 38
é -24 -10ù
\ C=ê ú
ë -28 -38û

é 3 -5ù 2 3
Q. 40 If A = ê ú, then find A - 5 A - 14 I . Hence, obtain A .
ë - 4 2 û
Sol. é 3 -5ù
We have, A=ê ú …(i)
ë -4 2 û
é 3 -5ù é 3 -5ù
\ A2 = A × A = ê úê ú
ë -4 2 û ë -4 2 û
é 29 -25ù
=ê ú …(ii)
ë -20 24 û
 é 29 -25ù é 15 -25ù é14 0 ù
\ A 2 - 5 A - 14I = ê ú-ê ú-ê ú
ë -20 24 û ë -20 10 û ë 0 14û
é 0 0ù
=ê ú
ë 0 0û
Now, A 2 - 5 A - 14 I = 0
2
Þ A × A - 5 A × A - 14 A I = 0
Þ A 3 - 5 A 2 - 14 A = 0 [Q A I = A]
3 2
Þ A = 5 A = 14 A
é 29 -25ù é 3 -5ù
= 5ê ú + 14 ê -4 2 ú [using Eqs. (i) and (ii)]
ë -20 24 û ë û
é 145 - 125ù é 42 - 70 ù
=ê ú+ ê ú
ë -100 120 û ë -56 28 û
é 187 -195ù
=ê ú
ë -156 148 û

Q. 41 Find the values of a, b, c and d, if

éa b ù é a 6 ù é 4 a + bù
3ê ú =ê ú +ê z.
ëc d û ë -1 2 d û ëc + d 3 úû
Sol. We have,
éa bù é a 6 ù é 4 a + bù
3ê = +
ëc d úû êë -1 2d úû êëc + d 3 úû
é3 a 3 bù é a + 4 6 + a + bù
Þ ê 3c =
ë 3d úû êëc + d - 1 3 + 2 d úû
Þ 3a = a + 4 Þ a = 2;
3b = 6 + a + b
Þ 3b - b = 8 Þ b = 4;
3d = 3 + 2d Þ d = 3
and Þ 3c = c + d - 1
Þ 2c = 3 - 1 c = 1
\ a = 2 , b = 4, c = 1and d = 3

é 2 -1ù é -1 -8 -10 ù
Q. 42 Find the matrix A such that ê 1 0 ú A = êê 1 -2 -5 úú.
ê ú
êë -3 4 úû êë 9 22 15 úû
é 2 -1ù é -1 -8 -10ù
ê 1 0ú A = ê 1 -2 -5 ú
Sol. We have,
ê ú ê ú
êë -3 4 úû 3 ´ 2 êë 9 22 15 úû 3 ´ 3
From the given equation, it is clear that order of A should be 2 ´ 3.
é a b cù
Let A=ê ú
ëd e f û
 é 2 -1ù é -1 -8 -10ù
\ ê 1 0 ú é a b c ù = ê 1 -2 -5 ú
ê ú êëd e f úû ê ú
êë -3 4 úû êë 9 22 15 úû
é2 a - d 2b - e 2c - f ù é -1 - 8 - 10ù
Þ ê a + 0d b + 0×e c + 0× f ú = ê 1 - 2 -5 ú
ê ú ê ú
êë -3 a + 4d - 3b + 4e - 3c + 4f úû êë 9 22 15 úû
é 2a - d 2b - e 2c - f ù é - 1 - 8 - 10ù
Þ ê a b c ú= ê 1 -2 - 5ú
ê ú ê ú
êë - 3a + 4d - 3b + 4e - 3c + 4f úû êë 9 22 15 úû
By equality of matrices, we get
a = 1, b = - 2 , c = - 5
and 2a - d = - 1 Þ d = 2 a + 1 = 3;
Þ 2b - e = - 8 Þ e = 2(-2 ) + 8 = 4
2c - f = - 10 Þ f = 2c + 10 = 0
é 1 -2 -5ù
\ A=ê
ë3 4 0 úû

é1 2ù 2
Q. 43 If A = ê ú, then find A + 2A + 7I.
ë 4 1û
é1 2 ù
Sol. We have, A = ê ú
ë 4 1û
é1 2 ù é1 2 ù
\ A2 = ê ú ê ú [Q A 2 = A × A]
ë 4 1û ë 4 1û
é1 + 8 2 + 2 ù é 9 4ù
=ê ú=ê
ë4 + 4 8 + 1 û ë8 9úû
é 9 4ù é2 4ù é7 0ù é18 8 ù
\ A2 + 2 A + 7 I = ê ú+ ê ú+ ê =
ë 8 9û ë 8 2 û ë 0 7 úû êë16 18úû

é cos a sin a ù -1
Q. 44 If A = ê ú and A = A¢, then find the value of a.
ë - sin a cos a û
Sol. We have,
é cos a sin a ù écos a - sin aù
A=ê ú and A¢= êsin a
ë - sin a cos aû ë cos a úû
Also, A -1 = A ¢
Þ AA -1 = AA¢
é cos a sin a ù écos a - sin a ù
Þ I=ê ú ê sin a cos a ú
ë - sin a cos a û ë û
0ù écos 2 a + sin2 a
é1 0 ù
Þ ê0 =ê ú
ë1úû ê 0 2 2
sin a + cos a úû
ë
By using equality of matrices, we get
cos 2 a + sin2 a = 1
which is true for all real values of a.
 é0 a 3 ù
Q. 45 If matrix êê2 b -1úú is a skew-symmetric matrix, then find the values
êëc 1 0 úû
of a, b and c.
K Thinking Process
We know that, a matrix A is skew-symmetric matrix, if A¢= - A, so by using this we can
get the values of a, b and c.
é0 a 3 ù
Sol. Let A= ê2 b -1ú
ê ú
êëc 1 0 úû
Since, A is skew-symmetric matrix.
\ A¢= - A
é0 2 c ù é0 a 3ù
Þ ê a b 1 = - ê2
ú b -1ú
ê ú ê ú
êë 3 -1 0úû êëc 1 0 úû
é0 2 c ù é 0 - a -3ù
Þ ê a b 1ú = ê -2 - b +1ú
ê ú ê ú
êë 3 -1 0úû êë - c -1 0 úû
By equality of matrices, we get
a = -2, c = -3 and b= - b Þ b=0
\ a = -2, b=0 and c= -3

écos x sin x ù
Q. 46 If P(x) = ê ú, then show that P (x ) × P ( y) = P (x + y)
ë - sin x cos x û
= P ( y) × P (x ).
Sol. We have,
écos x sin x ù
P(x ) = ê ú
ë - sin x cos x û
écos y sin y ù
\ P( y) = ê ú
ë - sin y cos yû
écos x sin x ù écos y sin y ù
Now, P(x ) × P( y) = ê ú ê ú
ë - sin x cos x û ë - sin y cos yû
écos x × cos y - sin x × sin y cos x × sin y + sin x × cos y ù
=ê
ë - sin x × cos y - cos x × sin y - sin x × sin y + cos x × cos yúû
écos (x + y) sin (x + y) ù
=ê ...(i)
ë - sin ( x + y) co s (x + y)úû
éQ cos (x + y) = cos x × cos y - sin x × sin y ù
êand sin (x + y) = sin x × cos y + cos x × sin yú
ë û
écos (x + y) sin (x + y)ù
and P (x + y) = ê ...(ii)
ë - sin (x + y) cos (x + y)úû
 écos y sin y ù écos x sin x ù
Also, P( y) × P (x ) = ê ú ê - sin x cos x ú
ë - sin y cos yû ë û
écos y × cos x - sin y × sin x cos y × sin x + sin y × cos x ù
=ê
ë - sin y × cos x - sin x × cos y - sin y × sin x + cos y × cos x úû
écos (x + y) sin (x + y) ù
=ê ú ...(iii)
ë - sin (x + y) cos (x + y)û
Thus, we see from the Eqs. (i), (ii) and (iii) that,
P(x ) × P( y) = P(x + y) = P( y) × P(x ) Hence proved.

Q. 47 If A is square matrix such that A 2 = A, then show that

(I + A) 3 = 7 A + I .
Sol. Since, A 2 = A and (I+A) × (I+A)= I 2 + IA + AI + A 2
= I 2 + 2 AI + A 2
= I + 2 A + A = I + 3A
and (I + A) × (I + A)(I + A) = (I + A)(I + 3 A)
= I 2 + 3 AI + AI + 3 A 2
= I + 4 AI + 3 A
= I + 7A = 7A + I Hence proved.

Q. 48 If A, B are square matrices of same order and B is a skew-symmetric

matrix, then show that A¢BA is skew-symmetric.
Sol. Since, A and B are square matrices of same order and B is a skew-symmetric matrix i.e.,
B¢ = - B.
Now, we have to prove that A¢ BA is a skew-symmetric matrix.
\ A¢BA¢= A¢BA¢= BA¢ A¢ [Q AB¢= B¢ A¢]
= A¢B¢ A = A'- BA = - A'BA
Hence, A ¢BA is a skew-symmetric matrix.

Long Answer Type Questions

Q. 49 If AB = BA for any two square matrices, then prove by mathematical
induction that (AB) n = A n B n .
Sol. Let P(n) : ( AB)n = A n Bn
\ P(1) : ( AB)1 = A1B1 Þ AB = AB
So, P(1) is true.
Now, P (k ) : ( AB)k = A k Bk , k Î N
So, P(K ) is true, whenever P(k+1) is true.
\ P(K+1 : AB)k + 1 = A k + 1B k + 1 ...(i)
Þ ABk × AB1 [Q AB = BA]
Þ A k B k × BA Þ A k B k +1 A
Þ A k × A × B k +1 Þ A k + 1B k + 1
Þ ( A × B) k + 1 = A k + 1Bk + 1
So, P(k + 1) is true for all n Î N, whenever P(k) is true.
By mathematical induction (AB) = A n B n is true for all n Î N.
 é0 2 y zù
Q. 50 Find x, y and z, if A = êêx y -z ú satisfies A ¢ = A -1 .
ú
êë x - y z úû
é0 2 y z ù é0 x xù
Sol. We have, A = ê x y - zú and A ¢ = ê2 y y - yú
ê ú ê ú
êë x - y z úû êë z - z z úû
By using elementary row transformations, we get
A=I A
é 0 2 y z ù é 1 0 0ù
Þ ê x y - zú = ê 0 1 0ú A
ê ú ê ú
êë x - y z úû êë 0 0 1úû
é0 2 y z ù é1 0 0ù
Þ êx y - zú = ê 0 1 0ú A [Q R 3 ® R 3 - R 2 ]
ê ú ê ú
êë 0 -2 y 2 z úû êë 0 -1 1úû
é 0 2 y z ù é1 0 0ù
ê x 3 y 0 ú = ê1 1 éQ R 3 ® R 3 + R1 ù
Þ 0ú A êand R ® R + R ú
ê ú ê ú ë 2 2 1û
êë 0 0 3 zúû êë1 -1 1úû

é-x -y zù é 0 -1 0 ù éQ R1 ® R1 - R 2 ù
ê ú
êx 3 y 0ú = ê 1 ê
Þ
ê ú
1 0ú A 1 ú
1 -1 1 ú êand R 3 ® R 3 ú
êë 0 0 zúû ê ë 3 û
êë 3 3 3 úû
-1 -2 -1
é ù
é - x - y 0ù ê 3 3 3ú
Þ ê x 3 y 0ú = ê 1 1 0 úA [Q R1 ® R1 - R 3 ]
ê ú ê ú
êë 0 0 zúû ê 1 -1 1 ú
ë3 3 3 û
é -1 -2 -1ù
ê3 3 3ú
é - x - y 0ù ê ú
ê 0 2 y 0ú = ê 2 1 -1ú
Þ A [Q R 2 ® R 2 + R1 ]
ê ú ê3 3 3ú
êë 0 0 zúû ê ú
ê1 -1 1 ú
ê 3 úû
ë3 3
é -1 -1ù
ê0 2 2ú
é - x 0 0ù ê ú
ê 0 2 y 0ú = ê 2 1 -1ú éQ R ® R + 1 R ù
Þ A
ê ú ê3 3 3ú êë 1 1
2 2 úû
êë 0 0 zúû ê ú
ê1 -1 1 ú
ê3 3 úû
ë 3
é 1 1ù -1
é ù
ê ú
0 2x 2x êQ R1 ® x R1, ú
é 1 0 0ù ê ú
ê ú
ê 0 1 0ú = 1 ê 1 -1 ú
Þ êR ® 1 R ú
ê ú ê 3y úA 2 2
6y 6y ú ê 2y ú
êë 0 0 1úû ê ê
ê1 1 ú
-1 1 ú êand R 3 ® R 3 ú
ê ú ë z û
ë 3z 3z 3z û
 é 1ù
1
ê 2x ú
0 2x
ê ú é0 x xù
-1 ê1 1 -1 ú ê2 y y - yú
\ A =ê ú= ê ú
ê 3y6y 6y ú
êë z - z z úû
ê1 -1 1 ú
ê ú
ë 3z3z 3z û
1 1
Þ =x Þ =±
2x 2
1 1
Þ = y Þ y=±
6y 6
1 1
and = z Þ z=±
3z 3
Alternate Method
We have,
é0 2 y z ù é0 x xù
A = ê x y - zú and A¢= ê2 y y - yú
ê ú ê ú
êë x - y z úû êë z - z z úû
Also, A¢= A -1
Þ AA¢= AA -1 [Q AA -1 = I]
Þ AA¢ = I
é0 2 y z ù é 0 x x ù é 1 0 0ù
Þ ê x y - zú ê2 y y - yú = ê 0 1 0ú
ê úê ú ê ú
êë x - y z úû êë z - z z úû êë 0 0 1úû
é 4 y2 + z2 2 y2 - z2 -2 y2 + z2 ù é 1 0 0ù
ê 2 ú
Þ ê 2y - z
2 2 2
x + y + z 2
x 2 - y2 - z2 ú = ê 0 1 0ú
ê ú
ê -2 y + z2
2 2
x -y -z 2 2
x + y + z úû êë 0 0 1úû
2 2 2
ë
Þ 2 y2 - z2 = 0 Þ 2 y2 = z2
Þ 4 y2 + z2 = 1
Þ 2 × z2 + z2 = 1
1
z=±
3
2 z2 1
\ y = Þ y=±
2 6
Also, x 2 + y2 + z2 = 1
1 1
Þ x 2 = 1 - y2 - z2 = 1 - -
6 3
3 1
= 1- =
6 2
1
Þ x=±
2
1 1
\ x = ±, ,y = ±
2 6
1
and z=±
3
Q. 51 If possible, using elementary row transformations, find the inverse of
the following matrices.
é 2 -1 3ù é 2 3 - 3ù
(i) ê-5 3 1ú (ii) ê-1 -2 2 ú
ê ú ê ú
êë-3 2 3úû êë 1 1 -1úû
é2 0 -1ù
(iii) ê5 1 0 ú
ê ú
êë0 1 3 úû
Sol. For getting the inverse of the given matrix A by row elementary operations we may write the
given matrix as
A = IA
é 2 -1 3ù é 1 0 0ù
(i) Q ê -5 3 1ú = ê 0 1 0ú A
ê ú ê ú
êë -3 2 3úû êë 0 0 1úû
é 2 -1 3ù é1 0 0ù
Þ ê -3 2 4ú = ê 1 1 0ú A [Q R 2 ® R 2 + R1 ]
ê ú ê ú
êë -3 2 3úû êë 0 0 1úû
é 2 -1 3 ù é1 0 0ù
Þ ê -3 2 4 ú = ê 1 1 0ú A [Q R 3 ® R 3 - R 2 ]
ê ú ê ú
êë 0 0 -1úû êë -1 -1 1úû
é -1 1 7 ù é2 1 0ù
Þ ê -3 2 4 ú = ê 1 1 0ú A [Q R1 ® R1 + R 2 ]
ê ú ê ú
êë 0 0 -1úû êë -1 -1 1úû
é -1 1 7 ù é2 1 0ù
Þ ê 0 -1 -17 ú = ê -5 -2 0ú A [Q R 2 ® R 2 - 3R1 ]
ê ú ê ú
êë 0 0 -1 úû êë -1 -1 1úû
é - 1 0 - 10 ù é -3 -1 0ù
ê 0 -1 -17 ú = ê -5 -2 éQ R1 ® R1 + R 2 ù
Þ 0ú A êand R ® -1× R ú
ê ú ê ú ë 3 3û
êë 0 0 1 úû êë 1 1 -1úû
é -1 0 0ù é 7 9 -10 ù
ê 0 -1 0ú éQ R1 ® R1 + 10R 3 ù
Þ = ê12 15 -17 ú A êand R ® R + 17 R ú
ê ú ê ú ë 2 2 3û
êë 0 0 1úû êë 1 1 -1 úû
é 1 0 0ù é -7 -9 10 ù
ê 0 1 0ú éQ R1 ® -1R1 ù
Þ = ê -12 -15 17 ú A êand R ® -1R ú
ê ú ê ú ë 2 2û
êë 0 0 1úû êë 1 1 -1úû
é -7 -9 10 ù
So, the inverse of A is ê -12 -15 17 ú .
ê ú
êë 1 1 -1úû
 é2 3 -3ù é10 0ù
(ii) \ ê -1 -2 2 ú = ê 0 1 0ú A
ê ú ê ú
êë 1 1 -1úû êë 0
0 1úû
é 0 1 -1ù é10 -2 ù
ê 0 -1 1 ú = ê 0 éQ R 2 ® R 2 + R 3 ù
Þ 1 1úA êand R ® R - 2 R ú
ê ú ê ú ë 1 1 3û
êë 1 1 -1úû êë 0
0 1 úû
é 0 1 -1ù é 1 0 -2 ù
Þ ê 0 0 0 ú = ê2 1 -2 ú A [Q R 2 ® R 2 + R1 ]
ê ú ê ú
êë 1 1 1 úû êë 00 1 úû
Since, second row of the matrix A on LHS is containing all zeroes, so we can
say that inverse of matrix A does not exist.
é2 0 -1ù é1 0 0ù
(iii) \ ê5 1 0 ú = ê0 1 0ú A
ê ú ê ú
êë 0 1 3 úû êë 0 0 1úû
é 2 0 - 1ù é1 0 0ù
Þ ê 3 1 1 ú = ê -1 1 0ú A [Q R 2 ® R 2 - R1 ]
ê ú ê ú
êë 0 1 3 úû êë 0 0 1úû
é2 0 -1ù é1 0 0ù
ê 1 1 2 ú = ê -2 éQ R 2 ® R 2 - R1 ù
Þ 1 0ú A êand R ® R + R ú
ê ú ê ú ë 3 3 1û
êë2 1 2 úû êë 1 0 1úû
é2 0 - 1ù é 1 0 0ù
ê ú ê ú éQ R 3 ® R 3 + R1 ù
Þ ê 0 1 5 ú = ê -5 1 0ú A ê 1 ú
ê 2ú ê2 ú êand R 2 ® R 2 - R1 ú
ê ú ê ú ë 2 û
ë 4 1 1 û ë2 0 1û
é2 0 -1ù é 1 0 0ù
ê 5ú ê -5 ú
Þ ê 0 1 = 1 0ú A [Q R 3 ® R 3 - 2 R1 ]
2ú ê2
êë 0 1 3 úû êë 0 0 1úû
é ù é ù
ê2 0 -1ú ê 1 0 0ú
ê 5ú ê -5 ú
Þ ê 0 1 = 1 0ú A [Q R 3 ® R 3 - R 2 ]
2ú ê2
ê 1ú ê 5 ú
ê0 0 ú ê -1 1ú
ë 2û ë2 û
é 1 0 -1ù é 1 0 0 ù
ê 2ú ê2 ú
ê éQ R ® 1 R ù
5 ú ê -5 ú
ê
Þ ê0 1 ú=ê 1 0ú A 1
2 1 ú
ê 0 0 1 ú ê 25
2 êand R ® 2 R ú
-2 2 ú ë 3 3û
ê ú ê ú
ë û ë û
é 1 0 0ù é 3 -1 1 ù éQ R ® R + 1 R ù
ê 0 1 0ú = ê -15 6 -5ú A ê 1 1
2 3 ú
Þ ê
ê ú ê ú 5 ú
êë 0 0 1úû êë 5 -2 2 úû êand R 2 ® R 2 - R 3 ú
ë 2 û
é 3 -1 1ù
Hence, ê -15 6 -5ú is the inverse of given matrix A.
ê ú
êë 5 -2 2 úû
 é2 3 1ù
Q. 52 Express the matrix êê1 -1 2úú as the sum of a symmetric and a
êë4 1 2úû
skew-symmetric matrix.
K Thinking Process
We know that, any square matrix A can be expressed as the sum of a symmetric matrix
A + A¢ A - A¢
and skew-symmetric matrix, i.e., A = + , where A + A¢ and A - A¢ are a
2 2
symmetric matrix and a skew-symmetric matrix, respectively.
é2 3 1ù
Sol. We have, A = ê 1 -1 2 ú
ê ú
êë 4 1 2 úû
é2 1 4ù
\ A ¢ = ê 3 -1 1ú
ê ú
êë 1 2 2 úû
é2 2 5 ù
é 4 4 5ù ê 2ú
A + A¢ 1 ê ê 3ú
Now, = 4 -2 3ú = ê 2 -1 ú
2 2 ê ú
ê5 3 2 ú
êë 5 3 4úû
ê 2ú
ë2 2 û
é 0 1 -3 ù
é 0 2 -3ù ê 2 ú
A - A¢ 1 ê ú ê 1ú
and = -2 0 1 = ê -1 0 ú
2 2ê ú 2
êë 3 -1 0 úû ê 3 -1 ú
ê 0ú
ë2 2 û
é -3 ù
é2 2 5 ù ê 0 1
2 ú
ê 2ú ê ú
A + A¢ A - A¢ ê ú ê 1ú
\ + = ê 2 -1 3 ú + ê -1 0
2 2 ê5 3 2ú ê 2 ú
ú
ê 2 ú ê 3 -1 ú
ë2 2 û ê 0ú
ë2 2 û
which is the required expression.
Objective Type Questions
é0 0 4 ù
Q. 53 The matrix P = êê0 4 0 úú is a
êë4 0 0 úû
(a) square matrix (b) diagonal matrix
(c) unit matrix (d) None of these
Sol. (a) We know that, in a square matrix number of rows are equal to the number of columns,
é 0 0 4ù
so the matrix P = ê 0 4 0ú is a square matrix.
ë 4 0 0û

Q. 54 Total number of possible matrices of order 3 ´ 3 with each entry 2 or 0

is
(a) 9 (b) 27
(c) 81 (d) 512
Sol. (d) Total number of possible matrices of order 3 ´ 3 with each entry 2 or 0 is 2 9 i.e., 512.

é2x + y 4x ù é 7 7 y - 13ù
Q. 55 ê5x - 7 4x ú = ê y x + 6 ú , then the value of x + y is
ë û ë û
(a) x = 3, y = 1 (b) x = 2, y = 3
(c) x = 2, y = 4 (d) x = 3, y = 3
Sol. (b) We have, 4x = x + 6 Þ x = 2
and 4x = 7 y - 13 Þ 8 = 7 y - 13
Þ 7 y = 21 Þ y = 3
\ x+ y=2+ 3= 5

é -1 æ x öù é æ x öù
sin (xp) tan -1 ç ÷ ú ê - cos -1 (xp) tan -1 ç ÷ ú
ê è p øú
Q. 56 If A = 1 ê è p ø ú and B = 1 ê ,
p êsin -1 æ x ö cot -1 (px)ú p ê -1 æ x ö -1 ú
ç ÷
êë èpø úû êsin çè p ÷ø - tan (px)ú
ë û
then A - B is equal to
1
(a) I (b) 0 (c) 2I (d) I
2
é 1 sin-1 xp 1
tan-1
x ù
êp p p ú
Sol. (d) We have, A = ê 1 x 1 ú
ê sin-1 cot -1 px ú
ëp p p û
é -1cos -1 xp 1
tan-1 ù
x
êp p p ú
and B= ê
1 x - 1 -1 ú
ê sin-1 tan px ú
ëp p p û
 é1 1 æ -1 x x ù
-1 -1
ê p (sin xp + co s x p) ç tan - tan-1 ö÷ ú
pè p pø
\ A-B= ê ú
ê 1 æ sin-1 x - sin-1 x ö 1
cot -1 px + tan-1 px ú
êë p çè p pø
÷
p úû
é1 p éQ sin-1 x + cos -1 x = p ù
× 0 ù
êp 2 ú ê 2ú
=ê
1 p ú ê pú
ê0 × ú êand tan-1 x + cot -1 x = ú
ë p 2 û ë 2û
1 é1 0ù
= ê
2 ë 0 1úû
1
= I
2

Q. 57 If A and B are two matrices of the order 3 ´ m and 3 ´ n, respectively

and m = n, then order of matrix (5A - 2B) is
(a) m ´3 (b) 3 ´ 3
(c) m ´ n (d) 3 ´n
Sol. (d) A3 ´m and B3 ´n are two matrices. If m = n, then A and B have same orders as 3 ´ n each,
so the order of (5A - 2B) should be same as 3 ´ n.

é0 1ù 2
Q. 58 If A = ê ú, then A is equal to
ë 1 0 û
é0 1ù é1 0 ù é0 1ù é1 0ù
(a) ê ú (b) ê ú (c) ê ú (d) ê ú
ë1 0û ë1 0 û ë0 1û ë0 1û
é 0 1ù é 0 1ù é 1 0ù
Sol. (d) Q A2 = A × A = ê ú× ê ú =ê ú
ë 1 0û ë 1 0û ë 0 1û

Q. 59 If matrix A = [aij ] 2´2 , where aij = 1, if i ¹ j = 0 and if i = j, then A 2 is

equal to
(a) I (b) A
(c) 0 (d) None of these
Sol. (a) We have, A = [aij ] 2 ´ 2 , where a ij = 1, if i ¹ j = 0 and if i = j
é0 1ù
\ A=ê
ë1 0úû
é0 1ù é 0 1ù é 1 0ù
and A2 = ê = =I
ë1 0úû êë 1 0úû êë 0 1úû

é1 0 0 ù
Q. 60 The matrix êê0 2 0 úú is a
êë0 0 4 úû
(a) identity matrix (b) symmetric matrix
(c) skew-symmetric matrix (d) None of these
 é 1 0 0ù
Sol. (b) Let A = ê 0 2 0ú
ê ú
êë 0 0 4úû
é 1 0 0ù
\ A¢= ê 0 2 0ú = A
ê ú
êë 0 0 4úû
So, the given matrix is a symmetric matrix.
[since, in a square matrix A, if A¢ =A, then A is called symmetric matrix]

é 0 -5 8 ù
Q. 61 The matrix ê 5 0 12úú is a
ê
êë -8 -12 0 úû
(a) diagonal matrix (b) symmetric matrix
(c) skew-symmetric matrix (d) scalar matrix
Sol. (c) We know that, in a square matrix, if bij = 0, when i ¹ j , then it is said to be a diagonal
matrix. Here, b12, b13, ... ¹ 0, so the given matrix is not a diagonal matrix.
é 0 -5 8 ù
Now, B= ê 5 0 12 ú
ê ú
êë -8 -12 0 úû
é 0 5 -8 ù é 0 -5 8 ù
\ B¢ = ê -5 0 -12 ú = - ê 5 0 12 ú = - B
ê ú ê ú
êë 8 12 0 úû êë -8 -12 0 úû
So, the given matrix is a skew-symmetric matrix, since we know that in a square matrix
B, if B¢ = - B, then it is called skew-symmetric matrix.

Q. 62 If A is matrix of order m ´ n and B is a matrix such that AB¢ and B¢ A are

both defined, then order of matrix B is
(a) m ´ m (b) n ´ n (c) n ´ m (d) m ´ n
Sol. (d) Let A =[aij ]m´ n and B = [bij]p ´q
\ B¢= [b ji ]q ´ p
Now, AB¢ is defined, so n = q
and B'A is also defined, so p= m
\ Order of B¢= [b ji ] n ´ m
and order of B = [bij ] m ´ n

Q. 63 If A and B are matrices of same order, then (AB¢ - BA¢) is a

(a) skew-symmetric matrix (b) null matrix
(c)symmetric matrix (d) unit matrix
Sol. (a) We have matrices A and B of same order.
Let P = ( AB¢ - BA¢)
Then, P¢ = ( AB¢ - BA¢)¢ = (AB¢)¢ - (BA¢)¢
= ( B¢)¢(A)¢ - (A¢)¢ B¢ = BA¢ - AB¢
= -(AB¢ - BA¢) = - P
Hence, (AB¢ - BA¢) is a skew-symmetric matrix.
Q. 64 If A is a square matrix such that A 2 = I, then
(A - I) 3 + ( A + I ) 3 - 7A is equal to
(a) A (b) I - A (c) I + A (d) 3A
2
Sol. (a) We have, A = I
\ ( A - I )3 + ( A + I )3 - 7 A = [( A - I ) + ( A + I ) {( A - I )2
+ ( A + I )2 - ( A - I ) ( A + I )}] - 7 A
[Q a3 + b 3 = (a + b ) (a2 + b 2 - ab )]
= [(2 A) { A + I - 2 AI + A + I + AI - ( A 2 - I 2 )}] - 7 A
2 2 2 2

= 2 A [I + I 2 + I + I 2 - A 2 + I 2 ] - 7 A [Q A 2 = AI ]
= 2 A [5I - I ] - 7 A
= 8 AI - 7 AI [Q A = AI ]
= AI = A

Q. 65 For any two matrices A and B, we have

(a) AB = BA (b) AB ¹ BA (c) AB = O (d) None of these
Sol. (d) For any two matrices A and B, we may have AB = BA = I, AB ¹ BA and AB = O but it is
not always true.

Q. 66 On using elementary column operations C 2 ® C 2 - 2C 1 in the

é1 -3ù é1 -1ù é3 1ù
following matrix equation ê ú =ê ú ê2 4 ú, we have
ë2 4 û ë0 1 û ë û
é 1 -5ù é 1 -1ù é3 -5 ù é 1 -5ù é 1 -1ù é 3 -5ù
(a) ê ú=ê ú ê 0 úû
(b) ê ú=ê ú ê ú
ë 0 4 û ë -2 2 û ë 2 ë 0 4 û ë 0 1 û ë -0 2 û
é 1 -5 ù é 1 -3 ù é 3 1ù é 1 -5ù é 1 -1ù é3 -5ù
(c) ê ú=ê ú ê 4úû
(d) ê ú=ê ú ê ú
ë 2 0 û ë 0 1 û ë -2 ë2 0 û ë0 1 û ë2 0 û

é1 - 3ù é1 - 1ù é 3 1 ù
Sol. (d) Given that, ê ú=ê úê ú
ë2 4 û ë 0 1û ë2 4û
é1 - 5ù é1 - 1ù é 3 - 5ù
On using C 2 ® C 2 - 2C1, ê ú=ê úê ú
ë2 0 û ë 0 1û ë2 0 û

Since, on using elementary column operation on X = AB, we apply these operations

simultaneously on X and on the second matrix B of the product AB on RHS.

Q. 67 On using elementary row operation R 1 ® R 1 - 3R2 in the following

é4 2ù é1 2ù é2 0ù
matrix equation ê ú =ê ú ê 1úû
, we have
ë3 3û ë0 3û ë1
é -5 - 7ù é1 - 7 ù é2 0 ù é -5 - 7ù é1 2 ù é -1 - 3ù
(a) ê = (b) ê ú=ê
ë 3 3 úû êë0 3úû ê1 1 ú
ë û
ú ê
ë 3 3 û ë0 3û ë 1 1 û
ú

é -5 - 7ù é1 2ù é2 0 ù é 4 2ù é 1 2 ù é2 0 ù
(c) ê = (d) ê ú=ê
ë 3 3 úû êë1 - 7úû ê1 1 ú
ë û
ú ê ú
ë -5 - 7 û ë -3 - 3û ë1 1 û
 é 4 2 ù é 1 2 ù é2 0ù
Sol. (a) We have, ê ú=ê úê ú
ë 3 3û ë 0 3û ë 1 1û
Using elementary row operation R1 ® R1 - 3R 2 ,
é -5 - 7 ù é1 - 7 ù é2 0ù
ê 3 3 ú = ê 0 3 ú ê1 1 ú
ë û ë ûë û
Since, on using elementary row operation on X = AB, we apply these operation
simultaneously on X and on the first matrix A of the product AB on RHS.

Fillers
Q. 68 ……… matrix is both symmetric and skew-symmetric matrix.
Sol. Null matrix is both symmetric and skew-symmetric matrix.

Q. 69 Sum of two skew-symmetric matrices is always ……… matrix.

Sol. Let A is a given matrix, then (– A) is a skew-symmetric matrix.
Similarly, for a given matrix - B is a skew-symmetric matrix.
Hence, - A - B = - ( A + B)Þ sum of two skew-symmetric matrices is always
skew-symmetric matrix.

Q. 70 The negative of a matrix is obtained by multiplying it by ……… .

Sol. Let A is a given matrix.
\ - A = - 1[ A]
So, the negative of a matrix is obtained by multiplying it by - 1.

Q. 71 The product of any matrix by the scalar ......... is the null matrix.
Sol. The product of any matrix by the scalar 0 is the null matrix. i.e., 0 × A = 0.
[where, A is any matrix]

Q. 72 A matrix which is not a square matrix is called a ……… matrix.

Sol. A matrix which is not a square matrix is called a rectangular matrix. For example a
rectangular matrix is A = [aij ]m x n , where m ¹ n.

Q. 73 Matrix multiplication is …… over addition.

Sol. Matrix multiplication is distributive over addition.
e.g., For three matrices A, B and C,
(i) A (B + C ) = AB + AC
(ii) ( A + B) C = AC + BC
Q. 74 If A is a symmetric matrix, then A 3 is a ……… matrix.
3
Sol. If A is a symmetric matrix, then A is a symmetric matrix.
Q A¢ = A
\ ( A 3 )¢ = A¢3
= A3 [Q( A¢)n = ( A n )¢]

Q. 75 If A is a skew-symmetric matrix, then A 2 is a ……… .

Sol. If A is a skew-symmetric matrix, then A 2 is a symmetric matrix.
Q A¢ = - A
\ ( A 2 )¢ = ( A¢)2
= (- A)2 [Q A¢ = - A]
2
= A
So, A 2 is a symmetric matrix.

Q. 76 If A and B are square matrices of the same order, then

(i) (AB)¢ = ………
(ii) (kA)¢ = ……… (where, k is any scalar)
(iii) [k (A - B)]¢ = ………
Sol. (i) ( AB)¢ = B¢ A¢
(ii) (kA)¢ = k A¢
(iii) [k( A - B)]¢ = k( A¢ - B¢)

Q. 77 If A is a skew-symmetric, then kA is a ……… (where, k is any scalar).

Sol. If A is a skew-symmetric, then kA is a skew-symmetric matrix (where, k is any scalar).
[Q A¢ = - A Þ (kA)¢ = k( A)¢= - (kA)]

Q. 78 If A and B are symmetric matrices, then

(i) AB - BA is a ………
(ii) BA - 2AB is a ………
Sol. (i) AB - BA is a skew-symmetric matrix.
Since, [ AB - BA]¢ = ( AB)¢ - (BA)¢
= B¢ A¢ - A¢B¢ [Q( AB)¢ = B¢ A¢]
= BA - AB [Q A¢ = A and B¢ = B ]
= - [ AB - BA]
So, [ AB - BA] is a skew-symmetric matrix.
(ii) [BA - 2 AB] is a neither symmetric nor skew-symmetric matrix.
\ (BA - 2 AB)¢ = (BA)¢ - 2( AB)¢
= A¢B¢ - 2 B¢ A¢
= AB - 2 BA
= - (2BA - AB)
So, [BA - 2 AB] is neither symmetric nor skew-symmetric matrix.
Q. 79 If A is symmetric matrix, then B¢AB is ………
Sol. If A is a symmetric matrix, then B¢ AB is a symmetric metrix.
Q [B¢ AB]¢ = [B¢( AB)]¢
= ( AB)¢ (B¢)¢ [Q( AB)¢ = B¢ A¢]
= B¢ A¢ B
= [B¢ A¢ B] [Q A¢ = A]
So, B¢ AB is a symmetric matrix.

Q. 80 If A and B are symmetric matrices of same order, then AB is symmetric

if and only if……… .
Sol. If A and B are symmetric matrices of same order, then AB is symmetric if and only if
AB = BA.
\ ( AB)¢
= B¢ A¢= BA
= AB [QAB = BA]

Q. 81 In applying one or more row operations while finding A -1 by

elementary row operations, we obtain all zeroes in one or more, then
A - 1 ……… .
Sol. In applying one or more row operations while finding A - 1 by elementary row operations, we
obtain all zeroes in one or more, then A -1 does not exist.

True/False
Q. 82 A matrix denotes a number.
Sol. False
A matrix is an ordered rectangular array of numbers or functions.

Q. 83 Matrices of any order can be added.

Sol. False
Two matrices are added, if they are of the same order.

Q. 84 Two matrices are equal, if they have same number of rows and same
number of columns.
Sol. False
If two matrices have same number of rows and same number of columns, then they are said
to be square matrix and if two square matrices have same elements in both the matrices,
only then they are called equal.

Q. 85 Matrices of different order cannot be subtracted.

Sol. True
Two matrices of same order can be subtracted
Q. 86 Matrix addition is associative as well as commutative.
Sol. True
Matrix addition is associative as well as commutative i.e.,
( A + B) + C = A + (B + C ) and A + B = B + A, where A, B and C are matrices of same order.

Q. 87 Matrix multiplication is commutative.

Sol. False
Since, AB ¹ BA is possible when AB and BA are both defined.

Q. 88 A square matrix where every element is unity is called an identity

matrix.
Sol. False
Since, in an identity matrix, the diagonal elements are all one and rest are all zero.

Q. 89 If A and B are two square matrices of the same order, then

A + B = B + A.
Sol. True
Since, matrix addition is commutative i.e., A + B = B + A, where A and B are two square
matrices.

Q. 90 If A and B are two matrices of the same order, then A - B = B - A.

Sol. False
Since, the addition of two matrices of same order are commutative.
\ A + (- B) = A - B = - [B - A] ¹ B - A

Q. 91 If matrix AB = 0, then A = 0 or B = 0 or both A and B are null matrices.

Sol. False
Since, for two non-zero matrices A and B of same order, it can be possible that A × B = 0 =
null matrix

Q. 92 Transpose of a column matrix is a column matrix.

Sol. False
Transpose of a column matrix is a row matrix.

Q. 93 If A and B are two square matrices of the same order, then AB = BA.
Sol. False
For two square matrices of same order it is not always true that AB = BA.

Q. 94 If each of the three matrices of the same order are symmetric, then
their sum is a symmetric matrix.
Sol. True
Let A, B and C are three matrices of same order
\ A¢ = A, B¢= B and C ¢= C
\ ( A + B + C )¢ = A¢ + B¢ + C ¢
= ( A + B + C)
Q. 95 If A and B are any two matrices of the same order, then ( AB)¢ = A ¢B ¢.
Sol. False
Q ( AB)¢ = B¢ A¢

Q. 96 If ( AB)¢ = B ¢ A ¢, where A and B are not square matrices, then number

of rows in A is equal to number of columns in B and number of
columns in A is equal to number of rows in B.
Sol. True
Let A is of order m ´n and B is of order p ´ q .
Since, ( AB)¢ = B¢ A¢
\ A( m ´n ) B ( p ´ q ) is defined Þ n = p ...(i)
and AB is of order m ´ q .
Þ ( AB)¢is of order q ´ m ...(ii)
Also, B¢ is of order q ´ p and A¢ is of order n ´ m
\ B¢ A¢is defined Þ p = n
and B¢ A¢is of order q ´ m. ...(iii)
Also, equality of matrices ( AB)¢ = B¢ A¢, we get the given statement as true.
e.g., If A is of order (3 ´ 1) and B is of order (1 ´ 3), we get
Order of ( AB)¢ = Order of (B¢ A¢) = 3 ´ 3

Q. 97 If A, B and C are square matrices of same order, then AB = AC always

implies that B = C .
Sol. False
If AB = AC = 0, then it can be possible that B and C are two non-zero matrices such that
B ¹ C.
\ A× B = 0 = A×C
é 1 0ù é 0 0ù
Let A=ê ú , B = ê 1 3ú
ë 0 0û ë û
é 0 0ù
and C=ê ú
ë 3 1û
é1 0ù é 0 0ù é 0 0ù
\ AB = ê =
ë0 0úû êë 1 3úû êë 0 0úû
é1 0ù é 0 0ù é 0 0ù
and AC = ê × =
ë0 0úû êë 3 1úû êë 0 0úû
Þ AB = AC but B ¹ C

Q. 98 AA¢ is always a symmetric matrix for any matrix A.

Sol. True
Q [ AA ¢]¢ = ( A ¢)¢ A ¢ = [ AA ¢]
 2 3
2 3 -1
Q. 99 If A= and B = 4 5 , then AB and BA are defined and
1 4 2
2 1
equal.
Sol. False
Since, AB is defined.
é2 3ù
é2 3 -1ù ê 4 5ú = é14 20ù
\ AB = ê ú
ë1 4 2 û ê ú êë22 25úû
êë2 1úû
Also, BA is defined.
é2 3ù
é2 3 -1ù
\ BA = ê 4 5ú ê
ê ú ë 1 4 2 úû
êë2 1úû
é 7 18 4ù
= ê13 32 6ú
ê ú
êë 5 10 0úû
\ AB ¹ BA

Q. 100 If A is skew-symmetric matrix, then A 2 is a symmetric matrix.

Sol. True
Q [ A 2 ]¢ = [ A¢]2
= [- A]2 [Q A ¢ = - A]
= A2
Hence, A 2 is symmetric matrix.

Q. 101 ( AB) -1 = A -1 × B -1 , where A and B are invertible matrices satisfying

commutative property with respect to multiplication.
Sol. True
We know that, if A and B are invertible matrices of the same order, then
( AB)-1 = (BA)-1 [QAB = BA]
-1 -1
Here, ( AB) = ( AB)
Þ B-1 A -1 = A -1B-1
[since, A and B are satisfying commutative property with respect to multiplications].
 4
Determinants
Short Answer Type Questions
x2 - x + 1 x - 1
Q. 1
x +1 x +1
x2 - x + 1 x - 1 x 2 - 2x + 2 x - 1
Sol. We have, = [Q C1 ® C1 - C 2 ]
x+1 x+1 0 x +1
= (x 2 - 2 x + 2 ) × (x + 1) - (x - 1) × 0
= x 3 - 2x 2 + 2x + x 2 - 2x + 2
= x3 - x2 + 2

a+x y z
Q. 2 x a + y z
x y a+z
a+ x y z a -a 0
é Q R1 ® R1 - R 2 ù
Sol. We have, x a+ y z = 0 a -a êand R ® R - R ú
ë 2 2 3û
x y a+ z x y a+ z
a 0 0
= 0 a -a [Q C 2 ® C 2 + C1 ]
x x + y a+ z
= a (a2 + az + ax + ay)
= a2 (a + z + x + y)
 0 xy 2 xz 2
Q. 3 x2 y 0 yz 2
x2 z zy 2 0

0 xy2 xz2 0 x x
2
Sol. We have, x y 0 yz2 = x 2 y2 z2 y 0 y
x 2 z zy2 0 z z 0
[taking x 2 , y2 and z2 common from C1, C 2 and C 3 , respectively]
0 0 x
= x 2 y2 z2 y - y y [Q C 2 ® C 2 - C 3 ]
z z 0
= x 2 y2 z2 [x ( yz + yz)]
= x 2 y2 z2 × 2 xyz = 2 x 3 y3 z3

3x -x + y -x + z
Q. 4 x - y 3y z-y
x-z y -z 3z
3x -x + y -x + z
Sol. We have, x- y 3y z- y
x - z y- z 3z
Applying, C1 ® C1 + C 2 + C 3 ,
x + y + z -x + y -x + z
= x + y+ z 3y z- y
x + y+ z y- z 3z
1 -x + y -x + z
= (x + y + z) 1 3y z- y
1 y- z 3z
[taking (x + y + z) common from column C1]
1 -x + y -x + z
= (x + y + z) 0 2 y + x x - y
0 x - z 2z + x
[Q R 2 ® R 2 - R1 and R 3 ® R 3 - R1 ]
Now, expanding along first column, we get
(x + y + z) × 1 [(2 y + x ) (2 z + x ) - (x - y) (x - z)]
= (x + y + z) (4 yz + 2 yx + 2 xz + x 2 - x 2 + xz + yx - yz )
= (x + y + z) (3 yz + 3 yx + 3xz)
= 3 (x + y + z)( yz + yx + xz)
 x +4 x x
Q. 5 x x + 4 x
x x x +4
x+ 4 x x 2x + 4 2x + 4 2x
Sol. We have, x x+ 4 x = x x+ 4 x [Q R1 ® R1 + R 2 ]
x x x+ 4 x x x+ 4
2x 2x 2x 4 4 0
= x x+ 4 x + x x+ 4 x
x x x+ 4 x x x+ 4
[here, given determinant is expressed in sum of two determinants]
1 1 1 1 1 0
= 2x x x + 4 x + 4 x x+ 4 x
x x x+ 4 x x x+ 4
[taking 2x common from first row of first determinant and 4 from first row of second
determinant]
Applying C1 ® C1 - C 3 and C 2 ® C 2 - C 3 in first and applying C1 ® C1 - C 2 in second, we
get
0 0 1 0 1 0
= 2x 0 4 x + 4 -4 x+ 4 x
-4 -4 x+ 4 0 x x+ 4
Expanding both the along first column, we get
2 x [- 4 (- 4)] + 4 [4 (x + 4 - 0)]
= 2 x ´ 16 + 16 (x + 4)
= 32 x + 16x + 64
= 16 (3x + 4)

a -b -c 2a 2a
Q. 6 2b b -c -a 2b
2c 2c c -a -b
Sol. Wehave, a - b - c 2a 2a
2b b-c - a 2b
2c 2c c-a-b
a+ b+c a+ b+c a+ b+c
= 2b b-c - a 2b [Q R1 ® R1 + R 2 + R 3 ]
2c 2c c-a-b
1 1 1
= (a + b + c ) 2 b b - c - a 2b
2c 2c c-a-b
[taking (a + b + c )common from the first row]
0 0 1
= (a + b + c ) 0 - (a + b + c ) 2b
(a + b + c ) (a + b + c ) ( c - a - b)
[Q C1 ® C1 - C 3 and C 2 ® C 2 - C 3 ]
 Expanding along R1,
= (a + b + c ) [1{0 + (a + b + c 2 }]
= (a + b + c ) [(a + b + c )2 ]
= (a + b + c )3

y 2z2 yz y +z
Q. 7 z 2 x 2 zx z + x =0
x2 y 2 xy x+y

Sol. We have to prove,

y2 z2 yz y+ z
z2 x 2 zx z+ x =0
x 2 y2 xy x+ y
2 2
y z yzy+ z x y2 z2 x yz x y + x z
1
\ LHS = z2 x 2 zx z + x = x 2 yz2 x yz yz + x y
xyz 2 2
x 2 y2 xy x + y x y z x yz x z + yz
[Q R1 ® x R1, R 2 ® y R 2 , R 3 ® z R 3 ]
yz 1 x y + x z
1
= (x yz)2 x z 1 yz + x y
x yz
x y 1 x z + yz
[taking (xyz) common from C1 and C 2 ]
yz 1 xy + yz + zx
= xyz xz 1 xy + yz + zx [C 3 ® C 3 + C1 ]
xy 1 xy + yz + zx
yz 1 1
= xyz (xy + yz + zx ) xz 1 1
xy 1 1
[taking (x y + yz + zx ) common from C 3 ]
=0 [since, C 2 and C 3 are identicals]
= RHS Hence proved.

y +z z y
Q. 8 z z+x x = 4 xyz
y x x+y
K Thinking Process
First in LHS use C 1® C 1 + C2 + C3 and then by using C 1® C 1 - C2 and R1® R1 - R3 , we
can get two zeroes in column 1 and then by simplification we will get the desired result.
Sol. We have to prove,
y+ z z y
z z+ x x = 4 x yz
y x x+ y
 y+ z z y
\ LHS = z z+ x x
y x x+ y
y+ z+ z+ y z y
= z+ z+ x + x z+ x x [Q C1 ® C1 + C 2 + C 3 ]
y+ x + x + y x x+ y
( y + z) z y
= 2 (z + x) z + x x [taking 2 common from C1]
(x + y) x x+ y
y z y
=2 0 z+ x x [Q C1 ® C1 - C 2 ]
y x x+ y
0 z-x -x
=2 0 z+ x x [Q R1 ® R1 - R 3 ]
y x x+ y
= 2 [ y(x z - x 2 + x z + x 2 )]
= 4x yz = RHS Hence proved.

a 2 + 2a 2a + 1 1
Q. 9 2a + 1 a + 2 1 = (a - 1) 3
3 3 1

K Thinking Process
Here, by using R1® R1 - R2 and R2 ® R2 - R3 in LHS, we can easily get the desired result.
Sol. We have to prove,
a2 + 2 a 2 a + 1 1
= 2 a + 1 a + 2 1 = (a - 1)3
3 3 1
a2 + 2 a 2 a + 1 1
\ LHS = 2 a + 1 a + 2 1
3 3 1
a2 + 2 a - 2 a - 1 2 a + 1 - a - 2 0
= 2a + 1 - 3 a+2-3 0
3 3 1
[Q R1 ® R1 - R 2 and R 2 ® R 2 - R 3 ]
(a - 1) (a + 1) (a - 1) 0 (a + 1) 1 0
= 2 (a - 1) (a - 1) 0 = (a - 1)2 2 1 0
3 3 1 3 3 1
[taking (a - 1) common from R1 and R 2 each]
= (a - 1)2 [1 (a + 1) - 2 ] = (a - 1)3
= RHS Hence proved.
 1 cos C cos B
Q. 10 If A + B + C = 0, then prove that cos C 1 cos A = 0.
cos B cos A 1
K Thinking Process
We have, given A + B + C = 0, so on solving the determinant by expansion, we can use
cos (A + B) = cos (- C) and similarly after simplification this expansion we will get the
desired result.
1 cos C cos B
Sol. We have to prove, cos C 1 cos A = 0
cos B cos A 1
1 cos C cos B
\ LHS = cos C 1 cos A
cos B cos A 1
= 1 (1 - cos 2 A) - cos C (cos C - cos A × cos B) + cos B (cos C × cos A - cos B)
= sin2 A - cos 2 C + cos A × cos B × cos C + cos A × cos B × cos C - cos 2 B
= sin2 A - cos 2 B + 2 cos A × cos B × cos C - cos 2 C
= - cos ( A + B) × cos ( A - B) + 2 cos A × cos B × cos C - cos 2 C
[Q cos 2 B - sin2 A = cos ( A + B) × cos ( A - B)]
= - cos (- C ) × cos ( A - B) + cos C (2 cos A × cos B - cos C ) [Q cos (- q) = cos q]
= - cos C (cos A × cos B + sin A × sin B - 2 cos A × cos B + cos C )
= cos C (cos A × cos B - sin A × sin B - cos C )
= cos C [cos ( A + B) - cos C ]
= cos C (cos C - cos C ) = 0 = RHS Hence proved.

Q. 11 If the coordinates of the vertices of an equilateral triangle with sides

of length ‘a’ are (x 1 , y 1 ),(x2 , y 2 ) and (x 3 , y 3 ), then
2
x1 y1 1
3a 4
x2 y2 1 = .
4
x3 y3 1
Sol. Since, we know that area of a triangle with vertices (x1, y1 ), (x 2 , y2 ) and (x 3 , y3 ), is given by
x1 y1 1
1
D= x 2 y2 1
2
x 3 y3 1
2
x1 y1 1
2 1
Þ D = x2 y2 1 ...(i)
4
x3 y3 1
We know that, area of an equilateral triangle with side a,
1 æ 3ö 2 3 2
D = çç ÷a = a
2 è 2 ÷ø 4
3 4
Þ D2 = a ...(ii)
16
 2
x1 y1 1
3 4 1
From Eqs. (i) and (ii), a = x2 y2 1
16 4
x3 y3 1
2
x1 y1 1
3 4
Þ x2 y2 1 = a Hence proved.
4
x3 y3 1

é 1 1 sin 3 q ù
ê
Q. 12 Find the value of q satisfying ê - 4 3 cos 2 q úú = 0
êë 7 - 7 - 2 úû
1 1 sin 3 q
Sol. We have, -4 3 cos 2 q = 0
7 -7 -2
0 1 sin 3 q
Þ -7 3 cos 2 q = 0 [Q C1 ® C1 - C 2 ]
14 - 7 -2
0 1 sin 3 q
Þ 7 -1 3 cos 2 q = 0 [taking 7 common from C1]
2 -7 -2
Þ 7 [0 - 1 ( 2 - 2 cos 2 q) + sin 3 q (7 - 6)] = 0 [expanding along R1]
Þ 7 [- 2 (1 - cos 2 q) + sin 3 q] = 0
Þ - 14 + 14 cos 2 q + 7 sin 3 q = 0
Þ 14 cos 2 q + 7 sin 3 q = 14
Þ 14 (1 - 2 sin2 q) + 7 (3 sin q - 4 sin3 q) = 14
Þ - 28 sin2 q + 14 + 21sin q - 28 sin3 q = 14
Þ - 28 sin2 q - 28 sin3 q + 21sin q = 0
Þ 28 sin3 q + 28 sin2 q - 21sin q = 0
Þ 4 sin3 q + 4 sin2 q - 3 sin q = 0
Þ sin q (4 sin2 q + 4 sin q - 3) = 0
Þ Either sin q = 0,
Þ q = np or 4 sin2 q + 4 sin q - 3 = 0
-4± 16 + 48 - 4 ± 64
\ sin q = =
8 8
- 4 ± 8 4 - 12
= = ,
8 8 8
1 -3
sin q = ,
2 2
1 p
If sin q = = sin , then
2 6
p
q = n p + (- 1)n
6
-3
Hence, sin q = [not possible because - 1 £ sin q £ 1]
2
 é4 - x 4 + x 4 + x ù
Q. 13 If êê4 + x 4 - x 4 + x úú = 0, then find the value of x.
êë4 + x 4 + x 4 - x úû
4-x 4+ x 4+ x
Sol. Given, 4+ x 4-x 4+ x =0
4+ x 4+ x 4-x

12 + x 12 + x 12 + x
Þ 4+ x 4-x 4+ x =0 [Q R1 ® R1 + R 2 + R 3 ]
4+ x 4+ x 4-x
1 1 1
Þ (12 + x ) 4 + x 4-x 4+ x =0 [taking (12 + x ) common from R1]
4+ x 4+ x 4-x
0 0 1
Þ (12 + x ) 0 8 4+ x =0 [Q C1 ® C1 - C 3 and C 2 ® C 2 + C 3 ]
2x 8 4-x
Þ (12 + x ) [1× (- 16x )] = 0
Þ (12 + x ) (- 16x ) = 0
\ x = - 12, 0

Q. 14 If a 1 , a2 , a 3 , ..., a r are in GP, then prove that the determinant

ar + 1 ar + 5 ar + 9
a r + 7 a r + 11 a r + 15 is independent of r.
a r + 11 a r + 17 a r + 21

K Thinking Process
We know that, nth term of a GP has value arn - 1, where a = first term and r = common
ratio. So, by using this result, we can prove the given determinant as independent of r.
Sol. We know that, ar +1 = AR (r + 1) - 1
= AR r
where r = r th term of a GP, A = First term of a GP and R = Common ratio of GP
ar + 1 ar + 5 ar + 9
We have, ar + 7 ar + 11 ar + 15
ar + 11 ar + 17 ar + 21

AR r AR r + 4 AR r + 8
r + 6
= AR AR r + 10 AR r + 14
+ 10
AR r AR r + 16 AR r + 20

1 AR 4 AR 8
r + 6 r + 10
r
= AR × AR × AR 1 AR 4 AR 8
1 AR 6 AR10
+ 6 + 10
[taking AR r , AR r and AR r common from R1, R 2 and R 3 , respectively]
= 0 [since, R1 and R 2 are identicals]
Q. 15 Show that the points (a + 5, a - 4), (a - 2, a + 3) and (a, a) do not lie on
a straight line for any value of a.
K Thinking Process
We know that, if three points lie in a straight line, then area formed by these points will
be equal to zero. So, by showing area formed by these points other than zero, we can
prove the result.
Sol. Given, the points are (a + 5, a - 4), (a - 2, a + 3) and (a, a).
a+ 5 a-4 1
1
\ D= a-2 a+ 3 1
2
a a 1
5 -4 0
1
=
-2 3 0 [Q R1 ® R1 - R 3 and R 2 ® R 2 - R 3 ]
2
a a 1
1
= [1 (15 - 8)]
2
7
Þ = ¹0
2
Hence, given points form a triangle i.e., points do not lie in a straight line.

Q. 16 Show that DABC is an isosceles triangle, if the determinant

1 1 1
D= 1 + cos A 1 + cos B 1 + cos C = 0.
2 2 2
cos A + cos A cos B + cos B cos C + cos C
1 1 1
Sol. We have, D = 1 + cos A 1 + cos B 1 + cos C =0
cos 2 A + cos A cos 2 B + cos B cos 2 C + cosC
0 0 1
D= cos A - cos C cos B - cos C 1 + cos C =0
cos 2 A + cos A - cos 2 C - cos C cos 2 B + cos B - cos 2 C - cos C cos 2 C + cos C
[\ C1 ® C1 - C 3 and C 2 ® C 2 - C 3 ]
Þ (cos A - cos C ) × (cos B - cos C )
0 0 1
1 1 1 + cos C =0
cos A + cos C + 1 cos B + cos C + 1 cos 2 C + cos C
[taking (cos A - cos C ) common from C1 and (cos B - cos C ) common from C 2 ]
Þ (cos A - cos C ) × (cos B - cos C ) [(cos B + cos C + 1) - (cos A + cos C + 1)] = 0
Þ (cos A - cos C ) × (cos B - cos C ) (cos B + cos C + 1 - cos A - cos C - 1) = 0
Þ (cos A - cos C ) × (cos B - cos C ) (cos B - cos A) = 0
i.e., cos A = cos C or cos B = cos C or cos B = cos A
Þ A = C or B = C or B = A
Hence, ABC is an isosceles triangle.
 0 1 1
A 2 - 3I
Q. -1
17 Find A , if A = 1 0 1 and show that A - 1 = .
2
1 1 0
0 1 1
Sol. We have, A= 1 0 1
1 1 0

\ A11 = - 1, A12 = 1, A13 = 1, A21 = 1, A22 = - 1, A23 = 1, A31 = 1, A32 = 1 and A33 = - 1
T
-1 1 1 -1 1 1
\ adj A = 1 - 1 1 = 1 -1 1
1 1 -1 1 1 -1
and | A| = - 1 (- 1) + 1× 1 = 2
é- 1 1 1ù
adj A 1 ê
\ A- 1 = = 1 -1 1ú ...(i)
| A| 2 ê ú
êë 1 1 - 1úû
é 0 1 1ù é 0 1 1ù é2 1 1ù
and A = ê 1 0 1ú × ê 1 0 1ú = ê 1 2 1ú
2
...(ii)
ê ú ê ú ê ú
êë 1 1 0úû êë 1 1 0úû êë 1 1 2 úû
ì2 1 1 3 0 0ü -1 1 1
A2 - 3 I 1 ï ï 1
\ = í 1 2 1 - 0 3 0ý= 1 -1 1
2 2ï 2
î 1 1 2 0 0 3 ïþ 1 1 -1
-1
= A [using Eq. (i)]
Hence proved.

Long Answer Type Questions

0 1 2
Q. 18 If A = - 2 - 1 - 2 , then find the value of A -1 .
0 -1 1
Using A -1 , solve the system of linear equations x - 2 y = 10,
2x - y - z = 8 and - 2 y + z = 7.
1 2 0
Sol. We have, A= -2 -1 -2 ...(i)
0 -1 1
\ A = 1 (- 3) - 2 (-2 ) + 0 = 1 ¹ 0
Now, A11 = - 3, A12 = 2, A13 = 2, A21 = - 2, A22 = 1, A23 = 1, A31 = - 4, A32 = 2 and A33 = 3
T
-3 2 2 -3 -2 -4
\ adj ( A) = - 2 1 1 = 2 1 2
-4 2 3 2 1 3
 adj A
\ A -1 =
| A|
-3 -2 -4
1
= 2 1 2
1
2 1 3
-3 -2 -4
Þ A- 1 = 2 1 2 ...(ii)
2 1 3
Also, we have the system of linear equations as
x - 2 y = 10,
2x - y - z = 8
and - 2y + z = 7
In the form of CX = D,
é1 2 0 ù é x ù é 10 ù
ê 2 - 1 - 1ú ê y ú = ê 8 ú
ê úê ú ê ú
êë 0 - 2 1 úû êë z úû êë 7 úû
é1 - 2 0ù é xù é 10ù
where, C = ê2 - 1 - 1ú, X = ê y ú and D = ê 8 ú
ê ú ê ú ê ú
êë 0 - 2 1 úû êë z úû êë 7 úû
T -1 -1 T
We know that, (A ) = (A )
1 2 0
\ CT = - 2 - 1 - 2 = A [using Eq. (i)]
0 -1 1
\ X = C- 1 D
é x ù é - 3 2 2 ù é 10ù
Þ ê y ú = ê - 2 1 1ú ê 8 ú
ê ú ê úê ú
êë z úû êë - 4 2 3úû êë 7 úû
é - 30 + 16 + 14ù é 0 ù
= ê - 20 + 8 + 7 ú = ê - 5ú
ê ú ê ú
êë - 40 + 16 + 21úû êë - 3úû
\ x = 0, y = - 5 and z = - 3

Q. 19 Using matrix method, solve the system of equations 3x + 2 y - 2z = 3,

x + 2 y + 3z = 6 and 2x - y + z = 2.
K Thinking Process
We know that, for given system of equations in the matrix form, we get AX = B Þ X = A- 1 B,
adj (A)
where A- 1 = and then by getting inverse of A and determinant of A, we can get
| A|
the desired result.
Sol. Given system of equations is
3x + 2 y - 2 z = 3,
x + 2 y + 3z = 6
and 2x - y + z = 2
 In the form of AX = B,
é 3 2 - 2 ù é x ù é3 ù
ê1 2 3 ú ê yú = ê6 ú
ê úê ú ê ú
êë 2 - 1 1 úû ëê z úû êë2 úû
For A - 1, | A| = |3 (5) - 2 (1 - 6) + (- 2 )(- 5)|
= |15 + 10 + 10| =|35| ¹ 0
\ A11 = 5, A12 = 5, A13 = - 5, A21 = 0, A22 = 7, A23 = 7, A31 = 10, A32 = - 11 and A33 = 4

T
55 -5 5 0 10
\ adj A = 0 7 7 = 5 7 -11
10 - 11 4 -5 7 4
5 0 10
adj A 1
Now, A- 1 = = 5 7 - 11
| A| 35
-5 7 4
For X = A - 1B,
éx ù é 5 0 10 ù é 3ù
ê y ú = 1 ê 5 7 - 11ú ê 6ú
ê ú 35 ê úê ú
êë z úû êë - 5 7 4 úû êë2 úû
é 15 + 20 ù é 35ù é 1ù
1 ê 1 ê ú
= 15 + 42 - 22 ú = 35 = ê 1ú
35 ê ú 35 ê ú ê ú
êë - 15 + 42 + 8úû êë 35úû êë 1úû
\ x = 1, y = 1 and z = 1

-42 2 1 -1 0
Q. 20 If A = - 4 2 - 4 and B = 2 3 4 , then find BA and use this
2 -1 5 0 1 2
to solve the system of equations y + 2z = 7, x - y = 3 and
2x + 3 y + 4 z = 17.
2 2 -4 1 -1 0
Sol. We have, A = -4 2 -4 and B = 2 3 4
2 -1 5 0 1 2
1 -1 0 2 2 -4 6 0 0
\ BA = 2 3 4 -4 2 -4 = 0 6 0 = 6 I
0 1 2 2 -1 5 0 0 6
2 2 -4
A 1 1
\ B-1 = = A= -4 2 - 4 ...(i)
6 6 6
2 -1 5
Also, x - y = 3, 2 x + 3 y + 4 z = 17 and y + 2 z = 7
é 1 -1 0ù é x ù é 3 ù
Þ ê 2 3 4ú ê y ú = ê 17 ú
ê úê ú ê ú
êë 0 1 2 úû êë z úû êë 7 úû
 -1
é x ù é 1 -1 0ù é 3 ù
\ ê y ú = ê 2 3 4ú ê 17 ú
ê ú ê ú ê ú
êë z úû êë 0 1 2 úû êë 7 úû
é 2 2 -4 ù é 3 ù
1
= ê -4 2 -4 ú ê 17 ú [using Eq. (i)]
6ê úê ú
êë 2 -1 5 úû êë 7 úû
é 6 + 34 - 28 ù é 12 ù é 2ù
1 1
= ê -12 + 34 - 28ú = ê -6ú = ê -1ú
6ê ú 6ê ú ê ú
êë 6 - 17 + 35 úû êë 24 úû êë 4 úû
\ x = 2, y = -1 and z = 4

a b c
Q. 21 If a + b + c ¹ 0 and b c a = 0 , then prove that a = b = c.
c a b

a b c
Sol. Let A= b c a
c a b
a+ b+c a+ b+c a+ b+c
= b c a [Q R1 ® R1 + R 2 + R 3 ]
c a b

1 1 1
= (a + b + c ) b c a
c a b
0 0 1
= (a + b + c ) b - a c - a a [Q C1 ® C1 - C 3 and C 2 ® C 2 - C 3 ]
c-b a-b b
Expanding along R1,
= (a + b + c ) [1 (b - a) (a - b ) - (c - a) (c - b ) ]
= (a + b + c ) (ba - b 2 - a2 + ab - c 2 + cb + ac - ab )
-1
= (a + b + c ) ´ (-2 ) (- a2 - b 2 - c 2 + ab + bc + ca)
2
-1
= (a + b + c )[a2 + b 2 + c 2 - 2 ab - 2 bc - 2c a + a2 + b 2 + c 2 ]
2
1
= - (a + b + c ) [a2 + b 2 - 2 ab + b 2 + c 2 - 2 bc + c 2 + a2 - 2 ac ]
2
-1
= (a + b + c ) [(a - b )2 + (b - c )2 + (c - a)2 ]
2
Also, A=0
-1
= (a + b + c ) [(a - b )2 + (b - c )2 + (c - a)2 ] = 0
2
(a - b )2 + (b - c )2 + (c - a)2 = 0 [Q a + b + c ¹ 0, given ]
Þ a- b= b-c =c - a= 0
a= b=c Hence proved.
 bc - a 2 ca - b 2 ab - c 2
Q. 22 Prove that ca - b 2 ab - c 2 bc - a 2 is divisible by (a + b + c ) and
ab - c 2 bc - a 2 ca - b 2
find the quotient.
bc - a2 ca - b 2 ab - c 2
Sol. Let D = ca - b 2 ab - c 2 bc - a2
ab - c 2 bc - a2 ca - b 2

bc - a2 - ca + b 2 ca - b 2 - ab + c 2 ab - c 2
2 2 2 2
= ca - b - ab + c ab - c - bc + a bc - a2
ab - c 2 - bc + a2 bc - a2 - ca + b 2 ca - b 2
[Q C1 ® C1 - C 2 and C 2 ® C 2 - C 3 ]
(b - a) (a + b + c ) (c - b ) (a + b + c ) ab - c 2
= (c - b ) (a + b + c ) (a - c ) (a + b + c ) bc - a2
(a - c ) (a + b + c ) (b - a) (a + b + c ) ca - b 2

b - a c - b ab - c 2
= (a + b + c )2 c - b a - c bc - a2
a - c b - a ca - b 2
[taking (a + b + c ) common from C1 and C 2 each]
0 0 ab + bc + ca - (a2 + b 2 + c 2 )
2
= (a + b + c ) c - b a - c bc - a2
a-c b- a ca - b 2
[Q R1 ® R1 + R 2 + R 3 ]
Now, expanding along R1,
= (a + b + c )2 [ab + bc + ca - (a2 + b 2 + c 2 )](c - b ) (b - a) - (a - c )2 ]
= (a + b + c )2 (ab + bc + ca - a2 - b 2 - c 2 )
(cb - ac - b 2 + ab - a2 - c 2 + 2 ac )
2 2 2 2
= (a + b + c ) (a + b + c - ab - bc - ca)
(a2 + b 2 + c 2 - ac - ab - bc )
1
= (a + b + c ) [(a + b + c ) (a2 + b 2 + c 2 - ab - bc - ca)]
2
[(a - b )2 + (b - c )2 + (c - a)2 ]
1
= (a + b + c ) (a3 + b 3 + c 3 - 3 abc ) [(a - b )2 + (b - c )2 + (c - a)2 ]
2
Hence, given determinant is divisible by (a + b + c ) and quotient is
(a3 + b 3 + c 3 - 3 abc ) [(a - b )2 + (b - c )2 + (c - a)2 ].
 xa yb zc a b c
Q. 23 If x + y + z = 0, then prove that yc za xb = xyz c a b .
zb xc ya b c a
K Thinking Process
We have, given x + y + z = 0 Þ x3 + y3 + z3 = 3 xyz. So, by using this in solving the given
determinant from both the sides, we can equate the obtained result from both the sides
to desired result.
Sol. Since, x + y + z = 0 , also we have to prove
x a yb zc a b c
yc za x b = xyz c a b
zb x c ya b c a
xa yb zc
\ LHS = yc za xb
zb xc ya
= x a ( za × ya - x b × x c ) - yb ( yc × ya - x b × zb ) + zc ( yc × xc - za × zb )
= x a (a2 yz - x 2 bc ) - yb ( y2 ac - b 2 xz) + zc (c 2 xy - z2 ab )
= x yza3 - x 3 abc - y3 abc + b 3 x yz + c 3 x yz - z3 abc
= x yz (a3 + b 3 + c 3 ) - abc (x 3 + y3 + z3 )
= x yz (a3 + b 3 + c 3 ) - abc (3 x yz)
[Q x + y + z = 0 Þ x 3 + y3 + z3 - 3xyz]
= x yz (a3 + b 3 + c 3 - 3 abc ) ...(i)
a b c a+ b+c b c
Now, RHS = x yz c a b = x yz a + b + c a b [Q C1 ® C1 + C 2 + C 3 ]
b c a a+ b+c c a
1 b c
= x yz (a + b + c ) 1 a b [taking (a + b + c ) common from C1]
1 c a
0 b-c c - a
= x yz (a + b + c ) 0 a - c b - a
1 c a
[Q R1 ® R1 - R 3 and R 2 ® R 2 - R 3 ]
Expanding along C1,
= x yz (a + b + c ) [1 (b - c ) (b - a) - (a - c ) (c - a)]
= x yz (a + b + c ) (b 2 - ab - bc + ac + a2 + c 2 - 2 ac )
= x yz (a + b + c ) (a2 + b 2 + c 2 - ab - bc - ca)
= x yz (a3 + b 3 + c 3 - 3 abc ) ...(ii)
From Eqs. (i) and (ii),
LHS = RHS
xa yb zc a b c
Þ yc za x b = x yz c a b Hence proved.
zb x c ya b c a
Objective Type Questions
2x 5 6 -2
Q. 24 If = , then the value of x is
8 x 7 3
(a) 3 (b) ± 3 (c) ± 6 (d) 6
2x 5 6 -2
Sol. (c) Q =
8 x 7 3
2
Þ 2 x - 40 = 18 + 14
Þ 2 x 2 = 32 + 40
72
Þ x2 = = 36
2
\ x=± 6

a -b b +c a
Q. 25 The value of b - a c + a b is
c -a a +b c
(a) a3 + b3 + c3 (b) 3bc
(c) a3 + b3 + c3 - 3abc (d) None of these
Sol. (d) We have,
a-b b+c a a+ c b+ c+ a a
b- a c + a b = b+ c c+ a+ b b [Q C1 ® C1 + C 2 and C 2 ® C 2 + C 3 ]
c-a a+ b c c+ b a+ b+c c
a+c 1 a
= (a + b + c ) b + c 1 b [taking (a + b + c ) common from C 2 ]
c+ b 1 c
a- b 0 a-c
= (a + b + c ) 0 0 b-c [Q R 2 ® R 2 - R 3 and R1 ® R1 - R 3 ]
c+ b 1 c
= (a + b + c ) [- (b - c ) × (a - b )] [expanding along R 2 ]
= (a + b + c ) (c - b ) (a - b )

Q. 26 If the area of a triangle with vertices (-3, 0), (3, 0) and (0, k) is
9 sq units. Then, the value of k will be
(a) 9 (b) 3 (c) -9 (d) 6
Sol. (b) We know that, area of a triangle with vertices (x1, y1 ), (x 2 , y2 ) and (x 3 , y3 ) is given by
x1 y1 1
1
D= x 2 y2 1
2
x 3 y3 1
-3 0 1
1
\ D= 3 0 1
2
0 k 1
 Expanding along R1,
1
9= [-3 (- k ) - 0 + 1 (3 k )]
2
Þ 18 = 3 k + 3 k = 6 k
18
\ k= =3
6

Q. 27 The determinant
b 2 - ab b - c bc - ac
ab - a 2 a - b b 2 - ab equals to
bc - ac c - a ab - a 2
(a) abc ( b - c)( c - a)( a - b) (b) ( b - c)( c - a)( a - b)
(c) ( a + b + c)( b - c)( c - a)( a - b) (d) None of these
Sol. (d) We have,
b 2 - ab b - c bc - ac b ( b - a) b - c c ( b - a)
ab - a2 a - b b 2 - ab = a (b - a) a - b b (b - a)
bc - ac c - a ab - a2 c ( b - a) c - a a ( b - a)
b b-c c
= (b - a)2 a a - b b
c c-a a
[on taking (b - a) common from C1 and C 3 each]
b-c b-c c
= (b - a)2 a - b a - b b [Q C1 ® C1 - C 3 ]
c-a c-a a
=0
[since, two columns C1 and C 2 are identical, so the value of determinant is zero]

sin x cos x cos x

Q. 28 The number of distinct real roots of cos x sin x cos x = 0 in the
cos x cos x sin x
p p
interval - £ x £ is
4 4
(a) 0 (b) 2 (c) 1 (d) 3
Sol. (c) We have,
sin x cos x cos x
cos x sin x cos x = 0
cos x cos x sin x
Applying C1 ® C1 + C 2 + C 3 ,
2 cos x + sin x cos x cos x
2 cos x + sin x sin x cos x = 0
2 cos x + sin x cos x sinx
 On taking ( 2 cos x + sin x ) common from C1, we get
1 cos x cos x
Þ ( 2 cos x + sin x ) 1 sin x cos x = 0
1 cos x sin x
1 cos x cos x
Þ ( 2 cos x + sin x ) 0 sin x - cos x 0 =0
0 0 (sin x - cos x )
[Q R 2 ® R 2 - R1 and R 3 ® R 3 - R1 ]
Expanding along C1,
( 2 cos x + sin x ) [1× (sin x - cos x ) 2 ] = 0
Þ ( 2 cos x + sin x ) (sin x - cos x )2 = 0
Either 2cos x = - sin x
1
Þ cos x = - sin x
2
Þ tanx = - 2 ...(i)
p p
But here for - £ x £ , we get -1 £ tanx £ 1 so, no solution possible
4 4
and for (sin x - cos x )2 = 0, sin x = cos x
p
Þ tan x = 1 = tan
4
p
\ x=
4
So, only one distinct real root exist.

Q. 29 If A, B and C are angles of a triangle, then the determinant

-1 cos C cos B
cos C -1
cos A is equal to
cos B cos A -1
(a) 0 (b) -1 (c) 1 (d) None of these
-1 cos C cos B
Sol. (a) We have, cos C -1 cos A
cos B cos A -1
Applying C1 ® a C1 + b C 2 + c C 3 ,
- a + b cos C + c cos B cos C cos B
acos C - b + c cos A -1 cos A
acos B + b cos A - c cos A -1
Also, by projection rule in a triangle, we know that
a = b cos C + c cos B, b = c cos A + acos C and c = acos B + b cos A
Using above equation in column first, we get
- a + a cos C cos B 0 cos C cos B
b-b -1 cos A = 0 -1 cos A = 0
c - c cos A -1 0 cos A -1
[since, determinant having all elements of any column or row gives value of
determinant as zero]
 é cos t t 1 ù
Q. 30 If f (t) = êê2 sin t t 2t úú, then lim f (2t) is equal to
t®0 t
êë sin t t t úû
(a) 0 (b) -1 (c) 2 (d) 3
Sol. (a) We have,
cos t t 1
f(t ) = 2 sin t t 2t
sin t t t
Expanding along C1,
= cos t (t 2 - 2t 2 ) - 2 sin t (t 2 - t ) + sin t ( 2t 2 - t )
= - t 2 cos t - (t 2 - t ) 2 sin t + ( 2t 2 - t ) sin t
= - t 2 cos t - t 2 × 2 sin t + t × 2 sin t + 2 t 2 sin t
= - t 2 cos t + 2 t sin t
f( t ) (- t 2 cos t ) 2t sin t
\ lim 2
= lim 2
+ lim
t ®0 t t ®0 t t ®0 t2
sin t
= - lim cos t + 2 × lim
t ®0 t ®0 t

é sin t ù
= -1+ 1 êëQ t lim = 1 and cos 0 = 1ú
®0 t û
=0

Q. 31 The maximum value of

1 1 1
D= 1 + sin q 1 is (where, q is real number)
1
1 + cos q 1 1
1 3 2 3
(a) (b) (c) 2 (d)
2 2 4
Sol. (a) Since,
1 1 1
D= 1 1 + sin q 1
1 + cos q 1 1
0 0 1
= 0 sin q 1 [Q C1 ® C1 - C 3 and C 2 ® C 2 - C 3 ]
cos q 0 1
= 1(sin q × cos q)
1 1
= × 2sin q cos q = sin 2 q
2 2
Since, the maximum value of sin 2 q is 1. So, for maximum value of q should be 45°.
1
\ D = sin 2 × 45°
2
1 1 1
= sin 90° = . 1 =
2 2 2
 0 x -a x -b
Q. 32 If f (x) = x + a 0 x - c , then
x +b x +c 0
(a) f ( a ) = 0 (b) f ( b) = 0 (c) f(0) = 0 (d) f (1) = 0
Sol. (c) We have,
0 x-a x-b
f( x ) = x + a 0 x -c
x+ b x+c 0
0 0 a-b
Þ f ( a) = 2a 0 a-c
a+ b a+c 0
= [(a - b ) {2 a × (a + c )}] ¹ 0
0 b-a 0
\ f (b) = b + a 0 b-c
2b b+c 0
= - (b - a) [2 b (b - c )]
= - 2b (b - a) (b - c ) ¹ 0
0 -a -b
\ f(0) = a 0 - c
b c 0
= a (bc ) - b (ac )
= abc - abc = 0

2 l -3
Q. 33 If A = 0 2 5 , then A -1 exists, if
1 1 3
(a) l = 2 (b) l ¹ 2
(c) l ¹ - 2 (d) None of these
Sol. (d) We have,
2 l -3
A= 0 2 5
1 1 3
Expanding along R1,
| A| = 2 (6 - 5) - l (-5) - 3 (-2 ) = 2 + 5l + 6
We know that, A -1 exists, if A is non-singular matrix i.e., A ¹ 0.
\ 2 + 5l + 6 ¹ 0
Þ 5l ¹ - 8
-8
\ l¹
5
-1 -8
So, A exists if and only if l ¹ .
5
Q. 34 If A and B are invertible matrices, then which of the following is not
correct?
(a) adj A = | A|× A-1 (b) det ( A) -1 = [det ( A) ]-1
(c) ( AB ) -1 = B-1 A-1 (d) ( A + B ) -1 = B-1 + A-1
Sol. (d) Since, A and B are invertible matrices. So, we can say that
( AB)-1 = B-1 A -1 ...(i)
1
Also, A -1 = (adj A)
| A|
Þ adj A = | A| × A -1 ...(ii)
Also, det ( A)-1 = [det ( A)]-1
1
Þ det ( A)-1 =
[det ( A)]
Þ det ( A) × det ( A)-1 = 1 ...(iii)
which is true.
1
Again, ( A + B)-1 = adj ( A + B )
( A + B)
Þ ( A + B )-1 ¹ B-1 + A -1 ...(iv)
So, only option (d) is incorrect.

1+x 1 1
Q. 35 If x, y and z are all different from zero and 1 1+ y 1 = 0,
1 1 1+z
then the value of x -1 + y -1 + z -1 is
(a) xyz (b) x -1y -1z -1 (c) - x - y - z (d) -1
1+ x 1 1
Sol. (d) We have, 1 1+ y 1 =0
1 1 1+ z
Applying C1 ® C1 - C 3 and C 2 ® C 2 - C 3 ,
x 0 1
Þ 0 y 1 =0
-z -z 1+ z
Expanding along R1,
x [ y (1 + z) + z] - 0 + 1 ( yz) = 0
Þ x ( y + yz + z) + yz = 0
Þ x y + x yz + x z + yz = 0
xy x yz xz yz
Þ + + + = 0 [on dividing (xyz) from both sides]
x yz x yz x yz x yz
1 1 1
Þ + + + 1= 0
x y z
1 1 1
Þ + + = -1
x y z
\ x -1 + y-1 + z-1 = - 1
 x x+y x + 2y
Q. 36 The value of x + 2y x x + y is
x+y x + 2y x
2
(a) 9x ( x + y) (b) 9y 2 ( x + y)
(c) 3y 2 ( x + y) (d) 7x 2 ( x + y)
x x+ y x + 2y
Sol. (b) We have, x + 2y x x+ y
x+ y x + 2y x
3 (x + y) x + y y
= 3 (x + y) x y [Q C1 ® C1 + C 2 + C 3 and C 3 ® C 3 - C 2 ]
3 (x + y) x + 2 y -2 y
1 (x + y) y
= 3 (x + y) 1 x y [taking 3 (x + y) common from first column]
1 (x + 2 y) -2 y
0 y 0
= 3 (x + y) 1 x y [Q R1 ® R1 - R 2 ]
1 (x + 2 y) -2 y
Expanding along R1,
= 3 (x + y) [- y (-2 y - y)]
= 3 y2 × 3 (x + y) = 9 y2 (x + y)

Q. 37 If there are two values of a which makes determinant,

1 -2 5
D= 2 a -1 = 86, then the sum of these number is
0 4 2a
(a) 4 (b) 5 (c) - 4 (d) 9
Sol. (c) We have,
1 -2 5
D = 2 a -1 = 86
0 4 2a
Þ 1 ( 2 a2 + 4) - 2 (- 4a - 20) + 0 = 86 [expanding along first column]
2
Þ 2 a + 4 + 8 a + 40 = 86
Þ 2 a2 + 8 a + 44 - 86 = 0
Þ a2 + 4 a - 21 = 0
Þ a2 + 7 a - 3 a - 21 = 0
Þ (a + 7 ) (a - 3) = 0
a = - 7 and 3
\ Required sum = - 7 + 3 = - 4
Fillers
Q. 38 If A is a matrix of order 3 ´ 3, then |3A | is equal to ......... .
Sol. If A is a matrix of order 3 ´ 3, then| 3 A| = 3 ´ 3 ´ 3| A| = 27 | A|

Q. 39 If A is invertible matrix of order 3 ´ 3, then | A -1| is equal to ......... .

1
Sol. If A is invertible matrix of order 3 ´ 3 , then| A -1| = . [since,| A|×| A -1| = 1]
| A|
(2 x + 2 - x )2 (2 x - 2 - x )2 1
Q. 40 If x, y, z Î R, then the value of (3 x + 3 - x )2 (3 x - 3 - x )2 1 is
(4 x + 4 - x )2 (4 x - 4 - x )2 1

Sol. We have,
(2 x + 2 - x )2 (2 x - 2 - x )2 1
(3x + 3- x )2 (3x - 3- x )2 1
(4x + 4- x )2 (4x - 4- x )2 1
(2 × 2 x ) (2 × 2 - x ) (2 x - 2 - x )2 1
= (2 × 3x ) (2 × 3- x ) (3x - 3- x )2 1 [Q (a + b )2 - (a - b )2 = 4 ab ]
(2 × 4x ) (2 × 4- x ) (4x - 4- x )2 1 [Q C1 ® C1 - C 2 ]

4 (2 x - 2 - x )2 1
= 4 (3x - 3- x )2 1 = 0 [since, C1 and C 3 are proportional to each other]
4 (4x - 4- x )2 1

2
0 cos q sin q
Q. 41 If cos2 q = 0 , then cos q sin q 0 is equal to ......... .
sin q 0 cos q

Sol. Since, cos2 q = 0

p p
Þ cos 2 q = cos Þ 2q=
2 2
p
Þ q=
4
p 1 p 1
\ sin = and cos =
4 2 4 2
2
1 1
0
2 2
1 1
\ 0
2 2
1 1
0
2 2
Expanding along R1,
2 2 2
é 1 æ 1ö 1 æ 1 öù é -2 ù æ -1 ö 1
= ê- ç ÷+ ç - ÷ú = ê ú =ç ÷ =
ë 2 è 2 ø 2 è 2 ø û ë 2 2 û è 2 ø 2
Q. 42 If A is a matrix of order 3 ´ 3 , then ( A 2 ) -1 is equal to ......... .
Sol. If A is a matrix of order 3 ´ 3 , then ( A 2 )-1 = ( A -1 )2 .

Q. 43 If A is a matrix of order 3 ´ 3 , then the number of minors in

determinant of A are ......... .
Sol. If A is a matrix of order 3 ´ 3 , then the number of minors in determinant of A are 9. [since, in
a 3 ´ 3 matrix, there are 9 elements]

Q. 44 The sum of products of elements of any row with the cofactors of

corresponding elements is equal to ......... .
Sol. The sum of products of elements of any row with the cofactors of corresponding elements
is equal to value of the determinant.
a11 a12 a13
Let D = a21 a22 a23
a31 a32 a33
Expanding along R1,
D = a11 A11 + a12 A12 + a13 A13
= Sum of products of elements of R1 with their
corresponding cofactors

x 3 7
Q. 45 If x = - 9 is a root of 2 x 2 = 0, then other two roots are ......... .
7 6 x

x 3 7
Sol. Since, 2 x 2 =0
7 6 x
Expanding along R1,
x (x 2 - 12 ) - 3 (2 x - 14) + 7 (12 - 7 x ) = 0
Þ x 3 - 12 x - 6x + 42 + 84 - 49x = 0
Þ x 3 - 67 x + 126 = 0 ...(i)
Here, 126 ´ 1 = 9 ´ 2 ´ 7
For x = 2, 2 3 - 67 ´ 2 + 126 = 134 - 134 = 0
Hence, x = 2 is a root.
For x = 7, 7 3 - 67 ´ 7 + 126 = 469 - 469 = 0
Hence, x = 7 is also a root.
 0 xyz x-z
Q. 46 y -x 0 y -z is equal to ......... .
z-x z-y 0

0 xyz x- z z-x xyz x- z

Sol. We have, y-x 0 y- z = z-x 0 y- z [Q C1 ® C1 - C 3 ]
z-x z- y 0 z-x z- y 0
1 xyz x - z
= (z - x) 1 0 y- z
1 z- y 0
[taking ( z - x ) common from column 1]
Expanding along R1,
= (z - x ) [1× {- ( y - z) ( z - y)} - xyz ( z - y) + (x - z) ( z - y)]
= (z - x ) ( z - y) (- y + z - xyz + x - z)
= (z - x ) ( z - y) (x - y - xyz)
= (z - x ) ( y - z) ( y - x + xyz)

(1 + x) 17 (1 + x) 19 (1 + x)23
Q. 47 If f (x) = (1 + x)23 (1 + x)29 (1 + x) 34
(1 + x) 41 (1 + x) 43 (1 + x) 47

= A + Bx + Cx 2 + ..., then A is equal to ......... .

Sol. Since,
1 (1 + x )2 (1 + x )6
17 23 41
f(x ) = (1 + x ) (1 + x ) (1 + x ) 1 (1 + x )6 (1 + x )11 = 0
1 (1 + x )2 (1 + x )6
[since, R1 and R 3 are identical]
\ A=0
True/False
Q. 48 ( A 3 ) -1 = ( A -1 ) 3 , where A is a square matrix and A ¹ 0 .
Sol. True
Since, ( A n )-1 = ( A -1 )n , where n Î N.

1 -1
Q. 49 (aA) -1 = A , where a is any real number and A is a square matrix.
a
Sol. False
Since, we know that, if A is a non-singular square matrix, then for any scalar a (non-zero), aA
is invertible such that
æ1 ö æ 1ö
(aA) ç A -1 ÷ = ç a × ÷ ( A × A -1 )
èa ø è aø
=I
æ 1 -1 ö 1
i.e., (aA) is inverse of ç A ÷ or (aA) = A -1, where a is any non-zero scalar.
-1
èa ø a
In the above statement a is any real number. So, we can conclude that above statement is
false.

Q. 50 | A -1| ¹ | A|-1 , where A is a non-singular matrix.

Sol. False
| A -1| = | A|-1, where A is a non-singular matrix.

Q. 51 If A and B are matrices of order 3 and | A | = 5, | B | = 3, then

|3 AB| = 27 ´ 5 ´ 3 = 405.
Sol. True
We know that, | AB| = | A||
× B|
\ |3 AB| = 27 | AB|
= 27| A| ×|B|
= 27 ´ 5 ´ 3 = 405

Q. 52 If the value of a third order determinant is 12, then the value of the
determinant formed by replacing each element by its cofactor will be
144.
Sol. True
Let A is the determinant.
\ | A| = 12
Also, we know that, if A is a square matrix of order n, then|adj A| = | A|n - 1
For n = 3,|adj A| = | A|3 - 1 = | A| 2
= (12 )2 = 144
 x +1 x +2 x +a
Q. 53 x + 2 x + 3 x + b = 0, where a, b and c are in AP.
x +3 x +4 x +c
Sol. True
Since, a, b and c are in AP, then 2b = a + c
x+1 x+2 x+ a
\ x+2 x+ 3 x+ b =0
x+ 3 x+ 4 x+c
2x + 4 2x + 6 2x + a + c
Þ x+2 x+ 3 x+ b =0 [Q R1 ® R1 + R 3 ]
x+ 3 x+ 4 x+c
2 (x + 2 ) 2 (x + 3) 2 (x + b )
Þ x+2 x+ 3 x+ b =0 [Q 2b = a + c]
x+ 3 x+ 4 x+c
Þ 0=0 [since, R1 and R 2 are in proportional to each other]
Hence, statement is true.

Q. 54 |adj A| = | A| 2, where A is a square matrix of order two.

Sol. False
If A is a square matrix of order n, then
|adj A| = | A|n - 1
Þ |adj A| = | A| 2 - 1 = | A| [Q n = 2 ]

sin A cos A sin A + cos B

Q. 55 The determinant sin B cos A sin B + cos B is equal to zero.
sin C cos A sin C + cos B
Sol. True
sin A cos A sin A + cos B sin A cos A sin A sin A cos A cos B
Since, sin B cos A sin B + cos B = sin B cos A sin B + sin B cos A cos B
sin C cos A sin C + cos B sin C cos A sin C sin C cos A cos B
sin A cos A cos B
= 0 + sin B cos A cos B
sin C cos A cos B
[since, in first determinant C1 and C 3 are identicals]
sin A 1 1
= cos A × cos B sin B 1 1
sin C 1 1
[taking cos A common from C 2 and cos B common from C 3 ]
=0 [since, C 2 and C 3 are identicals]
 x +a p +u l + f
Q. 56 If the determinant y +b q + v m + g splits into exactly k
z +c r +w n +h
determinants of order 3, each element of which contains only one
term, then the value of k is 8.
Sol. True
x + a p+ u l + f
Since, y+ b q + v m+ g
z+c r+ w n+ h
x p l a u f
= y+ b q + v m+ g + y+ b q + v m+ g [splitting first row]
z+c r+ w n+ h z+c r+ w n+ h
x p l x p l
= y q m + b v g
z+c r+ m n+ h z+c r+ w n+ h
a u f a u f
+ y q m + b v g [splitting second row]
z+c r+ w n+ h z+c r+ w n+ h
Similarly, we can split these 4 determinants in 8 determinants by splitting each one in two
determinants further. So, given statement is true.

a p x p+x a+x a+p

Q. 57 If D = b q y = 16, then D 1 = q + y b + y b + q = 32.
c r z r + z c + z c +r
Sol. True
a p x
We have, D = b q y = 16
c r z
p+ x a+ x a+ p
and we have to prove, D1 = q + y b + y b + q = 32
r+ z c+ z c+ r
2 p + 2x + 2a a + x a + p
D1 = 2q + 2 y + 2 b b + y b + q [Q C1 ® C1 + C 2 + C 3 ]
2r + 2 z + 2c c+ z c+ r
p x - p a+ p
=2 q y-q b+ q
r z-r c + r
[taking 2 common from C1 and then C1 ® C1 - C 2 , C 2 ® C 2 - C 3 ]
é p x a + pù é p p a + pù
= 2 ê q y b + q ú - êq q b + q ú
ê ú ê ú
êë r z c + r úû êë r r c + r ûú
 p x a+ p
=2 q y b+q -0
r z c+ r
[since, two columns C1 and C 2 are identicals]
p x a p x p
=2 q y b +2 q y q
r z c r z r
a p x
=2 b q y + 0
c r z
[since, C1 and C 3 are identical in second determinant and in first determinant, C1 « C 2
and then C1 « C 3 ]
= 2 ´ 16 [QD = 16]
= 32 Hence proved.

1 1 1
1
Q. 58 The maximum value of 1 1 + sin q 1 is .
2
1 1 1 + cos q
Sol. True
1 1 1
Since, 0 sin q 0 [Q R 2 ® R 2 - R1 and R 3 ® R 3 - R1]
0 0 cos q
On expanding along third row, we get the value of the determinant
1 1
= cos q × sin q = sin 2 q =
2 2
[when q is 45° which gives maximum value]
 5
Continuity and
Differentiability
Short Answer Type Questions
Q. 1 Examine the continuity of the function f (x) = x 3 + 2x 2 - 1 at x = 1.
K Thinking Process
We know that, function f will be continuous at x = a, if lim f (x) = lim f (x) = f (a) .
x®a- x ® a+

Sol. We have, f(x ) = x 3 + 2 x 2 - 1 at x = 1.

\ lim f(x ) = lim (1 + h)3 + 2 (1 + h)2 - 1 = 2
x ® 1+ h®0

and lim f(x ) = lim (1 - h)3 + 2(1 - h)2 - 1 = 2

x ® 1- h®0

\ lim f(x ) = lim f(x )

x ® 1+ x ® 1-

and f(1) = 1 + 2 - 1 = 2
So, f(x ) is continuous at x = 1.
Note Every polynomial function is continuous at any real point.

ì3x + 5, if x ³ 2
Q. 2 f (x) = í 2 at x = 2.
îx , if x < 2
ì3x + 5, if x ³ 2
Sol. We have, f( x ) = í 2 at x = 2.
îx , if x < 2
At x = 2, LHL = lim (x )2
x ® 2-

= lim (2 - h)2 = lim (4 + h2 - 4h) = 4

h®0 h®0

and RHL = lim (3x + 5)

x ® 2+
= lim [3 (2 + h) + 5] = 11
h®0

Since, LHL ¹ RHL at x = 2

So, f(x ) is discontinuous at x = 2.
 ì 1 - cos 2x
ï , if x ¹ 0
Q. 3 f (x) = í x2 at x = 0.
ïî 5, if x = 0
ì1 - cos 2 x , if x ¹ 0
ï
Sol. We have, f( x ) = í x2 at x = 0 .
ïî5, if x = 0
1 - cos 2 x
At x = 0, LHL = lim
x ® 0- x2
1 - cos 2(0 - h)
= lim
h®0 (0 - h)2
1 - cos 2 h
= lim [Q cos (- q) = cos q]
h®0 h2
1 - 1 + 2 sin2 h
= lim [Q cos 2 q = 1 - 2 sin2 q]
h®0 h2
2 (sin h)2 é sin h ù
= lim êëQ hlim = 1ú
h®0 (h)2 ®0 h û
=2
1 - cos 2 x
RHL = lim
x ® 0+ x2
1 - cos 2 (0 + h)
= lim
h®0 (0 + h)2
2 sin2 h é sin h ù
= lim 2
=2 êëQ hlim = 1ú
h®0 h ®0 h û
and f(0) = 5
Since, LHL = RHL ¹ f(0)
Hence, f(x ) is not continuous at x = 0.

ì 2x 2 - 3x - 2
ï , if x ¹ 2
Q. 4 f (x) = í x -2 at x = 2.
ï 5, if x = 2
î
ì2 x 2 - 3x - 2
ï , if x ¹ 2
Sol. We have, f(x ) = í x -2 at x = 2.
ï5, if x = 2
î
2 x 2 - 3x - 2
At x = 2, LHL = lim
x ®2 - x -2
2(2 - h)2 - 3 (2 - h) - 2
= lim
h®0 (2 - h) - 2
8 + 2 h2 - 8 h - 6 + 3 h - 2
= lim
h®0 -h
2 h2 - 5 h h ( 2 h - 5)
= lim = lim =5
h®0 -h h®0 -h
2 x 2 - 3x - 2
RHL = lim
x ® 2+ x -2
 2 ( 2 + h)2 - 3 ( 2 + h) - 2
= lim
h®0 (2 + h) - 2
8 + 2 h2 + 8 h - 6 - 3 h - 2
= lim
h®0 h
2 h2 + 5 h h( 2 h + 5)
= lim = lim =5
h®0 h h®0 h
and f(2 ) = 5
\ LHL = RHL = f ( 2 )
So, f(x ) is continuous at x = 2.

ì | x - 4|
ï , if x ¹ 4
Q. 5 f (x) = í 2 (x - 4) at x = 4.
ï 0, if x = 4
î
ì|x - 4| if
ï , x¹4
Sol. We have, f(x ) = í2(x - 4) at x = 4.
ï 0, if x = 4
î
|x - 4|
At x = 4, LHL = lim
x ® 4- 2(x - 4)
|4 - h - 4| |0 - h|
= lim = lim
h ® 0 2[( 4 - h) - 4] h ® 0 ( 8 - 2 h - 8)
h -1
= lim = and f(4) = 0 ¹LHL
h ® 0 -2 h 2
So, f(x ) is discontinuous at x = 4.

ì| x | cos 1 , if x ¹ 0
ï
Q. 6 f (x) = í x at x = 0.
ïî 0, if x = 0
ì 1
ï|x| cos , if x ¹ 0
Sol. We have, f( x ) = í x at x = 0
ïî0, if x = 0

1 1
At x = 0, LHL = lim |x| cos = lim |0 - h| cos
x ® 0- x h®0 0-h
-1
= lim h cos æç ö÷
h®0 è hø
= 0 ´[an oscillating number between -1 and 1 ] = 0
1
RHL = lim |x | cos
+ x
x ®0
1
= lim |0 + h| cos
h®0 (0 + h)
1
= lim h cos
h®0 h
= 0 ´ [an oscillating number between -1 and 1] = 0
and f(0) = 0
Since, LHL = RHL = f(0)
So, f(x ) is continuous at x = 0.
 ì| x - a| sin 1 , if x ¹ 0
ï
Q. 7 f (x) = í x -a at x = a.
ï 0, if x = a
î
ì 1
ï|x - a| sin , if x ¹ 0
Sol. We have, f( x ) = í x-a at x = a
ï 0, if x = a
î
1
At x = a, LHL = lim |x - a| sin
x ® a- x-a
æ 1 ö
= lim |a - h - a| sin çç ÷÷
h®0
èa - h - aø
æ1ö
= lim - h sin çç ÷÷ [Q sin (- q) = - sin q]
h®0
è hø
= 0 ´ [an oscillating number between -1 and 1 ] = 0
æ 1 ö
RHL = lim |x - a| sin çç ÷÷
èx - aø
+
x ®a

æ 1 ö 1
= lim |a + h - a| sin çç ÷÷ = lim h sin
h®0
è a + h - a ø h®0 h
= 0 ´ [an oscillating number between -1 and 1 ] = 0
and f ( a) = 0
\ LHL = RHL = f(a)
So, f(x ) is continuous at x = a.

ì e 1/ x
ï , if x ¹ 0
Q. 8 f (x) = í 1 + e 1/ x at x = 0.
ï 0, if x = 0
î

ì e1/ x
ï , if x ¹ 0
Sol. We have, f(x ) = í1 + e1/ x at x = 0
ï0, if x = 0
î
e1/ x e1/ 0 - h
At x = 0, LHL = lim = lim
x ®0 -
1+ e 1/ x h®0 1 + e1/ 0 - h
e -1/ h 1
= lim -1/ h
= lim
h®0 1+ e e (1 + e -1/ h )
h®0 1/ h

1 1 1
= lim = ¥ = [Q e ¥ = ¥ ]
h®0 e1/ h +1 e +1 ¥ +1
1
= =0
1
0
e1/ x
RHL = lim
x ®0 +
1 + e1/ x
e1/ 0 + h
e1/ h
= lim = lim
h®0 1 + e1/ 0 + h h®0 1 + e1/ h
 1 1
= lim =
h®0 e -1/ h + 1 e -¥ + 1
1
= =1 [Q e - ¥ = 0]
0+1
Hence, LHL ¹ RHL at x = 0.
So, f(x ) is discontinuous at x = 0.

ì x2
ïï , if 0 £ x £ 1
Q. 9 f (x) = í 2 at x = 1.
ï2x 2 - 3x + 3 , if 1 < x £ 2
ïî 2

ì x2
ïï , if 0 £ x £ 1
Sol. We have, f( x ) = í 2 at x = 1
ï 2 x 2 - 3x + 3 , if 1 < x £ 2
ïî 2
x2 (1 - h)2
At x = 1, HL = lim = lim
- 2 ® 0 2
x ®1 h

1 + h2 - 2 h 1
= lim =
h®0 2 2
æ 2
RHL = lim ç 2 x - 3x +
3 ö
÷
x ® 1+ è 2 ø

= lim é2(1 + h)2 - 3 (1 + h) + ù

3
h®0 ê ë 2 úû
= lim æç 2 + 2 h2 + 4 h - 3 - 3 h + ö÷ = - 1 + =
3 3 1
h®0 è 2ø 2 2
12 1
and f(1) = =
2 2
\ LHL = RHL = f(1)
Hence, f(x ) is continuous at x = 1.

Q. 10 f (x) = | x | + | x - 1| at x = 1.
Sol. We have, f(x ) = |x| + |x - 1| at x = 1
At x = 1, LHL = lim [|x| + |x - 1|]
x ® 1-

= lim [|1 - h| + |1 - h - 1|] = 1 + 0 = 1

h®0
and RHL = lim [|x| + |x - 1|]
x ® 1-

= lim [|1 + h| + |1 + h - 1|] = 1 + 0 = 1

h®0

and f(1) = |1| + |0| = 1

\ LHL = RHL = f(1)
Hence, f(x ) is continuous at x = 1.
Note Every modulus function is a continuous function at any real point.
 ì3x - 8, if x £ 5
Q. 11 f (x) = í at x = 5.
î 2k, if x > 5
ì3x - 8, if x £ 5
Sol. We have, f( x ) = í at x = 5
î 2 k, if x > 5
Since, f(x ) is continuous at x = 5.
\ LHL = RHL = f(5)
Now, LHL = lim (3x - 8) = lim [3 ( 5 - h) - 8]
x ®5 - h ®0

= lim [15 - 3h - 8] = 7
h®0

RHL = lim 2 k = lim 2 k = 2 k = 7 [Q LHL = RHL ]

x ®5 + h®0
and f(5) = 3 ´ 5 - 8 = 7
7
\ 2k = 7 Þ k=
2

ì 2 x + 2 - 16
ï , if x¹ 2
Q. 12 f ( x ) = í 4 x - 16 at x = 2.
ïî k, if x=2
ì2 x + 2 - 16
ï , if x ¹ 2
Sol. We have, f(x ) = í 4x - 16 at x = 2
ïî k, if x = 2
Since, f(x ) is continuous at x = 2.
\ LHL = RHL = f(2 )
2x × 22 - 24 4 × (2 x - 4)
At x = 2, lim 2
= lim
x ®2 x
4 -4 x ®2 (2 x )2 - (4)2
4 × (2 x - 4)
= lim [Q a2 - b 2 = (a + b )(a - b )]
x ®2 (2 - 4) (2 x + 4)
x

4 4 1
= lim x = =
x ®2 2 + 4 8 2
But f(2 ) = k
1
\ k=
2

ì 1 + kx - 1 - kx
ï , if - 1 £ x < 0
Q.13 f (x) = í x at x = 0.
2x + 1 if 0 £ x £ 1
ï ,
î x-1
ì 1 + kx - 1 - kx
ï , if - 1 £ x < 0
Sol. We have, f(x ) = í x at x = 0.
2x + 1 if 0 £ x £ 1
ï ,
î x-1
 1 + kx - 1 - kx
\ LHL = lim
- x
x ®0
æ 1 + kx - 1 - kx ö æ 1 + kx + 1 - kx ö
= lim çç ÷×ç
÷ ç 1 + kx +
÷
x ®0 - è x ø è 1 - kx ÷ø
1 + kx - 1 + kx
= lim
x ®0 - x [ 1 + kx + 1 - kx ]
2 kx
= lim
x ®0 - x 1 + kx + 1 - kx
2k
= lim
h ®0 1 + k ( 0 - h) + 1 - k ( 0 - h)

2k 2k
= lim = =k
h ®0 1 - kh + 1 + kh 2
2 ´0+ 1
and f(0) = = -1
0-1
Þ k = -1 [Q LHL = RHL = f(0)]

ì 1 - cos kx
ïï x sin x , if x ¹ 0
Q. 14 f (x) = í at x = 0.
ï 1, if x = 0
ïî 2
ì1 - cos kx
ïï x sin x , if ¹ 0
x
Sol. We have, f( x ) = í at x = 0
ï 1, if x = 0
ïî2
1 - cos kx 1 - cos k (0 - h)
At x = 0, LHL = lim = lim
x ®0 - x sin x h ®0 ( 0 - h) sin ( 0 - h)
1 - cos (- kh)
= lim
h ®0 - h sin (- h)
1 - cos kh
= lim [Q cos (- q) = cos q, sin(- q) = - sin q]
h ®0 h sin h
kh
1 - 1 + 2 sin2
= lim 2 éQ cos q = 1 - 2 sin2 q ù
h ®0 h sin h êë 2 úû
kh
2 sin2
= lim 2
h ®0 h sin h

kh kh
2 sin sin 2
= lim 2 . 2 . 1 . k h/ 4
h ®0 kh kh sin h h
2 2 h
2 k2 k2 éQ lim sin h = 1ù
= =
4 2 êë h ®0 h úû
2
1 k 1
Also, f(0) = Þ = Þ k = ±1 p
2 2 2
 ì x
, if x ¹ 0
ï
Q. 15 Prove that the function f defined by f (x) = í| x | + 2x 2
ï k, if x = 0
î
remains discontinuous at x = 0, regardless the choice of k.
ì x
ï , if x ¹ 0
Sol. We have, f(x ) = í |x| + 2 x 2
ï k, if x = 0
î
x (0 - h)
At x = 0, LHL = lim 2
= lim
x ®0 - | x| + 2 x h ®0 | 0 - h| + 2( 0 - h)2

-h -h
= lim = lim = -1
h ®0 h + 2 h2 h ®0 h (1 + 2 h)

x 0+ h
RHL = lim 2
= lim
x ®0 + | x| + 2 x h ®0 |0 + h| + 2( 0 + h)2

h h
= lim = lim =1
h ® 0 h + 2 h2 h ® 0 h (1 + 2 h)

and f(0) = k
Since, LHL ¹ RHL for any value of k.
Hence, f(x ) is discontinuous at x = 0 regardless the choice of k.

Q. 16 Find the values of a and b such that the function f defined by

ì x -4
ï x - 4 + a, if x < 4
ïï
f (x) = í a + b, if x = 4
ï x -4
ï + b, if x > 4
ïî| x - 4|
is a continuous function at x = 4.
ì x-4
ï|x - 4| + a, if x < 4
ïï
Sol. We have, f(x ) = í a + b, if x = 4
ï x-4
ï + b, if x > 4
ïî|x - 4|
x-4
At x = 4, LHL = lim + a
- | x - 4|
x ®4
4-h-4 -h
= lim + a = lim + a
h ®0 |4 - h - 4| h ®0 h

= - 1+ a
x-4
RHL = lim + b
x ®4 + | x - 4|
4+ h-4 h
= lim + b = lim + b = 1 + b
h ®0 |4 + h - 4| h ®0 h

f(4) = a + b Þ -1 + a = 1 + b = a + b
Þ -1 + a = a + b and 1 + b = a + b
\ b = - 1 and a = 1
 1
Q. 17 If the function f (x) =
, then find the points of discontinuity of
x +2
the composite function y = f { f (x)}.
Sol. 1
We have, f( x ) =
x+2
\ y = f{f(x )}
æ 1 ö 1
= f çç ÷÷ =
è x +2 ø 1
+2
x +2
1 (x + 2 )
= × (x + 2 ) =
1 + 2x + 4 (2 x + 5)
So, the function y will not be continuous at those points, where it is not defined as it is a
rational function.
x +2
Therefore, y = is not defined, when 2 x + 5 = 0
(2 x + 5)
-5
\ x=
2
-5
Hence, y is discontinuous at x = .
2

1
Q. 18 Find all points of discontinuity of the function f (t) = 2
, where
t + t -2
1
t= .
x -1
1 1
Sol. We have, f (t ) = and t =
t2 + t - 2 x -1
1
\ f( t ) =
æ 1 ö æ 1 ö 2
ç ÷ ç ÷
ç x 2 + 1 - 2 x ÷ + ç x -1 ÷ - 1
è ø è ø
1
=
æ 1 + x - 1 + [-2(x - 1)2 ] ö
ç ÷
ç (x 2 + 1 - 2 x ) ÷
è ø
x 2 + 1 - 2x
=
x - 2 x 2 - 2 + 4x
x 2 + 1 - 2x
=
- 2 x 2 + 5x - 2
(x - 1)2
=
- ( 2 x 2 - 5x + 2 )
(x - 1)2
=
( 2 x - 1) (2 - x )
So, f( t ) is discontinuous at 2 x - 1 = 0 Þ x = 1/ 2
and 2-x = 0 Þ x = 2.
Q. 19 Show that the function f (x) = |sin x + cos x | is continuous at x = p.
Sol. We have, f(x ) = |sin x + cos x| at x = p
Let g (x ) = sin x + cos x
and h(x ) = |x |
\ hog (x ) = h[g (x )]
= h (sin x + cos x )
= |sin x + cos x |
Since, g (x ) = sin x + cos x is a continuous function as it is forming with addition of two
continuous functions sin x and cos x.
Also, h(x ) = |x | is also a continuous function. Since, we know that composite functions of
two continuous functions is also a continuous function.
Hence, f(x ) = |sin x + cos x| is a continuous function everywhere.
So, f(x ) is continuous at x = p.

Q. 20 Examine the differentiability of f, where f is defined by

ì x [ x], if 0 £ x < 2
f (x) = í at x = 2.
î (x - 1)x, if 2 £ x < 3
K Thinking Process
We know that, a function f is differentiable at a point a in its domain, if both Lf ¢(a) and
f(a - h) - f(a)
Rf ¢(a) are finite and equal, where Lf ¢(c) = lim and
h ®0 -h
f(a + h) - f(a)
Rf ¢(c) = lim .
h ®0 h
ìx[x ], if 0 £ x < 2
Sol. We have, f( x ) = í at x = 2.
î( x - 1 ) x if 2 £x<3
f(2 - h) - f(2 )
At x = 2, Lf ¢(2 ) = lim
h ®0 -h
(2 - h) [2- h] - (2 - 1) 2
= lim
h ®0 -h
{Q [a - h] = [a - 1], where a is any positive number}
(2 - h) (1) - 2
= lim
h ®0 -h
2 - h-2 -h
= lim = lim =1
h ®0 -h h ®0 - h
f(2 + h) - f(2 )
Rf ¢(2 ) = lim
h ®0 h
(2 + h - 1) (2 + h) - (2 - 1) × 2
= lim
h ®0 h
(1 + h) (2 + h) - 2
= lim
h ®0 h
2 + h + 2 h + h2 - 2
= lim
h ®0 h
h2 + 3h h (h + 3)
= lim = lim =3
h ®0 h h ®0 h
\ Lf ¢(2 ) ¹ Rf ¢(2 )
So, f(x ) is not differentiable at x = 2.
 ì x 2 sin 1 , if x ¹ 0
ï
Q. 21 f (x) = í x at x = 0.
ïî 0, if x = 0
ì 2 1
ïx sin , if x ¹ 0
Sol. We have, f(x ) = í x at x = 0
îï 0, if x = 0
For differentiability at x = 0,
1
x 2 sin -0
f(x ) - f(0) x
Lf ¢(0) = lim = lim
x ®0 - x-0 x ®0 - x-0
æ 1 ö -1
(0 - h)2 sin çç ÷ h2 sin æç ö÷
è 0 - h ÷ø è hø
= lim = lim
h ®0 0-h h ®0 -h

= lim + h sin æç ö÷
1
[Q sin (- q) = - sin q ]
h ®0 è hø
= 0 ´[an oscillating number between -1and1] = 0
1
x 2 sin - 0
f(x ) - f(0) x
Rf ¢(0) = lim = lim
x ®0 + x-0 x ®0 + x-0
æ 1 ö
(0 + h)2 sin çç ÷
è 0 + h ÷ø h2 sin (1 / h)
= lim = lim
h ®0 0+ h h ®0 h
= lim h sin (1 / h)
h ®0

= 0 ´ [an oscillating number between -1 and 1] = 0

Q Lf ¢(0) = Rf ¢(0)
So, f(x ) is differentiable at x = 0.

ì1 + x, if x £ 2
Q. 22 f (x) = í at x = 2.
î5 - x, if x > 2
ì1 + x, if x £ 2
Sol. We have, f(x ) = í at x = 2.
î5 - x, if x > 2
For differentiability at x = 2,
f(x ) - f(2 ) (1 + x ) - (1 + 2 )
Lf ¢(2 ) = lim = lim
x ®2 - x -2 x ®2 - x -2
(1 + 2 - h) - 3 -h
= lim = lim =1
h ®0 2 - h-2 h ®0 - h
f(x ) - f(2 ) (5 - x ) - 3
Rf ¢(2 ) = lim = lim
x ®2 + x -2 x ®2 + x -2
5 - (2 + h) - 3
= lim
h ®0 2 + h-2
5-2 - h- 3 -h
= lim = lim
h ®0 h h ®0 + h

= -1
Q Lf ¢(2 ) ¹ Rf ¢(2 )
So, f(x ) is not differentiable at x = 2.
Q. 23 Show that f (x) = | x - 5| is continuous but not differentiable at x = 5.
Sol. We have, f(x ) = |x - 5|
ì- (x - 5), if x < 5
\ f( x ) = í
îx - 5, if x ³ 5
For continuity at x = 5,
LHL = lim (- x + 5)
x ®5 -

= lim [- (5 - h) + 5] = lim h = 0
h ®0 h ®0

RHL = lim (x - 5)
x ®5 +
= lim (5 + h - 5) = lim h = 0
h ®0 h ®0
\ f(5) = 5 - 5 = 0
Þ LHL = RHL = f(5)
Hence, f(x ) is continuous at x = 5.
f(x ) - f(5)
Now, Lf ¢(5) = lim
x ®5 - x-5
-x + 5 - 0
= lim = -1
x ®5 - x-5
f(x ) - f(5)
Rf ¢(5) = lim
x ®5 + x-5
x -5-0
= lim =1
x ®5 + x- 5
\ Lf ¢(5) ¹ Rf ¢(5)
So, f(x ) = |x - 5| is not differentiable at x = 5.

Q. 24 A function f : R ® R satisfies the equation f (x + y) = f (x) × f ( y) for

all x, y Î R , f (x) ¹ 0. Suppose that the function is differentiable at
x = 0 and f ¢ (0) = 2, then prove that f ¢ (x) = 2 f (x).
Sol. Let f : R ® R satisfies the equation f (x + y) = f(x ) × f( y), " x, y Î R, f(x ) ¹ 0.
Let f(x ) is differentiable at x = 0 and f¢ (0) = 2.
f(x ) - f(0)
Þ f ¢(0) = lim
x ®0 x-0
f(x ) - f(0)
Þ 2 = lim
x ®0 x
f(0 + h) - f(0)
Þ 2 = lim
h ®0 0+ h
f(0) × f(h) - f(0)
Þ 2 = lim
h ®0 h
f(0) [f(h) - 1]
Þ 2 = lim [Q f(0) = f(h)] ...(i)
h ®0 h
f(x + h) - f(x )
Also, f ¢(x ) = lim
h ®0 h
f(x ) × f(h) - f(x )
= lim [Q f(x + y) = f(x ) × f( y)]
h ®0 h
f(x ) [f(h) - 1]
= lim = 2f(x ) [using Eq. (i)]
h ®0 h
\ f ¢(x ) = 2 f(x. )
 2
Q. 25 2 cos x

2
Sol. Let y = 2cos x

2
\ log y = log 2cos x
= cos 2 x × log 2
On differentiating w.r.t. x, we get
d dy d
log y. = log 2 × cos 2 x
dy dx dx
1 dy d
Þ × = log 2 (cos x )2
y dx dx
1 dy d
Þ × = log 2 × [2 cos x ] × cos x
y dx dx
= log 2 × 2 cos x × (- sin x )
= log 2 × [ - (sin 2 x )]
dy
\ = - y × log 2 (sin 2 x )
dx
2
= - 2cos x
× log 2 (sin2 x )

x
Q. 26 8 8
x
8x 8x
Sol. Let y= 8
Þ log y = log
x x8
d dy d
Þ log y × = [log 8x - log x 8 ]
dy dx dx
1 dy d
Þ × = [x × log 8 - 8 × log x ]
y dx dx
On differentiating w.r.t. x, we get
1 dy 1
× = log 8 × 1 - 8 ×
y dx x
1 dy 8
Þ × = log 8 -
y dx x
8x
= y æç log 8 - ö÷ = 8 æç log 8 - ö÷
dy 8 8
\
dx è xø x è xø

Q. 27 log (x + x2 + a )
Sol. Let y = log (x + x 2 + a)
dy d
\ = log (x + x 2 + a)
dx dx
1 d
= . [x + x 2 + a ]
(x + x 2 + a) dx
1 é1 + 1 (x 2 + a)-1/ 2 × 2 x ù
=
ê ûú
(x + x 2 + a) ë 2

1 æ x ö
= × ç1 + ÷
(x + x 2 + a ) çè x 2 + a ÷ø
( x2 + a + x) 1
= =
2 2 2
(x + x + a) ( x + a) ( x + a)
Q. 28 log [log (log x 5 )]
Sol. Let y = log [log (log x 5 )]
dy d
\ = [log (log log x 5 )]
dx dx
1 d
= 5
× (log × log x 5 )
log log x dx
1 æ 1 ö d
= ×ç ÷. log x 5
log log x 5 çè log x 5 ÷ø dx
1 1 d 5
= × × (5 log x ) =
log log x 5 log x 5 dx x × log (log x 5 ) × log (x 5 )

Q. 29 sin x + cos2 x
Sol. Let y = sin x + (cos x )2
dy d d
\ = sin(x 1/ 2 ) + [cos (x 1/ 2 )]2
dx dx dx
d 1/ 2 d
= cos x 1/ 2 × x + 2 cos (x 1/ 2 ) [cos (x 1/ 2 )]
dx dx
= cos (x 1/ 2 ) x -1/ 2 + 2 × cos (x 1/ 2 ) × é - sin (x 1/ 2 ) .
1 d 1/ 2 ù
x
2 ëê dx ûú
1 1/ 2 1/ 2 1
= cos x × [- 2 cos (x )] × sin x ×
2 x 2 x
1
= [cos ( x ) - sin (2 x )]
2 x

Q. 30 sin n (ax 2 + bx + c )
Sol. Let y = sin (ax 2 + bx + c )
n

dy d
\ = [sin (ax 2 + bx + c )]n
dx dx
d
= n × [sin (ax 2 + bx + c )]n -1 × sin (ax 2 + bx + c )
dx
d
= n × sinn -1 (ax 2 + bx + c ) × cos (ax 2 + bx + c ) . (ax 2 + bx + c )
dx
= n × sinn - 1 (ax 2 + bx + c ) × cos (ax 2 + bx + c ) × (2 ax + b )
= n × (2 ax + b ) × sinn - 1 (ax 2 + bx + c ) × cos (ax 2 + bx + c )

Q. 31 cos(tan x + 1)
Sol. Let y = cos (tan x + 1)
dy d d
\ = cos (tan x + 1) = - sin (tan x + 1) × (tan x + 1)
dx dx dx
= - sin (tan x + 1) × sec 2 x + 1 ×
d
(x + 1)1/ 2 éQ d (tan x ) = sec 2 x ù
dx êë dx úû
2 1 -1/ 2 d
= - sin (tan x + 1) × (sec x + 1) × (x + 1) × (x + 1)
2 dx
-1
= × sin (tan x + 1) × sec 2 ( x + 1)
2 x+1
Q. 32 sin x 2 + sin 2 x + sin 2 (x 2 )
Sol. Let y = sin x 2 + sin2 x + sin2 (x 2 )
dy d d d
\ = sin (x 2 ) + (sin x )2 + (sin x 2 )2
dx dx dx dx
d d d
= cos (x 2 ) (x 2 ) + 2 sin x × sin x + 2 sin x 2 × sin x 2
dx dx dx
d 2
= cos x 2 2 x + 2 × sin x × cos x + 2 sin x 2 cos x 2 × x
dx
= 2 x cos (x )2 + 2 × sin x × cos x + 2 sin x 2 × cos x 2 × 2 x
= 2 x cos (x )2 + sin 2 x + sin 2 (x )2 × 2 x
= 2 x cos (x 2 ) + 2 x × sin 2 (x 2 ) + sin 2 x

1
Q. 33 sin -1
x+1
1
Sol. Let y = sin-1
x+1
dy d 1
\ = sin-1
dx dx x+1
1 d 1 é d 1 ù
= × êQ (sin-1 x ) = ú
æ 1 ö
2 dx (x + 1)1/ 2 ë dx 1- x 2 û
1- ç ÷
è x + 1ø
1 d
= × × (x + 1)-1/ 2
x + 1 - 1 dx
x+1
1
x + 1 -1 - -1 d
= × (x + 1) 2 × (x + 1)
x 2 dx
(x + 1)1/ 2 æ 1 ö -1 æ 1 ö
= × ç - ÷ (x + 1)-3 / 2 = ×ç ÷
x 1/ 2 è 2ø 2 x çè x + 1 ÷ø

Q. 34 (sin x) cos x
Sol. Let y = (sin x )cos x
Þ log y = log(sin x )cos x = cos x log sin x
d dy d
\ log y × = (cos x × logsin x )
dy dx dx
1 dy d d
Þ × = cos x × log sin x + log sin x × cos x
y dx dx dx
1 d
= cos x × × sin x + log sin x × (- sin x )
sin x dx
é cos x ù
= cot x × cos x - log (sin x ) × sin x êQ cot x = sin x ú
ë û
dy é cos 2 x ù
\ = yê - sin x × log (sin x )ú
dx ë sin x û
é cos 2
x ù
= sin x cos x ê - sin x × log (sin x )ú
ë sin x û
Q. 35 sin m x × cos n x
Sol. Let y = sinm x × cos n x
dy d
\ = [(sin x )m × (cos x )n ]
dx dx
d d
= (sin x )m × (cos x )n + (cos x )n × (sin x )m
dx dx
d d
= (sin x )m × n (cos x )n -1 × cos x + (cos x )n m (sin x )m -1 × sinx
dx dx
m n -1 n m -1
= (sin x ) × n(cos x ) (- sin x ) + (cos x ) × m (sin x ) cos x
= - n sinm x × cos n -1 x × (sin x ) + m cos n x × sinm -1 x × cos x
1 1
= - n × sinm x × sin x × cos n x × + m × sinm x. × cos n x × cos x
cos x sin x
= - n × sinm x × cos n x × tan x + m sinm x × cos n x × cot x
= sinm x × cos n x [- n tan x + m cot x ]

Q. 36 (x + 1)2 (x + 2) 3 (x + 3) 4
Sol. Let y = (x + 1)2 (x + 2 )3 (x + 3)4
\ log y = log {(x + 1)2 × (x + 2 )3 (x + 3)4 }
= log (x + 1)2 + log(x + 2 )3 + log (x + 3)4
d dy d d d
and log y × = [2 log (x + 1) ] + [3log (x + 2 )] + [4 log (x + 3)]
dy dx dx dx dx
1 dy 2 d 1 d
× = × (x + 1) + 3 × × (x + 2 )
y dx (x + 1) dx (x + 2 ) dx
1 d éQ d (log x ) = 1 ù
+ 4× × (x + 3)
(x + 3) dx êë dx x úû
é 2 3 4 ù
=ê + +
ë x + 1 x + 2 x + 3 úû
dy é 2 3 4 ù
\ = yê + + ú
dx ë (x + 1) (x + 2 ) (x + 3)û
é 2 3 4 ù
= (x + 1)2 × (x + 2 )3 × (x + 3)4 ê + + ú
ë (x + 1) (x + 2 ) (x + 3)û
= (x + 1)2 × (x + 2 )3 × (x + 3)4
é 2 (x + 2 ) (x + 3) + 3 (x + 1) (x + 3) + 4(x + 1) (x + 2 )ù
ê (x + 1) (x + 2 ) (x + 3) ú
ë û
2 3 4
(x + 1) (x + 2 ) (x + 3)
=
(x + 1) (x + 2 ) (x + 3)
[2 (x 2 + 5x + 6) + 3 (x 2 + 4x + 3) + 4 (x 2 + 3x + 2 )]
= (x + 1) (x + 2 )2 (x + 3)3
[2 x 2 + 10x + 12 + 3x 2 + 12 x + 9 + 4x 2 + 12 x + 8]
= (x + 1) (x + 2 )2 (x + 3)3 [9x 2 + 34x + 29]
 sin x + cos x ö p p
Q. 37 cos -1 æç ÷, - < x <
è 2 ø 4 4
æ sin x + cos x ö
Sol. Let y = cos -1 ç ÷
è 2 ø
dy d æ sin x + cos xö
\ = cos -1 ç ÷
dx dx è 2 ø
-1 d æ sin x + cos x ö
= . ç ÷
2 dx è 2 ø
æ sin x + cos x ö
1- ç ÷
è 2 ø
é d 1 ù
êQ (cos x ) = - ú
ë dx 1- x 2 û
-1 1
= × (cos x - sin x )
2 2
(sin x + cos x + 2 sin x × cos x ) 2
4-
2
- 1× 2 1
= × (cos x - sin x )
1 - sin2 x 2
[ Q 1 - sin 2 x = (cos x - sin x )2 = cos 2 x + sin2 x - 2 sin x cos x ]
-1 (cos x - sin x )
= = -1
(cos x - sin x )

1 - cos x p p
Q. 38 tan -1 , - < x<
1 + cos x 4 4
1 - cos x
Sol. Let y = tan-1
1 + cos x
dy d 1 - cos x
\ = tan-1
dx dx 1 + cos x
d é 1 - cos x ù1
/2
1 é d -1 1 ù
= × êëQ dx (tan x ) = 1 + x 2 úû
æ 1 - cos x ö
2 dx êë 1 + cos x úû
1+ ç ÷
è 1 + cos x ø
-1/ 2
1 1 é 1 - cos x ù d æ 1 - cos x ö
= . . çç ÷÷
1 - cos x 2 êë 1 + cos x úû dx è 1 + cos x ø
1+
1 + cos x
-1/ 2
1 1 é (1 - cos x ) (1 - cos x ) ù
= . ×
1 + cos x + 1 - cos x 2 êë (1 + cos x ) (1 - cos x ) úû
1 + cos x
(1 + cos x ) × sin x + (1 - cos x ) × sin x
×
(1 + cos x )2
-1/ 2
(1 + cos x ) 1 é (1 - cos x )2 ù é sin x (1 + cos x + 1 - cos x )ù
= × ê 2 ú ê ú
2 2 ë (1 - cos x )û ë (1 + cos x )2 û
-1/ 2
(1 + cos x ) 1 é (1 - cos x )2 ù é sin x (1 + cos x + 1 - cos x )ù
= × ê 2 ú ê ú
2 2 ë (1 - cos x )û ë (1 + cos x )2 û
 -1/ 2
(1 + cos x ) 1 é (1 - cos x )2 ù 2 sin x
= × ê ú ×
2 2 ë sin x û (1 + cos x )2
(1 + cos x ) 1 sin x 2 sin x
= × × ×
2 2 (1 - cos x ) (1 + cos x )2
2 sin2 x 1 sin2 x
= = ×
4 (1 + cos x ) (1 - cos x ) 2 (1 - cos 2 x )
1 sin2 x 1
=× =
2 sin2 x 2
Alternate Method
æ 1 - cos x ö
Let y = tan-1 çç ÷÷
è 1 + cos x ø
æ x ö
ç 1 - 1 + 2 sin2 ÷
= tan -1 ç 2 ÷ éQ cos x = 1 - 2 sin2 x = 2 cos 2 x - 1ù
ç 1 + 2 cos 2 x - 1 ÷ ëê 2 2 úû
ç 2 ÷
è ø
-1 æ xö x
= tan ç tan ÷ =
è 2ø 2
On differentiating w.r.t. x, we get
dy 1
=
dx 2

-p p
Q. 39 tan -1 (sec x + tan x), < x<
2 2
Sol. Let y = tan-1 (sec x + tan x )
dy d
\ = tan-1 (sec x + tan x )
dx dx
1 d é d -1 1 ù
= . (sec x + tan x ) êëQ dx (tan x ) = 1 + x 2 úû
1 + (sec x + tan x )2 dx
1
= × [sec x × tan x + sec 2 x ]
1 + sec 2 x + tan2 x + 2 sec x × tan x
1
= × sec x × (sec x + tan x )
(sec 2 x + sec 2 x + 2 sec x × tan x )
1 1
= × sec x (sec x + tan x ) =
2 sec x (tan x + sec x ) 2

æ a cos x - b sin x ö -p p a
Q. 40 tan -1 çç ÷÷, < x < and tan x > - 1.
è b cos x + a sin x ø 2 2 b
æ a cos x - b sin x ö
Sol. Let y = tan-1 çç ÷
è b cos x + a sin x ÷ø
é a cos x b sin x ù é a - tan x ù
-
ê
-1 b cos x b cos x ú -1 ê b ú
= tan ê ú = tan ê ú
ê b cos x +
a sin x ú a
ê 1 + tan x ú
ëê b cos x b cos x úû ë b û
a é æ x - y öù
= tan-1 - tan-1 tan x -1 -1
êQ tan x - tan y = tan
-1
çç ÷÷ ú
b ë è 1 + xy ø û
 a
= tan-1-x
b
dy d æ -1 a ö d
\ = ç tan ÷- (x )
dx dx è b ø dx
é d æaö ù
= 0-1 êëQ dx çè b ÷ø = 0úû
= -1

æ 1ö
Q. 41 sec -1 çç ÷, 0 < x < 1
3 ÷
è 4 x - 3x ø 2
æ 1 ö
Sol. Let y = sec -1 çç 3 ÷
÷ ...(i)
è 4x - 3 x ø
On putting x = cos q in Eq. (i), we get
1
y = sec -1
4cos 3 q - 3 cos q
1
= sec -1
cos 3 q
= sec -1 (sec 3 q) = 3 q
= 3 cos -1 x [Q q = cos -1 x ]
dy d
\ = (3 cos -1 x )
dx dx
-1
= 3×
1 - x2

æ 3a 2 x - x 3 ö -1 x 1
Q. 42 tan -1 çç 3 2 ÷
÷, < <
è a - 3ax ø 3 a 3
æ 3 a2 x - x 3 ö
Sol. Let y = tan-1 çç 3 2
÷
÷
è a - 3ax ø
x
Put x = a tan q Þ q = tan-1
a
-1 é 3 tan q - tan q ù é 3 tan q - tan3 q ù
3
\ y = tan ê 2 ú êQ tan 3 q = ú
ë 1 - 3 tan q û ë 1 - 3 tan2 q û
= tan-1 (tan 3 q) = 3 q

= 3 tan-1
x éQ q = tan-1 x ù
a êë a úû
é ù
ê 1 ú d
× æç ÷ö
dy d -1 x x
\ = 3× tan = 3× ê ú×
dx dx a ê x2 ú dx è a ø
êë + a2
1
úû
a2 1 3a
= 3× × =
a + x2
2
a a2 + x 2
 é 1 + x2 + 1 - x2 ù
Q. 43 tan -1 ê ú , - 1 < x < 1, x ¹ 0
êë 1 + x 2 - 1 - x 2 úû
é 1 + x2 + 1 - x2 ù
Sol. Let y = tan-1 ê ú
êë 1 + x 2 - 1 - x 2 úû
Put x 2 = cos 2 q
æ 1 + cos 2 q + 1 - cos 2 q ö
\ y = tan-1 çç ÷
è 1 + cos 2 q - 1 - cos 2 q ÷ø
æ 1 + 2 cos 2 q - 1 + 1 - 1 + 2 sin2 q ö÷
= tan-1 ç
ç 1 + 2 cos 2 q -1 - 1 - 1 + 2 sin2 q ÷ø
è
æ 2 cos q + 2 sin q ö é 2 (cos q + sin q)ù
= tan-1 çç ÷ = tan-1 ê ú
è 2 cos q - 2 sin q ÷ø ë 2 (cos q - sin q) û
æ cos q + sin q ö
ç ÷
æ cos q + sin q ö cos q
= tan-1 çç ÷÷ = tan-1 ç ÷
è cos q - sin q ø ç cos q - sin q ÷
ç cos q ÷
è ø
-1 æ 1 + tan q ö
= tan çç ÷÷
è 1 - tan q ø
p é tan a + tan b ù
= tan-1 tan æç + q ö÷ êQ tan (a + b ) = 1 - tan a × tan b ú
è4 ø ë û
p p 1 éQ 2 q = cos -1 x 2 Þ q = 1 cos -1 x 2 ù
= + q = + cos -1 x 2
4 4 2 ëê 2 ûú
dy d æpö d æ1 -1 2 ö
\ = ç ÷+ ç cos x ÷
dx dx è 4 ø dx è 2 ø
1 -1 d 2 1 -2 x -x
= 0+ × × x = × =
2 1- x 4 dx 2 1- x 4
1 - x4

dy
Find of each of the functions expressed in parametric form.
dx

Q. 44 x = t + 1 , y =t -
1
t t
1 1
Sol. Q x =t + and y = t -
t t
dx d æ 1ö dy d æ 1ö
\ = çt + ÷ and = çt - ÷
dt dt è tø dt dt è tø
dx dy
Þ = 1 + ( - 1 ) t -2 and = 1 - (-1)t -2
dt dt
dx 1 dy 1
Þ = 1- 2 and = 1+ 2
dt t dt t
dx t 2 - 1 dy t 2 + 1
Þ = and =
dt t2 dt t2
dy dy / dt t 2 + 1/ t 2 2
t +1
\ = = = 2
dx dx / dt t 2 - 1/ t 2 t -1
 y = e -q æç q - ö÷
Q. 45 x = e q æç q + 1 ö÷, 1
è qø è qø
x = e q æç q + ö÷ and y = e -q æç q - ö÷
1 1
Sol. Q
è q ø è qø
dx d é q æ 1 öù
\ = e × ç q + ÷ú
dq dq êë è q øû
q d æ 1ö æ 1ö d q
=e × çq + ÷ + çq + ÷ × e
dq è qø è q ø dq
æ 1 ö
= e q ç 1 - 2 ÷ + æç q + ö÷ e q
1
è q ø è qø
æ 1 1ö
= e q ç1 - 2 + q + ÷
è q qø
æ q2 - 1 + q3 + q ö
= e q çç ÷
÷ ...(i)
è q2 ø
dy d é - q æ 1 öù
and = e × ç q - ÷ú
dq dq êë è q øû
= e -q ×
d æ 1 ö d e -q æ q - 1 ö
çq - ÷ + ç ÷
dq è q ø dq è qø
æ 1 ö
= e - q ç 1 + 2 ÷ + æç q - ö÷ e - q ×
1 d
(- q)
è q ø è qø dq
é q2 + 1 q2 - 1ù -q é q + 1 - q + q ù
2 3
= e -q ê 2 - ú=e ê ú ...(ii)
ë q q û ë q 2
û
-q æ q2 + 1 - q3 + q ö
e çç ÷
÷
dy dy / dq è q2 ø
\ = =
dx dx / dq q æ q2 - 1 + q3 + q ö
e çç ÷
÷
è q2 ø
-2 q æç
- q3 + q2 + q + 1 ö
÷
=e ç q3 + q2 + q - 1 ÷
è ø

Q. 46 x = 3 cos q - 2 cos 3 q, y = 3 sin q - 2 sin 3 q

Sol. Q x = 3 cos q - 2 cos 3 q and y = 3sin q - 2 sin3 q
dx d d
\ = (3 cos q) - (2 cos 3 q)
dq dq dq
d
= 3 × (- sin q) - 2 × 3 cos 2 q × × cos q
dq
= - 3 sin q + 6 cos 2 q sin q
dy d
and = 3 cos A - 2 × 3 sin2 q × × sin q
dq dq
= 3 cos q - 6 sin2 q × cos q
dy dy / dq 3cos q - 6 sin2 q cos q
Now, = =
dx dx / dq - 3 sin q + 6cos 2 q sin q
3cos q (1 - 2 sin2 q) cos 2 q
= = cot q × = cot q
3sin q ( - 1 + 2 cos 2 q) cos 2 q
 2t 2t
Q. 47 sin x = 2
, tan y =
1+t 1 - t2
2t
Sol. Q sin x = ...(i)
1+ t2
2t
and tan y = ...(ii)
1- t2
d dx d æ 2t ö
\ sin x × = ç ÷
dx dt dt çè 1 + t 2 ÷ø
d d
(1 + t 2 ) × (2t ) - (2t ) × (1 + t 2 )
dx dt dt
Þ cos x =
dt (1 + t 2 )2
2
2 (1 + t 2 ) - 2t × 2t 2 + 2t 2 - 4t
= =
(1 + t 2 )2 (1 + t 2 )2
2
dx 2(1 - t ) 1
Þ = ×
dt (1 + t 2 )2 cos x
dx 2(1 - t 2 ) 1 2(1 - t 2 ) 1
Þ = 2 2
× = 2 2
×
dt (1 + t ) 1 - sin x (1 + t )
2
æ 2t ö
2

1 - çç 2
÷
÷
è1 + t ø
dx 2(1 - t 2 ) (1 + t 2 ) 2
Þ = × = ...(iii)
dt (1 + t 2 )2 (1 - t 2 ) 1 + t 2
d dy d æ 2t ö
Also, tan y × = ç ÷
dy dt dt ç1 - t 2 ÷
è ø
2 d d
(1 - t ) × (2t ) - 2t × (1 - t 2 )
dy dt dt
sec 2 y =
dt (1 - t 2 )2
dy 2 - 2t 2 + 4t 2 1
= ×
dt (1 - t 2 )2 sec 2 y
2(1 + t 2 ) 1 2(1 + t 2 ) 1
= 2 2
× 2
= ×
(1 - t ) (1 + tan y) (1 - t 2 )2 4t 2
1+
(1 - t 2 )2
2 2 2
2(1 + t ) (1 - t ) 2
= × = ...(iv)
(1 - t 2 )2 (1 + t 2 )2 1 + t 2
dy dy / dt 2 /1+t2
\ = = =1 [from Eqs. (iii) and (iv)]
dx dx / dt 2 / 1 + t 2

1 + log t 3 + 2 log t
Q. 48 x = 2
,y=
t t
1 + log t 3 + 2 log t
Sol. Q x= and y =
t2 t
d d 2
t 2. (1 + log t ) - (1 + log t ). t
dx dt dt
\ =
dt ( t 2 )2
 1
t2 × - (1 + log t ) × 2t
t t - (1 + log t ) × 2t
= 4
=
t t4
t - 1 - 2 log t
= 4 [1 - 2 (1 + log t ) = ... (i)
t t3
d d
t. (3 + 2 log t ) - (3 + 2 log t ). t
dy dt dt
and = 2
dt t
1
t × 2 × - (3 + 2 log t ) × 1
= t
t2
2 - 3 - 2 log t - 1 - 2 log t
= = ...(ii)
t2 t2
dy dy / dt - 1 - 2 log t / t 2
\ = = =t
dx dx / dt - 1 - 2 log t / t 3

dy y log x
Q. 49 If x = e cos2 t and y = e sin 2t , then prove that =- .
dx x log y
Sol. Q x = ecos 2 t and y = e sin 2 t
dx d cos 2 t d
\ = e = ecos 2 t . cos 2 t
dt dt dt
d
= ecos 2 t × (- sin 2 t ) × (2 t )
dt
dx
= - 2 ecos 2 t × sin 2 t ...(i)
dt
dy d sin 2 t d
and = e = e sin 2 t × sin 2 t
dt dt dt
d
= esin 2 t cos 2 t × 2 t
dt
= 2e sin 2 t × cos 2t ...(ii)
dy dy / dt 2esin 2 t × cos 2 t
\ = =
dx dx / dt -2ecos 2 t × sin 2 t
esin 2 t × cos 2 t
= ...(iii)
ecos 2 t × sin 2 t
We know that, log x = cos 2 t × log e = cos 2 t ...(iv)
and log y = sin2 t × log e = sin2 t ...(v)
dy - y log x
\ =
dx x log y
[using Eqs. (iv) and (v) in Eq. (iii) and x = ecos 2 t , y = esin 2 t ]
Hence proved.

Q. 50 If x = a sin 2 t (1 + cos 2 t) and y = b cos 2 t (1 - cos 2 t), then show that

æ dy ö b
= .
ç ÷
è ø t = p /4 a
d x
Sol. Q x = a sin2 t (1 + cos 2 t ) and y = b cos 2 t (1 - cos 2 t )

= a ésin2 t × sin 2 t ù
dx d d
\ (1 + cos 2 t ) + (1 + cos 2 t ) ×
dt êë dt dt úû
 = a ésin2 t × (- sin 2 t ) × 2 t + (1 + cos 2 t ) × cos 2 t × 2t ù
d d
ëê dt dt úû
= - 2 a sin2 2 t + 2 a cos 2 t (1 + cos 2 t )
dx
Þ = - 2 a [sin2 2 t - cos 2 t (1 + cos 2 t )] ...(i)
dt
= b écos 2 t × cos 2 t ù
dy d d
and (1 - cos 2 t ) + (1 - cos 2 t ) ×
dt êë dt dt úû
é d
= b cos 2 t × (sin 2 t ) 2 t + (1 - cos 2 t ) (- sin 2 t ) ×
d
2t ù
ëê dt dt ûú
= b [2 sin 2t × cos 2t + 2 (1 - cos 2t ) (- sin 2t )]
= 2 b [sin 2t × cos 2t - (1 - cos 2t ) sin 2t ] …(ii)
dy dy / dt -2 b [- sin 2t × cos 2t + (1 - cos 2t ) sin 2t ]
\ = =
dx dx / dt -2 a [sin2 2t - cos 2t (1 + cos 2t )]
é p p p pù
- sin cos + æç 1 - cos ö÷ sin ú
æ dy ö b êë 2 2 è 2ø 2û
Þ ç ÷ =
è dx øt = p/ 4 a é 2 p p æ p öù
êësin 2 - cos 2 çè 1 + cos 2 ÷ø úû
b (0 + 1) éQ sin p = 1 and cos p = 0ù
= ×
a (1 - 0) êë 2 2 úû
b
= Hence proved.
a

dy p
Q. 51 If x = 3 sin t - sin 3 t, y = 3 cos t - cos 3t, then find at t = .
dx 3
Sol. Q x = 3 sin t - sin 3t and y = 3 cos t - cos 3t
dx d d
\ = 3× sin t - sin 3t
dt dt dt
d
= 3 cos t - cos 3t × 3t = 3 cos t - 3 cos 3t ...(i)
dt
dy d d
and = 3× cos t - cos 3 t
dt dt dt
d
= - 3 sin t + sin 3 t × 3t
dt
dy
= 3 sin 3 t - 3 t sin t ...(ii)
dt
dy dy / dt 3 (sin 3 t - sin t )
\ = =
dx dx / dt 3 (cos t - cos 3 t )
3p p
sin - sin
æ dy ö 3 3 0 - 3 /2
Now, ç ÷ = =
è dx øt = p/ 3 æ cos p - cos 3 p ö 1
- (-1)
ç ÷
è 3 3ø 2
- 3/2 - 3 -1
= = =
3/2 3 3
 x
Q. 52 Differentiate w.r.t. sin x.
sin x
x
Sol. Let u= and v = sin x
sin x
d d
sin x × x - x× sin x
du dx dx
\ = 2
dx (sin x )
sin x - x cos x
= ...(i)
sin2 x
dv d
and = sin x = cos x ...(ii)
dx dx
du du / dx sin x - x cos x / sin2 x
\ = =
dv dv / dx cos x
sin x - x cos x
sin x - x cos x cos x
= =
sin2 x cos x sin2 x cos x
cos x
[dividing by cos x in both numerator and denominator]
tan x - x
=
sin2 x

1 + x2 - 1
Q. 53 Differentiate tan -1 w.r.t. tan -1 x, when x ¹ 0.
x
1 + x2 -1
Sol. Let u = tan-1 and v = tan-1 x
x
\ x = tan q
1 + tan2 q - 1
Þ u = tan-1
tan q
-1 (sec q - 1) cos q
= tan
sin q
æ 1 - cos q ö
= tan-1 çç ÷÷
è sin q ø
é 1 - 1 + 2 sin2 q/ 2 ù
= tan-1 ê ú [Q cos q = 1 - 2 sin2 q]
ë 2 sin q / 2 × cos q/ 2 û
q
= tan-1 é tan ù
êë 2 úû
q 1
= = tan-1 x
2 2
du 1 d 1 1
\ = tan-1 x = × ...(i)
dx 2 dx 2 1 + x2
dv d 1
and = tan-1 x = ...(ii)
dx dx 1 + x2
du du / dx
\ =
dv dv / dx
1 / 2 (1 + x 2 ) (1 + x 2 ) 1
= 2
= 2
=
1 / (1 + x ) 2(1 + x ) 2
 dy
Find when x and y are connected by the relation given.
dx
x
Q. 54 sin (xy) + = x2 - y
y
x
Sol. We have, sin (xy) + = x2 - y
y
On differentiating both sides w.r.t. x, we get
d d æx ö d 2 d
(sin xy) + ç ÷= x - y
dx dx è y ø dx dx
d d
y x - x× y
d dx = 2 x - dy
Þ cos xy × (xy) + dx 2
dx y dx
dy
y-x
cos xy × é x × × xù +
d d dx = 2 x - dy
Þ y + y×
ëê dx dx ûú y2 dx
dy y x dy dy
Þ x cos xy × + ycos xy + 2 - 2 = 2x -
dx y y dx dx
dy é x ù y
Þ x cos xy - 2 + 1ú = 2 x - y cos x y - 2
dx êë y û y
dy é 2 x y - y2 cos x y - 1ù é y2 ù
\ =ê úê 2 2ú
dx ë y û ë x y cos x y - x + y û
(2 x y - y2 cos x y - 1) y
=
(x y2 cos x y - x + y2 )

Q. 55 sec (x + y) = xy
Sol. We have, sec (x + y) = x y
On differentiating both sides w.r.t. x, we get
d d
sec (x + y ) = (xy)
dx dx
d d d
Þ sec (x + y) × tan (x + y) × (x + y) = x × y + y× x
dx dx dx
sec (x + y) × tan (x + y) × æç 1 +
dy ö dy
Þ ÷=x + y
è dx ø dx
dy dy
Þ sec (x + y) tan (x + y) + sec (x + y) × tan (x + y) . =x + y
dx dx
dy
Þ [sec (x + y) × tan (x + y) - x ] = y - sec (x + y) × tan (x + y)
dx
dy y - sec(x + y) × tan (x + y)
\ =
dx sec (x + y) × tan (x + y) - x
Q. 56 tan -1 (x 2 + y2) = a
Sol. We have, tan-1 (x 2 + y2 ) = a
On differentiating both sides w.r.t. x, we get
d d
tan-1 (x 2 + y2 ) = ( a)
dx dx
1 d
Þ × ( x 2 + y2 ) = 0
1 + (x 2 + y2 )2 dx
d 2 dy
Þ 2x + y × =0
dy dx
dy
Þ 2y× = - 2x
dx
dy -2 x - x
\ = =
dx 2y y

Q. 57 (x 2 + y 2 )2 = xy
Sol. We have, (x 2 + y2 )2 = xy
On differentiating both sides w.r.t. x, we get
d d
(x 2 + y2 )2 = (xy)
dx dx
d d d
Þ 2 (x 2 + y 2 ) × (x 2 + y 2 ) = x × y + y× x
dx dx dx
2(x 2 + y2 ) × æç 2 x + 2 y
dy ö dy
Þ ÷=x + y
è dx ø dx
dy dy dy
Þ 2x 2 × 2x + 2x 2 × 2 y + 2 y2 × 2 x + 2 y2 × 2 y =x + y
dx dx dx
dy
Þ [4x 2 y + 4 y3 - x ] = y - 4x 3 - 4xy2
dx
dy ( y - 4x 3 - 4xy2 )
\ =
dx (4x 2 y + 4 y3 - x )

dy dx
Q. 58 If ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0, then show that . = 1.
dx dy
Sol. We have, ax 2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0 ...(i)
On differentiating both sides w.r.t. x, we get
d d d d d d
(ax 2 ) + (2 hxy) + (by2 ) + (2 gx ) + (2 fy) + (c) = 0
dx dx dx dx dx dx
2 ax + 2 h æç x × + y × 1ö÷ + b × 2 y
dy dy dy
Þ + 2g + 2f + 0=0
è dx ø dx dx
dy
Þ [2 hx + 2 by + 2 f ] = - 2 ax - 2 hy - 2 g
dx
dy - 2 (ax + hy + g )
Þ =
dx 2 (hx + by + f )
- (ax + hy + g )
= ...(ii)
(hx + by + f )
 Now, differentiating Eq. (i) w.r.t. y, we get
d d d d d d
(ax 2 ) + (2 hxy) + (by2 ) + (2 gx ) + (2 fy) + (c) = 0
dy dy dy dy dy dy
dx æ d d ö dx
Þ a × 2x × + 2h × ç x × y+ y× x ÷ + b × 2 y + 2g × + 2f + 0 = 0
dy è dy dy ø dy
dx
Þ [2 ax + 2 hy + 2 g ] = - 2 hx - 2 by - 2 f
dy
dx - 2 (hx + by + f ) - (hx + by + f )
Þ = = ... (iii)
dy 2 (ax + hy + g ) (ax + hy + g )
dy dx - (ax + hy + g ) - (hx + by + f )
\ × = × [using Eqs. (ii) and (iii)]
dx dy (hx + by + f ) (ax + hy + g )
= 1= RHS Hence proved.

x-y
Q. 59 If x = e x / y , then prove that dy = .
dx x log x
Sol. We have, x = ex/ y
d d x/ y
\ x= e
dx dx
d
Þ 1 = ex/ y . (x / y)
dx
é y × 1 - x × dy / dx ù
Þ 1 = ex/ y × ê ú
ë y2 û
2 x/ y dy x/ y
Þ y = y×e - x× ×e
dx
dy x / y
Þ x× ×e = ye x / y - y2
dx
dy y (e x / y - y)
\ =
dx x ×ex/ y
(e x / y - y) é x/ y xù
= êQ x = e Þ log x = ú
x ë yû
ex/ y ×
y
x- y
= Hence proved.
x × log x

2
dy (1 + log y)
Q. 60 If y x = e y - x , then prove that = .
dx log y
Sol. We have, y x = ey - x
Þ log y x = log e y - x
Þ x log y = y - x × loge = ( y - x ) [Q loge = 1]
(y - x)
Þ log y = ...(i)
x
Now, differentiating w.r.t. x, we get
d dy d (y - x)
log y × =
dy dx dx x
 d d
x× ( y - x ) - ( y - x )× ×x
1 dy dx dx
Þ × =
y dx x2
x æç - 1ö÷ - ( y - x )
dy
1 dy
= è ø
dx
Þ
y dx x2
x 2 dy dy
Þ × =x -x - y+ x
y dx dx
dy æ x 2 ö
Þ ç - x ÷÷ = - y
ç
dx è y ø
dy - y2 - y2
\ = 2 =
dx x - xy x(x - y)
y2 x y2 1
= × = 2 ×
x( y - x ) x x (y - x)
x
(1 + log y)2 é y-x
log y = - 1 Þ 1 + log y = ù
y y
= Q log y =
log y êë x x x úû
Hence proved.

(cos x )... ¥ dy y 2 tan x

Q. 61 If y = (cos x)(cos x) , then show that = .
dx y log cos x - 1
(cosx )... ¥
Sol. We have, y = (cos x )( cos x )
y
Þ y = (cos x )
y
\ log y = log (cos x )
Þ log y = ylog cos x
On differentiating w.r.t. x, we get
1 dy d dy
× = y × log cos x + log cos x ×
y dx dx dx
1 dy y d dy
Þ × = × cos x + log cos x ×
y dx cos x dx dx
dy é 1 ù - ysin x
Þ - log cos x ú = = - y tan x
dx êë y û cos x
dy - y2 tan x
\ =
dx (1 - ylog cos x )
y2 tan x
= Hence proved.
y log cos x - 1

Q. 62 If x sin (a + y) + sin a × cos(a + y) = 0, then prove that

2
dy sin (a + y)
= .
dx sin a
Sol. We have,
x sin (a + y) + sin a × cos(a + y) = 0
Þ x sin (a + y) = - sin a × cos(a + y)
- sin a × cos(a + y)
Þ x=
sin (a + y)
 Þ x = - sin a × cot(a + y)
dx d
\ = - sin a × [-cosec 2 (a + y)]× (a + y)
dy dy
1
= sin a × 2 ×1
sin (a + y)
sin2 (a + y)
= Hence proved.
sin a

dy 1 - y2
Q. 63 If 1 - x 2 + 1 - y 2 = a(x - y), then prove that = .
dx 1 - x2
Sol. We have,
1 - x2 + 1 - y2 = a(x - y)
On putting x = sin a and y = sin b, we get
1 - sin2 a + 1 - sin2 b = a(sin a - sin b )
Þ cos a + cos b = a(sin a - sin b )
a+b a -b a+b a -bö
Þ 2 cos .cos = aæç 2 cos .sin ÷
2 2 è 2 2 ø
a -b a -b
Þ cos = asin
2 2
a -b
Þ cot =a
2
a -b
Þ = cot -1 a
2
Þ a - b = 2 cot -1 a
Þ sin-1 x - sin-1 y = 2 cot -1 a [Qx = sin a and y = sin b]
On differentiating both sides w.r.t. x, we get
1 1 dy
- =0
2 2 dx
1- x 1- y
dy 1 - y2 1 - y2
\ = = Hence proved.
dx 1- x 2 1 - x2

2
Q. 64 If y = tan -1 x, then find d y
in terms of y alone.
dx 2
Sol. We have, y = tan-1 x [on differentiating w.r.t. x]
dy 1
\ = [again differentiating w.r.t. x]
dx 1 + x 2
d2y d
Now, = (1 + x 2 )-1
dx 2 dx
d
= - 1 (1 + x 2 )-2 . (1 + x 2 )
dx
1
=- × 2x
(1 + x 2 )2
- 2 tan y
= [Q y = tan-1 x Þ tan y = x ]
(1 + tan2 y)2
 - 2 tan y
=
(sec 2 y)2
sin y
= -2 × cos 2 y × cos 2 y
cos y
= - sin 2 y × cos 2 y [Q sin 2 x = 2 sin x cos x ]
Verify the Rolle's theorem for each of the functions in following questions.

Q. 65 f (x) = x(x - 1)2 in [0,1]

K Thinking Process
We know that, Rolle’s theorem states that, if f be a real valued function, defined in the
closed interval [a, b], such that (i) f is continuous on [a, b]. (ii) f is differentiable on ]a,
b[.(iii) f(a) = f(b).
Then, there exists a real number c in the open interval ] a, b [, such that f ¢(c) = 0 ¢. Here,
we shall verify the Rolle’s theorem for the given function.
Sol. We have, f(x ) = x(x - 1)2 in [0, 1].
(i) Since, f(x ) = x(x - 1)2 is a polynomial function.
So, it is continuous in [0, 1].
d d
(ii) Now, f ¢(x ) = x × (x - 1)2 + (x - 1)2 x
dx dx
= x × 2(x - 1) × 1 + (x - 1)2
= 2x 2 - 2x + x 2 + 1 - 2x
= 3x 2 - 4x + 1, which exists in (0, 1).
So, f(x ) is differentiable in (0, 1).
(iii) Now, f(0) = 0 and f(1) = 0 Þ f(0) = f(1)
f satisfies the above conditions of Rolle’s theorem.
Hence, by Rolle’s theorem $ c Î (0, 1) such that
f ¢(c ) = 0
Þ 3c 2 - 4c + 1 = 0
Þ 3c 2 - 3c - c + 1 = 0
Þ 3c(c - 1) - 1(c - 1) = 0
Þ (3c - 1)(c - 1) = 0
1 1
Þ c = , 1Þ Î(0, 1)
3 3
Thus, we see that there exists a real number c in the open interval (0, 1).
Hence, Rolle’s theorem has been verified.
 p
Q. 66 f (x) = sin 4 x + cos 4 x in é0, ù
êë 2 úû
p
Sol. We have, f(x ) = sin4 x + cos 4 x in é 0, ù ...(i)
êë 2 úû
p
(i) f(x ) is continuous in é 0, ù
êë 2 úû
[since, sin4 x and cos 4 x are continuous functions and we know that, if g and h be
continuous functions, then (g + h) is a continuous function.]
(ii) f ¢(x ) = 4(sin x )3 × cos x + 4(cos x )3 × (- sin x )
= 4sin3 x × cos x - 4 sin x × cos 3 x
p
= 4sin x cos x (sin2 x - cos 2 x ) which exists in æç 0, ö÷ ...(ii)
è 2ø
p
Hence, f(x ) is differentiable in æç 0, ö÷.
è 2ø
p
(iii) Also, f(0) = 0 + 1 = 1 and f ¢æç ö÷ = 1 + 0 = 1
è ø
2
p
Þ f(0) = f æç ö÷
è ø
2
Conditions of Rolle’s theorem are satisfied.
p
Hence, there exists atleast one c Î æç 0, ö÷ such that f ¢(c ) = 0.
è 2 ø
\ 4 sin c cos c (sin2 c - cos 2 c ) = 0
Þ 4 sinc cos c (- cos 2 c ) = 0
Þ - 2 sin 2c × cos 2 c = 0
Þ - sin4 c = 0
Þ sin 4 c = 0
Þ 4c = p
p
Þ c=
4
p æ pö
and Î ç 0, ÷
4 è 2ø
Hence, Rolle’s theorem has been verified.

Q. 67 f (x) = log (x 2 + 2) - log 3 in [ - 1, 1]

Sol. We have, f(x ) = log (x 2 + 2 ) - log 3.
(i) Logarithmic functions are continuous in their domain.
Hence, f(x ) = log (x 2 + 2 ) - log 3 is continuous in [- 1 ,1.]
1
(ii) f ¢(x ) = × 2x - 0
x2 + 2
2x
= , which exists in (- 1, 1.)
x2 + 2
Hence, f(x ) is differentiable in (- 1, 1.)
 (iii) f(-1) = log [(-1)2 + 2 ] - log 3 = log 3 - log 3 = 0 and
f(1) = log (12 + 2 ) - log 3 = log 3 - log 3 = 0
Þ f(-1) = f(1)
Conditions of Rolle’s theorem are satisfied.
Hence, there exists a real number c such that
f ¢(c ) = 0.
2c
Þ =0
c2 + 2
Þ c = 0 Î (- 1, 1)
Hence, Rolle’s theorem has been verified.

Q. 68 f (x) = x (x + 3) e - x /2 in [ - 3, 0]
Sol. We have, f(x ) = x (x + 3) e - x / 2
(i) f(x ) is a continuous function. [since, it is a combination of polynomial functions x(x + 3)
and an exponential function e - x / 2 which are continuous functions]
So, f(x ) = x (x + 3) e - x / 2 is continuous in [- 3, 0 ].
d - x/ 2 d
(ii) \ f ¢(x ) = (x 2 + 3x ) × e + e -x / 2 × (x 2 + 3x )
dx dx
= (x 2 + 3x ) × e - x / 2 × æç - ö÷ + e - x / 2 × (2 x + 3)
1
è 2ø
= e - x / 2 é2 x + 3 - × (x 2 + 3x )ù
1
êë 2 úû
é 4x + 6 - x 2 - 3x ù
= e -x / 2 ê ú
ë 2 û
-x / 2 1 2
=e × [- x + x + 6]
2
- 1 -x / 2 2
= e [x - x - 6]
2
- 1 -x / 2 2
= e [x - 3x + 2 x - 6]
2
- 1 -x / 2
= e [(x + 2 ) (x - 3)], which exists in (- 3, 0).
2
Hence, f(x ) is differentiable in (- 3, 0).
(iii) \ f(- 3) = - 3 (- 3 + 3)e -3 / 2 = 0
and f(0) = 0(0 + 3) e -0 / 2 = 0
Þ f(- 3) = f (0)
Since, conditions of Rolle’s theorem are satisfied.
Hence, there exists a real number c such that f ¢(c ) = 0
1
Þ - e - c / 2 (c + 2 ) (c - 3) = 0
2
Þ c = - 2, 3, where - 2 Î (- 3, 0)
Therefore, Rolle’s theorem has been verified.
Q. 69 f (x) = 4 - x 2 in [ - 2, 2]
Sol. We have, f(x ) = 4 - x 2 = (4 - x 2 )1/ 2

(i) f(x ) = 4 - x 2 is a continuous function.

[since every polynomial function is a continuous function]
Hence, f(x ) is continuous in [- 2, 2 ] .
1
(ii) f ¢(x ) = (4 - x 2 )-1/ 2 × (- 2 x )
2
1
= - x. , which exists everywhere except at x = ± 2.
4 - x2
Hence, f(x ) is differentiable in (- 2, 2 ).
(iii) f(- 2 ) = (4 - 4) = 0 and f(2 ) = (4 - 4) = 0
Þ f(- 2 ) = f(2 )
conditions of Rolle’s theorem are satisfied.
Hence, there exists a real number c such that f ¢ (c ) = 0.
1
Þ -c =0
4 - c2
Þ c = 0 Î (- 2, 2 )
Hence, Rolle’s theorem has been verified.

Q. 70 Discuss the applicability of Rolle’s theorem on the function given by

ì x 2 + 1, if 0 £ x £ 1
f (x) = í
î3 - x, if 1 £ x £ 2
ìx 2 + 1, if 0 £ x £ 1
Sol. We have, f( x ) = í
î3 - x, if 1 £ x £ 2
We know that, polynomial function is everywhere continuous and differentiability.
So, f(x ) is continuous and differentiable at all points except possibly at x = 1.
Now, check the differentiability at x = 1,
At x = 1,
f(x ) - f(1)
LDH = lim
x ® 1- x -1
(x 2 + 1) - (1 + 1)
= lim [Q f(x ) = x 2 + 1, " 0 £ x £ 1]
x ®1 x -1
x2 -1 (x + 1)(x - 1)
= lim = lim
x ®1 x - 1 x ®1 x -1
=2
f(x ) - f(1) (3 - x ) f(1 + 1)
and RDH = lim = lim
x ®1+ x - 1 x ® 1 (x - 1)
3- x -2 -(x - 1)
= lim = lim = -1
x ®1 x - 1 x ®1 x - 1

\ LHD ¹ RHD
So, f(x ) is not differentiable at x = 1.
Hence, polle’s theorem is not applicable on the interval [0, 2].
Q. 71 Find the points on the curve y = (cos x - 1) in [0, 2p], where the
tangent is parallel to X-axis.
K Thinking Process
We know that, if f be a real valued function defined in the closed interval [a , b] such that
it follows all the three conditions of Rolle’s theorem, then f ¢(c) = 0 shows that the
tangent to the curve at x = c has a slope 0, i.e., it is parallel to the X-axis. So, by getting
the value of c¢ we can get the required point.
Sol. The equation of the curve is y = cos x - 1.
Now, we have to find a point on the curve in [0, 2p ],
where the tangent is parallel to X-axis i.e., the tangent to the curve at x = c has a slope o,
where c Î] 0, 2p[.
Let us apply Rolle’s theorem to get the point.
(i) y = cos x - 1is a continuous function in [0, 2p ].
[since it is a combination of cosine function and a constant function]
(ii) y¢ = - sin x , which exists in (0, 2p).
Hence, y is differentiable in (0, 2p).
(iii) y (0) = cos 0 - 1 = 0 and y (2 p ) = cos 2 p - 1 = 0,
\ y (0) = y (2 p)
Since, conditions of Rolle’s theorem are satisfied.
Hence, there exists a real number c such that
f ¢(c ) = 0
Þ - sinc = 0
Þ c = p or 0, where p Î(0, 2 p)
Þ x=p
\ y = cos p - 1 = - 2
Hence, the required point on the curve, where the tangent drawn is parallel to the X-axis is
(p, - 2 ).

Q. 72 Using Rolle’s theorem, find the point on the curve

y = x (x - 4), x Î [0, 4], where the tangent is parallel to X-axis.
Sol. We have, y = x (x - 4), x Î [0,4]
(i) y is a continuous function since x(x - 4) is a polynomial function.
Hence, y = x (x - 4) is continuous in [0, 4].
(ii) y¢ = (x - 4) × 1 + x × 1 = 2 x - 4 which exists in (0,4).
Hence, y is differentiable in (0,4).
(iii) y(0) = 0 (0 - 4) = 0
and y(4) = 4 (4 - 4) = 0
Þ y(0) = y(4)
Sicne, conditions of Rolle’s theorem are satisfied.
Hence, there exists a point c such that
f ¢(c ) = 0 in (0,4) [Q f ¢(x ) = y¢]
Þ 2c - 4 = 0
Þ c =2
Þ x = 2; y = 2(2 - 4) = - 4
Thus, (2, - 4) is the point on the curve at which the tangent drawn is parallel to X-axis.
Verify mean value theorem for each of the functions.
1
Q. 73 f (x) = in [1, 4]
4x - 1
K Thinking Process
We know that, mean value theorem states that, if f be a real function such that
(i) f (x) is continuous on [a,b]
(ii) f (x) is differentiable on ]a,b[
f (b) - f (a)
Then, there exists a real number c Î] a,b [ such that f ¢(c) = , thus we can
verify it for given function. b-a

Sol. 1
We have, f(x ) = in [1, 4]
4x - 1
(i) f(x ) is continuous in [1, 4].
1
Also, at x = , f(x ) is discontinuous.
4
Hence, f(x ) is continuous in [1, 4].
4
(ii) f ¢(x ) = - , which exists in (1, 4).
(4x - 1)2
Since, conditions of mean value theorem are satisfied.
Hence, there exists a real number c Î] 1, 4 [ such that
f (4) - f (1)
f ¢(c ) =
4-1
1 1 1 1
- -
-4 16 - 1 4 - 1 15 3
Þ = =
(4c - 1)2 4-1 3
-4 1- 5 - 4
Þ = =
(4 c - 1)2 45 45
Þ (4c - 1)2 = 45
Þ 4c - 1 = ± 3 5
3 5+1
Þ c= Î (1, 4) [neglecting (- ve) value]
4
Hence, mean value theorem has been verified.

Q. 74 f (x) = x 3 - 2x 2 - x + 3 in [0, 1]
Sol. We have, f(x ) = x 3 - 2 x 2 - x + 3 in [0, 1]
(i) Since, f(x ) is a polynomial function.
Hence, f(x ) is continuous in [0, 1].
(ii) f ¢(x ) = 3x 2 - 4x - 1, which exists in (0,1).
Hence, f(x ) is differentiable in (0,1).
Since, conditions of mean value theorem are satisfied.
Therefore, by mean value theorem $ c Î (01, , ) such that
f(1) - f(0)
f ¢(c ) =
1- 0
 [1 - 2 - 1 + 3] - [0 + 3]
Þ 3c 2 - 4c - 1 =
1- 0
2 -2
Þ 3c - 4c - 1 =
1
Þ 3c 2 - 4c + 1 = 0
Þ 3c 2 - 3c - c + 1 = 0
Þ 3c (c - 1) - 1(c - 1) = 0
Þ (3c - 1) (c - 1) = 0
1
Þ c = 1 / 3 , 1, where Î(0, 1)
3
Hence, the mean value theorem has been verified.

Q. 75 f (x) = sin x - sin 2x in [0, p]

Sol. We have, f(x ) = sin x - sin2 x in [0,p]
(i) Since, we know that sine functions are continuous functions hence f(x ) = sin x - sin2 x is
a continuous function in [0,p].
(ii) f ¢(x ) = cos x - cos2 x × 2 = cos x - 2 cos 2 x , which exists in (0, p).
So, f(x ) is differentiable in (0, p). Conditions of mean value theorem are satisfied.
f (p) - f (0)
Hence, $ c Î (0, p ) such that, f ¢( c ) =
p-0
sin p - sin 2 p - sin 0 + sin 2 × 0
Þ cos c - 2 cos 2c =
p-0
0
Þ 2 cos 2 c - cos c =
p
Þ 2 × (2 cos 2 c - 1) - cos c = 0
Þ 4 cos 2 c - 2 - cos c = 0
Þ 4 cos 2 c - cos c - 2 = 0
1 ± 1 + 32
1 ± 33
Þ cos c = =
8 8
-1 æ 1 ± 33 ö
\ c = cos çç ÷÷
è 8 ø
æ 1 ± 33 ö
Also, cos -1 çç ÷÷ Î (0, p)
è 8 ø
Hence, mean value theorem has been verified.

Q. 76 f (x) = 25 - x 2 in [1, 5]
Sol. We have, f(x ) = 25 - x 2 in [1, 5]
(i) Since, f(x ) = (25 - x 2 )1/ 2 , where 25 - x 2 ³ 0
Þ x 2 £ ± 5 Þ -5 £ x £ 5
Hence, f(x ) is continuous in [1, 5].
1 -x
(ii) f ¢(x ) = (25 - x 2 )-1/ 2 × - 2 x = , which exists in (1, 5).
2 25 - x 2
Hence, f ¢(x ) is differentiable in (1, 5).
 Since, conditions of mean value theorem are satisfied.
By mean value theorem $ c Î (1, 5) such that
f(5) - f(1) -c 0 - 24
f ¢(c ) = Þ =
5-1 25 - c 2 4

c2 24
Þ =
25 - c 2 16
Þ 16 c 2 = 600 - 24 c 2
600
Þ c2 = = 15
40
\ c = ± 15
Also, c = 15 Î (1, 5)
Hence, the mean value theorem has been verified.

Q. 77 Find a point on the curve y = (x - 3)2 , where the tangent is parallel to

the chord joining the points (3, 0) and (4, 1).
K Thinking Process
We know that, if y = f (x) be a function defined on [a, b] which follows mean value
theorem, then there exists atleast one point c in (a, b) such that the tangent at the point
[c , f (c)] is parallel to the secant joining the points [a , f (a)] and [b, f (b)] . So, we shall use
this concept.
Sol. We have, y = (x - 3)2 , which is continuous in x1 = 3 and x 2 = 4 i.e., [3, 4].
Also, y¢ = 2(x - 3) × 1 = 2(x - 3) which exists in (3, 4).
Hence, by mean value theorem there exists a point on the curve at which
tangent drawn is parallel to the chord joining the points (3,0) and (4,1).
f(4) - f(3)
Thus, f ¢(c ) =
4-3
(4 - 3)2 - (3 - 3)2
Þ 2 (c - 3) =
4-3
1- 0 7
Þ 2c - 6 = Þ c=
1 2
2 2
7 æ 7 ö æ 1ö 1
For x = , y = ç - 3÷ = ç ÷ =
2 è2 ø è2 ø 4
So, æç , ö÷ is the point on the curve at which tangent drawn is parallel to the chord joining
7 1
è2 4ø
the points (3, 0) and (4, 1).

Q. 78 Using mean value theorem, prove that there is a point on the curve
y = 2x 2 - 5x + 3 between the points A(1, 0) and B(2, 1), where
tangent is parallel to the chord AB. Also, find that point.
Sol. We have, y = 2 x 2 - 5x + 3, which is continuous in [1, 2] as it is a polynomial function.
Also, y¢ = 4x - 5, which exists in (1, 2).
By mean value theorem, $ c Î (1, 2) at which drawn tangent is parallel to the chord AB,
where A and B are (1, 0) and (2,1), respectively.
f(2 ) - f(1)
\ f ¢(c ) =
2 -1
 (8 - 10 + 3) - (2 - 5 + 3)
Þ 4c - 5 =
1
Þ 4c - 5 = 1
6 3
\ c = = Î (1 , 2 )
4 2
3 3 2
y = 2 æç ö÷ - 5 æç ö÷ + 3
3
For x = ,
2 è2 ø è2 ø
9 15 9 - 15 + 6
=2 ´ - + 3= =0
4 2 2
Hence, æç , 0 ö÷ is the point on the curve y = 2 x 2 - 5x + 3 between the points A (1, 0) and
3
è2 ø
B (2, 1), where tangent is parallel to the chord AB.

Long Answer Type Questions

ì x 2 + 3x + p, if x £ 1
Q. 79 Find the values of p and q, so that f (x) = í is
î qx + 2, if x > 1
differentiable at x = 1.
Sol. ì x 2 + 3x + p, if x £ 1
We have, f (x ) = í is differentiable at x = 1.
î qx + 2, if x > 1
f(x ) - f(1)
\ Lf ¢ (1) = lim
x ®1- x -1
(x 2 + 3 x + p) - (1 + 3 + p)
= lim
x ®1- x -1
[(1 - h)2 + 3(1 - h) + p] - [1 + 3 + p]
= lim
h ®0 (1 - h) - 1
[1 + h2 - 2 h + 3 - 3h + p] - [4 + p]
= lim
h®0 -h
[h2 - 5 h + p + 4 - 4 - p] h [h - 5]
= lim = lim
h®0 -h h ® 0 -h
= lim - [h - 5] = 5
h®0
f(x ) - f(1) (qx + 2 ) - (1 + 3 + p)
R f ¢ (1) = lim = lim
x ®1 +x - 1 x ®1+ x -1
[q (1 + h) + 2 ] - (4 + p)
= lim
h®0 1+ h - 1
[q + qh + 2 - 4 - p] qh + (q - 2 - p)
= lim = lim
h®0 h h®0 h
Þ q - 2 - p = 0Þ p - q = - 2 ...(i)
qh + 0
Þ lim =q [for existing the limit]
h®0 h
If Lf ¢(1) = Rf ¢(1,) then 5 = q
Þ p- 5 = - 2Þ p= 3
\ p = 3 and q = 5
Q. 80 If x m × y n = (x + y) m + n , prove that
dy y d2 y
(i) = and (ii) =0
dx x dx 2
Sol. We have, x m × y n = (x + y)m + n
...(i)
(i) Differentiating Eq. (i) w.r.t. x, we get
d d
(x m × y n ) = (x + y)m + n
dx dx
d n dy n d d
Þ xm × y × + y × x m = (m + n) (x + y)m + n - 1 (x + y)
dy dx dx dx

+ y × mx m - 1 = (m + n) (x + y)m + n -1 æç 1 +
dy n dy ö
Þ x m × nyn - 1 ÷
dx è dx ø
dy m n
Þ [x × ny -1 - (m + n) × (x + y)m + n - 1 ] = (m + n) (x + y)m + n - 1 - yn mx m-1
dx
n
dy n -1 y -1 × y × mx m
Þ [nx m y - (m + n)(x + y)m + n -1 ] = (m + n) × (x + y)m + n -1 -
dx x
(m + n) (x + y)m + n y n -1× y × mx m
-
dy (x + y) x
\ =
dx n x m yn m+ n 1
- (m + n) (x + y)
y (x + y)
x (m + n) (x + y)m + n - (x + y) × y × n - 1 y × mx m
(x + y) × x
= n
(x + y) n x m y - y (m + n) (x + y)m + n
(x + y) × y
x (m + n) × x m × yn - m (x + y) yn x m
(x + y) × x n
= [Q (x + y)m + n
= xm × y ]
(x + y) nx m × yn - y (m + n) × x m × y n
(x + y) × y
x m y n [mx + nx - mx - my]× (x + y) y
=
x m y n [nx + ny - my - ny]× (x + y) × x
y
= …(ii)
x
Hence proved.
dy y
(ii) Further, differentiating Eq. (ii) i.e., = on both the sides w.r.t. x, we get
dx x
dy
x× - y× 1
d2y dx
=
dx 2 x2
y
x× - y
= x éQ dy = y ù
x2 ëê dx x ûú
=0 Hence proved.
Q. 81 If x = sin t and y = sin pt, then prove that
2
d y dy
(1 - x 2 ) 2 - x + p 2 y = 0.
dx dx
Sol. We have, x = sin t and y = sin pt
dx dy
\ = cos t and = cos pt×p
dt dt
dy dy / dt p × cos pt
Þ = = …(i)
dx dx / dt cos t
Again, differentiating both sides w.r.t. x, we get
d dt d dt
cos t × ( p × cos pt ) - pcos pt × cos t ×
d2y dt dx dt dx
=
dx 2 cos 2 t
dt
[cos t × p × (- sin pt ) × p - p cos pt × (- sin t )]
= dx
cos 2 t
1
[- p2 sin pt × cos t + p sin t × cos pt ] ×
cos t
=
cos 2 t
d2y - p2 sin pt × cos t + p cos pt × sin t
Þ = ...(ii)
dx 2 cos 3 t
Since, we have to prove
d2y dy
(1 - x 2 ) -x + p2 y = 0
dx 2 dx
[- p2 sin pt × cos t + pcos pt × sin t ]
\ LHS = (1 - sin2 t )
cos 3 t
pcos pt
- sin t × + p2 sin pt
cos t
1 é(1 - sin t ) (- p sin pt × cos t + pcos pt × sin t )ù
2 2
= 3 ê ú
cos t êë - p cos pt × sin t × cos 2 t + p2 sin pt × cos 3 t úû

1 é - p2 sin pt × cos 3 t + pcos pt × sin t × cos 2 t ù 2 2

= ê
3 ú [Q 1 - sin t = cos t ]
cos t êë - pcos pt × sin t × cos 2 t + p2 sin pt × cos 3 t úû
1
= ×0
cos 3 t
=0 Hence proved.

dy x2 + 1
Q. 82 Find the value of , if y = x tan x + .
dx 2
x2 + 1
Sol. We have, y = x tan x + ...(i)
2
x2 + 1
Taking u = x tan x and v = ,
2
log u = tan x log x ...(ii)
x2 + 1
and v2 = ...(iii)
2
 On, differentiating Eq. (ii) w.r.t. x, we get
1 du 1
× = tan x × + log x × sec 2 x
u dx x
=u é
tan x
+ log x × sec 2 x ù
du
Þ
dx ëê x ûú
tan x é tan x
=x + log x × sec 2 x ù …(iv)
êë x úû
Also, differentiating Eq. (iii) w.r.t. x, we get
dv 1 dv 1
2 v× = (2 x )Þ = × (2 x )
dx 2 d x 4v
dv 1 x× 2
Þ = × 2x =
dx 2
x +1 2 x2 + 1
4×
2
dv x
Þ = ...(v)
dx 2 ( x 2 + 1)
Now, y=u + v
d y du d v
\ = +
dx dx dx

= x tan x é
tan x
+ log x × sec 2 x ù +
x
ëê x ûú 2(x 2 + 1)

Objective Type Questions

x2
Q. 83 If f (x) = 2x and g(x) = + 1, then which of the following can be a
2
discontinuous function?
(a) f ( x) + g ( x) (b) f ( x) - g ( x)
g ( x)
(c) f ( x) × g ( x) (d)
f ( x)
Sol. (d) We know that, if f and g be continuous functions, then
(a) f + g is continuous (b) f - g is continuous.
f
(c) fg is continuous (d) is continuous at these points, where g (x ) ¹ 0.
g
x2
+1
g(x ) x2 + 2
Here, = 2 =
f( x ) 2x 4x
which is discontinuous at x = 0.
 4 - x2
Q. 84 The function f (x) = is
4x - x 3
(a) discontinuous at only one point
(b) discontinuous at exactly two points
(c) discontinuous at exactly three points
(d) None of the above
4 - x2 (4 - x 2 )
Sol. (c) We have, f( x ) = 3
=
4x - x x (4 - x 2 )
(4 - x 2 ) 4 - x2
= =
2
x (2 - x ) 2
x (2 + x ) (2 - x )
Clearly, f(x ) is discontinuous at exactly three points x = 0, x = - 2 and x = 2.

Q. 85 The set of points where the function f given by f (x) = | 2x - 1| sin x is

differentiable is
1
(a) R (b) R - æç ö÷
è2ø
(c) (0 , ¥ ) (d) None of these
Sol. (b) We have, f(x ) = |2 x - 1| sin x
1
At x = , f(x ) is not differentiable.
2
Hence, f(x ) is differentiable in R - æç ö÷.
1
è2 ø
f æç + h ö÷ - f æç ö÷
1 1
Q æ 1ö
Rf ¢ç ÷ = lim è 2 ø è 2ø
è2 ø h ® 0 h

2 ç + h ö÷ - 1 sin æç + h ö÷ - 0
æ 1 1
è2 ø è2 ø
= lim
h®0 h
æ 1 + 2h ö
|2 h|× sin ç ÷
= lim è 2 ø = 2 × sin 1
h®0 h 2
f æç - h ö÷ - f æç ö÷
1 1
L f ¢æç ö÷ = lim è
1 2 ø è2 ø
and
è2 ø h ® 0 -h
-1
2 æç - h ö÷
1
- sin æç - h ö÷ - 0
1
è2 ø è2 ø
= lim
h®0 -h

|0 - 2 h|- sin æç - h ö÷
1
= lim è 2 ø = - 2 sin æ 1 ö
ç ÷
h®0 -h è2 ø

Q æ 1ö æ 1ö
Rf ¢ç ÷ ¹ Lf ¢ç ÷
è2 ø è2 ø
1
So, f(x ) is not differentiable at x = .
2
Q. 86 The function f (x) = cot x is discontinuous on the set
(a) { x = np : n Î Z } (b) { x = 2np : n Î Z }
p np
(c) ìí x = (2n + 1) ; n Î Z üý (d) ìíx = ; n Î Z üý
î 2 þ î 2 þ
Sol. (a) We know that, f(x ) = cot x is continuous in R - {n p : n Î Z}.
cos x
Since, f(x ) = cot x = [since, sin x = 0 at n p , n Î Z ]
sin x
Hence, f(x ) = cot x is discontinuous on the set {x = np : n Î Z}.

Q. 87 The function f (x) = e |x| is

(a) continuous everywhere but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) None of the above
Sol. (a) Let u (x ) = |x| and v (x ) = e x
\ f(x ) = vou(x ) = v[u (x )]
= v | x | = e | x|
Since, u(x ) and v (x ) are both continuous functions.
So, f(x ) is also continuous function but u (x ) = |x| is not differentiable at x = 0, whereas
v(x ) = e x is differentiable at everywhere.
Hence, f(x ) is continuous everywhere but not differentiable at x = 0.

1
Q. 88 If f (x) = x 2 sin , where x ¹ 0, then the value of the function f at
x
x = 0, so that the function is continuous at x = 0, is
(a) 0 (b) - 1
(c) 1 (d) None of these
Q f(x ) = x 2 sin æç ö÷, where x ¹ 0
1
Sol. (a)
èx ø
Hence, value of the function f at x = 0, so that it is continuous at x = 0 is 0.

émx + 1, p
if x £
ê 2 is continuous at x = p , then
Q. 89 If f (x) = ê
p 2
êsin x + n, if x >
ë 2
np
(a) m = 1, n = 0 (b) m = +1
2
mp p
(c) n = (d) m = n =
2 2
ìmx + 1, p
if x £
ï 2 p
Sol. (c) We have, f(x ) = í p
is continuous at x =
ï(sin x + n), if x > 2
î 2
 é p ù mp
\ LHL = lim (mx + 1) = lim ê m æç - h ö÷ + 1ú = +1
x®
p- h®0 ë è 2 ø û 2
2
é p ù
and RHL = lim (sin x + n) = lim êsin æç + h ö÷ + nú
x®
p+ h®0 ë è 2 ø û
2

= lim cos h + n = 1 + n
h®0

\ LHL = RHL é to be continuous at x = p ù

êë 2 úû
p
Þ m× + 1= n + 1
2
p
\ n = m×
2

Q. 90 If f (x) = |sin x |, then

(a) f is everywhere differentiable
(b) f is everywhere continuous but not differentiable at x = np, n Î Z
p
(c) f is everywhere continuous but not differentiable at x = (2n + 1) , n Î Z
2
(d) None of the above
Sol. (b) We have, f(x ) = |sin x|
Let f(x ) = vou (x ) = v [u(x )] [where, u (x ) = sin x and v (x ) = |x|]
= v (sin x ) = |sin x|
where, u(x ) and v (x ) are both continuous.
Hence, f(x ) = vo u(x ) is also a continuous function but v(x ) is not differentiable at x = 0.
So, f(x ) is not differentiable where sin x = 0Þ x = n p, n Î Z
Hence, f(x ) is continuous everywhere but not differentiable at x = np, n Î Z.

æ 1 - x2 ö
Q. 91 If y = log çç ÷, then dy is equal to
2 ÷
è1 + x ø dx
4 x3 - 4x 1 - 4 x3
(a) (b) (c) (d)
1- x4 1- x4 4 - x4 1- x4
æ 1 - x2 ö
Sol. (b) We have, y = log çç ÷
2 ÷
è1 + x ø
dy 1 d æ 1 - x2 ö
\ = . ç ÷
dx 1 - x 2 dx çè 1 + x 2 ÷ø
1 + x2
(1 + x 2 ) (1 + x 2 ) × (-2 x ) - (1 - x 2 ) × 2 x
= ×
(1 - x 2 ) (1 + x 2 )2
-2 x[1 + x 2 + 1 - x 2 ] - 4x
= =
(1 - x 2 ) × (1 + x 2 ) 1 - x4
 dy
Q. 92 If y = sin x + y, then is equal to
dx
cos x cos x sin x sin x
(a) (b) (c) (d)
2y - 1 1 - 2y 1 - 2y 2y - 1

Sol. (a) Q y = (sin x + y )1/ 2

dy 1 d
\ = (sin x + y)-1/ 2 × (sin x + y)
dx 2 dx
dy 1 1
× æç cos x +
dy ö
Þ = × ÷
dx 2 (sin x + y)1/ 2 è dx ø
dy 1 æ dy ö
Þ = ç cos x + ÷ [Q (sin x + y)1/ 2 = y]
dx 2 y è dx ø
dy æ 1 ö cos x
Þ ç1 - ÷=
dx è 2yø 2y
dy cos x 2y cos x
\ = × =
dx 2y 2y - 1 2y - 1

Q. 93 The derivative of cos -1 (2x 2 - 1) w.r.t. cos -1 x is

-1
(a) 2 (b)
2 1- x2
2
(b) (d) 1 - x 2
x
Sol. (a) Let u = cos -1 (2 x 2 - 1) and v = cos -1 x
dv + -1 - 4x
\ = × 4x =
dx 2
1 - (2 x - 1) 2
1 - (4x 4 + 1 - 4x 2 )
- 4x - 4x
= =
4 2
- 4x + 4x 4x (1 - x 2 )
2

-2
=
1 - x2
du -1
and =
dx 1 - x2
2
dx du / dx - 2 / 1 - x
\ = = =2
dv dv / dx -1 / 1 - x 2

2
Q. 94 If x = t 2 and y = t 3 , then d y
is equal to
dx 2
3 3 3 3
(a) (b) (c) (d)
2 4t 2t 2t

Sol. (b) We have, x = t 2 and y = t 3

dx dy
\ = 2 t and = 3t2
dt dt
dy dy / dt 3t 2 3
\ = = = t
dx dx / dt 2t 2
 On further differentiating w.r.t. x, we get
d 2 y 3 d dt
= × t×
dx 2 2 dt dx
3 1 éQ dt = 1 ù
= ×
2 2t êë dx 2t úû
3
=
4t

Q. 95 The value of c in Rolle’s theorem for the function f (x) = x 3 - 3x in the

interval [0, 3] is
3 1
(a) 1 (b) - 1 (c) (d)
2 3
Sol. (a) Q f ¢(c ) = 0 [Q f ¢(x ) = 3x 2 - 3]
Þ 3c 2 - 3 = 0
3
Þ c2 = = 1
3
Þ c = ± 1, where 1 Î(0, 3 )
\ c =1

1
Q. 96 For the function f (x) = x + , x Î [1, 3], the value of c for mean value
x
theorem is
(a) 1 (b) 3
(c) 2 (d) None of these
f ( b ) – f ( a)
Sol. (b) Q f ¢ (c ) =
b– a
é 3 + 1 ù – é1 + 1ù
ê éQ f ¢ (x ) = 1 – 1 ù
1 3 ûú êë 1úû
Þ 1– 2 = ë ê x2 ú
c 3–1 ê ú
ëand b = 3, a = 1û
10
–2
c2 – 1
Þ 2
= 3
c 2
c2 - 1 4 2
Þ = =
c2 3 ´2 3
Þ 3(c 2 - 1) = 2c 2
Þ 3 c2 - 2c2 = 3
Þ c2 = 3 Þ c = ± 3
Q c = 3 Î(1, 3)
Fillers
Q. 97 An example of a function which is continuous everywhere but fails to
be differentiable exactly at two points is ......... .
Sol. | x | + | x – 1| is continuous everywhere but fails to be differentiable exactly at two points
x = 0 and x = 1.
So, there can be more such examples of functions.

Q. 98 Derivative of x 2 w.r.t. x 3 is ......... .

Sol. 2
Derivative of x 2 w.r.t. x 3 is .
3x
Let u = x 2 and v = x 3
du dv
\ = 2 x and = 3x 2
dx dx
du 2x 2
Þ = =
dv 3x 2 3x

p
Q. 99 If f (x) = |cos x |, then f ¢ æç ö÷ is equal to ......... .
è4ø
p
Sol. If f(x ) = |cos x |, then f¢ æç ö÷
è 4ø
p
Q 0< x < , cos x > 0.
2
f (x ) = + cos x
\ f ¢ (x ) = (– sin x )
p p –1 é p 1 ù
Þ f¢ æç ö÷ = – sin = êëQ sin 4 = 2 úû
è ø
4 4 2

p
Q. 100 If f (x) = |cos x – sin x |, then f ¢ æç ö÷ is equal to ......... .
è3ø
Sol. Q f(x ) = |cos x – sin x|,
p 3+1
\ f¢ æç ö÷ =
è 3ø 2
p p
We know that, < x < , sin x > cos x
4 2
\cos x – sin x £ 0 i .e., f(x ) = – (cos x – sin x )
f ¢(x ) = - [– sin x – cos x ]
p æ - 3 1ö æ 3+ 1ö
\ f¢æç ö÷ = – çç – ÷÷ = çç ÷÷
è 3ø è 2 2ø è 2 ø
 at æç , ö÷ is ......... .
dy 1 1
Q. 101 For the curve x+ y = 1,
dx è4 4ø
at æç , ö÷ is - 1.
dy 1 1
Sol. For the curve x + y = 1,
dx è 4 4ø
We have, x + y =1
1 1 dy
Þ + =0
2 x 2 y dx
dy y
Þ =–
dx x
1
–
\ æ dy ö = 2 = –1
ç ÷æ 1 1
è dx ø ç ,
1ö
÷
è4 4ø
2

True/False
Q. 102 Rolle’s theorem is applicable for the function f (x) = | x – 1| in [ 0, 2 ].
Sol. False
Hence, f(x ) = x – 1 in [0, 2 ]is not differentiable at x = 1 Î (0, 2 ).

Q. 103 If f is continuous on its domain D, then | f | is also continuous on D.

Sol. True

Q. 104 The composition of two continuous function is a continuous

function.
Sol. True

Q. 105 Trigonometric and inverse trigonometric functions are differentiable

in their respective domain.
Sol. True

Q. 106 If f × g is continuous at x = a, then f and g are separately continuous

at x = a.
Sol. False
Let f(x ) = sin x and g (x ) = cot x
cos x
\ f(x ) × g (x ) = sin x × = cos x
sin x
which is continuous at x = 0 but cot x is not continuous at x = 0.
 6
Application of Derivatives
Short Answer Type Questions
Q. 1 A spherical ball of salt is dissolving in water in such a manner that the
rate of decrease of the volume at any instant is propotional to the
surface. Prove that the radius is decreasing at a constant rate.
K Thinking Process
First, let V be the volume of the ball and S be the surface area of the ball and then by
dV
using µ S , we can prove the required result.
dt

Sol. We have, rate of decrease of the volume of spherical ball of salt at any instant is µ surface.
Let the radius of the spherical ball of the salt be r.
4
\ Volume of the ball (V ) = pr 3
3
and surface area (S) = 4pr 2
dV d æ4 3ö 2
Q µS Þ ç pr ÷ µ 4pr
dt dt è 3 ø
4 dr dr 4pr 2
Þ p × 3 r2. µ 4pr 2 Þ µ
3 dt dt 4pr 2
dr
Þ = k × 1 [where, k is the proportionality constant]
dt
dr
Þ =k
dt
Hence, the radius of ball is decreasing at a constant rate.

Q. 2 If the area of a circle increases at a uniform rate, then prove that

perimeter varies inversely as the radius.
Sol. Let the radius of circle = r And area of the circle, A = pr 2
d d
\ A= pr 2
dt dt
dA dr
Þ = 2pr × ...(i)
dt dt
 Since, the area of a circle increases at a uniform rate, then
dA
=k ...(ii)
dt
where, k is a constant.
dr
From Eqs. (i) and (ii), 2pr × =k
dt
dr k k æ 1ö
Þ = = ×ç ÷ ...(iii)
dt 2 pr 2 p è r ø
Let the perimeter, P = 2pr
dP d dP dr
\ = × 2pr Þ = 2p ×
dt dt dt dt
k 1 k
= 2p × . = [using Eq. (iii)]
2p r r
dP 1
Þ µ Hence proved.
dt r

Q. 3 A kite is moving horizontally at a height of 151.5 m. If the speed of kite

is 10 m/s, how fast is the string being let out, when the kite is 250 m
away from the boy who is flying the kite, if the height of boy is 1.5 m?
Sol. We have , height (h) = 151. 5 m, speed of kite (v)= 10 m/s F C
Let CD be the height of kite and AB be the height of boy.
Let DB = x m = EA and AC = 250 m
dx
\ = 10 m/s
dt
From the figure, we see that 151.5 m
EC = 151. 5 - 1. 5 = 150 m A E
and AE = x G
Also, AC = 250 m 1.5 m
In right angled DCEA, B
H D
AE 2 + EC 2 = AC 2
Þ x 2 + (150)2 = y2 ... (i)
2 2 2
Þ x + (150) = (250)
Þ x 2 = (250)2 - (150)2
= (250 + 150) (250 - 150)
= 400 ´ 100
\ x = 20 ´ 10 = 200
From Eq. (i), on differentiating w.r.t. t, we get
dx dy
2x × + 0 = 2y
dt dt
dy dx
Þ 2y = 2x
dt dt
dy x dx
\ = ×
dt y dt
200 éQ dx = 10 m / sù
= × 10 = 8 m/s
250 ëê dt ûú
So, the required rate at which the string is being let out is 8 m/s.
Q. 4 Two men A and B start with velocities v at the same time from the
junction of two roads inclined at 45° to each other. If they travel by
different roads, then find the rate at which they are being separated.
K Thinking Process
By drawing figure such that men start moving at a point C, A and B are separating points,
then draw perpendicular from that point C to AB to get D. Now, get the value of ÐACD in
dy
terms of x and y, then by using get desired result. [ let AC = BC = x and AB = y]
dt

Sol. Let two men start from the point C with velocity v each at the same C
time.
Also, ÐBCA = 45°

°
45
Since, A and B are moving with same velocity v, so they will cover
same distance in same time.
Therefore, DABC is an isosceles triangle with AC = BC.
90°
Now, draw CD ^ AB. A B
Let at any instant t, the distance between them is AB.
Let AC = BC = x and AB = y
In DACD and DDCB,
ÐCAD = Ð CBD [Q AC = BC]
ÐCDA = Ð CDB = 90°
\ ÐACD = ÐDCB
1
or Ð ACD = ´ Ð ACB
2
1
Þ ÐACD = ´ 45°
2
p
Þ ÐACD =
8
p AD
\ sin =
8 AC
p y/2
Þ sin = [Q AD = y / 2]
8 x
y p
Þ = x sin
2 8
p
Þ y = 2 x × sin
8
Now, differentiating both sides w.r.t. t, we get
dy p dx
= 2 × sin ×
dt 8 dt
p éQ v = dx ù
= 2 × sin . v
8 ëê dt ûú
2- 2 é p 2- 2ù
= 2v × êQ sin = ú
2 êë 8 2 úû
= 2 - 2 v unit/s
which is the rate at which A and B are being separated.
Q. 5 Find an angle q, where 0 < q < p , which increases twice as fast as its sine.
2
Sol. Let q increases twice as fast as its sine.
Þ q = 2 sin q
Now, on differentiating both sides w.r.t. t, we get
dq dq
= 2 × cos q × Þ 1 = 2 cos q
dt dt
1 p
Þ = cos q Þ cos q = cos
2 3
p
\ q=
3
p
So, the required angle is .
3

Q. 6 Find the approximate value of (1.999) 5 .

Sol. Let x =2
and D x = - 0.001 [Q 2 - 0. 001 = 1. 999]
Let y = x5
On differentiating both sides w.r.t. x, we get
dy
= 5x 4
dx
dy
Now, Dy = × Dx = 5x 4 ´ D x
dx
= 5 ´ 2 4 ´ [-0.001]
= - 80 ´ 0.001 = - 0.080
\ (1999
. )5 = y + D y
= 2 5 + (- 0.080)
= 32 - 0.080 = 31. 920

Q. 7 Find the approximate volume of metal in a hollow spherical shell whose

internal and external radii are 3 cm and 3.0005 cm, respectively.
Sol. Let internal radius = r and external radius = R
4
\Volume of hollow spherical shell, V = p (R 3 - r 3 )
3
4
Þ V = p [(3.0005)3 - (3)3 ] ... (i)
3
Now, we shall use differentiation to get approximate value of (3.0005)3 .
Let (3.0005)3 = y + D y
and x = 3, D x = 0.0005
Also, let y = x3
On differentiating both sides w.r.t. x, we get
dy
= 3x 2
dx
dy
\ Dy= ´ D x = 3x 2 ´ 0.0005
dx
= 3 ´ 32 ´ 0.0005
= 27 ´ 0.0005 = 0.0135
 Also, (3.0005)3 = y + Dy
= 33 + 0.0135 = 27.0135
4
\ V = p [27.0135 - 27.000] [using Eq. (i)]
3
4
= p [0.0135] = 4p ´(0.0045)
3
= 0.0180 p cm3

2
Q. 8 A man, 2 m tall, walks at the rate of 1 m/s towards a street light
1 3
which is 5 m above the ground. At what rate is the tip of his shadow
3
moving
1 and at what rate is the length of the shadow changing when he
is 3 m from the base of the light?
3
Sol. Let AB be the street light post and CD be the height of man i.e., CD = 2 m.
A
D
5 13 m
2m

B E
x C y
dx - 5
Let BC = x m, CE = y m and = m/s
dt 3
From DABE and DDCE, we see that
DABE ~ DDCE [by AAA similarity]
16
AB BE x+ y
\ = Þ 3 =
DC CE 2 y
16 x + y
Þ =
6 y
Þ 16 y = 6x + 6 y Þ 10 y = 6x
3
Þ y= x
5
On differentiating both sides w.r.t. t, we get
dy 3 dx 3 æ 2 ö
= × = × ç -1 ÷
dt 5 dt 5 è 3 ø
[since, man is moving towards the light post]
3 æ -5ö
= ×ç ÷ = - 1m/s
5 è 3 ø
Let z=x + y
Now, differentiating both sides w.r.t. t, we get
5
= - æç + 1ö÷
dz dx dy
= +
dt dt dt è3 ø
8 2
= - = - 2 m/s
3 3
2
Hence, the tip of shadow is moving at the rate of 2 m/s towards the light source and
3
length of the shadow is decreasing at the rate of 1 m/s.
Q. 9 A swimming pool is to be drained for cleaning. If L represents the
number of litres of water in the pool t seconds after the pool has been
plugged off to drain and L = 200 (10 - t)2 . How fast is the water running
out at the end of 5 s and what is the average rate at which the water
flows out during the first 5 s?
Sol. Let L represents the number of litres of water in the pool t seconds after the pool has been
plugged off to drain, then
L = 200 (10 - t )2
dL
\ Rate at which the water is running out = -
dt
dL
= - 200 × 2 (10 - t ) × (-1)
dt
= 400 (10 - t )
Rate at which the water is running out at the end of 5 s
= 400 (10 - 5)
= 2000 L/s = Final rate
initial rate = - æç ö÷
dL
Since, = 4000 L/s
è dt øt = 0
Initial rate + Final rate
\ Average rate during 5 s =
2
4000 + 2000
=
2
= 3000 L/s

Q. 10 The volume of a cube increases at a constant rate. Prove that the

increase in its surface area varies inversely as the length of the side.
Sol. Let the side of a cube be x unit.
\ Volume of cube (V ) = x 3
On differentiating both side w.r.t. t, we get
dV dx
= 3x 2 =k [constant]
dt dt
dx k
Þ = ... (i)
dt 3x 2
Also, surface area of cube, S = 6x 2
On differentiating w.r.t. t, we get
dS dx
= 12 x ×
dt dt
dS k
Þ = 12 x × 2 [using Eq. (i)]
dt 3x
dS 12 k
= 4 æç ö÷
k
Þ =
dt 3x èx ø
dS 1
Þ µ
dt x
Hence, the surface area of the cube varies inversely as the length of the side.
Q. 11 If x and y are the sides of two squares such that y = x - x 2 , then find
the rate of change of the area of second square with respect to the
area of first square.
K Thinking Process
First, let A 1 and A 2 be the areas of two squares and get their values in one variable and
then by using dA1/ dt and dA 2 / dt get the value of dA 2 / dA1.

Sol. Since, x and y are the sides of two squares such that y = x - x 2 .
\ Area of the first square ( A1) = x 2
and area of the second square ( A2 ) = y2 = (x - x 2 )2
dA2
(x - x 2 )2 = 2 (x - x 2 ) æç
d dx dx ö
\ = - 2x × ÷
dt dt è dt dt ø
dx
= (1 - 2 x ) 2 (x - x 2 )
dt
dA1 d 2 dx
and = x = 2x ×
dt dt dt
dx
dA2 dA2 / dt × (1 - 2 x ) (2 x - 2 x 2 )
\ = = dt
dA1 dA1 / dt dx
2x ×
dt
(1 - 2 x ) 2 x (1 - x )
=
2x
= (1 - 2 x ) (1 - x )
= 1 - x - 2x + 2x 2
= 2 x 2 - 3x + 1

Q. 12 Find the condition that curves 2x = y 2 and 2xy = k intersect

orthogonally.
K Thinking Process
First, get the intersection point of the curve and then get the slopes of both the curves at
that point. Then, by using m1× m 2=-1, get the required condition.

Sol. Given, equation of curves are 2 x = y2 ... (i)

and 2xy = k ... (ii)
k
Þ y= [from Eq. (ii)]
2x
k ö2
From Eq. (i), 2 x = æç ÷
è 2x ø
3 2
Þ 8x = k
1
Þ x 3 = k2
8
1
Þ x = k2/ 3
2
k k
\ y= = = k1/ 3
2 x 2 × 1 k2/ 3
2
1
Thus, we get point of intersection of curves which is æç k 2 / 3 , k1/ 3 ö÷.
è2 ø
 From Eqs. (i) and (ii),
dy
2 = 2y
dx
2 éx × + y × 1ù = 0
dy
and
ëê dx ûú
dy 1
Þ =
dx y
and æ dy ö = - 2 y = - y
ç ÷
è dx ø 2x x
æ dy ö 1
Þ ç ÷æ 1 = [say m1 ]
è dx ø ç k 2 / 3, k1 / 3 ö÷ k1/ 3
è2 ø

æ dy ö - k1/ 3
and ç ÷æ 1 = = - 2 k - 1/ 3 [say m2 ]
è dx ø ç k 2 / 3, k1 / 3 ö÷ 1 k 2 / 3
è2 ø
2
Since, the curves intersect orthogonally.
i.e., m1 × m2 = - 1
1
Þ × (- 2 k - 1/ 3 ) = - 1
k1/ 3
Þ - 2 k- 2/ 3 = - 1
2
Þ =1
k2/ 3
Þ k2/ 3 = 2
\ k2 = 8
which is the required condition.

Q. 13 Prove that the curves xy = 4 and x 2 + y 2 = 8 touch each other.

K Thinking Process
First, find the intersection points of curves and then equate the slopes of both the curves
at the obtained point.

Sol. Given equation of curves are

xy = 4 ...(i)
and x 2 + y2 = 8 ...(ii)
dy
Þ x× + y=0
dx
dy
and 2x + 2 y =0
dx
dy - y
Þ =
dx x
dy -2 x
and =
dx 2y
dy - y
Þ = = m1 [say]
dx x
dy - x
and = = m2 [say]
dx y
Since, both the curves should have same slope.
- y -x
\ = Þ - y2 = - x 2
x y
Þ x 2 = y2 ...(iii)
 Using the value of x 2 in Eq. (ii), we get
y2 + y2 = 8
Þ y2 = 4 Þ y = ± 2
4
For y = 2, x = = 2
2
4
and for y = - 2, x = = -2
-2
Thus, the required points of intersection are (2, 2) and (- 2, - 2 ).
- y -2
For (2, 2), m1 = = = -1
x 2
- x -2
and m2 = = = -1
y 2
Q m1 = m2
- y - (- 2 )
For (-2,-2), m1 = = = -1
x -2
- x - (- 2 )
and m2 = = = -1
y -2
Thus, for both the intersection points, we see that slope of both the curves are same.
Hence, the curves touch each other.

Q. 14 Find the coordinates of the point on the curve x+ y = 4 at which

tangent is equally inclined to the axes.
Sol. We have, x + y=4 ...(i)
Þ x 1/ 2 + y1/ 2 = 4
1 1 1 1 dy
Þ × + × × =0
2 x 1/ 2 2 y1/ 2 dx
dy 1
\ = - × x - 1/ 2 2 × y1/ 2
dx 2
y
=-
x
Since, tangent is equally inclined to the axes.
dy
\ =±1
dx
y
Þ - =±1
x
y
Þ = 1Þ y = x
x
From Eq. (i), y+ y=4
Þ 2 y=4
Þ 4 y = 16
\ y = 4 and x = 4
When y=4, then x = 4
So, the required coordinates are (4, 4).
Q. 15 Find the angle of intersection of the curves y = 4 - x 2 and y = x 2 .

Sol. We have, y = 4 - x2 ...(i)

and y = x2 ... (ii)
dy
Þ = - 2x
dx
dy
and = 2x
dx
Þ m1 = - 2 x
and m2 = 2 x
From Eqs. (i) and (ii), x2 = 4 - x2
Þ 2x 2 = 4
Þ x2 = 2
Þ x=± 2
\ y = x 2 = ( ± 2 )2 = 2
So, the points of intersection are ( 2 , 2 ) and (- 2 , 2 ).
For point (+ 2 , 2), m1 = - 2 x = - 2 × 2 = - 2 2
and m2 = 2 x = 2 2
m - m2 -2 2 -2 2 -4 2
and for point ( 2 , 2 ), tan q = 1 = =
1 + m1m2 1-2 2 ×2 2 -7
æ4 2 ö
\ q = tan- 1 çç ÷÷
è 7 ø

Q. 16 Prove that the curves y 2 = 4 x and x 2 + y 2 - 6x + 1 = 0 touch each

other at the point (1, 2).
Sol. We have, y2 = 4x and x 2 + y2 - 6x + 1 = 0
Since, both the curves touch each other at (1, 2) i.e., curves are passing through (1, 2).
dy
\ 2y× =4
dx
dy
and 2x + 2 y =6
dx
dy 4
Þ =
dx 2 y
dy 6 - 2 x
and =
dx 2y
æ dy ö 4
Þ ç ÷ = =1
è dx ø( 1, 2 ) 4
æ dy ö 6 -2 ×1 4
and ç ÷ = = =1
è ø( 1, 2 )
dx 2 ×2 4
Þ m1 = 1 and m2 = 1
Thus, we see that slope of both the curves are equal to each other i.e., m1 = m2 = 1 at the
point (1, 2).
Hence, both the curves touch each other.
Q. 17 Find the equation of the normal lines to the curve 3x 2 - y 2 = 8 which
are parallel to the line x + 3 y = 4.
Sol. Given equation of the curve is
3x 2 - y2 = 8 ...(i)
On differentiating both sides w.r.t. x, we get
dy
6x - 2 y =0
dx
dy 6x 3x
Þ = =
dx 2 y y
3x
Þ m1 = [say]
y
-1 - y
and slope of normal (m2 ) = = ...(ii)
m1 3x
Since, slope of normal to the curve should be equal to the slope of line x + 3 y = 4, which is
parallel to curve.
4 - x -x 4
For line, y= = +
3 3 3
-1
Þ Slope of the line (m3 ) =
3
\ m2 = m3
-y 1
Þ =-
3x 3
Þ - 3 y = - 3x
Þ y=x ...(iii)
On substituting the value of y in Eq. (i), we get
3x 2 - x 2 = 8
Þ x2 = 4
Þ x=±2
For x = 2, y=2 [using Eq. (iii)]
and for x = - 2, y= -2 [using Eq. (iii)]
Thus,the points at which normal to the curve are parallel to the line x + 3 y = 4 are (2, 2) and
(- 2, - 2 ).
Required equations of normal are
y - 2 = m2 (x - 2 ) and y + 2 = m2 (x + 2 )
-2 -2
Þ y-2 = (x - 2 ) and y + 2 = (x + 2 )
6 6
Þ 3 y - 6 = - x + 2 and 3 y + 6 = - x - 2
Þ 3y + x = + 8 and 3 y + x = - 8
So, the required equations are 3 y + x = ± 8.

Q. 18 At what points on the curve x 2 + y 2 - 2x - 4 y + 1 = 0, the tangents

are parallel to the Y-axis?
Sol. Given, equation of curve which is
x 2 + y2 - 2 x - 4 y + 1 = 0 ... (i)
dy dy
Þ 2x + 2 y -2 - 4 =0
dx dx
 dy
Þ (2 y - 4) = 2 - 2 x
dx
dy 2(1 - x )
Þ =
dx 2( y - 2 )
dy
Since, the tangents are parallel to the Y-axis i.e., tan q = tan 90° = .
dx
1- x 1
\ =
y-2 0
Þ y-2 = 0
Þ y=2
For y = 2 from Eq. (i), we get
x 2 + 22 - 2x - 4 ´ 2 + 1 = 0
Þ x 2 - 2x - 3 = 0
Þ x 2 - 3x + x - 3 = 0
Þ x(x - 3) + 1(x - 3) = 0
Þ (x + 1)(x - 3) = 0
\ x = - 1, x = 3
So, the required points are (- 1, 2 ) and (3, 2).

x y
Q. 19 Show that the line + = 1, touches the curve y = b × e -x/ a at the
a b
point, where the curve intersects the axis of Y.
Sol. x y
We have the equation of line given by + = 1 ,which touches the curve y = b × e - x / a at
a b
the point, where the curve intersects the axis of Y i.e., x = 0.
\ y = b × e -0/ a = b [Q e 0 = 1]
So, the point of intersection of the curve with Y-axis is (0,b).
Now, slope of the given line at (0, b) is given by
1 1 dy
×1 + × =0
a b dx
dy -1
Þ = ×b
dx a
dy 1 -b
Þ = - ×b = = m1 [say]
dx a a
Also, the slope of the curve at (0, b ) is
dy -1
= b × e -x / a ×
dx a
dy - b - x / a
= e
dx a
æ dy ö - b -0 - b
ç ÷ = e = = m2 [say]
è dx ø( 0, b) a a
-b
Since, m1 = m2 =
a
Hence, the line touches the curve at the point, where the curve intersects the axis of Y.
Q. 20 Show that f (x) = 2x + cot -1 x + log( 1 + x 2 - x) is increasing in R.
K Thinking Process
If f ¢(x) ³ 0, then we can say that f(x) is increasing function. Use this condition to show
the desired result.

Sol. We have, f(x ) = 2 x + cot -1 x + log( 1 + x 2 - x )

æ -1 ö 1 æ 1 ö
\ f ¢(x ) = 2 + çç ÷+ ç × 2 x - 1÷
2 ÷
è 1 + x ø ( 1 + x - x ) çè 2 1 + x ÷
2 2
ø
1 1 (x - 1 + x 2 )
=2 - 2
+ ×
1+ x 2
( 1 + x - x) 1 + x2
1 1
=2 - -
1 + x2 1 + x2
2 + 2x 2 - 1 - 1 + x 2 1 + 2 x 2 `- 1 + x 2
= 2
=
1+ x 1 + x2
For increasing function, f ¢(x ) ³ 0
1 + 2x - 1 + x 2
2
Þ ³0
1 + x2
Þ 1 + 2x 2 ³ 1 + x 2
Þ (1 + 2 x 2 )2 ³ 1 + x 2
Þ 1 + 4x 4 + 4x 2 ³ 1 + x 2
Þ 4x 4 + 3x 2 ³ 0
Þ x 2 (4x 2 + 3) ³ 0
which is true for any real value of x.
Hence, f(x ) is increasing in R.

Q. 21 Show that for a ³ 1, f (x) = 3 sin x - cos x - 2ax + b is decreasing in R.

K Thinking Process
If f ¢(x) £ 0, then we can say that f(x) is a decreasing function. So, use this condition to
show the result.

Sol. We have, a ³ 1, f(x ) = 3 sin x - cos x - 2 ax + b

\ f ¢(x ) = 3 cos x - (- sin x ) - 2 a
= 3 cos x + sin x - 2 a
é 3 1 ù
= 2ê × cos x + × sin x ú - 2 a
ë 2 2 û
p p
= 2 cos × cos x + sin × sin x ù - 2 a
é
ëê 6 6 ûú
æ p ö
= 2 ç cos - x ÷ - 2 a
è 6 ø
[Qcos( A - B) = cos A × cos B + sin A × sin B ]
éæ p ö ù
= 2 ê ç cos - x ÷ - aú
ëè 6 ø û
 We know that, cos x Î [- 1, 1]
and a ³1
é p ù
So, 2 êcosæç - x ö÷ - aú £ 0
ë è 6 ø û
\ f ¢(x ) £ 0
Hence, f(x ) is a decreasing function in R.

Q. 22 Show that f (x) = tan -1 (sin x + cos x) is an increasing function in æç 0, p ö÷.

è 4ø
Sol. We have, f(x ) = tan-1(sin x + cos x )
1
\ f ¢(x ) = × (cos x - sin x )
1 + (sin x + cos x )2
1
= (cos x - sin x )
1 + sin2 x + cos 2 x + 2 sin x × cos x
1
= (cos x - sin x )
(2 + sin 2 x )
[Q sin 2 x = 2 sin x cos x and sin2 x + cos 2 x = 1 ]
For f ¢(x ) ³ 0,
1
× (cos x - sin x ) ³ 0
(2 + sin2 x )

Þ cos x - sin x ³ 0
é æ p öù
êëQ ( 2 + sin2 x ) ³ 0 in çè 0, 4 ÷ø úû
Þ cos x ³ sin x
p
which is true, if x Î æç 0, ö÷.
è 4ø
p
Hence, f(x ) is an increasing function in æç 0, ö÷.
è 4ø

Q. 23 At what point, the slope of the curve y = - x 3 + 3x 2 + 9x - 27 is

maximum? Also, find the maximum slope.
Sol. We have, y = - x 3 + 3x 2 + 9x - 27
dy
\ = - 3x 2 + 6x + 9 = Slope of tangent to the curve
dx
d2y
Now, = - 6x + 6
dx 2
d æ dy ö
For ç ÷ = 0,
dx è dx ø
- 6x + 6 = 0
-6
Þ x= =1
-6
d æd2y ö
\ ç ÷ = - 6< 0
dx çè dx 2 ÷ø
So, the slope of tangent to the curve is maximum, when x = 1.
For x = 1, æ dy ö = - 3 × 12 + 6 × 1 + 9 = 12,
ç ÷
è dx ø( x = 1)
which is maximum slope.
 Also, for x = 1, y = - 13 + 3 × 12 + 9 × 1 - 27
= - 1 + 3 + 9 - 27
= - 16
So, the required point is (1, - 16).

p
Q. 24 Prove that f (x) = sin x + 3 cos x has maximum value at x = .
6
Sol. We have, f(x ) = sin x + 3 cos x
\ f ¢(x ) = cos x + 3 (- sin x )
= cos x - 3 sin x
For f ¢(x ) = 0, cos x = 3sin x
1 p
Þ tan x = = tan
3 6
p
Þ x=
6
Again, differentiating f ¢(x ), we get
f ¢¢(x ) = - sin x - 3 cos x
p p p
At x = , f ¢¢(x ) = - sin - 3 cos
6 6 6
1 3
= - - 3×
2 2
1 3
= - - = -2<0
2 2
p p
Hence, at x = , f(x ) has maximum value at is the point of local maxima.
6 6

Long Answer Type Questions

Q. 25 If the sum of lengths of the hypotenuse and a side of a right angled
triangle is given, then show that the area of triangle is maximum,
p
when the angle between them is .
3
Sol. Let ABC be a triangle with AC = h, AB = x and BC = y.
Also, ÐCAB = q
Let h+ x =k ...(i)
C

h y

q
A x B
x
\ cos q =
h
Þ x = hcos q
Þ h + hcos q = k [using Eq. (i)]
Þ h (1+ cos q) = k
k
Þ h= …(ii)
(1 + cos q)
1
Also, area of DABC = ( AB × BC )
2
1
A = ×x × y
2
1 éQ sin q = y ù
= hcos q × hsin q
2 ëê h úû
1
= h2 sin q × cos q
2
2 h2
= sin q × cos q
4
1
= h2 sin 2 q ...(iii)
4
k
Since, h=
1 + cos q
2
1æ k ö
\ A= ç ÷ × sin 2 q
4 çè 1 + cos q ÷ø
k2 sin 2 q
Þ A= × ...(iv)
4 (1 + cos q)2
dA k 2 é (1 + cos q)2 × cos 2 q × 2 - sin2 q × 2(1 + cos q) × (0 - sin q)ù
\ = ê ú
dq 4 ë (1 + cos q)4 û
k 2 ì2(1 + cos q)[(1 + cos q) × cos 2 q + sin 2 q (sin q)ü
= í ý
4î (1 + cos q)4 þ
k2 2 2
= × [(1 + cos q) × cos 2 q + 2 sin q × cos q]
4 (1 + cos q)3
k2
= [(1 + cos q)(1 - 2 sin2 q) + 2 sin2 q × cos q]
2(1 + cos q)3
k2
= [1 + cos q - 2 sin2 q - 2 sin2 q × cos q + 2 sin2 q × cos q]
2(1 + cos q)3
k2
= [(1 + cos q) - 2 sin2 q]
2(1 + cos q)3
k2
= [1 + cos q - 2 + 2 cos 2 q]
2(1 + cos q)3
k2
= (2 cos 2 q + cos q - 1) ...(v)
2(1 + cos q)3
dA
For = 0,
dq
k2
(2 cos 2 q + cos q - 1) = 0
2 (1 + cos q)3
Þ 2 cos 2 q + cos q - 1 = 0
2
Þ 2 cos q + 2 cos q - cos q - 1 = 0
Þ 2 cos q (cos q + 1) - 1 (cos q + 1) = 0
Þ (2 cos q - 1)(cos q + 1) = 0
 1
Þ cos q =
or cos q = - 1
2
p
Þ q= [possible]
3
or q = 2np ± p [not possible]
p
\ q=
3
Again, differentiating w.r.t. q in Eq. (v), we get
d æ dA ö d é k2 ù
ç ÷= ê (2 cos 2 q + cos q - 1)ú
dq è dq ø dq ë 2(1 + cos q)3
û
d2A d é k 2 (2 cos q - 1)(1 + cos q)ù d é k 2 (2 cos q - 1)ù
\ = ú= ê ×
dq dq êë
2
2(1 + cos q)3 û dq ë 2 (1 + cos q) û
2ú

k 2 é (1 + cos q)2 × (- 2 sin q) - 2(1 + cos q) × (- sin q)(2 cos q - 1)ù

=
2 êë (1 + cos q)4
ú
û
é (1 + cos q) × [1 + cos q](-2 sin q) + 2 sin q (2 cos q - 1)ù
k2
= ê ú
2ë (1 + cos q)4 û
k é -2 sin q - 2 sin q × cos q + 4sin q × cos q - 2 sin q ù
2
=
2 êë (1 + cos q)3
ú
û
k 2 é - 4sin q - sin2 q + 2 sin2 q ù k 2 é sin2 q - 4sin q ù
= ê ú = ê ú
2 ë (1 + cos q)3 û 2 ë (1 + cos q)3 û
é 2p pù é 3 4 3ù
- 4sin ú -
k2 ê 2 ê ú
æd2A ö sin
3 3ú= k
\ ç 2 ÷ = ê ê 2 2 ú
ç dq ÷ p 2 ê æ 3
pö ú 2 ê æ 1ö ú3
è ø at q =
3 êë çè 1 + cos 3 ÷ø úû êë çè 1 + 2 ÷ø úû
k2 é - 3 3 × 8ù æ2 3 ö
= = - k 2 çç ÷÷
2 êë 2 × 27 úû è 9 ø
which is less than zero.
p
Hence, area of the right angled triangle is maximum, when the angle between them is .
3

Q. 26 Find the points of local maxima, local minima and the points of
inflection of the function f (x) = x 5 - 5x 4 + 5x 3 - 1. Also, find the
corresponding local maximum and local minimum values.
Sol. Given that, f(x ) = x 5 - 5x 4 + 5x 3 - 1
On differentiating w.r.t. x, we get
f ¢(x ) = 5x 4 - 20x 3 + 15x 2
For maxima or minima, f ¢(x ) = 0
Þ 5x 4 - 20x 3 + 15x 2 = 0
Þ 5x 2 (x 2 - 4x + 3) = 0
Þ 5x (x 2 - 3x - x + 3) = 0
2

Þ 5x 2 [x(x - 3) - 1(x - 3)] = 0

Þ 5x 2 [(x - 1)(x - 3)] = 0
\ x = 0, 1, 3
 dy
Sign scheme for = 5x 2 (x - 1)(x - 3)
dx
–¥+ + – + +¥
0 1 3
So, y has maximum value at x = 1and minimum value at x = 3.
At x = 0 ,y has neither maximum nor minimum value.
\ Maximum value of y = 1 - 5 + 5 - 1= 0
and minimum value = (3)5 - 5(3)4 + 5(3)3 - 1
= 243 - 81 ´ 5 - 27 ´ 5 - 1= - 298

Q. 27 A telephone company in a town has 500 subscribers on its list and

collects fixed charges of ` 300 per subscriber per year. The company
proposes to increase the annual subscription and it is believed that for
every increase of ` 1 per one subscriber will discontinue the service.
Find what increase will bring maximum profit?
Sol. Consider that company increases the annual subscription by ` x.
So, x subscribes will discontinue the service.
\ Total revenue of company after the increment is given by
R (x ) = (500 - x ) (300 + x )
= 15 ´ 104 + 500x - 300x - x 2
= - x 2 + 200x + 150000
On differentiating both sides w.r.t. x, we get
R ¢(x ) = - 2 x + 200
Now, R ¢ (x ) = 0
Þ 2 x = 200 Þ x = 100
\ R ¢¢ (x ) = - 2 < 0
So, R (x ) is maximum when x = 100 .
Hence, the company should increase the subscription fee by ` 100, so that it has maximum
profit.

Q. 28 If the straight line x cos a + y sin a = p touches the curve

2
x y2
2
+ 2
= 1, then prove that a 2 cos2 a + b 2 sin 2 a = p 2 .
a b
Sol. Given, line is x cos a + y sin a = p ... (i)
x2 y2
and curve is 2 + 2 = 1
a b
Þ b 2 x 2 + a 2 y2 = a 2 b 2 ...(ii)
Now, differentiating Eq. (ii) w.r.t. x, we get
dy
b 2 × 2 x + a2 × 2 y × =0
dx
dy - 2 b 2 x - xb 2
Þ = = ... (iii)
dx 2 a2 y ya2
From Eq. (i), y sin a = p - x cos a
p
Þ y = - x cot a +
sin a
Thus, slope of the line is (- cot a ).
 æ x b2 ö
So, the given equation of line will be tangent to the Eq. (ii), if çç - × 2 ÷ = (- cot a)
÷
è y a ø
x y
Þ = =k [say]
a2 cos a b 2 sin a
Þ x = ka2 cos a
and y = b 2 k sin a
x2 y2
So, the line x cos a + y sin a = p will touch the curve 2
+ at point
a b2
(ka2 cos a, kb 2 sin a).
From Eq. (i), ka2 cos 2 a + k b 2sin2 a = p
p
Þ a2 cos 2 a + b 2 sin2 a =
k
p2
Þ (a2 cos 2 a + b 2 sin2 a )2 = 2 ...(iv)
k
From Eq. (ii), b 2 k 2 a4 cos 2 a + a2 k 2 b 4 sin2 a = a2 b 2
Þ k 2 (a2 cos 2 a + b 2 sin2 a ) = 1
1
Þ (a2 cos 2 a + b 2 sin2 a ) = 2 ... (v)
k
On dividing Eq. (iv) by Eq. (v), we get
a2 cos 2 a + b 2 sin2 a = p2 Hence proved.

Alternate Method
x2 y2
We know that, if a line y = mx + c touches ellipse 2
+ = 1, then
a b2
the required condition is c 2 = a2 m2 + b 2
Here, given equation of the line is
x cos a + y sin a = p
p - x cos a
Þ y=
sin a
p
= - x cot a +
sin a
p
Þ c=
sin a
and m = - cot a
2
æ p ö
\ çç ÷÷ = a2 (- cot a) 2 + b 2
è sin a ø
p2 cos 2 a
Þ 2
= a2 + b2
sin a sin2 a
Þ p2 = a2 cos 2 a + b 2 sin2 a Hence proved.
Q. 29 If an open box with square base is to be made of a given quantity of
card board of area c 2, then show that the maximum volume of the box
c3
is cu units.
6 3
K Thinking Process
dV dV
First, let the sides of box in x and y then find in terms c and x. Also, for = 0 get
dx dx
d 2V
the value of x and if < 0 at the value of x, then by putting that value of x in the
dx2
equation of V, get the desired result.

Sol. Let the length of side of the square base of open box be x units and its height be y units.
\ Area of the metal used = x 2 + 4xy
Þ x 2 + 4xy = c 2 [given]
c2 - x 2
Þ y= ...(i)
4x
Now, volume of the box (V ) = x 2 y
æ c2 - x 2 ö
Þ V = x 2 × çç ÷
÷
è 4x ø y
1
= x (c 2 - x 2 ) x
4 x
1
= (c 2 x - x 3 )
4
On differentiating both sides w.r.t. x, we get
dV 1 2
= (c - 3x 2 ) ... (ii)
dx 4
dV
Now, = 0 Þ c 2 = 3x 2
dx
c2
Þ x2 =
3
c
Þ x= [using positive sign]
3
Again, differentiating Eq. (ii) w.r.t. x, we get
d 2V 1 -3
= (- 6x ) = x<0
dx 2 4 2
æ d 2v ö
= - × æç
3 c ö
\ ç 2÷ ÷<0
ç dx ÷ 2 è 3ø
è ø at x = c
3
c
Thus, we see that volume (V) is maximum at x = .
3
1æ c c3 ö
\ Maximum volume of the box, (V ) = çç c 2 . - ÷
3 3 3 ÷ø
c
x= 4è
3
1 (3c 3 - c 3 ) 1 2 c 3
= × = ×
4 3 3 4 3 3
c3
= cu units
6 3
Q. 30 Find the dimensions of the rectangle of perimeter 36 cm which will
sweep out a volume as large as possible, when revolved about one of
its sides. Also, find the maximum volume.
Sol. Let breadth and length of the rectangle be x and y, respectively.

x
Q Perimeter of the rectangle = 36 cm
Þ 2 x + 2 y = 36
Þ x + y = 18
Þ y = 18 - x ... (i)
Let the rectangle is being revolved about its length y.
Then, volume (V) of resultant cylinder = p x 2 × y
Þ V = px 2 × (18 - x ) [Q V = pr 2 h] [using Eq. (i)]
= 18px 2 - px 3 = p [18x 2 - x 3 ]
On differentiating both sides w.r.t. x, we get
dV
= p (36x - 3x 2 )
dx
dV
Now, =0
dx
Þ 36x = 3x 2
Þ 3x 2 - 36x = 0
Þ 3 (x 2 - 12 x ) = 0
Þ 3x (x - 12 ) = 0
Þ x = 0, x = 12
\ x = 12 [Q, x ¹ 0]
Again, differentiating w.r.t. x, we get
d 2V
= p (36 - 6x )
dx 2
æ d 2V ö
Þ ç 2÷ = p(36 - 6 ´ 12 ) = - 36p < 0
ç dx ÷
è ø x = 12
At x = 12, volume of the resultant cylinder is the maximum.
So, the dimensions of rectangle are 12 cm and 6 cm, respectively. [using Eq. (i)]
\ Maximum volume of resultant cylinder,
(V )x = 12 = p [18 × (12 )2 - (12 )3 ]
= p [12 2 (18 - 12 )]
= p ´ 144 ´ 6
= 864 p cm3
Q. 31 I the sum of the surface areas of cube and a sphere is constant, what is
the ratio of an edge of the cube to the diameter of the sphere, when
the sum of their volumes is minimum?
Sol. Let length of one edge of cube be x units and radius of sphere be r units.
\ Surface area of cube = 6x 2
and surface area of sphere = 4pr 2
Also, 6x 2 + 4pr 2 = k [constant, given]
Þ 6x 2 = k - 4pr 2
k - 4pr 2
Þ x2 =
6
1/ 2
é k - 4pr 2 ù
Þ x=ê ú ... (i)
ë 6 û
Now, volume of cube = x 3
4
and volume of sphere = pr 3
3
Let sum of volume of the cube and volume of the sphere be given by
3/ 2
4 3 é k - 4pr 2 ù 4 3
S = x3 + pr = ê ú + pr
3 ë 6 û 3
On differentiating both sides w.r.t. r, we get
1/ 2
dS 3 é k - 4pr 2 ù - 8pr ö 12 2
= ê ú × æç ÷+ pr
dr 2 ë 6 û è 6 ø 3
1/ 2
é k - 4pr 2 ù 2
= - 2 pr ê ú + 4pr ... (ii)
ë 6 û
éì k - 4pr 2 ü1/ 2 ù
= - 2 pr êí ý - 2 rú
êëî 6 þ úû
dS
Now, =0
dr
1/ 2
æ k - 4pr 2 ö
Þ r = 0 or 2 r = çç ÷
÷
è 6 ø
2 k - 4pr 2
Þ 4r = Þ 24r = k - 4pr 2
2
6
Þ 24r 2 + 4pr 2 = k Þ r 2 [24 + 4p ] = k
k 1 k
\ r = 0 or r= =
24 + 4p 2 6+ p
We know that, r¹0
1 k
\ r=
2 6+ p
Again, differentiating w.r.t. r in Eq. (ii), we get

d 2S é ìæ k - 4pr 2 ö
1/ 2
üïù
ê - 2 pr ïíç
d
2
= ç
÷
÷ + 4pr 2 ýú
dr dr ê ïîè 6 ø ïþúû
ë
 é 1 æ k - 4pr 2 ö - 1/ 2 - 8pr 2 1/ 2 ù
= -2 p ê r × çç ÷ × æ ö + æç k - 4pr ö÷ × 1ú + 4p × 2 r
÷ ç ÷ ç ÷
êë 2 è 6 ø è 6 ø è 6 ø úû
é ù
ê ú
1 æ - 8pr ö k - 4pr 2
= - 2 p êr × ×ç ÷ + ú + 8pr
ê k - 4pr 2 è 6 ø 6 ú
ê 2 ú
ë 6 û
é æ 4p r 2
ö ù
ê - 8pr + 12 çç k - ÷ú
2

ê è 6 ÷ø ú
= - 2p + 8pr
ê k - 4pr 2 ú
ê 12 ú
êë 6 úû
é ù é ù
ê ú ê ú
- 48pr 2 + 72 k - 48pr 2 - 96pr 2 + 72 k
= -2 p ê ú + 8pr = - 2 p ê ú + 8pr > 0
ê k - 4pr 2 ú ê k - 4pr 2 ú
ê 72 ú ê 72 ú
ë 6 û ë 6 û
1 k
For r = , then the sum of their volume is minimum.
2 6+ p
é k - 4p × 1 k ù1/ 2
1 k ê 4 (6 + p) ú
For r = , x=ê ú
2 6+ p ê 6 ú
êë úû
1/ 2 1/ 2
é (6 + p) k - pk ù é k ù
=ê ú = ê 6 + p ú = 2r
ë 6 ( 6 + p) û ë û
Since, the sum of their volume is minimum when x = 2 r.
Hence, the ratio of an edge of cube to the diameter of the sphere is 1:1.

Q. 32 If AB is a diameter of a circle and C is any point on the circle, then

show that the area of DABC is maximum, when it is isosceles.
Sol. We have, AB = 2 r
and ÐACB = 90° [since, angle in the semi-circle is always 90°]
Let AC = x and BC = y
\ (2 r )2 = x 2 + y2
Þ y2 = 4r 2 - x 2
Þ y= 4r 2 - x 2 ... (i)
1
Now, area of DABC, A = ´ x ´ y
2
1
= ´ x ´ (4r 2 - x 2 )1/ 2 [using Eq. (i)]
2
Now, differentiating both sides w.r.t. x, we get
dA 1 é 1 2
= x × (4r - x 2 )- 1/ 2 × (0 - 2 x ) + (4r 2 - x 2 )1/ 2 × 1ù
dx 2 êë 2 úû

1 é - 2x 2 ù
= ê + (4r 2 - x 2 )1/ 2 ú
2 ê 2 4r 2 - x 2 úû
ë
 1 é - x2 ù
ê= + 4r 2 - x 2 ú
2 ê 4r 2 - x 2 úû
ë
1 é - x + 4r - x ù 1 é - 2 x 2 + 4r 2 ù
2 2 2
= ê ú= ê ú
2 ê 4r 2 - x 2 úû 2 êë 4r 2 - x 2 úû
ë
dA é (- x 2 + 2 r 2 )ù
Þ =ê ú B
dx ê 4r 2 - x 2 ú
ë û
dA
Now, =0
dx
2 2
Þ - x + 2r = 0 C
1
Þ r2 = x 2
2
1
Þ r= x A x C
2
\ x=r 2
Again, differentiating both sides w.r.t. x, we get
1
4r 2 - x 2 × (- 2 x ) + (2 r 2 - x 2 ) × (4r 2 - x 2 )- 1/ 2 (- 2 x )
d2A 2
=
dx 2 ( 4r 2 - x 2 )2
é 1 ù
- 2 x ê 4r 2 - x 2 + (2 r 2 - x 2 ) × ú
2 2
êë 2 4r - x úû
=
( 4r 2 - x 2 )2
2
- 4x × çæ 4r 2 - x 2 ÷ö + (2 r 2 - x 2 ) (- 2 x )
= è ø
2 × (4r 2 - x 2 )3 / 2
- 4x (4r 2 - x 2 ) + (2 r 2 - x 2 )× (- 2 x )
=
2 × (4r 2 - x 2 )3 / 2
- 16xr 2 + 4x 3 + (2 r 2 - x 2 ) (- 2 x )
=
2 × (4r 2 - x 2 )3 / 2
æd2A ö - 16 × r 2 × r 2 + 4 × (r 2 )3 + [2 r 2 - (r 2 )2 ] × (- 2 × r 2 )
ç 2÷ = [Q x = r 2 ]
ç dx ÷ 2 × (4r 2 - 2 r 2 )3 / 2
è øx = r 2

- 16 2 × r 3 + 8 2 r 3 8 2 r 2 [r - 2 r ]
= 2 3/ 2
=
2 (2 r ) 4r 3
- 8 2 r3
= = - 2 2< 0
4r 3
For x = r 2, the area of triangle is maximum.
For x = r 2, y= 4r 2 - (r 2 )2 = 2 r 2 = r 2
Since, x=r 2=y
Hence, the triangle is isosceles.
Q. 33 A metal box with a square base and vertical sides is to contain
1024 cm 3 . If the material for the top and bottom costs ` 5per cm 2 and
the material for the sides costs ` 2.50 per cm 2 . Then, find the least
cost of the box.
Sol. Since, volume of the box = 1024 cm3
Let length of the side of square base be x cm and height of the box be y cm.

y
x
x
2
\ Volume of the box (V ) = x × y = 1024
1024
Since, x 2 y = 1024 Þ y =
x2
Let C denotes the cost of the box.
\ C = 2 x 2 ´ 5 + 4xy ´ 2.50
= 10x 2 + 10xy = 10x (x + y)
1024 ö
= 10x æç x + ÷
è x2 ø
10x
= 2 (x 3 + 1024)
x
10240
Þ C = 10x 2 + ... (i)
x
On differentiating both sides w.r.t. x, we get
dC
= 20x + 10240 (- x )- 2
dx
10240
= 20x - ...(ii)
x2
dC
Now, =0
dx
10240
Þ 20x =
x2
3
Þ 20x = 10240
Þ x 3 = 512 = 83 Þ x=8
Again, differentiating Eq. (ii) w.r.t. x, we get
d 2C 1
2
= 20 - 10240 (- 2 ) × 3
dx x
20480
= 20 + > 0
x3
æd C ö
2
20480
\ ç 2÷ = 20 + = 60 > 0
ç dx ÷ 512
è øx = 8
For x = 8, cost is minimum and the corresponding least cost of the box,
10240
C(8) = 10 × 82 +
8
= 640 + 1280 = 1920
\ Least cost = ` 1920
Q. 34 The sum of surface areas of a rectangular parallelopiped with sides x,
x
2x and and a sphere is given to be constant. Prove that the sum of
3
their volumes is minimum, if x is equal to three times the radius of the
sphere. Also, find the minimum value of the sum of their volumes.
Sol. We have given that, the sum of the surface areas of a rectangular parallelopiped with sides
x
x, 2x and and a sphere is constant.
3
Let S be the sum of both the surface area.
S = 2 æç x × 2 x + 2 x × + × x ö÷ + 4pr 2 = k
x x
\
è 3 3 ø
é 2 2x 2 x2 ù 2
k = 2 ê2 x + + ú + 4pr
ë 3 3 û
= 2 [3x 2 ] + 4pr 2 = 6x 2 + 4pr 2
Þ 4pr 2 = k - 6x 2
k - 6x 2
Þ r2 =
4p
k - 6x 2
Þ r= ... (i)
4p
Let V denotes the volume of both the parallelopiped and the sphere.
x 4 2 4
Then, V = 2 x × x × + pr 3 = x 3 + pr 3
3 3 3 3
3/ 2
2 3 4 æ k - 6x 2 ö
=x + p çç ÷
3 3 è 4p ÷ø
2 4 1
= x 3 + p × 3 / 2 (k - 6x 2 )3 / 2
3 3 8p
2 1
= x3 + (k - 6x 2 )3 / 2 ...(ii)
3 6 p
On differentiating both sides w.r.t. x, we get
dV 2 1 3
= × 3x 2 + × (k - 6x 2 )1/ 2 × (- 12 x )
dx 3 6 p 2
12 x
= 2x 2 - k - 6x 2
4 p
3x
= 2x 2 - (k - 6x 2 )1/ 2 ...(iii)
p
dV
Q =0
dx
3x
Þ 2x 2 = (k - 6x 2 )1/ 2
p
9x 2
Þ 4x 4 = (k - 6x 2 )
p
Þ 4px 4 = 9 k x 2 - 54x 4
Þ 4px 4 + 54x 4 = 9 k x 2
Þ x 4 [4p + 54] = 9 × k × x 2
9k
Þ x2 =
4p + 54
k
Þ x = 3× ...(iv)
4p + 54
Again, differentiating Eq. (iii) w.r.t. x, we get
d 2V 3 é x × 1 (k - 6x 2 )- 1/ 2 × (- 12 x ) + (k - 6x 2 )1/ 2 × 1ù
= 4x -
dx 2 p ëê 2 ûú
3
= 4x - [- 6x 2 × (k - 6x 2 )- 1/ 2 + (k - 6x 2 )1/ 2 ]
p
3 é - 6x 2 + k - 6x 2 ù
= 4x - ê ú
p ê k - 6x 2 úû
ë
3 é k - 12 x ù
2
= 4x - ê ú
p ê k - 6x 2 ú
ë û
é k ù
æ d 2V ö ê k - 12 × 9 × ú
k 3 4p + 54
Now, çç 2 ÷÷ = 4× 3 - ê ú
è dx ø x = 3 × k 4p + 54 p ê 6× 9× k ú
4 p + 54 ê k - ú
ë 4p + 54 û
é 108k ù
ê k- ú
k 3 4p + 54
= 12 - ê ú
4p + 54 p ê 54 k ú
ê k -
ë 4p + 54 úû
k 3 é 4 kp + 54 k - 108 k / 4p + 54 ù
= 12 - ê ú
4p + 54 p ë 4 kp + 54 k - 54 k / 4p + 54 û
k 3 é 4 kp - 54k ù
= 12 - ê ú
4p + 54 p ë 4 kp 4p + 54 û

k 6 é k (2 p - 27 ) ù
= 12 - ê ú
4p + 54 p ê k 16p 2 + 216p ú
ë û
é d 2V ù
êsince, (2 p - 27) < 0 Þ 2
> 0; k > 0ú
ë dx û
k
For x = 3 , the sum of volumes is minimum.
4p + 54
k k - 6x 2
For x = 3 , then r= [using Eq. (i)]
4p + 54 4p
1 9k
= k - 6×
2 p 4p + 54
1 4kp + 54 k - 54 k
= ×
2 p 4p + 54
1 4kp k 1
= = = x
2 p 4p + 54 4p + 54 3
Þ x = 3r Hence proved.
\ Minimum sum of volume,
3
2 3 4 3 2 3 4 æ1 ö
Væ ö
= x + pr = x + p × ç x ÷
ç x = 3×
k
÷ 3 3 3 3 è3 ø
ç 4 p + 54 ÷ø
è
2 3 4 x3 2 3 æ 2p ö
= x + p× = x ç1 + ÷
3 3 27 3 è 27 ø
Objective Type Questions
Q. 35 If the sides of an equilateral triangle are increasing at the rate of 2
cm/s then the rate at which the area increases, when side is 10 cm, is
(a) 10 cm2 / s (b) 3 cm2 / s
2 10
(c) 10 3 cm / s (d) cm2 / s
3
Sol. (c) Let the side of an equilateral triangle be x cm.
3 2
\ Area of equilateral triangle, A = x …(i)
4
dx
Also, = 2cm/s
dt
On differentiating Eq. (i) w.r.t. t, we get
dA 3 dx
= × 2x ×
dt 4 dt
3 é Q x = 10 and dx = 2 ù
= × 2 × 10 × 2
4 êë dt úû
= 10 3 cm2 /s

Q. 36 A ladder, 5 m long, standing on a horizontal floor, leans against a

vertical wall. If the top of the ladder slides downwards at the rate of
10 cm/s, then the rate at which the angle between the floor and the
ladder is decreasing when lower end of ladder is 2 m from the wall is
1 1
(a) rad/s (b) rad/s
10 20
(c) 20 rad/s (d) 10 rad/s

Sol. (b) Let the angle between floor and the ladder be q.
Let AB = x cm and BC = y cm
x y A
\ sin q = and cos q =
500 500
Þ x = 500 sin q and y = 500 cos q
dx
Also, = 10 cm/s
cm

dt x
0
50

dq
Þ 500 × cos q × = 10 cm/s
dt
dq 10 1
Þ = = q
dt 500cos q 50 cos q C B
y
For y = 2 m = 200 cm,
dq 1 10
= =
dt y y
50 ×
500
10 1
= = rad/s
200 20
Q. 37 The curve y = x 1/ 5 has at (0, 0)
(a) a vertical tangent (parallel to Y-axis)
(b) a horizontal tangent (parallel to X-axis)
(c) an oblique tangent
(d) no tangent

Sol. (a) We have, y = x 1/ 5

1
dy 1 5 - 1 1 -4 / 5
Þ = x = x
dx 5 5
æ dy ö 1
\ ç ÷ = ´ (0)-4 / 5 = ¥
è dx ø( 0, 0 ) 5
So, the curve y = x 1/ 5 has a vertical tangent at (0, 0), which is parallel to Y-axis.

Q. 38 The equation of normal to the curve 3x 2 - y 2 = 8 which is parallel to

the line x + 3 y = 8 is
(a) 3x - y = 8 (b) 3x + y + 8 = 0
(c) x + 3y ± 8 = 0 (d) x + 3y = 0

Sol. (c) We have, the equation of the curve is 3x 2 - y2 = 8 ...(i)

Also, the given equation of the line is x + 3 y = 8.
Þ 3y = 8 - x
x 8
Þ y=- +
3 3
1
Thus, slope of the line is - which should be equal to slope of the equation of normal
3
to the curve.
On differentiating Eq. (i) w.r.t. x, we get
dy
6x - 2 y =0
dx
dy 6x 3x
Þ = = = Slope of the curve
dx 2 y y
1
Now, slope of normal to the curve = -
æ dy ö
ç ÷
è dx ø
1 y
=- =-
æ 3x ö 3x
ç ÷
è y ø
1
- æç
y ö
\ ÷=-
è 3x ø 3
Þ - 3 y = - 3x
Þ y=x
On substituting the value of the given equation of the curve, we get
3x 2 - x 2 = 8
8
Þ x2 =
2
Þ x=±2
 For x = 2, 3 (2 )2 - y2 = 8
Þ y2 = 4
Þ y=±2
and for x = - 2, 3 (- 2 )2 - y2 = 8
Þ y=±2
So, the points at which normals are parallel to the given line are (± 2, ± 2).
Hence, the equation of normal at (± 2, ± 2 ) is
1
y - (± 2 ) = - [x - (± 2 )]
3
Þ 3[ y - ( ± 2 )] = - [x - (± 2 )]
Þ x + 3y ± 8 = 0

Q. 39 If the curve ay + x 2 = 7 and x 3 = y, cut orthogonally at (1, 1), then

the value of a is
(a) 1 (b) 0 (c) - 6 (d) 6

Sol. (d) We have, ay + x 2 = 7 and x 3 = y

On differentiating w.r.t. x in both equations, we get
dy dy
a× + 2x = 0 and 3x 2 =
dx dx
dy 2x dy
Þ =- and = 3x 2
dx a dx
æ dy ö -2
Þ ç ÷ = = m1
è dx ø( 1, 1) a

and æ dy ö = 3 × 1 = 3 = m2
ç ÷
è dx ø( 1, 1)
Since, the curves cut orthogonally at (1, 1).
\ m1 × m2 = - 1

Þ æ -2 ö × 3 = - 1
ç ÷
è a ø
\ a=6

Q. 40 If y = x 4 - 10 and x changes from 2 to 1.99, then what is the change in

y?
(a) 0.32 (b) 0.032 (c) 5.68 (d) 5.968
dy
Sol. (a) We have, y = x 4 - 10 Þ = 4x 3
dx
and Dx = 2.00 - 199
. = 0.01
dy
\ Dy = ´ Dx
dx
= 4x 3 ´ Dx
= 4 ´ 2 3 ´ 0.01
= 32 ´ 0.01 = 0.32
So, the approximate change in y is 0.32.
Q. 41 The equation of tangent to the curve y(1 + x 2 ) = 2 - x, where it
crosses X-axis, is
(a) x + 5y = 2 (b) x - 5y = 2
(c) 5x - y = 2 (d) 5x + y = 2

Sol. (a) We have, equation of the curve y(1 + x 2 ) = 2 - x ...(i)

dy
\ y × (0 + 2 x ) + (1 + x 2 ) × = 0-1 [on differentiating w.r.t. x]
dx
dy
Þ 2 xy + (1 + x 2 ) = -1
dx
dy - 1 - 2 xy
Þ = ...(ii)
dx 1 + x2
Since, the given curve passes through X-axis i.e., y = 0.
\ 0 (1 + x 2 ) = 2 - x [using Eq. (i)]
Þ x =2
So, the curve passes through the point (2, 0).
æ dy ö - 1- 2 ´ 0 1
\ ç ÷ = = - = Slope of the curve
è dx ø( 2, 0 ) 1 + 22 5
1
\Slope of tangent to the curve = -
5
\Equation of tangent of the curve passing through (2, 0) is
1
y - 0 = - (x - 2 )
5
x 2
Þ y+ =+
5 5
Þ 5y + x = 2

Q. 42 The points at which the tangents to the curve y = x 3 - 12x + 18 are

parallel to X-axis are
(a) (2, - 2), ( - 2, - 34) (b) (2, 34), ( - 2, 0)
(c) (0 , 34), ( - 2, 0) (d) (2, 2), ( - 2, 34)

Sol. (d) The given equation of curve is

y = x 3 - 12 x + 18
dy
\ = 3x 2 - 12 [on differentiating w.r.t. x]
dx
So, the slope of line parallel to the X-axis.
\ æ dy ö = 0
ç ÷
è dx ø
Þ 3x 2 - 12 = 0
12
Þ x2 = =4
3
\ x=±2
For x = 2, y = 2 3 - 12 ´ 2 + 18 = 2
and for x = - 2, y = (- 2 )3 - 12(- 2 ) + 18 = 34
So, the points are (2, 2 ) and (- 2, 34).
Q. 43 The tangent to the curve y = e2x at the point (0, 1) meets X-axis at
(b) æç - , 0 ö÷
1
(a) (0, 1) (c) (2, 0) (d) (0 , 2)
è 2 ø
Sol. (b) The equation of curve is y = e 2 x
Since, it passes through the point (0, 1).
dy
\ = e 2x × 2 = 2 × e 2x
dx
Þ æ dy ö = 2 × e 2 ×0 = 2 = Slope of tangent to the curve
ç ÷
è dx ø( 0, 1)
\ Equation of tangent is y - 1 = 2(x - 0)
Þ y = 2x + 1
Since, tangent to curve y = e 2 x at the point (0, 1) meets X-axis i.e., y = 0.
1
\ 0 = 2 x + 1Þ x = -
2
-1
So, the required point is æç , 0 ö÷.
è2 ø

Q. 44 The slope of tangent to the curve x = t 2 + 3t - 8 and y = 2 t 2 - 2 t - 5

at the point (2, - 1) is
22 6 6
(a) (b) (c) - (d) - 6
7 7 7
Sol. (b) Equation of curve is given by
x = t 2 + 3 t - 8 and y = 2 t 2 - 2 t - 5.
dx dy
\ = 2t + 3 and = 4t - 2
dt dt
dy
dy dt 4t - 2
Þ = = ...(i)
dx dx 2 t + 3
dt
Since, the curve passes through the point (2, - 1).
\ 2 = t 2 + 3t - 8
and - 1 = 2t 2 - 2t - 5
2
Þ t + 3t - 10 = 0
and 2t 2 - 2t - 4 = 0
2
Þ t + 5t - 2 t - 10 = 0
and 2 t 2 + 2 t - 4t - 4 = 0
Þ t ( t + 5) - 2 ( t + 5) = 0
and 2 t ( t + 1) - 4( t + 1) = 0
Þ ( t - 2 ) ( t + 5) = 0
and ( 2 t - 4)( t + 1) = 0
Þ t = 2, - 5 and t = - 1, 2
Þ t =2
\ Slope of tangent,
æ dy ö 4 ´2 - 2 6
ç ÷ = = [using Eq. (i)]
è dx ø at t = 2 2 ´ 2 + 3 7
Q. 45 Two curves x 3 - 3xy 2 + 2 = 0 and 3x 2 y - y 3 - 2 = 0 intersect at an
angle of
p p p p
(a) (b) (c) (d)
4 3 2 6
Sol. (c) Equation of two curves are given by
x 3 - 3xy2 + 2 = 0
and 3x 2 y - y3 - 2 = 0 [on differentiating w.r.t. x]

3x 2 - 3 é x × 2 y + y2 × 1ù + 0 = 0
dy
Þ
êë dx úû

3 éx 2 + y × 2 x ù - 3 y2
dy dy
and -0=0
êë dx úû dx
dy
Þ 3x × 2 y + 3 y2 = 3x 2
dx
dy dy
and 3 y2 = 3x 2 + 6xy
dx dx
dy 3x 2 - 3 y2
Þ =
dx 6xy
dy 6xy
and =
dx 3 y2 - 3x 2
2 2
Þ æ dy ö = 3(x - y )
ç ÷
è dx ø 6xy
æ dy ö = -6xy
and ç ÷
è dx ø 3(x 2 - y2 )
( x 2 - y2 )
Þ m1 =
2 xy
-2 xy
and m2 =
x 2 - y2
x 2 - y2 - (2 xy)
\ m1m2 = × 2 = -1
2 xy x - y2
p
Hence, both the curves are intersecting at right angle i.e., making with each other.
2

Q. 46 The interval on which the function f (x) = 2x 3 + 9x 2 + 12x - 1 is

decreasing is
(a) [ - 1, ¥ ) (b) [ - 2, - 1] (c) ( - ¥ , - 2] (d) [ - 1, 1]

Sol. (b) We have, f(x ) = 2 x 3 + 9x 2 + 12 x - 1

\ f ¢(x ) = 6x 2 + 18x + 12
= 6 (x 2 + 3x + 2 ) = 6 (x + 2 )(x + 1)
So, f ¢(x ) £ 0, for decreasing.
On drawing number lines as below
+ve – +ve
–2 –1
We see that f ¢(x ) is decreasing in [- 2, - 1 ].
Q. 47 If f : R ® R be defined by f (x) = 2x + cos x, then f
(a) has a minimum at x = p (b) has a maximum at x = 0
(c) is a decreasing function (d) is an increasing function

Sol. (d) We have, f(x ) = 2 x + cos x

\ f ¢(x ) = 2 + (- sin x ) = 2 - sin x
Since, f ¢(x ) > 0, " x
Hence, f(x ) is an increasing function.

Q. 48 If y = x (x - 3)2 decreases for the values of x given by

3
(a) 1 < x < 3 (b) x < 0 (c) x > 0 (d) 0 < x <
2
Sol. (a) We have, y = x(x - 3)2
dy
\ = x × 2(x - 3) × 1 + (x - 3)2 × 1
dx
= 2 x 2 - 6x + x 2 + 9 - 6x = 3x 2 - 12 x + 9
= 3 (x 2 - 3x - x + 3) = 3 (x - 3)(x - 1)
+ – +
1 3
So, y = x(x - 3)2 decreases for (1, 3).
[since, y¢ < 0 for all x Î(1, 3), hence y is decreasing on (1, 3)]

Q. 49 The function f (x) = 4 sin 3 x - 6 sin 2 x + 12 sin x + 100 is strictly

3p p
(a) increasing in æç p , ö÷ (b) decreasing in æç , p ö÷
è 2ø è 2 ø
-p p ù pù
(c) decreasing in é é
(d) decreasing in 0 ,
êë 2 , 2 úû êë 2 úû

Sol. (b) We have, f(x ) = 4sin3 x - 6sin2 x + 12 sin x + 100

\ f ¢(x ) = 12 sin2 x × cos x - 12 sin x × cos x + 12 cos x
= 12 [sin2 x × cos x - sin x × cos x + cos x ]
= 12 cos x [sin2 x - sin x + 1]
Þ f ¢(x ) = 12 cos x [sin2 x + (1 - sin x )] ...(i)
2
Q 1 - sinx ³ 0 and sin x ³ 0
\ sin2 x + 1 - sin x ³ 0
p p
Hence, f ¢(x ) > 0, when cos x > 0 i.e., x Î æç - , ö÷.
è 2 2ø
p p p 3p ö
So, f(x ) is increasing when x Î æç - , ö÷ and f ¢(x ) < 0, when cos x < 0 i.e., x Î æç , ÷
è 2 2ø è2 2 ø
p 3p ö
Hence, f(x ) is decreasing when x Î æç , ÷
è 2 2 ø

Since, æ p , p ö Î æ p , 3p ö
ç ÷ ç ÷
è2 ø è2 2 ø
p
Hence, f(x ) is decreasing in æç , p ö÷.
è 2 ø
Q. 50 Which of the following functions is decreasing on æç 0, p ö÷?
è 2ø
(a) sin 2x (b) tan x (c) cos x (d) cos 3x
p
Sol. (c) In the interval æç 0, ö÷, f(x ) = cos x
è 2ø
Þ f ¢(x ) = - sin x
p
which gives f ¢(x ) < 0 in æç 0, ö÷
è 2ø
p
Hence, f(x ) = cos x is decreasing in æç 0, ö÷.
è 2ø

Q. 51 The function f (x) = tan x - x

(a) always increases
(b) always decreases
(c) never increases
(d) sometimes increases and sometimes decreases

Sol. (a) We have, f(x ) = tan x - x

\ f ' (x ) = sec2 x - 1
Þ f ¢(x ) > 0, " x Î R
So, f (x ) always increases.

Q. 52 If x is real, then the minimum value of x 2 - 8 x + 17 is

(a) - 1 (b) 0 (c) 1 (d) 2

Sol. (c) Let f(x ) = x 2 - 8x + 17

\ f ¢(x ) = 2 x - 8
So, f ¢(x ) = 0, gives x = 4
Now, f ¢¢(x ) = 2 > 0, " x
So, x = 4 is the point of local minima.
\ Minimum value of f(x ) at x = 4,
f(4) = 4 ´ 4 - 8 ´ 4 + 17 = 1

Q. 53 The smallest value of polynomial x 3 - 18 x 2 + 96x in [0, 9] is

(a) 126 (b) 0 (c) 135 (d) 160

Sol. (b) We have, f(x ) = x 3 - 18x 2 + 96x

\ f ¢(x ) = 3x 2 - 36x + 96
So, f ¢(x ) = 0
Gives, 3x 2 - 36x + 96 = 0
Þ 3(x 2 - 12 x + 32 ) = 0
Þ (x - 8)(x - 4) = 0
Þ x = 8, 4 Î [0, 9]
 We shall now evaluate the value of f at these points and at the end points of the interval
[0, 9] i.e., at x = 4 and x = 8 and at x = 0 and at x = 9.
\ f(4) = 43 - 18 × 42 + 96 × 4
= 64 - 288 + 384 = 160
f(8) = 83 - 18 × 82 + 96 × 8 = 128
f(9) = 93 - 18 × 92 + 96 × 9
= 729 - 1458 + 864 = 135
and f(0) = 03 - 18 × 02 + 96 × 0 = 0
Thus, we conclude that absolute minimum value of f on [0, 9] is 0 occurring at x = 0.

Q. 54 The function f (x) = 2x 3 - 3x 2 - 12x + 4, has

(a) two points of local maximum
(b) two points of local minimum
(c) one maxima and one minima
(d) no maxima or minima

Sol. (c) We have f(x ) = 2 x 3 - 3x 2 - 12 x + 4

\ f ¢(x ) = 6 x 2 - 6x - 12
Now, f ¢(x ) = 0 Þ 6 (x 2 - x - 2 ) = 0
Þ 6 (x + 1)(x - 2 ) = 0
Þ x = - 1and x = + 2
On number line for f ¢(x ), we get
+ – +
–1 2
Hence x = - 1is point of local maxima and x = 2 is point of local minima.
So, f(x ) has one maxima and one minima.

Q. 55 The maximum value of sin x × cos x is

1 1
(a) (b) (c) 2 (d) 2 2
4 2
1
Sol. (b) We have, f(x ) = sin x × cos x = sin 2 x
2
1
\ f ¢(x ) = × cos 2 x × 2 = cos 2 x
2
Now, f ¢(x ) = 0Þ cos2 x = 0
p p
Þ cos 2 x = cos Þ x =
2 4
d
Also f ¢¢(x ) = cos 2 x = - sin2 x × 2 = - 2 sin2 x
dx
p p
\ [f ¢¢(x )]at x = p/ 4 = - 2 × sin 2 × = - 2 sin = - 2 < 0
4 2
p p
At , f(x ) is maximum and is point of maxima.
4 4
æ pö 1 p 1
\ f ç ÷ = × sin 2 × =
è 4ø 2 4 2
Q. 56 At x = 5p , f (x) = 2 sin 3x + 3 cos 3x is
6
(a) maximum (b) minimum
(c) zero (d) neither maximum nor minimum

Sol. (d) We have, f(x ) = 2 sin 3x + 3cos 3x

\ f ¢(x ) = 2 × cos 3x × 3 + 3 (- sin 3x ) × 3
Þ f ¢ (x ) = 6 cos 3x - 9sin 3x ...(i)
Now, f ¢¢(x ) = - 18sin 3x - 27 cos 3x
= - 9 (2 sin 3x + 3cos 3x )
æ 5p ö æ 5p ö - 9 sin æ 3 × 5p ö
\ f ¢ç ÷ = 6 cosç 3 × ÷ ç ÷
è 6 ø è 6 ø è 6 ø
5p 5p
= 6 cos - 9sin
2 2
pö p
æ
= 6 cos ç 2 p + ÷ - 9 sin æç 2 p + ö÷
è 2ø è 2ø
=0-9¹0
5p
So, x = cannot be point of maxima or minima.
6
5p
Hence, f(x ) at x = is neither maximum nor minimum.
6

Q. 57 The maximum slope of curve y = - x 3 + 3x 2 + 9x - 27 is

(a) 0 (b) 12 (c) 16 (d) 32

Sol. (b) We have, y = - x 3 + 3x 2 + 9x - 27

dy
\ = - 3x 2 + 6x + 9 = Slope of the curve
dx
d2y
and = - 6x + 6 = - 6 (x - 1)
dx 2
d2y
\ =0
dx 2
Þ - 6(x - 1) = 0 Þ x = 1 > 0
d3y
Now, = - 6< 0
dx 3
So, the maximum slope of given curve is at x = 1.
\ æ dy ö = - 3 × 12 + 6 × 1 + 9 = 12
ç ÷
è dx ø( x = 1)

Q. 58 The functin f (x) = x x has a stationary point at

1
(a) x = e (b) x = (c) x = 1 (d) x = e
e
Sol. (b) We have, f( x ) = x x
Let y = xx
and log y = x log x
1 dy 1
\ × = x × + log x × 1
y dx x
 dy
Þ = (1 + log x ) × x x
dx
dy
\ =0
dx
Þ (1 + log x ) × x x =0
Þ log x = -1
Þ log x = log e -1
Þ x = e -1
1
Þ x=
e
1
Hence, f(x ) has a stationary point at x = .
e

x
Q. 59 The maximum value of æç 1 ö÷ is
èxø
1 1/ e
(a) e (b) ee (c) e1/ e (d) æç ö÷
è eø
x
1
Sol. (c) Let y = æç ö÷
è ø
x
1
Þ log y = x × log
x
1 dy 1 1 1
\ × = x × × æç - 2 ö÷ + log × 1
y dx 1 è x ø x
x
1
= - 1 + log
x
x
1 1
= æç log - 1ö÷ × æç ö÷
dy
\
dx è x ø èx ø
dy
Now, =0
dx
1
Þ log = 1 = log e
x
1
Þ =e
x
1
\ x=
e
1
Hence, the maximum value of f æç ö÷ = (e )1/ e .
èe ø
Fillers
Q. 60 The curves y = 4 x 2 + 2x - 8 and y = x 3 - x + 13 touch each other at
the point ......... .
Sol. The curves y = 4x 2 + 2 x - 8 and y = x 3 - x + 13 touch each other at the point (3, 34).
Given, equation of curves are y = 4x 2 + 2 x - 8 and y = x 3 - x + 13
dy
\ = 8x + 2
dx
dy
and = 3x 2 - 1
dx
So, the slope of both curves should be same
\ 8x + 2 = 3x 2 - 1
Þ 3x 2 - 8x - 3 = 0
Þ 3x 2 - 9x + x - 3 = 0
Þ 3x(x - 3) + 1 (x - 3) = 0
Þ (3x + 1)(x - 3) = 0
1
\ x = - and x = 3,
3
2
1 1 -1
For x = - , y = 4 × æç - ö÷ + 2 × æç ö÷ - 8
3 è 3ø è 3ø
4 2 4 - 6 - 72
= - -8=
9 3 9
74
=-
9
2
and for x = 3, y = 4 × (3) + 2 × (3) - 8
= 36 + 6 - 8 = 34
1 -74 ö
So, the required points are (3, 34) and æç - , ÷.
è 3 9 ø

Q. 61 The equation of normal to the curve y = tan x at (0, 0) is ......... .

Sol. The equation of normal to the curve y = tan x at (0, 0) is x + y = 0 .
dy
Q y = tan x Þ = sec2 x
dx
æ dy ö 1 1
Þ ç ÷ = sec 2 0 = 1 and - =-
è dx ø( 0, 0 ) æ dy ö 1
ç ÷
è dx ø
\Equation of normal to the curve y = tan x at (0, 0) is
y - 0 = - 1( x - 0 )
Þ y+ x =0
Q. 62 The values of a for which the function f (x) = sin x - ax + b increases
on R are ......... .
Sol. The values of a for which the function f(x ) = sin x - ax + b increases on R are (- ¥, - 1.)
Q f ¢(x ) = cos x - a
and f ¢(x ) > 0 Þ cos x > a
Since, cos x Î [- 1, 1]
Þ a < - 1 Þ a Î (- ¥, - 1)

2x 2 - 1
Q. 63 The function f (x) = , (where, x > 0) decreases in the interval
x4
......... .
2x 2 - 1
Sol. The function f(x ) = , where x > 0, decreases in the interval (1, ¥ ).
x4
x 4 × 4x - (2 x 2 - 1) × 4x 3 4x 5 - 8x 5 + 4x 3
Q f ¢(x ) = =
x8 x8
5 3 3 2
- 4x + 4x 4x (- x + 1)
= =
x8 x8
Also, f ¢(x ) < 0
4x 3 (1 - x 2 )
Þ < 0 Þ x 2 >1
x8
Þ x >± 1
\ x Î (1, ¥ )

b
Q. 64 The least value of function f (x) = ax + (where ,a > 0, b > 0, x > 0) is
x
......... .
b
Sol. The least value of function f(x ) = ax + (where, a > 0, b > 0, x > 0) is 2 ab.
x
b
Q f ¢(x ) = a - and f ¢(x ) = 0
x2
b
Þ a=
x2
b b
Þ x2 = Þ x = ±
a a
(- 2 ) 2b
Now, f ¢¢(x ) = - b × 3 = + 3
x x
b 2b + 2 b × a3 / 2
At x = , f ¢¢(x ) = + 3/ 2
=
a æbö b3 / 2
ç ÷
èaø
a3
= + 2 b -1/ 2 × a3 / 2 = + 2 >0 [Q a, b > 0 ]
b
æ bö b b
\ Least value of f(x ), f çç ÷÷ = a × +
è a ø a b
a
= a × a-1/ 2 × b1/ 2 + b × b -1/ 2 × a1/ 2
= ab + ab = 2 ab
 7
Integrals
Short Answer Type Questions
Verify the following
2x - 1
Q. 1 ò dx = x - log|(2 x + 3)2 | + C
2x + 3
2x - 1 2x + 3 - 3 - 1
Sol. Let I =ò dx = ò dx
2x + 3 2x + 3
1 4
= ò 1dx - 4ò dx = x - ò dx
2x + 3 æ 3ö
2çx + ÷
è 2ø
æ 3ö ½ æ 2 x + 3 ö½
= x - 2 log + ç x + ÷ C ¢ = x - 2 log½ç ÷½ + C ¢
è 2ø ½è 2 ø½
é m ù
= x - 2 log|(2 x + 3)| + 2 log 2 + C ¢ êëQ log n = log m - log núû
= x - log|(2 x + 3)2| + C [Q C = 2 log 2 + C ¢]

2x + 3
Q. 2 ò 2
dx = log| x 2 + 3x | + C
x + 3x
2x + 3
Sol. Let I= ò x 2 + 3x dx
Put x 2 + 3x = t
Þ (2 x + 3) dx = dt
1
\ I = ò dt = log|t| + C
t
= log|(x 2 + 3x )| + C
 (x 2 + 2)d
Q. 3 ò x
x +1
K Thinking Process
1
First of all divided numerator by denominator, then use the formula ò dx = log | x| to
get the solution. x
x2 + 2
Sol. Let I= ò
x+1
dx

æ 3 ö
= ò çç x - 1 + ÷ dx
è x + 1 ÷ø
1
= ò (x - 1) dx + 3ò dx
x+1
x2
= - x + 3log|(x + 1)| + C
2

e 6 log x - e 5 log x
Q. 4 ò dx
e 4 log x - e 3 log x
æ e 6 log x - e 5 log x ö
Sol. Let I = ò çç 4 log x ÷ dx
èe - e 3 log x ÷ø
æ elog x 6 - elog x 5 ö
= òç ÷ dx [Q alog b = log b a ]
ç log x 4 log x 3 ÷
è e - e ø
æ x6 - x5 ö
= ò çç 4 ÷ dx
3 ÷
[Q elog x = x ]
èx - x ø
æ x3 - x2 ö 2
= ò çç ÷ dx = x (x - 1)dx
÷ ò
è x -1 ø x -1
x3
= ò x 2dx = +C
3

(1 + cos x)
Q. 5 ò dx
x + sin x
(1 + cos x )
Sol. Consider that, I= ò (x + sin x ) dx
Let x + sin x = t Þ (1 + cos x ) dx = dt
1
\ I = ò dt = log|t| + C
t
= log|(x + sin x )| + C
 dx
Q. 6 ò
1 + cos x
K Thinking Process
x
cos x = 2 cos2 - 1 and also use formula i.e., ò sec x = tanx + C to solve
2
the above
2
problem.
dx dx
Sol. Let I= ò 1 + cos x = ò x
-1 1 + 2 cos 2
2
1 1 1 x
= ò dx = ò sec 2 dx
2 cos 2 x 2 2
2
1 x x
= × tan × 2 + C = tan + C [Q ò sec 2 x dx = tan x ]
2 2 2

Q. 7 ò tan 2 x sec 4 x dx
K Thinking Process
Use the formula sec2x = 1 + tan2 x and put tanx = t to solve this problem.
Sol. Let I = ò tan2 x sec 4 x dx
Put tanx = t Þ sec 2 x dx = dt
\ I = ò t 2 (1 + t 2 ) dt = ò (t
2
+ t 4 ) dt
3 5 5
t t tan x tan3 x
= + +C= + +C
3 5 5 3

sin x + cos x
Q. 8 ò dx
1 + sin 2x
sin x + cos x (sin x + cos x )
Sol. Let I= ò 1 + sin2 x
dx = ò sin2 x + cos 2 x + 2 sin x cos x
dx

sin x + cos x
= ò (sin x + cos x )2
dx = ò 1dx = x + C

Q. 9 ò 1 + sin x dx
Sol. Let I= ò 1 + sinx dx
x x x x é 2 x 2 x ù
= ò sin2
2
+ cos 2 + 2 sin cos dx
2 2 2 êëQ sin 2 + cos 2 = 1úû

æ x x ö2 æ x xö
= òç sin + cos ÷ dx = ò ç sin + cos ÷ dx
è 2 2ø è 2 2ø
x x x x
= - cos × 2 + sin × 2 + C = - 2 cos + 2 sin + C
2 2 2 2
 x
Q. 10 ò dx
x +1
x
Sol. Let I= ò x +1
dx

1
Put x =t Þ dx = dt
2 x
Þ dx = 2 x dt
æx x ö t 2 ×t t3
\ I = 2 ò çç ÷ dt = 2 ò
÷ dt = 2 ò dt
è t + 1ø t +1 t +1
t3 + 1- 1 ( t + 1) (t 2 - t + 1) 1
= 2ò dt = 2 ò dt - 2 ò dt
t +1 t +1 t +1
1
= 2 ò ( t 2 - t + 1) dt - 2 ò dt
t +1
ét 3 t 2 ù
= 2ê - + t - log|(t + 1)|ú + C
ë3 2 û
éx x x ù
= 2ê - + x - log|( x + 1)|ú + C
ë 3 2 û

a+x
Q. 11 ò dx
a-x
K Thinking Process
Here, put x = a cos2 q and also use the formula i.e., cos 2 q = 2 cos2 q - 1 = 1 - 2 sin2 q, to
get the solution.
a+ x
Sol. Let I= òa-x
dx

Put x = acos2 q
Þ dx = - a × sin2 q × 2 × dq
a + acos 2 q
\ I = - 2ò × asin2 qdq
a - acos 2 q
é x -1 x 1 -1 x ù
êëQ cos 2 q = a Þ 2 q = cos a Þ q = 2 cos a úû

1 + cos 2 q 2 cos 2 q
= - 2 aò sin2 qdq = - 2 a ò sin2 qdq
1 - cos 2 q 2 sin2 q
cos q
= - 2 aò cot q × sin2 qdq = - 2 aò × 2 sin q × cos qdq
sin q
= - 4aò cos 2 qdq = - 2 aò (1 + cos 2 q)dq
é sin2 q ù
= - 2 aê q + +C
ë 2 úû
é1 x 1 x2 ù
= - 2 aê cos -1 + 1- 2 ú + C
êë 2 a 2 a úû
é æx ö x ù
2
= - aêcos -1 ç ÷ + 1 - 2 ú + C
êë è ø
a a úû
 Alternate Method
a+ x (a + x )(a + x )
Let I= ò a-x
dx = ò
(a - x )(a + x )
dx

(a + x )
= ò a2 - x 2
dx

a x
I= ò 2
a -x 2
+ ò
a - x2
2
dx

\ I = I1 + I 2 …(i)
a æx ö
Now, I1 = ò = asin-1 ç ÷ + C1
2
a -x 2 èaø
x
and I2 = ò a2 - x 2
dx

Put a2 - x 2 = t 2 Þ - 2 x dx = 2t dt
t
\ I 2 = - ò dte = - ò 1dt
t
= - t + C 2 = - a2 - x 2 + C 2
æx ö
\ I = asin-1 ç ÷ + C1 - a2 - x 2 + C 2 [Q t 2 = a2 - x 2 ]
èaø
æx ö
I = asin-1 ç ÷ - a2 - x 2 + C [Q C = C1 + C 2 ]
èaø

x 1/2
Q. 12 ò dx
1 + x 3/ 4 x1 2
Sol. Let I= ò 1 + x 3 4 dx
Put x = t4 Þ dx = 4t 3dt
t 2 (t 3 ) æ t2 ö
\ I = 4ò 3
dt = 4ò çç t 2 - ÷ dt
÷
1+ t è 1+ t3 ø
t2
I = 4ò t 2dt - 4ò dt
1+ t3
I = I1 - I 2
t3 4 34
I1 = 4ò t 2dt = 4 × + C1 = x + C1
3 3
t2
Now, I 2 = 4ò dt
1+ t3
Again, put 1 + t 3 = z Þ 3t 2dt = dz
1 4 1
Þ t 2dt = dz = ò dz
3 3 z
4 4
= log| z| + C 2 = log|(1 + t 3 )| + C 2
3 3
4 34
= log|(1 + x )| + C 2
3
4 4
\ I = x 3 4 + C1 - log|(1 + x 3 4 )| - C 2
3 3
4
= x 3 4 - log|(1 + x 3 4 ) + C [Q C = C1 - C 2 ]
3
 1 + x2
Q. 13 ò dx
x4
1 + x2 1 + x2 1
Sol. Let I= ò x 4
dx = ò x
×
x3
dx

1 + x2 1 1 1
= ò x2
× 3 dx = ò
x x2
+ 1 × 3 dx
x
1 -2
Put 1+ = t2 Þ 3 dx = 2t dt
x2 x
1
Þ - = t dt
x3
3/ 2
t3 1æ 1 ö
\ I = - ò t 2dt = - + C = - ç1 + 2 ÷ +C
3 3è x ø

dx
Q. 14 ò
16 - 9x 2
K Thinking Process
1
First of all concert the expression in form of , then use the formula,
a - x2
2

1 æ xö
ò dx = sin-1ç ÷ + C .
2
a -x 2 è aø
dx dx 1 -1 æ 3x ö
Sol. Let I= ò 16 - 9x 2
= ò 2
(4) - (3x ) 2
dx =
3
sin ç ÷+C
è 4 ø

dt
Q. 15 ò
3t - 2t 2
dt 1 dt
Sol. Let I= ò 3t - 2 t 2
=
2 ò æ 3 ö
- çt 2 - t ÷
è 2 ø
1 dt
=
2 ò éæ 1 3 ö æ 3 ö2 æ 3 ö2 ù
- êçt 2 - 2 × × t ÷ + ç ÷ - ç ÷ ú
êë è 2 2 ø è 4ø è 4 ø ûú
1 dt
=
2 ò éæ æ 3ö ù
2 2
3ö
- êçt - ÷ - ç ÷ ú
êë è 4ø è 4 ø úû
1 dt
=
2 ò2
æ 3ö æ 3 ö2
ç ÷ - çt - ÷
è 4ø è 4ø
æ 3ö
çt - ÷
=
1
sin-1 ç 4 ÷ + C = 1 sin-1 æç 4t - 3 ö÷ + C
2 ç 3 ÷ 2 è 3 ø
ç ÷
è 4 ø
 3x - 1
Q. 16 ò dx
x2 + 9
K Thinking Process
First of all convert to the given integral into two parts, then by using formula i.e.,
1
ò a2 + x2 = log| x + a + x | + C, get the desired result.
2 2

3x - 1
Sol. Let I= ò x2 + 9
dx

3x 1
I= ò x2 + 9
dx - ò x2 + 9
dx

I = I1 - I 2
3x
Now, I1 = ò
x2 + 9
Put x 2 + 9 = t 2 Þ 2 x dx = 2t dt Þ xdx = tdt
t
\ I1 = 3ò dt
t
= 3ò dt = 3t + C1 = 3 x 2 + 9 + C1
1 1
and I2 = ò x2 + 9
dx = ò x 2 + (3)2
dx

= log|x + x 2 + 9| + C 2
\ I = 3 x 2 + 9 + C1 - log|x + x 2 + 9| - C 2
= 3 x 2 + 9 - log|x + x 2 + 9| + C [Q C = C1 - C 2 ]

Q. 17 ò 5 - 2x + x 2 dx

K Thinking Process
First of all convert the given expression into x2 + a2 form, then use the formula i.e.,
1 a2
ò x2 + a2 dx = x x2 + a2 + log| x + x2 + a2 | + C .
2 2
Sol. Let I=ò 5 - 2 x + x 2 dx = ò x 2 - 2x + 1 + 4d x

=ò (x - 1)2 + (2 )2 dx = ò (2 )2 + (x - 1)2 dx
x -1 2
= 2 + (x - 1)2 + 2 log|x - 1 + 2 2 + (x - 1)2| + C
2
x -1
= 5 - 2 x + x 2 + 2 log|x - 1 + 5 - 2 x + x 2| + C
2
 x
Q. 18 ò 4
dx
x -1
x
Sol. Let I= ò x 4 - 1dx
1
Put x 2 = t Þ 2xdx = dt Þ xdx = dt
2
1 dt 1 1 ½ t - 1½ é dx 1 ½ x - a½ ù
\ I= ò = × log½
2 t - 1 2 2 ½t + 1½
2
½+ C êQ ò 2
êë x -a 2
= log ½
2 a ½ x + a½
½+ Cú
úû
1
= [log|x 2 - 1| - log|x 2 + 1|] + C
4

x2
Q. 19 ò dx
1 - x4
K Thinking Process
1 1 1 ½1 + x½
Here, use ò dx = tan-1x + C and ò a2 - x2 dx = 2 a log½½1 - x½+ C, to solve this
1 + x2 ½
problem.
x2
Sol. Let I= ò 1 - x 4 dx
æ 1 x2 1 x2 ö
ç + - + ÷
ç2 2 2 2 ÷ø
=ò è dx [Q a2 - b 2 = (a + b )(a - b )]
(1 - x 2 )(1 + x 2 )
1 1
(1 + x 2 ) - (1 - x 2 )
=ò 2 2 dx
(1 - x 2 )(1 + x 2 )
1
(1 + x 2 )
2 1 (1 - x 2 )
=ò 2 2
dx - ò dx
(1 - x )(1 + x ) 2 (1 - x 2 )(1 + x 2 )
1 1 1 1 1 1 ½1 + x ½ 1
½ + C1 - tan-1 x + C 2
2 ò 1 - x2
= dx - ò dx = × log ½
2 1 + x2 2 2 ½1 - x ½ 2
1 ½1 + x½ 1
= log½ ½ - tan-1 x + C [Q C = C1 + C 2 ]
4 ½1 - x½ 2

Q. 20 ò 2ax - x 2 dx

Sol. Let I= ò 2 ax - x 2 dx = ò - (x 2 - 2 ax )dx

=ò - (x 2 - 2 ax + a2 - a2 ) dx = ò - (x - a)2 - a2 dx

=ò a2 - (x - a)2 dx
x-a 2 a 2 -1 æ x - a ö
= a - (x - a)2 + sin ç ÷+C
2 2 è a ø
x-a a 2 -1 æ x - a ö
= 2 ax - x 2 + sin ç ÷+C
2 2 è a ø
 sin -1 x
Q. 21 ò dx
(1 - x 2 ) 3 4
sin-1 x sin-1 x
Sol. Let I= ò (1 - x 2 )3 4 dx = ò (1 - x 2 ) 1 - x2
dx

1
Put sin-1 x = t Þ dx = dt
1 - x2
and x = sin t Þ 1 - x 2 = cos 2 t
Þ cost = 1 - x 2
t
\ I=ò dt = ò t × sec 2 tdt
cos 2 t
æd ö
= t × ò sec 2 tdt - ò ç t × ò sec 2 t dt ÷ dt
è dt ø
= t × tan t - ò 1× tan t dt
= t tan t + log|cos t| + C [Q ò tan x dx = - log|cos x| + C ]
x
= sin-1 x × + log| 1 - x 2| + C
2
1- x

(cos 5x + cos 4 x)
Q. 22 ò dx
1 - 2 cos 3x
9x x
2 cos × cos
cos 5x + cos 4x 2 2 dx
Sol. Let I=ò dx = ò
1 - 2 cos 3x æ 3x ö
1 - 2 ç 2 cos 2 - 1÷
è 2 ø
é C+D C-D 2 ù
êëQ cos C + cos D = 2 cos 2 . cos 2 and cos 2 x = 2 cos x - 1úû
9x x 9x x
2 cos × cos 2 cos × cos
\ I=ò 2 2 dx = - ò 2 2 dx
3x 3x
3 - 4cos 2 4cos 2 -3
2 2
9x x 3x
2 cos × cos × cos
2 2 2 dx é 3x ù
= -ò êmultiply and divide by cos 2 ûú
3 3x 3x ë
4cos - 3cos
2 2
9x x 3x
2 cos × cos × cos
= -ò 2 2 2 dx = - 2 cos 3x × cos x dx
3x ò 2 2
cos 3 ×
2
ì æ 3x xö æ 3x x ö ü
= - ò ícos ç + ÷ + cos ç - ÷ ý dx
î è 2 2 ø è 2 2 øþ
= - ò (cos 2x + cos x ) dx
é sin2 x ù
= -ê + sin x ú + C
ë 2 û
1
= - sin2 x - sin x + C
2
 sin 6 x + cos6 x
Q. 23 ò dx
sin 2 x cos2 x
K Thinking Process
Use a3 + b3 = (a + b)(a2 - ab + b2) and sec2 x = 1 + tan2 x, cosec 2x = 1 + cot2 x, to solve
the above problem.
sin6 x + cos 6 x (sin2 x )3 + (cos 2 x )3
Sol. Let I= ò sin2 x cos 2 x
dx = ò sin2 x × cos 2 x
dx

(sin2 x + cos 2 x )(sin4 x - sin2 x cos 2 x + cos 4 x )

= ò sin2 x × cos 2 x
dx

sin4 x cos 4 x sin2 x cos 2 x

= ò sin2 x cos2 x dx + ò sin2 x × cos2 x dx - ò sin2 x × cos2 x dx
= ò tan2 x dx + ò cot 2 x dx - ò 1dx
= ò (sec 2 x - 1) dx + ò (cos ec 2 x - 1) dx - ò 1dx
= ò sec 2 x dx + ò cos ec 2 x dx - 3ò dx
I = tan x - cot x - 3x + C

x
Q. 24 ò dx
a - x3
3

x x
Sol. Let I= ò
a3 - x 3
dx = ò
(a3 2 )2 - (x 3 2 )2
3
Put x 3 2 = t Þ x 1/ 2dx = dt
2
2 dt 2 t
\ I= ò = sin-1 3 2 + C
3 (a3 2 )2 - t 2 3 a

2 -1 x 3 2 2 -1 x 3
= sin + C = sin +C
3 a3 2 3 a3

cos x - cos 2x
Q. 25 ò dx
1 - cos x
K Thinking Process
C+D D -C x
Apply the formula, cos C - cos D = 2 sin × sin and cos x = 1 - 2 sin2 to solve
2 2 2
it.
3x x
2 sin × sin
cos x - cos 2 x 2 2
Sol. Let I=ò dx = ò dx
1 - cos x x
1 - 1 + 2 sin2
2
3x x 3x
sin × sin sin
= 2ò 2 2 dx = 2 dx
2 x
ò x
2 sin sin
2 2
 x x
3sin - 4sin3
=ò 2 2 dx [Q sin 3x = 3sin x - 4sin3 x ]
x
sin
2
x 1 - cos x
= 3ò dx - 4ò sin2 dx = 3ò dx - 4ò dx
2 2
= 3ò dx - 2 ò dx + 2 ò cos x dx
= ò dx + 2 ò cos x dx = x + 2sin x + C = 2sinx + x + C

dx
Q. 26 ò
x x4 - 1
dx
Sol. Let I= òx x4 - 1
Put x 2 = sec q Þ q = sec -1 x 2
Þ 2x dx = sec q × tan q dq
1 sec q × tan q 1 1
2 ò sec q tan q
\ I= dq = ò dq = q + C
2 2
1
= sec -1(x 2 ) + C
2

2
Q. 27 ò0 (x 2 + 3)dx

K Thinking Process
b b-a
òa f(x)dx = lim h [ f (a) + f (a + h) + ¼+ f {a + (n - 1) h}], where h =
h ®0 n
® 0 as
n® ¥ ×
2
ò0 (x
2
Sol. Let I= + 3) dx
b-a 2-0
Here, a = 0, b = 2 and h = =
n n
2
Þ h= Þ nh = 2 Þ f(x ) = (x 2 + 3)
n
2
Now, ò0 (x 2 + 3) dx = lim h [f(0) + f (0 + h) + f(0 + 2 h) + ¼ + f {0 + (n - 1) h}]
h ®0
...(i)
Q f(0) = 3
Þ f(0 + h) = h2 + 3, f(0 + 2 h) = 4h2 + 3 = 2 2 h2 + 3
f [0 + (n - 1)h] = (n2 - 2 n + 1)h + 3 = (n - 1)2 h + 3
From Eq. (i),
2
ò0 (x 2 + 3) dx = lim h[3 + h2 + 3 + 2 2 h2 + 3 + 32 h2 + 3 + ¼ + (n - 1)2 h2 + 3]
h ®0

= lim h[3n + h2 {12 + 2 2 + ¼ + (n - 1)2 }]

h ®0
é æ (n - 1) (2 n - 2 + 1) (n - 1 + ) öù é n(n + 1) (2 n + 1)ù
= lim h ê 3n + h2 ç 2
÷ ú êQ S n = úû
h ®0 ë è 6 øû ë 6
é æ ( n2
- n)(2 n - 1 ) ö÷ ù
= lim h ê 3n + h2 çç ÷ú
h ®0 êë è 6 ø úû
 é h2 ù
= lim h ê 3n + (2 n3 - n2 - 2 n2 + n)ú
h ®0
ë 6 û
é 2 n h - 3n h × h + nh × h2 ù
3 3 2 2
= lim ê 3nh + ú
h ®0 ê 6 úû
ë
é 2 × 8 - 3×2 × h + 2 × h ù
2 2 é 16 - 12 h + 2 h2 ù
= lim ê 3 × 2 + ú = lim ê 6 + ú
h ®0 ê 6 úû h ®0 êë 6 úû
ë
16 8 26
=6+ =6+ =
6 3 3

2
Q. 28 ò0 e x dx
2
ò0 e
x
Sol. Let I= dx
Here, a = 0 and b=2
b-a
\ h=
n
Þ nh = 2 and f(x ) = e x
2
ò0 e dx = hlim
x
Now, h[f(0) + f(0 + h) + f(0 + 2 h) + ¼ + f{0 + (n - 1) h}]
®0

\ I = lim h[1 + e h + e 2 h + ¼ + e( n -1) h]

h ®0

é 1× (e h )n - 1ù æ e nh - 1 ö
= lim h ê h ú = lim hçç h ÷
÷
h ®0 ê e - 1 ú
ë û h ®0 è e - 1 ø
æ e2 - 1ö
= lim hçç h ÷
÷
h ®0
è e - 1ø
h h é h ù
= e 2 lim h
- lim h êQ lim h = 1ú
h ®0 e -1 h ®0 e -1 êë h ®0 e - 1 úû
2 2
=e -1 =e -1
Evaluate the following questions.
1 dx
Q. 29 ò0
e + e -x
x

1 dx 1 ex
Sol. Let I= ò0 e + e -x
x
= ò0 1 + e 2x
dx

Put ex = t
Þ e x dx = dt
e dt
\ I= ò1 = [tan-1 t ]e1
1+ t2
= tan-1 e - tan-1 1
p
= tan-1 e -
4
 p2 tan x
Q. 30 ò0 dx
1 + m2 tan 2 x
p/ 2 tan x dx
Sol. Let I= ò0 1 + m2 tan2 x
dx

sin x
p/ 2 cos x
=ò 2
dx
0 sin x
2
1+ m ×
cos 2 x
sin x
p/ 2 cos x
=ò dx
0 cos 2 x + m2 sin2 x
cos 2 x
p/ 2 sin x cos x dx
= ò0 1 - sin2 x + m2 sin2 x
dx

p/ 2 sin x cos x
= ò0 1 - sin2 x(1 - m2 )
dx

Put sin2 x = t
Þ 2sin x cos x dx = dt
1 1 dt
\ I= ò
2 0 1 - t (1 - m2 )
1
1é 1 ù
= ê - log|1 - t (1 - m2 )|× ú
2 êë 1 - m2 úû 0
1é 2 1 1 ù
= ê - log|1 - 1 + m | × 2
+ log|1|× ú
2 êë 1+ m 1 - m2 úû
1é 2 1 ù 2 log m
= ê - log|m | × ú= ×
2 êë 1 - m2 úû 2 (m2 - 1)
m
= log
m2 - 1

2 dx
Q. 31 ò 1
(x - 1)(2 - x)
K Thinking Process
1
First of all convert the given function into form, then apply the formula i.e.,
a2 - x2
1 x
ò dx = sin-1 + C .
2 2 a
a -x
2 dx 2 dx
Sol. Let I= ò1 (x - 1)(2 - x )
= ò1 2x - x 2 - 2 + x
2 dx
= ò1 - (x 2 - 3x + 2 )
 2 dx
= ò1 é 3 æ 3ö
2
9ù
- êx 2 - 2 × x + ç ÷ + 2 - ú
êë 2 è2 ø 4 úû
2 dx
= ò1 ìïæ 3 ö 2 æ 1 ö 2 üï
- íç x - ÷ - ç ÷ ý
îïè 2ø è 2 ø þï
2
é æ 3 öù
2 dx ê -1 ç x - 2 ÷ ú
= ò1 2
= êsin ç
ç 1 ÷ú
÷ú
æ 1ö æ 3 ö2 ê ç ÷
ç ÷ - çx - ÷ êë è 2 ø úû1
è2 ø è 2ø
= [sin-1(2 x - 3)]12 = sin-1 1 - sin-1(-1)
p p é p ù
= + êëQ sin 2 = 1 and sin (- q) = - sin qúû
2 2
=p

1 x
Q. 32 ò0 dx
1 + x2
1 x
Sol. Let I= ò0 1 + x2
dx

Put 1 + x2 = t 2
Þ 2 x dx = 2tdt
Þ x dx = tdt
2 tdt
\ I= ò1 t
2
= [t ]1 = 2 -1

p
Q. 33 ò0 x sin x cos2 x dx

K Thinking Process
a a
Here, use the property i.e., ò f (x)dx = ò (a - x)dx and
0 0
sin (p - x) = sin x, cos(p - x) = cos x .
p
ò0 x sin x cos x dx
2
Sol. Let I= …(i)
p
and I = ò (p - x )sin(p - x )cos 2 (p - x )dx
0
p
Þ I = ò (p - x )sin x cos 2 x dx …(ii)
0
On adding Eqs. (i) and (ii), we get
p
ò0 p sin x cos
2
2I = x dx
Put cos x = t
Þ - sinx dx = dt
 As x ® 0, then t ® 1
and x ® p, then t ® -1
-1
-1 ét 3 ù
\ I = - p ò t 2 dt Þ I = - p ê ú
ë 3 û1
1

p 2p
Þ 2 I = - [-1 - 1] Þ 2 I =
3 3
p
\ I=
3

1/2 dx
Q. 34 ò0
(1 + x 2 ) 1 - x 2
1/ 2 dx
Sol. Let I= ò0 (1 + x ) 1 - x 2
2

Put x = sin q
Þ dx = cos q dq
As x ® 0, then q ® 0
1 p
and x ® , then q ®
2 6
p/ 6 cos q p/ 6 1
\ I=ò dq = ò dq
0 (1 + sin2 q) cos q 0 1 + sin2 q

p/ 6 1
=ò dq
0 cos 2 q (sec 2 q + tan2 q)

p/ 6 sec 2 q
= ò0 sec q + tan2 q
2
dq

p/ 6 sec 2 q
= ò0 1 + tan2 q + tan2 q
dq

p/ 6 sec 2 q
= ò0 1 + 2 tan2 q
dq

Again, put tan q = t

Þ sec 2 q dq = dt
As q® 0, then t ® 0
p 1
and q ® , then t ®
6 3
1/ 3 dt 1 1/ 3 dt
\ I= ò0 1 + 2t 2
=
2 ò0 æ 1 ö
2
ç ÷ + t2
è 2ø
1/ 3
é ù
1 1 ê -1 t ú 1
= × ê = [tan-1( 2 t )]10/ 3
1 ú
tan
2 1/ 2 ê ú 2
êë 2 úû 0
1 é -1 2 ù 1 æ 2ö
= ê tan - 0ú = tan-1 çç ÷
÷
2ë 3 û 2 è 3ø
Long Answer Type Questions
x2
Q. 35 ò dx
x 4 - x 2 - 12
K Thinking Process
px + q A B
Use = + , where a ¹ b , then compare the coefficient of x to
(x - a)(x - b) (x - a) x - b
get the value of A and B.
x2
Sol. Let I= ò x 4 - x 2 - 12 dx
x2
= ò x 4 - 4x 2 + 3x 2 - 12 dx
x 2dx
= ò x 2 (x 2 - 4) + 3 (x 2 - 4)
x 2dx
= ò (x 2 - 4)(x 2 + 3)
x2
Now, [ let x 2 = t ]
(x - 4)(x 2 + 3)
2

t A B
Þ = +
(t - 4)(t + 3) t - 4 t + 3
Þ t = A (t + 3) + B (t - 4)
On comparing the coefficient of t on both sides, we get
A+ B=1 …(i)
Þ 3 A - 4B = 0 …(ii)
Þ 3(1 - B) - 4B = 0
Þ 3 - 3B - 4B = 0
Þ 7B = 3
3
Þ B=
7
3 3
If B = , then A + = 1
7 7
3 4
Þ A = 1- =
7 7
x2 4 3
= +
(x - 4)(x 2 + 3)
2
7(x 2 - 4) 7(x 2 + 3)
4 1 3 1
7ò 7 ò x 2 + ( 3 )2
\ I= dx + dx
x 2 - ( 2 )2

4 1 ½x - 2 ½ 3 1 x
= × log½ ½+ × tan-1 +C
7 2 × 2 ½x + 2 ½ 7 3 3
1 ½x - 2½ 3 x
= log½ ½+ tan-1 +C
7 ½x + 2½ 7 3
 x2
Q. 36 ò dx
(x 2 + a 2 )(x 2 + b 2 )
x2
Sol. Let I= ò (x 2 + a2 )(x 2 + b2 )dx
x2
Now, [let x 2 = t ]
(x + a )(x 2 + b 2 )
2 2

t A B
= = +
(t + a2 )(t + b 2 ) (t + a2 ) (t + b 2 )
t = A (t + b 2 ) + B (t + a2 )
On comparing the coefficient of t, we get
A+ B=1 …(i)
b 2 A + a2 B = 0 …(ii)
Þ b 2 (1 - B) + a2 B = 0
Þ b 2 - b 2 B + a2 B = 0
Þ b 2 + ( a2 - b 2 ) B = 0
- b2 b2
Þ B= 2 2
=
a -b b - a2
2

2
b
From Eq. (i), A+ =1
b 2 - a2
b 2 - a2 - b 2 - a2
Þ A= 2 2
=
b -a b - a2
2

2
-a b2 1
\ I= ò (b2 - a2 )(x 2 + a2 )dx + ò b2 - a2 × x 2 + b2 dx
- a2 1 b2 1
= 2
(b - a 2 ò
) x 2 + a2
dx +
b - a2 x 2 + b 2 2 ò dx

- a2 1 x b 2
1 x
= 2 × tan-1 + 2 × tan-1
b - a2 a a b - a2 b b
1 é -1 x -1 x ù
= 2 - a tan + b tan
b - a2 êë a b úû
1 é x xù
= 2 a tan-1 - b tan-1 ú
a - b 2 êë a bû

p x
Q. 37 ò0
1 + sin x
px
Sol. Let I= ò0
1 + sin x
dx ...(i)

p p-x p p-x
and I=ò dx = ò0 dx ...(ii)
0 1 + sin( p - x ) 1 + sin x
On adding Eqs. (i) and (ii), we get
p1
2I = pò dx
1 + sinx
0

p (1 - sin x ) dx
= pò
0 (1 + sin x )(1 - sin x )
 p (1 - sin x ) dx
= pò
0 cos 2 x
p
= p ò (sec 2 x - tan x × sec x ) dx
0
p p
= p ò sec 2 x dx - p ò sec x x × tan x dx
0 0
= p[tan x ]p0 - p[sec x ]p0
p
= p[tan x - sec x ]
0
= p [tan p - sec p - tan 0 - sec 0]
Þ 2 I = p[0 + 1 - 0 + 1]
2I = 2p
\ I=p

2x - 1
Q. 38 ò dx
(x - 1)(x + 2)(x - 3)
K Thinking Process
px + q A B C
Apply = + + , then get the values of A, B and
(x - a)(x - b)(x - c) (x - a) (x - b) (x - c)
1
C and use ò dx = log | x | + C .
x
Sol. (2 x - 1)
Let I= ò (x - 1)(x + 2 )(x - 3)dx
2x - 1 A B C
Now, = + +
(x - 1)(x + 2 )(x - 3) (x - 1) (x + 2 ) (x - 3)
Þ 2 x - 1 = A(x + 2 )(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x + 2 )
Put x = 3, then
6 - 1 = C(3 - 1)(3 + 2 )
1
Þ 5 = 10C Þ C =
2
Again, put x = 1, then
2 - 1 = A (1 + 2 )(1 - 3)
1
Þ 1 = - 6A Þ A=-
6
Now, put x = - 2, then
-4 - 1 = B (-2 - 1)(-2 - 3)
1
Þ -5 = 15B Þ B = -
3
1 1 1 1 1 1
6ò x - 1
\ I=- dx - ò dx + ò dx
3 x+2 2 x-3
1 1 1
= - log| (x - 1)| - log| (x + 2 )| + log| (x - 3)| + C
6 3 2
= - log| (x - 1)|1 6 - log| (x + 2 )|1 3 + log| (x - 3)|1 2 + C

½ x-3 ½
= log½ 16 13 ½
+C
½ (x - 1) (x + 2 ) ½
 -1
xæ 1 + x + x2 ö
Q. 39 ò e tan ç ÷dx
ç 1 + x 2 ÷ø
è
æ
-1 x + x 2 ö÷
x ç1 +
Sol. Let I = ò e tan ç 1 + x 2 ÷ dx
è ø
-1 æ 1 + x ö
2
x
= ò e tan x çç 2
+ ÷ dx
÷
è1 + x 1 + x2 ø
-1
-1 x e tan x
= ò e tan x
dx + ò 2
dx
1+ x
I = I1 + I 2 …(i)
-1
x e tan x
Now, I2 = ò 1 + x2
dx

Put tan-1 x = t Þ x = tan t

1
Þ dx = dt
1 + x2
\ I = ò tan t × et dt
I II

= tan t × et - ò sec 2 t × et dt + C
= tan t × et - ò (1 + tan2 t ) et dt + C [Q sec 2 q = 1 + tan2 q]
-1 x
e tan
I 2 = tan t × et - ò (1 + x 2 ) dx + C
1 + x2
-1
I 2 = tan t × et - ò e tan x
dx + C
tan -1 x -1
\ I = òe dx + tan t × et - ò e tan x
dx + C
t
= tan t × e + C
-1
= x e tan x
+C

x
Q. 40 ò sin -1 dx
a+x
K Thinking Process
First of all put x = tan2 q and convert the given expression into two parts, then use the
æ d ö
formulae for integration by part i.e., ò I× IIdx = Iò IIdx - ò ç Iò IIdx÷ d x.
è dx ø
-1 x
Sol. Let I= ò sin a+ x
dx
Put
x = a tan2 q
Þ dx = 2 a tan q sec 2 q d q
a tan2 q
\ I = ò sin-1 (2 a tan q × sec 2 q) dq
a + a tan2 q
æ tan q ö
= 2 aò sin-1 çç ÷ tan q × sec 2 q dq
÷
è sec q ø
-1
= 2 aò sin (sin q)tan q × sec 2 q dq
 = 2 aò q × tan q sec 2 q d q
I II
é æd ö ù
= 2 a ê q × ò tan q × sec 2 q dq - ò ç q × ò tan q × sec 2 q dq ÷ dqú
ë è dq ø û
é ù
êPut tan q = t ú
êÞ sec q × tan q × dq = dt ú
ê ú
êëÞ ò tan q sec 2 q dq = ò t dt úû
é tan2 q tan2 q ù
= 2 a êq × -ò dqú
ë 2 2 û
= aq tan2 q - aò ( sec 2 q - 1) dq
= aq × tan2 q - a tan q + aq + C
éx x xù
= aê tan-1 + tan-1 ú+C
ë a a aû

p /2 1 + cos x
Q. 41 òp / 3 dx
(1 - cos x) 5/2
p/ 2 1 + cos x
Sol. Let I= òp/ 3 (1 - cos x )5 / 2
dx

p/ 2 1 + cos x
= òp/ 3 (1 - cos x )2 1 + cos x
dx

p2 1 p2 1
= òp3 (1 - cos 2 x )
dx = ò
p3 sin2 x
dx
p2
= òp3 cosec
2
x dx = [- cot x ]pp// 23
é p pù é 1 ù 1
= - êcot - cot ú = - ê 0 - =+
ë 2 3û ë 3 úû 3

Alternate Method
1/ 2
æ 2 xö
ç 2 cos ÷
p/2 1 + cos x è 2øp/ 2
Let I=ò dx = òp/ 3 æ dx
p/ 3 (1 - cos x )5 / 2 5/ 2
2 xö
ç 2 sin ÷
è 2ø
æx ö æx ö
cosç ÷ cosç ÷
2 p/2 è 2 ø dx = 1 p/ 2 è 2 ø dx
= ò
4 2 p/3 5æ x ö
ò
4 p/ 3 5æ x ö
sin ç ÷ sin ç ÷
è2 ø è2 ø
x
Put sin = t
2
x 1
Þ cos × dx = dt
2 2
x
Þ cos dx = 2dt
2
 p 1
As x® , then t ®
3 2
p 1
and x ® , then t ®
2 2
1/ 2
2 1 2 dt 1 é t -5 + 1 ù
\ I= ò
4 1/ 2 t 5
= ê ú
2 ë -5 + 1û1/ 2
é ù
ê ú
1ê 1 1 ú
=- -
8 ê æ 1 ö4 æ 1 ö4 ú
êç ÷ ç ÷ ú
ëè 2 ø è2 ø û
1 12 3
= - (4 - 16) = =
8 8 2
Note If we integrate the trigonometric function in different ways [using different identities]
then, we can get different answers.

Q. 42 ò e -3x cos 3 x dx
Sol. Let I = ò e -3 x cos 3 x dx
II I
æd ö
= cos 3 x ò e -3 x dx - ò ç cos 3 x ò e -3 x dx ÷ dx
è dx ø
-3 x -3 x
e e
= cos 3 x × - ò (- 3cos 2 x ) sin x × dx
-3 -3
1
= - cos 3 x e -3 x - ò cos 2 x sin x e -3 xdx
3
1
= - cos 3 x e -3 x - ò (1 - sin2 x )sin x e -3 xdx
3
1
= - cos 3 x e -3 x - ò sin x e -3 xdx + ò sin3 x e -3 xdx
3
I II
1 e -3 x e -3 x
= - cos 3 x e -3 x - ò sin x e -3 xdx + sin3 x × - ò 3sin2 x cos x × dx
3 -3 -3
1 1
= - cos 3 x e -3 x - ò sin x e -3 xdx - sin3 x e -3 x + ò (1 - cos 2 x )cos x e -3 xdx
3 3
1 -3 x -3 x 1 3
I = - cos x e3
- ò sin x e - sin x e -3 x + ò cos x e -3 xdx - ò cos 3 x e -3 xdx
3 3
I II
e -3 x é e -3 x e -3 x ù
2I = [cos 3 x + sin3 x ] - êsin x × - ò cos x × dx ú + ò cos x e -3 xdx
3 ë -3 -3 û
e -3 x 1 -3 x 1
2I = 3 3
[cos x + sin x ] + sin x × e - ò cos x × e dx + ò cos x e -3 xdx
-3 x
-3 3 3

e -3 x 1 2
2I = [cos 3 x + sin3 x ] + sin x e -3 x + ò cos x e -3 xdx
-3 3 3
 -3 x
Now, let I1 = ò cos x e dx
I II
e -3 x e -3 x
I1 = cos x × - ò (- sin x ) × dx
-3 -3
-1 1
cos x × e -3 x - sin x × e -3 xdx
3 ò
I1 =
3
1 1é e -3 x e -3 x ù
= - cos x × e -3 x - êsin x × - ò cos x × dx ú
3 3ë -3 -3 û
1 -3 x 1 -3 x 1 -3 x
= - cos x × e + sin x × e - ò cos x × e dx
3 9 9
1 1 1
I1 + I1 = - e -3 x × cos x + sin x × e -3 x
9 3 9
æ 10 ö 1 -3 x 1
ç ÷ I1 = - e × cos x + sin x × e -3 x
è 9ø 3 9
- 3 -3 x 1 -3 x
I1 = e × cos x + e sin x
10 10
1 1 3
2 I = - e -3 x [sin3 x + cos 3 x ] + sin x × e -3 x - e -3 x × cos x
3 3 10
1 -3 x
+ e × sin x + C
10
1 -3 x 13 3
\ I = - e [sin3 x + cos 3 x ] + e -3 x × sin x - e -3 x × cos x + C
6 30 10
éQ sin 3x = 3sin x - 4sin3 x ù
ê ú
êëand cos 3x = 4cos 3 x - 3cos x úû
e -3 x 3e -3 x
= [sin 3x - cos 3x ] + [sin x - 3cos x ] + C
24 40

Q. 43 ò tan x dx
Sol. Let I= ò tan x dx
Put tanx = t Þ sec 2 x dx = 2t dt
2

2t t2
\ I = òt × 2
dt = 2 ò dt
sec x 1+ t4
( t 2 + 1) + (t 2 - 1)
= ò (1 + t 4 )
dt

t2 + 1 t2 - 1
= ò 1 + t 4 dt + ò 1 + t 4 dt
1 1
1+ 1-
t 2 dt + t 2 dt
= ò 1 ò 1
t2 + 2
t2 +
t t2
æ 1ö æ 1ö
1 - ç - 2 ÷ dt 1+ ç- 2 ÷
=ò è t ø + è t ø
æ 1ö
2 ò æ 1 ö 2 dt
çt - ÷ + 2 çt + ÷ - 2
è tø è tø
 1 æ 1ö
Put u =t - Þ du = ç 1 + 2 ÷ dt
t è t ø
1 æ 1ö
and v = t + Þ dv = ç 1 - 2 ÷ dt
t è t ø
du dv
\ I=ò 2 2
+ ò 2
u + ( 2) v - ( 2 )2
1 u 1 ½v- 2½
= tan-1 + log ½ ½+ C
2 2 2 2 ½v + 2½
æ tan x - 1 ö
=
1
tan-1 çç ÷ + 1 log ½
½
tan x - 2 tan x + 1½
½+ C
2 2 tan x ÷ 2 2 tan x + 2 tanx + 1½
è ø ½

p /2 dx
Q. 44 ò0
(a cos x + b 2 sin 2 x)2
2 2

p/ 2 dx
Sol. Let I= ò0 (a2 cos 2 x + b 2 sin2 x )2
Divide numerator and denominator by cos 4 x, we get
p/ 2 sec 4 x dx
I=ò
0 2
(a + b 2 tan2 x )2
p/ 2 (1 + tan2 x ) sec 2 x dx
= ò0 (a2 + b 2 tan2 x )2
Put tanx = t
Þ sec 2 x dx = dt
As x ® 0, then t ® 0
p ¥ (1 + t 2 )
and x ® , then t ® ¥
2
I= ò0 (a2 + b 2 t 2 )2
1+ t2
Now, [let t 2 = u]
(a + b 2 t 2 )2
2

1+ u A B
= +
(a2 + b 2u )2 (a2 + b 2u ) (a2 + b 2u )2
Þ 1 + u = A(a2 + b 2u ) + B
On comparing the coefficient of x and constant term on both sides, we get
a2 A + B = 1 …(i)
2
and b A=1 …(ii)
1
\ A= 2
b
a2
Now, + B=1
b2
a2 b 2 - a2
Þ B = 1- 2
=
b b2
2
¥ (1 + t )
\ I= ò0 (a2 + b 2t 2 )2
1 ¥ dt b 2 - a2 ¥ dt
=
b 2 ò0 2
a + bt 2 2
+
b2
ò0 (a2 + b 2t 2 )2
 1 ¥ dt b 2 - a2 ¥ dt
=
b 2 ò0 æa 2 ö
+
b 2 ò0 (a2 + b 2t 2 )2
b ç 2 + t 2 ÷÷
2ç

è b ø
¥
1 é -1 æ tb ö ù b 2 - a2 æ p 1 ö
= 3 ê
tan ç ÷ ú + ç × ÷
ab ë è a øû0 b 2 è 4 a3 b ø
1 p b 2 - a2
= 3
[tan-1 ¥ - tan-1 0] + ×
ab 4 ( a3 b 3 )
p p b 2 - a2
= + ×
2 ab 3 4 ( a3 b 3 )
æ 2 a2 + b 2 - a2 ö p æ a2 + b 2 ö
= p çç ÷= ç
÷ 4 ç a3 b 3
÷
÷
è 4a3 b 3 ø è ø

1
Q. 45 ò0 x log(1 + 2x) dx
K Thinking Process
æ d ö
Use formula for integration by part i.e., ò I× IIdx = Iò IIdx - ò ç Iò Idx÷ dx and also
è dx ø
1
use ò = log | x | + C.
x
1
Sol. Let I= ò0 x log(1 + 2 x )dx1
é x2 ù 1 x2
= êlog(1 + 2 x ) ú - ò ×2 × dx
ë 2 û0 1 + 2x 2
1 2 x2
= [x log(1 + 2 x )]10 - ò dx
2 1 + 2x
é æ x ö ù
1 ê 1ç x 2
÷ ú
= [1log 3 - 0] - ê ò ç - ÷dx ú
2 ê çç 2 1 + 2 x ÷÷ ú
0
êë è ø úû
1 1 1 1 1 x
= log 3 - ò x dx + ò dx
2 2 0 2 0 1 + 2x
1
1 (2 x + 1 - 1)
1 1 éx2 ù 1 1 2
= log 3 - ê ú + ò dx
2 2 ë 2 û0 2 0 (2 x + 1)
1 1 é1 ù 1 1 1 1 1
= log 3 - ê - 0ú + ò dx - ò dx
2 2 ë2 û 4 0 4 0 1 + 2x
1 1 1 1
= log 3 - + [x ]10 - [log|(1 + 2 x )|]10
2 4 4 8
1 1 1 1
= log 3 - + - [log 3 - log 1]
2 4 4 8
1 1
= log 3 - log 3
2 8
3
= log 3
8
 p
Q. 46 ò0 x logsin x dx

K Thinking Process
a a
First of all use property of definite integral i.e., ò0 f(x) dx = ò0 f(a - x) dx, then use
2a a
ò0 f(x) dx = 2ò f(x) dx.
0
p
Sol. Let I= ò0 x logsin x dx ...(i)
p
I = ò (p - x )logsin(p - x ) dx
0
p
= ò (p - x )logsin x dx ...(ii)
0
p
2 I = p ò logsinx dx ...(iii)
0

2I = 2pò
p/2 éQ f(x ) dx = 2 ò f (x ) dx ù
2a a

ëê ò0
logsinx dx
0 0 ûú
p/2
I = pò logsinx dx ...(iv)
0
p/ 2
Now, I=p ò0 logsin(p / 2 - x ) dx ...(v)
On adding Eqs. (iv) and (v), we get
p/2
2I = p ò0 (logsin x + logcos x ) dx
p/ 2
2I = p ò0 logsin x cos x dx
p / 22 2 sin x cos x
=p ò0 log
2
dx
p/2
2I = p ò0 (logsin2 x - log 2 ) dx
p/ 2 p/2
2I = p ò0 logsin2 x dx - p ò0 log 2 dx
1
Put 2x = t Þ dx = dt
2
As x ® 0, then t ® 0
p
and x ® , then t ® p
2
p p p2
\ 2I =
2 ò0 logsint dt -
2
log 2
2
p p p
Þ 2I =
2 ò0 logsin x dx - 2 log2
p2
Þ 2I = I - log 2 [from Eq. (iii)]
2
2
p p2 æ 1ö
\ I=- log 2 = log ç ÷
2 2 è2 ø
 p /4
Q. 47 òp / 4 log(sin x + cos x) dx
p/ 4
Sol. Let I= ò- p / 4 log(sin x + cos x ) dx ...(i)
p/ 4 ì æp p ö æp p öü
I= ò- p/ 4 log íîsin çè 4 - 4
- x ÷ + cos ç - - x ÷ ý dx
ø è4 4 øþ
p/ 4
= ò- p/ 4 log{sin(- x ) + cos(- x )} dx
p4
and I= ò- p/ 4 log(cos x - sin x ) dx ...(ii)
From Eqs. (i) and (ii),
p4
2I = ò- p/ 4 logcos2 x dx
p/ 4
2I = ò0 logcos 2 x dx ...(iii)
éQ f(x ) dx = 2
ò0 f(x ), if f(- x ) = f(x )ùûú
a a

ëê ò- a
dt
Put 2x = t Þ dx =
2
As x ® 0, then t ® 0
p p
and x ® , then t ®
4 2
1 p2
2 ò0
2I = logcost dt ...(iv)

1 p2 æp ö éQ af(x ) dx = af(a - x ) dx ù
2 I = ò logcosç - t ÷ dt
2 0 è2 ø ëê ò0 ò0 ûú
1 p2
Þ 2 I = ò logsin t dx ...(v)
2 0
On adding Eqs. (iv) and (v), we get
1 p2
4I = ò logsin t cos t dt
2 0
1 p2 sin2 t
Þ 4I = ò log dt
2 0 2
1 p 2 1 p2
Þ 4I = ò logsin2 x dx - ò log 2 dx
2 0 2 0
1 p2 æp ö p
Þ 4I = ò logsinç - 2 x ÷ dx - log 2 .
2 0 è2 ø 4
1 p2 p
Þ 4I = ò logcos 2 x dx - log 2
2 0 4
p/4 p éQ 2 a f(x ) dx = 2 af(x ) dx ù
4I = ò logcos 2 x dx - log 2
êë ò0 ò0
Þ
0 4 úû
p
Þ 4I = 2 I - log 2 [from Eq. (iii)]
4
p p æ 1ö
\ I = - log 2 = log ç ÷
8 8 è2 ø
Objective Type Questions
cos 2x - cos 2 q
Q. 48 ò cos x - cos q
dx is equal to

(a) 2 (sin x + x cos q) + C (b) 2 (sin x - x cos q) + C

(c) 2 (sin x + 2x cos q) + C (d) 2 (sin x - 2x cos q) + C
K Thinking Process
Use formula cos 2 q = 2 cos2 q - 1 to get simplest form, then apply ò cos x dx = sin x + C .
cos 2 x - cos 2 q
Sol. (a) Let I= ò cos x - cos q
dx

(2 cos 2 x - 1 - 2 cos 2 q + 1)
= ò cos x - cos q
dx

(cos x + cos q) (cos x - cos q)

= 2ò dx
(cos x - cos q)
= 2 ò (cos x + cos q) dx
= 2(sin x + x cos q) + C

dx
Q. 49 is equal to
sin(x - a)sin(x - b)
sin( x - b) sin( x - a)
(a) sin( b - a) log +C (b) cosec ( b - a) log +C
sin( x - a) sin( x - b)
sin( x - b) sin( x - a)
(c) cosec( b - a) log +C (d) sin( b - a) log +C
sin( x - a) sin( x - b)
dx
Sol. (c) Let I= ò sin(x - a)sin(x - b)
1 sin(b - a)
sin(b - a) ò sin(x - a)sin(x - b )
= dx

1 sin(x - a - x + b )
sin(b - a) ò sin(x - a)sin(x - b )
= dx

1 sin{(x - a) - (x - b )}
sin(b - a) ò sin(x - a)sin(x - b )
= dx

1 sin(x - a)cos(x - b ) - cos(x - a)sin(x - b )

sin(b - a) ò
= dx
sin(x - a)sin(x - b )
1
sin(b - a) ò
= [cot(x - b ) - cot(x - a)] dx

1
= [log|sin(x - b )| - log|sin(x - a)|] + C
sin(b - a)
sin(x - b )
= cosec(b - a) log +C
sin(x - a)
Q. 50 ò tan -1 x dx is equal to
(a) ( x + 1) tan -1 x - x + C (b) x tan -1 x - x + C
(c) x - x tan -1 x + C (d) x - ( x + 1) tan -1 x + C
K Thinking Process
æ d ö
Use formula for integration by part i.e., ò I× II dx = Iò II dx - ò ç Iò II dx÷ dx.
è dx ø
Sol. (a) Let I = ò 1 × tan-1 x dx
1 1 2
= tan-1 x × x - ò × dx
2 (1 + x ) x
1 2
= x tan-1 x - ò dx
2 x (1 + x )
Put x = t 2 Þ dx = 2t dt
t
\ I = x tan-1 x - ò t (1 + t 2 )dt
t2
= x tan-1 x - ò 1 + t 2 dt
æ 1 ö
= x tan-1 x - ò çç 1 - ÷ dt
÷
è 1+ t2 ø
= x tan-1 x - x + tan-1 t + C
-1
= x tan x - x + tan-1 x + C
= (x + 1)tan-1 x - x +C

x9
Q. 51 ò dx is equal to
(4 x 2 + 1)6
1 æ 1 ö -5 1æ 1 ö -5
(a) ç4 + 2÷ + C (b) ç4 + 2÷ + C
5x è x ø 5è x ø
-5
1 1 æ 1 ö
(c) (1 + 4) -5 + C (d) ç 2 + 4÷ + C
10 x 10 è x ø
x9 x9
Sol. (d) Let I= ò (4x 2 + 1)6 dx = ò æ 1 ö
6
dx
x 12 ç 4 + 2 ÷
è x ø
dx
= ò æ 1 ö
6
3
x ç4 + 2 ÷
è x ø
1 -2
Put 4+ = t Þ 3 dx = dt
x2 x
1 1
Þ dx = - dt
x3 2
1 dt 1 é t -6 + 1 ù
\ I =- ò 6 =- ê ú+C
2 t 2 ë -6 + 1û
-5
1 é1ù 1 æ 1 ö
= +C= ç4 + 2 ÷ +C
10 êë t 5 úû 10 è x ø
 dx 1
Q. 52 If ò 2
= a log|1 + x 2| + b tan -1 x + log| x + 2| + C , then
(x + 2) (x + 1) 5
-1 -2 1 2
(a) a = ,b= (b) a = ,b=-
10 5 10 5
-1 2 1 2
(c) a = , b = (d) a = ,b=
10 5 10 5
K Thinking Process
1 A Bx + C
Use method of partial fraction i.e., = +
(x - a) (x2 + bx + c) (x - a) (x2 + bx + c)
to solve the above problem.
dx 1
Sol. (c) Given that, ò = alog|1 + x 2| + b tan-1 x + log|x + 2| + C
(x + 2 ) (x 2 + 1) 5
dx
Now, I= ò (x + 2 ) (x 2 + 1)
1 A Bx + C
= + 2
(x + 2 ) (x 2 + 1) x+2 x +1
Þ 1 = A(x 2 + 1) + (Bx + C ) (x + 2 )
Þ 1 = Ax 2 + A + Bx 2 + 2 Bx + Cx + 2C
Þ 1 = ( A + B)x 2 + (2 B + C )x + A + 2C
Þ A + B = 0, A + 2 C = 1, 2 B + C = 0
1 1 2
We have, A = , B = - and C =
5 5 5
1 2
- x+
dx 1 1
\ ò (x + 2 ) (x 2 + 1) = 5 ò x + 2 dx + ò x 2 + 15 dx
5

1 1 1 x 1 2
5ò x + 2 5ò 1 + x 2
=dx - dx + ò dx
5 1 + x2
1 1 2
= log|x + 2| - log|1 + x 2| + tan-1 x + C
5 10 5
2 -1
\ b = and a =
5 10

3
Q. 53 ò x is equal to
x +1
x 2 x3 x2 x3
(a) x + + - log|1 - x| + C (b) x + - - log|1 - x| + C
2 3 2 3
2 3 2 3
x x x x
(c) x - - - log|1 + x| + C (d) x - + - log|1 + x| + C
2 3 2 3
x3
Sol. (d) Let I= ò x + 1 dx
æ 1 ö
= ò çç (x 2 - x + 1) - ÷ dx
è ( x + 1) ÷ø
x3 x2
= - + x - log|x + 1|+ C
3 2
 x + sin x
Q. 54 ò dx is equal to
1 + cos x
(a) log|1 + cos x| + C (b) log| x + sin x| + C
x x
(c) x - tan +C (d) x × tan + C
2 2
x + sin x
Sol. (d) Let I= ò 1 + cos x dx
x sin x
= ò 1 + cos x dx + ò 1 + cos x dx
x 2 sin x / 2 cos x / 2
= ò 2 cos2 x / 2 dx + ò 2 cos 2 x / 2
dx

1
2ò
= x sec 2 x / 2 dx + ò tan x / 2 dx

1é x ù x
= ê x × tan x / 2 × 2 - ò tan × 2 dx ú + ò tan 2 dx
2ë 2 û
x
= x × tan +C
2

x 3dx
Q. 55 If = a(1 + x 2 ) 3/2 + b 1 + x 2 + C , then
2
1+x
1 -1
(a) a = , b = 1 (b) a = , b =1
3 3
-1 1
(c) a = , b = -1 (d) a = , b = -1
3 3
x3
Sol. (d) Let I= ò 1 + x2
dx = a(1 + x 2 )3 / 2 + b 1 + x 2 + C

x3 x2 × x
Q I= ò 1+ x 2
dx = ò 1 + x2
dx

Put 1 + x2 = t 2
Þ 2 x dx = 2t dt
t (t 2 - 1) t3
\ I=ò dt = -t + C
t 3
1
= (1 + x 2 )3 / 2 - 1 + x 2 + C
3
1
\ a = and b = - 1
3

p /4 dx
Q. 56 ò-p / 4 is equal to
1 + cos 2x
(a) 1 (b) 2 (c) 3 (d) 4
p/4dx p/4 dx
Sol. (a) Let I=ò =ò
- p / 4 1 + cos 2 x - p / 4 2 cos 2 x

1 p/4 p/4
= ò sec 2 x dx = ò sec 2 x dx = [tan x ]0p/ 4 = 1
2 -p / 4 0
 p /2
Q. 57 ò0 1 - sin 2x dx is equal to
(a) 2 2 (b) 2 ( 2 + 1)
(c) 2 (d) 2 ( 2 - 1)
p/ 2
Sol. (d) Let I= ò0 1 - sin2 x dx
p/4 p/ 2
= ò0 (cos x - sin x )2 dx + òp / 4 (sin x - cos x )2 dx

= [sin x + cos x ]p0 / 4 + [- cos x - sin x ]pp // 24

1 1 æ 1 1 ö
= + - 0 - 1+ ç- 0 - 1+ + ÷
2 2 è 2 2ø
= 2 2 - 2 = 2( 2 - 1)

p /4
Q. 58 ò0 cos xe sin x dx is equal to
(a) e + 1 (b) e - 1 (c) e (d) - e

p/ 2
Sol. (b) Let I= ò0 cos x esin xdx
Put sinx = t Þ cos x dx = dt
As x ® 0, then t ® 0
and x ® p / 2, then t ® 1
1 t
ò0e dt = [e ]0
t 1
\ I=
= e1 - e 0 = e - 1

x +3
Q. 59 ò 2
e x dx is equal to
(x + 4)
æ 1 ö æ 1 ö
(a) ex çç ÷+C
÷ (b) e- x çç ÷+C
÷
è x + 4ø è x + 4ø
æ 1 ö æ 1 ö
(c) e- x çç ÷+C
÷ (d) e2x çç ÷+C
÷
è x - 4ø è x - 4ø
x+ 3
ò (x + 4)2 e
x
Sol. (a) Let I= dx

ex ex
= ò (x + 4) - ò (x + 4)2 dx
æ 1 1 ö
= ò e x çç - ÷ dx
÷
è ( x + 4 ) (x + 4)2 ø
æ 1 ö
= e x çç ÷+C
÷ [Q ò e x {f(x ) + f ¢(x )} dx = e x f(x ) + C ]
è x + 4ø
Fillers
a 1 p
Q. 60 If ò0 dx = , then a = ......... .
2
1 + 4x 8
a 1 p
Sol. Let I = ò0 1 + 4x 2 dx = 8
a 1 2
Now, ò0 dx = [tan-1 2 x ]a0
æ1 2ö 4
4ç + x ÷
è4 ø
1
= tan-1 2 a - 0 = p/ 8
2
1 p
tan-1 2 a =
2 8
Þ tan-1 2 a = p/ 4
Þ 2a = 1
1
\ a=
2

sin x
Q. 61 ò dx = ......... .
3 + 4 cos2 x
Sol. sin x
Let I= ò 3 + 4cos2 x dx
Put cos x = t Þ - sinx dx = dt
dt 1 dt
\ I = -ò 2
=- ò
3 + 4t 4 æ 3 ö2
ç ÷ 2
ç 2 ÷ +t
è ø
1 2 2t
=- × tan-1 +C
4 3 3
1 æ 2 cos x ö
=- tan-1 ç ÷+C
2 3 è 3 ø

p
Q. 62 The value of ò-p sin 3 x cos2 x dx is ......... .
p
ò- psin x cos x dx
3 2
Sol. We have, f( x ) =
p
f(- x ) = ò sin3 (-2 ) - cos 2 (- x ) dx
-p

= - f( x )
Since, f(x ) is an odd function.
p
ò- psin
3
\ x cos 2 x dx = 0
 8
Application o f Integrals
Short Answer Type Questions
Q. 1 Find the area of the region bounded by the curves y 2 = 9x and y = 3x.
K Thinking Process
On solving both the equation of curves, get the values of x and then at those values, find
the area of the shaded region.
Sol. We have, y2 = 9x and y = 3x
Þ (3x )2 = 9x
2
Þ 9x - 9x = 0
Þ 9x(x - 1) = 0
Þ x = 1, 0
Y
y=3x

(0, 0) X
(1, 0)

y 2 = 9x

1 1
\ Required area, A = ò0 9x dx - ò0 3x dx
1 1
= 3ò x 1/ 2 dx - 3ò x dx
0 0
1 1
é x 3/ 2 ù éx2 ù
=3ê ú -3ê ú
ë 3 / 2 û0 ë 2 û0

= 3 æç - 0 ö÷ - 3 æç - 0 ö÷
2 1
è3 ø è2 ø
3 1
= 2 - = sq units
2 2
Q. 2 Find the area of the region bounded by the parabola y 2 = 2 px
x 2 = 2 py.
Sol. We have, y2 = 2 px and x 2 = 2 py
\ y = 2 px
Þ x 2 = 2 p × 2 px
Þ x 4 = 4 p2 × (2 px )
Þ x 4 = 8 p3 x
Þ x - 8 p3 x = 0
4

Þ x (x - 8 p3 ) = 0
3

Þ x = 0, 2 p
Y
x 2 = 2py

X
O 2p

y 2 = 2 px

2p x2 2p
\ Required area = ò0 2 px dx -
2p ò0
dx

2p 1 2p 2
= 2 pò x 1/ 2dx -
2 p ò0
x dx
0

2p 2p
é 2(x )3 / 2 ù 1 éx3 ù
= 2p ê ú - ê ú
ë 3 û0 2 p ë 3 û0

= 2 p é × (2 p)3 / 2 - 0ù -
1 é1
(2 p)3 - 0ù
2
ëê 3 úû 2 p ëê 3 úû

= 2 p æç × 2 2 p3 / 2 ö÷ -
2 1 æ1 3ö
ç 8p ÷
è3 ø 2p è3 ø
æ 4 2 3/ 2 ö 1 æ8 3ö
= 2 p çç p ÷÷ - ç p ÷
è 3 ø 2 p è3 ø
4 2 8
= × 2 p2 - p2
3 6
(16 - 8)p2 8 p2
= =
6 6
4 p2
= sq units
3
Q. 3 Find the area of the region bounded by the curve y = x 3 , y = x + 6 and
x = 0.
Sol. We have, y = x 3 , y = x + 6 and x = 0
Y
y=x 3 y=x + 6

X
O (0, 0) (2, 0)

\ x3 = x + 6
3
Þ x -x=6
Þ x3 - x - 6 = 0
Þ x 2 (x - 2 ) + 2 x (x - 2 ) + 3 (x - 2 ) = 0
Þ (x - 2 )(x 2 + 2 x + 3) = 0
Þ x = 2, with two imaginary points
2
ò0 (x + 6 - x
3
\ Required area of shaded region = ) dx
2
éx2 x4 ù
=ê + 6x - ú
ë2 4 û0

= é + 12 - - 0ù
4 16
ëê 2 4 ûú
= [2 + 12 - 4] = 10 sq units

Q. 4 Find the area of the region bounded by the curve y 2 = 4 x and x 2 = 4 y.

K Thinking Process
First, by using both the equation get the values of x and then find the shaded region by
using these value of x in the equation of curve in x only.
Sol. Given equation of curves are
y2 = 4x …(i)
2
and x = 4y …(ii)
2
æx ö
2
Þ ç ÷
ç 4 ÷ = 4x
Y
è ø x 2 = 4y
x4
Þ = 4x
4× 4
Þ x 4 = 64x (4, 0)
X
(0, 0)
4
Þ x - 64x = 0
Þ x(x 3 - 43 ) = 0
Þ x = 4, 0 y 2 = 4x
 4æ x2 ö
\ Area of shaded region, A = ò0 çç 4x -
4
÷ dx
÷
è ø
4
4æ x2 ö é2 x 3 / 2 × 2 1 x 3 ù
= ò0 çç 2 x - 4 ÷÷ dx = ê 3 - 4 × 3 ú
è ø ë û0
2 ×2 1 64 32 16 16
= ×8 - × - 0= - = sq units
3 4 3 3 3 3

Q. 5 Find the area of the region included between y 2 = 9x and y = x.

2
Sol. We have, y = 9x and y = x
Þ x 2 = 9x
2
Þ x - 9x = 0
Þ x(x - 9) = 0
Þ x = 0, 9
Y
y=x

X
(0, 0) (9, 0)
O

y 2 = 9x

9 9 9
ò0 ( ò0 3x ò0
1/ 2
\ Area of shaded region, A = 9x - x ) dx = dx - x dx
9 2 ù9
é x 3/ 2 ù éx
= ê3 × × 2ú - ê ú
ë 3 û0 ë 2 û0
é 3
´2 ù
ê 3 × 32 ú
× 2 - 0ú - é - 0ù
81
=ê
3 ê
ë2 úû
ê ú
ë û
81 108 - 81 27
= 54 - = = sq units
2 2 2

Q. 6 Find the area of the region enclosed by the parabola x 2 = y and the line
y = x + 2.
Sol. We have, x 2 = y and y = x + 2 Y
2
Þ x =x+2
Þ x2 - x - 2 = 0
y
Þ x 2 - 2x + x - 2 = 0 x=
2

Þ x(x - 2 ) + 1(x - 2 ) = 0 X¢ X
–1 2
Þ (x + 1)(x - 2 ) = 0
Þ x = - 1, 2 y=x+2

Y¢
 2
2 éx2 x3 ù
ò-1(x + 2 - x
2
\ Required area of shaded region = ) dx = ê + 2x - ú
ë2 3 û -1

=é + 4- - +2 - ù
4 8 1 1
êë 2 3 2 3 úû
3 9 36 + 9 - 18 27 9
=6+ - = = = sq units
2 3 6 6 2

Q. 7 Find the area of the region bounded by line x = 2 and parabola y 2 = 8 x.

Sol. We have, y2 = 8x and x = 2
Y
y2 = 8x

X
(0, 0) (2, 0)

x=2
2 2
\ Area of shaded region, A = 2 ò 8x dx = 2 ×2 2 ò x 1/ 2dx
0 0
2
é x 3/ 2 ù é2 ù
= 4 × 2 × ê2 × ú = 4 2 ê × 2 2 - 0ú
ë 3 û0 ë3 û
32
= sq units
3

Q. 8 Sketch the region {(x, 0) : y = 4 - x 2 } and X-axis. Find the area of the
region using integration.
Sol. Given region is {(x, 0) : y = 4 - x 2 } and X-axis.
We have, y= 4 - x 2 Þ y2 = 4 - x 2Þ x 2 + y2 = 4
Y

X
(–2, 0) (2, 0)

2 2
\ Area of shaded region, A = ò-2 4 - x 2 dx = ò-2 2 2 - x 2 dx
2
éx 22 xù
= ê 22 - x 2 + × sin-1 ú
ë 2 2 2 û -2
2 p 2 -1 p p
= × 0 + 2 . + × 0 - 2 sin (-1) = 2 . + 2 .
2 2 2 2 2
= 2p sq units
Q. 9 Calculate the area under the curve y = 2 x included between the lines
x = 0 and x = 1.
Sol. We have, y = 2 x , x = 0 and x = 1
Y
y=2Öx

X
(0, 0) O (1,0)

x=1
1
\ Area of shaded region, A = ò0 (2 x ) dx
1
é x 3/ 2 ù
= 2×ê × 2ú
ë 3 û0
æ 2 ö 4
= 2 ç × 1 - 0 ÷ = sq units
è3 ø 3

Q. 10 Using integration, find the area of the region bounded by the line
2 y = 5x + 7, X-axis and the lines x = 2 and x = 8.
Sol. We have, 2 y = 5x + 7
5x 7
Þ y= +
2 2
+7
5x
=
2y

X
(0, 0) (2, 0) (8, 0)
x=2 x=8
8
1 8 1é x ù 2

2 ò2
\ Area of shaded region = (5x + 7 ) d x = ê 5 × + 7x ú
2ë 2 û2
1 1
= [5× 32 + 7 × 8 - 10 - 14] = [160 + 56 - 24]
2 2
192
= = 96 sq units
2
Q. 11 Draw a rough sketch of the curve y = x - 1 in the interval [1, 5].
Find the area under the curve and between the lines x = 1 and x = 5.
Sol. Given equation of the curve is y = x - 1.
Þ y2 = x - 1
Y

X
O 1 5

y=Öx – 1
x=5
x=1
5
5 é 2 × (x - 1)3 / 2 ù
ò1 (x - 1)
1/ 2
\ Area of shaded region, A = dx = ê ú
ë 3 û1

= é 2
× (5 - 1)3/ 2 ù
-0 =
16
sq units
ëê 3 ûú 3

Q. 12 Determine the area under the curve y = a 2 - x 2 included between

the lines x = 0 and x = a.
Sol. Given equation of the curve is y = a2 - x 2 .
Þ y2 = a 2 - x 2 Þ y2 + x 2 = a 2
Y

X
–a O a

x=0 x=a
a
\ Required area of shaded region, A = ò0 a2 - x 2 dx
a
éx 2 a 2 -1 x ù
=ê a - x2 + sin ú
ë2 2 a û0

é a 2 -1 a 2 -1 ù
= ê0 + sin (1) - 0 - sin 0ú
ë 2 2 û
a2 p p a2
= × = sq units
2 2 4
Q. 13 Find the area of the region bounded by y = x and y = x.
Sol. Given equation of curves are y= x and y = x .
Þ x= x Þ x2 = x
Þ x 2 - x = 0 Þ x(x - 1) = 0
Þ x = 0, 1
Y
y=Öx

(0, 0)
X
(1, 0)

y=x
1 1
\ Required area of shaded region, A = ò0 ( x )dx - ò0 xdx
1 1
é x 3/ 2 ù éx2 ù
= ê2 × ú -ê ú
ë 3 û0 ë 2 û0
2 1 2 1 1
= × 1 - = - = sq units
3 2 3 2 6

Q. 14 Find the area enclosed by the curve y = - x 2 and the straight line
x + y + 2 = 0.
Sol. We have, y = - x 2 and x + y + 2 = 0
Y

–1 2
X

–x 2=y

Þ -x - 2 = - x 2 Þ x 2 - x - 2 = 0
2
Þ x + x - 2 x - 2 = 0 Þ x(x + 1) - 2(x + 1) = 0
Þ (x - 2 )(x + 1) = 0Þ x = 2, - 1
2 2
\ Area of shaded region, A = ò-1(-x - 2 + x
2
) dx = ½ò (x 2 - x - 2 ) dx½
½ -1 ½
2
éx3 x2 ù
- 2 x ú = ½é - - 4 + + - 2 ù½
8 4 1 1
= ê -
ê úû½
ë 3 2 û - 1 ½ë 3 2 3 2
16 - 12 - 24 + 2 + 3 - 12½ ½ 27½ 9
=½ = - = sq units
½ 6 ½ ½ 6½ 2
Q. 15 Find the area bounded by the curve y = x, x = 2 y + 3 in the first
quadrant and X-axis.
Sol. Given equation of the curves are y = x and x = 2 y + 3 in the first quadrant.

On solving both the equations for y, we get

3
2y+
Y
y = 2y + 3

x=
Þ y2 = 2 y + 3 y = Öx
3
2
Þ y - 2y - 3 = 0
2
Þ y - 3y + y - 3 = 0 X
Þ y( y - 3) + 1( y - 3) = 0 –1
Þ ( y + 1)( y - 3) = 0
Þ y = - 1, 3
\ Required area of shaded region,
3
3 é 2 y2 y3 ù
ò0 (2 y + 3 - y )dy = ê 2 + 3 y - 3 ú
2
A=
ë û0
= é 18 ù
+ 9 - 9 - 0 = 9 sq units
ëê 2 ûú

Long Answer Type Questions

Q. 16 Find the area of the region bounded by the curve y 2 = 2x and
x 2 + y 2 = 4x.
Sol. We have, y2 = 2 x and x 2 + y2 = 4x
Y
y2 = 2x

(0, 0) (2, 0) X

(x – 2)2 + y2 = 4

Þ x 2 + 2 x = 4x
Þ x 2 - 2x = 0
Þ x(x - 2 ) = 0
Þ x = 0, 2
Also, x 2 + y2 = 4x
Þ x 2 - 4x = - y2
Þ x - 4x + 4 = - y2 + 4
2

Þ (x - 2 )2 - 2 2 = - y2
 \ Required area = 2 × ò é 2 2 - (x - 2 )2 - 2 x ù dx
2
0 ë û
éé x - 2 2 2 -1 æ x - 2 ö ù
2
é x 3/ 2 ù ù
2

= 2 êê × 2 2 - (x - 2 )2 + sin ç ÷ú - ê 2 × ú ú
êë ë 2 2 è 2 øû0 ë 3 / 2 û0 ú
û
éæ p ö 2 2 3/ 2 ù
= 2 ê ç 0 + 0 - 1× 0 + 2 × ÷ - (2 - 0)ú
ëè 2ø 3 û
4p 8 × 2 16 æ 8ö
= - = 2p - = 2 ç p - ÷ sq units
2 3 3 è 3ø

Q. 17 Find the area bounded by the curve y = sin x between x = 0 and x = 2p.
K Thinking Process
We know that, sin x curve has positive region from [0 , p] and negative region in [p , 2 p].
2p p 2p
Sol. Required area = ò0 sin x dx = ò0 sin x dx + ½½òp sin x dx½
½
= - [cos x ] p0 + [- cos x ] 2p p
= - [cos p - cos 0] + - [cos 2 p - cos p ]
Y

y=sin x

2p
p X
O

= - [- 1 - 1] + - (1 + 1)
= 2 + 2 = 4 sq units

Q. 18 Find the area of region bounded by the triangle whose vertices are
(- 1, 1), (0, 5) and (3, 2), using integration.
Sol. Let we have the vertices of a DABC as A (- 1, 1,) B (0, 5) and C (3, 2 ).
Y
B (0, 5)

C (3, 2)
A
(–1, 1)
X¢ X
–1 3

Y¢
æ 5 - 1ö
\ Equation of AB is y - 1 = çç ÷÷ (x + 1)
è 0 + 1ø
Þ y - 1 = 4x + 4
Þ y = 4x + 5 …(i)
æ2 - 5ö
and equation of BC is y - 5 = çç ÷÷(x - 0)
è 3 - 0ø
 -3
Þ y-5= (x )
3
Þ y=5-x …(ii)
æ 2 - 1ö
Similarly, equation of AC is y - 1 = çç ÷÷ (x + 1)
è 3 + 1ø
1
Þ y - 1 = (x + 1)
4
Þ 4y = x + 5 …(iii)
0 3
\ Area of shaded region = ò-1 ( y1 - y2 )dx + ò0 ( y1 - y2 )dx
0 é x + 5ù 3é x + 5ù
= ò-1 êë 4x + 5 - 4 úû
dx + ò0 êë 5 - x - 4 úû
dx

0 3
é 4x 2 x 2 5x ù é x 2 x 2 5x ù
=ê + 5x - - ú + ê 5x - - - ú
ë 2 8 4 û -1 ë 2 8 4 û0
é 1 5 ù é ù
= ê 0 - æç 4 × + 5 (-1) - + ö÷ ú + ê æç 15 - - -
1 9 9 15 ö
÷ - 0ú
ë è 2 8 4 øû ëè 2 8 4ø û
é 1 5
= - 2 + 5 + - + 15 - - -
9 9 15 ù
êë 8 4 2 8 4 úû
æ 1 - 10 - 36 - 9 - 30 ö
= 18 + ç ÷
è 8 ø
æ 84 ö
= 18 + ç - ÷ = 18 -
21 15
= sq units
è 8 ø 2 2

Q. 19 Draw a rough sketch of the region {(x, y) : y 2 £ 6 ax and

x 2 + y 2 £ 16a 2 }. Also, find the area of the region sketched using
method of integration.
Sol. We have, y2 = 6ax and x 2 + y2 = 16a2
Þ x 2 + 6ax = 16a2
Þ x + 6ax - 16a2 = 0
2

Þ x + 8ax - 2 ax - 16a2 = 0
2

Þ x(x + 8a) - 2 a(x + 8a) = 0

Þ (x - 2 a)(x + 8a) = 0
Þ x = 2 a, - 8a
Y

X
–4a O 2a 4a
–1

x 2+y 2 = (4a)2
y 2 = 6a x
 \ Area of required region = 2 é ò (4 a)2 - x 2 dx ù
2a 4a
6ax dx + ò
êë 0 2a úû
= 2 éò (4 a)2 - x 2 dx ù
2 a 4 a
6a x 1/ 2dx + ò
ëê 0 2a ûú
é é x 3/ 2 ù
2a
æx (4 a)2 -1 x ö ù
4a

= 2 ê 6a ê ú + ç ( 4 a )2
- x 2
+ sin ÷ ú
êë ç2 4 a ÷ø 2 a ú
ë 3 / 2 û0 è 2
û
é 2 4a 16 a 2
p 2 a 16 a2 2aù
= 2 ê 6a × ((2 a)3 / 2 - 0) + ×0+ × - 16 a2 - 4 a2 - × sin-1 ú
ë 3 2 2 2 2 2 4a û
é 2 3/ 2 2 2a 2 pù
= 2 6a ×2 2 a + 0 + 4p a - ×2 3 a - 8a ×
êë 3 2 6 úû
é 4 4a p ù
2
= 2 ê 12 × a2 + 4p a2 - 2 3 a2 - ú
ë 3 3 û
é 8 3 a2 + 12 p a2 - 6 3 a2 - 4 a2 p ù
=2 ê ú
ë 3 û
2
= a2 [8 3 + 12 p - 6 3 - 4p ]
3
2 2 4
= a [2 3 + 8p ] = a2 [ 3 + 4p ]
3 3

Q. 20 Compute the area bounded by the lines x + 2 y = 2, y - x = 1 and

2x + y = 7.
Sol. We have, x + 2y = 2 …(i)
y- x =1 …(ii)
and 2x + y = 7 …(iii)
On solving Eqs. (i) and (ii), we get
y - (2 - 2 y) = 1 Þ 3 y - 2 = 1 Þ y = 1
Y

2x+y=7

(0, 3)

(0, 1)

X¢ X
O x+
1 (0, –1)
2y =
x= 2
y–

Y¢
2( y - 1) + y = 7
 On solving Eqs. (ii) and (iii), we get
Þ 2y - 2 + y = 7
Þ y=3
On solving Eqs. (i) and (iii), we get
2(2 - 2 y) + y = 7
Þ 4 - 4y + y = 7
Þ -3 y = 3
Þ y = -1
1 3 (7 - y) 3
\ Required area = ò (2 - 2 y) dy + ò dy - ò ( y - 1) dy
-1 -1 2 1
1 3 3
é 2 y2 ù é7 y y2 ù é y2 ù
= ê -2 y + ú + ê - ú -ê - yú
ë 2 û -1 ë 2 2 × 2 û -1 ë 2 û1
é 2
= -2 + - 2 -
2 ù é 21 9 7 1 ù é 9
+ - + + - - 3 - + 1ù
1
êë 2 2 úû êë 2 4 2 4 úû êë 2 2 úû
42 - 9 + 14 + 1ù é 9 - 6 - 1 + 2 ù
= [-4] + é -
ëê 4 ûú ëê 2 ûú
= - 4 + 12 - 2 = 6 sq units

Q. 21 Find the area bounded by the lines y = 4 x + 5, y =5- x and

4 y = x + 5.
Sol. Y y = 4x + 5

(0, 5)

X¢ X
–1 3
x +5
4 y= y=5–x

Y¢
Given equations of lines are
y = 4x + 5 …(i)
y=5-x …(ii)
and 4y = x + 5 …(iii)
On solving Eqs. (i) and (ii), we get
4x + 5 = 5 - x
Þ x=0
On solving Eqs. (i) and (iii), we get
4(4x + 5) = x + 5
Þ 16x + 20 = x + 5
Þ 15x = - 15
Þ x = -1
On solving Eqs. (ii) and (iii), we get
4 (5 - x ) = x + 5
Þ 20 - 4x = x + 5
 Þ x=3
0 3 1 3
\ Required area = ò-1 (4x + 5)dx + ò0 (5 - x )dx -
4 ò-1
(x + 5)dx
0 3 3
é 4x 2 ù é x2 ù 1 éx2 ù
=ê + 5x ú + ê 5x - ú - ê + 5x ú
ë 2 û -1 ë 2 û0 4 ë 2 û -1
é
= [0 - 2 + 5] + 15 - - 0 -
9 ù 1 é9
+ 15 - + 5ù
1
êë 2 úû 4 êë 2 2 úû
21 1
=3+ - × 24
2 4
21 15
=-3+ = sq units
2 2

Q. 22 Find the area bounded by the curve y = 2cos x and the X-axis from
x = 0 to x = 2p.
2p
Sol. Required area of shaded region = ò0 2 cos xdx
p/ 2 3 p/ 2 2p
=ò 2 cos x dx + ½ò 2 cos x dx½ + ò3 p/ 2 2 cos x dx
0 ½ p/ 2 ½
Y
y=cos x

p
O p/2 3p/2 X
2p

= 2[sin x ]0p/ 2 + 2(sin x )3p/p2/ 2 + 2[sin x ]23 pp/ 2

= 2 + 4 + 2 = 8 sq units

Q. 23 Draw a rough sketch of the given curve y = 1 + | x + 1|, x = - 3, x = 3,

y = 0 and find the area of the region bounded by them, using
integration.
Sol. We have, y = 1 + |x + 1|, x = - 3, x = 3 and y = 0
Y

–3 –1 0 3 X
 ì - x, if x < - 1
Q y=í
îx + 2, if x ³ - 1
-1 3
\ Area of shaded region, A = ò-3 - x dx + ò-1(x + 2 ) dx
-1 3
éx2 ù éx2 ù
=-ê ú + ê + 2x ú
2
ë û -3 ë 2 û -1
=- é 1 9ù é 9
- + + 6- +
1
2ù
êë 2 2 úû êë 2 2 úû
= - [- 4] + [8 + 4]
= 12 + 4 = 16 sq units

Objective Type Questions

Q. 24 The area of the region bounded by the Y-axis y = cos x and y = sin x,
p
where 0 £ x £ , is
2
(a) 2 sq units (b) ( 2 + 1) sq units
(c) ( 2 - 1) sq units (d) (2 2 - 1) sq units
p
Sol. (c) We have, Y-axis i.e., x = 0, y = cos x and y = sin x , where 0 £ x £ .
2
Y
y=sin x
1

X¢ X
O p/4 p/2

y=cos x
Y¢
p/ 4
\ Required area = ò0 (cos x - sin x )dx
= [sin x ]p0/ 4 + [cos x ]p0/ 4
p p
= æç sin - sin 0 ö÷ + æç cos - cos 0 ö÷
è 4 ø è 4 ø
æ 1 ö æ 1 ö
= ç - 0÷ + ç - 1÷
è 2 ø è 2 ø
1 1
= + -1
2 2
2 - 2 +2
= - 1+ =
2 2
-2 + 2 2
= = ( 2 - 1) sq units
2
Q. 25 The area of the region bounded by the curve x 2 = 4 y and the straight
line x = 4 y - 2 is
3 5
(a) sq unit (b) sq unit
8 8
7 9
(c) sq unit (d) sq units
8 8
Sol. (d) Given equation of curve is x 2 = 4 y and the straight line x = 4 y - 2.
Y x 2 = 4y

2
4 y–
x=
(–1, 1/4)
(2, 1)

O X

Y¢
For intersection point, put x = 4 y - 2 in equation of curve, we get
(4 y - 2 )2 = 4 y
Þ 16 y2 + 4 - 16 y = 4 y
Þ 16 y2 - 20 y + 4 = 0
Þ 4 y2 - 5 y + 1 = 0
Þ 4 y2 - 4 y - y + 1 = 0
Þ 4 y ( y - 1) - 1( y - 1) = 0
Þ (4 y - 1)( y - 1) = 0
1
\ y = 1,
4
For y = 1, x = 4 × 1 = 2 [since, negative value does not satisfy the equation of line]
1 1
For y = , x = 4 × = - 1 [positive value does not satisfy the equation of line]
4 4
So, the intersection points are (2, 1) and æç - 1, ö÷.
1
è 4ø
2
2 æx +2ö 2 x
\ Area of shaded region = ò ç ÷ dx - ò dx
-1 è 4 ø -1 4
2 2
1 éx2 ù 1 x3
= ê + 2x ú -
4ë 2 û -1 4 3 -1

= - é + 4- - +2 ù- é + ù
1 4 1 1 8 1
4 êë 2 2 úû 4 êë 3 3 úû
1 15 1 9 45 - 18
= × - × =
4 2 4 3 24
27 9
= = sq units
24 8
Q. 26 The area of the region bounded by the curve y = 16 - x 2 and X-axis is
(a) 8p sq units (b) 20 p sq units
(c) 16p sq units (d) 256p sq units
Sol. (a) Given equation of curve is y = 16 - x 2 and the equation of line is X-axis i.e., y = 0.
Y

X¢ X
(–4, 0) O (4, 0)

Y¢
\ 16 - x 2 = 0 …(i)
2
Þ 16 - x = 0
Þ x 2 = 16
Þ x=± 4
So, the intersection points are (4, 0) and (-4, 0).
4
ò-4 (16 - x
2 1/ 2
\ Area of curve, A = ) dx
4
= ò-4 (42 - x 2 ) dx
4
éx 2 42 -1 x ù
=ê 4 - x2 + sin ú
ë2 2 4 û -4
é 4 2 4 ù
= é 42 - 42 + 8sin-1 ù - ê - 4 - (- 4 )2 + 8sin-1 æç - ö÷ ú
4 4
ëê 2 4 ûú ë 2 è 4 øû
p p
= é2 × 0 + 8 × - 0 + 8 × ù = 8p sq units
ëê 2 2 ûú

Q. 27 Area of the region in the first quadrant enclosed by the X-axis, the
line y = x and the circle x 2 + y 2 = 32 is
(a) 16p sq units (b) 4p sq units (c) 32p sq units (d) 24 p sq units
Sol. (b) We have, area enclosed by X-axis i.e., y = 0, y = x and the circle x 2 + y2 = 32 in first
quadrant.
Since, x 2 + (x )2 = 32 [Q y = x ]
Þ 2 x 2 = 32
Þ x=±4
So, the intersection point of circle x 2 + y2 = 32 and line y = x are (4, 4) or (- 4, 4).
and x 2 + y2 = (4 2 )2
Since, y=0
\ x 2 + (0)2 = 32
Þ x=± 4 2
 So, the circle intersects the X-axis at (±4 2 , 0).
Y

y=x

(4, 4)

(0, 0)
X¢ X
(4, 0)

32
2 +
y2 =
(–4, –4) x

Y¢
4 4 2
Area of shaded region = ò xdx + ò (4 2 )2 - x 2 dx
0 4

2 4 4 2
x x (4 2 )2 -1 x
= + (4 2 )2 - x 2 + sin
2 0
2 2 4 2 4
16 é4 2 (4 2 ) 4 4 ù
= + ê × 0 + 16 sin-1 - (4 2 ) - 16 - 16 sin-1
2
2 ë 2 (4 2 ) 2 4 2 úû
p p
= 8 + é 16 × - 2 × 16 - 16 × ù
êë 2 4 úû
= 8 + [8p - 8 - 4p] = 4p sq units

Q. 28 Area of the region bounded by the curve y = cos x between x = 0 and

x = p is
(a) 2 sq units (b) 4 sq units
(c) 3 sq units (d) 1 sq unit
Sol. (a) Required area enclosed by the curve y = cos x , x = 0 and x = p is
Y

O p/2 p
X

y=cos x
p/ 2 p
A= ò0 cos x dx + ½ò cos x dx½
½ p/ 2 ½
é p ù ½ p ½
= sin - sin 0 + sin - sin p
êë 2 úû ½ 2 ½
= 1 + 1 = 2 sq units
Q. 29 The area of the region bounded by parabola y 2 = x and the straight
line 2y = x is
4 2 1
(a) sq units (b) 1 sq unit (c) sq unit (d) sq unit
3 3 3
Sol. (a) We have to find the area enclosed by parabola y2 = x and the straight line 2 y = x .
Y
(4, 2)
2y = x

(0, 0)
X
(4, 0)

y2 = x

2
\ æx ö = x
ç ÷
è2 ø
Þ x 2 = 4x Þ x(x - 4) = 0
Þ x = 4Þ y = 2 and x = 0 Þ y = 0
So, the intersection points are (0, 0) and (4, 2).
Area enclosed by shaded region,
A = ò é x - ù dx
4 x
0 êë 2 úû
4
é 1+1 ù 4
êx 2 1 x2 ú é x 3/ 2 x 2 ù
=ê - × ú = ê2 × - ú
1 2 2 3 4 û0
ê +1 ú ë
ë2 û0
2 3 / 2 16 2 1
= 4 - - ×0 + ×0
3 4 3 4
16 16 64 - 48 16 4
= - = = = sq units
3 4 12 12 3

Q. 30 The area of the region bounded by the curve y = sin x between the
p
ordinates x = 0, x = and the X-axis is
2
(a) 2 sq units (b) 4 sq units (c) 3 sq units (d) 1 sq unit
p
Sol. (d) Area of the region bounded by the curve y = sin x between the ordinates x = 0, x =
and the X-axis is 2
Y

X
O p/2 p

y = sin x
 p/ 2
A= ò0 sin x dx
p
= - [cos x ] p0/ 2 = - écos - cos 0ù
êë 2 úû
= - [0 - 1] = 1 sq unit

2
y2
Q. 31 The area of the region bounded by the ellipse x + = 1 is
25 16
(a) 20 p sq units (b) 20 p 2 sq units (c) 16p 2 sq units (d) 25p sq units
x2 y2
Sol. (a) We have, + =1
52 42
Here, a = ± 5 and b = ± 4
y2 x2
and 2
= 1- 2
4 5
æ x2 ö
Þ y = 16 çç 1 -
2 ÷
è 25 ÷ø
Y
(0, 4)

(–5, 0) (5, 0)
X

(0, –4) x2 y2
+ =1
25 16

16
Þ y= (25 - x 2 )
25
4
Þ (52 - x 2 )
y=
5
4 5
\ Area enclosed by ellipse, A = 2 × ò 52 - x 2 dx
5 -5
8 5
= 2 × ò 52 - x 2 dx
5 0
5
8 éx 2 52 -1 x ù
=2× ê 5 - x2 + sin ú
5 ë2 2 5 û0
8 é5 2 2 52 -1 5 25 ù
=2× ê 5 -5 + sin -0- × 0ú
5 ë2 2 5 2 û
8 é 25 p ù
=2× ×
5 êë 2 2 úû
16 25p
= ×
5 4
= 20p sq units
Q. 32 The area of the region bounded by the circle x 2 + y 2 = 1 is
(a) 2p sq units (b) p sq units
(c) 3p sq units (d) 4p sq units
Sol. (b) We have, x 2 + y2 = 12 [Q r = ± 1]
2 2 2
Þ y = 1- x Þ y = 1- x
Y
(0, 1)

X
(–1, 0) (0, 0) (1, 0)

(0, –1)

1 1
\ Area enclosed by circle = 2 ò 12 - x 2 dx = 2 × 2 ò 12 - x 2 dx
-1 0
1
éx 12 xù
= 2 × 2 ê 12 - x 2 + sin-1 ú
ë 2 2 1 û0
1 p
= 4 é × 0 + × - 0 - × 0ù
1 1
ëê 2 2 2 2 ûú
p
= 4 × = p sq units
4

Q. 33 The area of the region bounded by the curve y = x + 1 and the lines
x = 2, x = 3, is
7 9 11 13
(a) sq units (b) sq units (c) sq units (d) sq units
2 2 2 2
3
3 éx2 ù
Sol. (a) Required area, A = ò2 (x + 1)dx = ê 2 + x ú
ë û2

= é 9 4
+ 3- -2 = ù é 5
+ 1ù = sq units
7
êë 2 2 úû êë 2 úû 2

y=x+1 x=2 x=3

Q. 34 The area of the region bounded by the curve x = 2 y + 3 and the lines
y = 1, y = - 1 is
3
(a) 4 sq units (b) sq units
2
(c) 6 sq units (d) 8 sq units
1
Sol. (c) Required area, A = ò-1(2 y + 3)dy
Y
y=1

y=–1
3
+
2y
x=

1
é 2 y2 ù
=ê + 3 yú
ë 2 û -1
= [ y2 + 3 y]1-1
= [1 + 3 - 1 + 3] = 6 sq units
 9
Differential Equations
Short Answer Type Questions
dy
Q. 1 Find the solution of = 2 y -x .
dx
dy
Sol. Given that, = 2 y -x
dx
dy 2 y é m - n am ù
Þ = êQ a = nú
dx 2 x ë a û
dy dx
Þ =
2y 2x
On integrationg both sides, we get
-y
ò2 dy = ò 2 - x dx
- 2- y - 2- x
Þ = +C
log 2 log 2
Þ - 2 - y + 2 - x = + C log 2
Þ 2 - x - 2 - y = - C log 2
Þ 2 -x - 2 -y = K [where, K = + C log2 ]

Q. 2 Find the differential equation of all non-vertical lines in a plane.

p
Sol. Since, the family of all non-vertical line is y = mx + c, where m ¹ tan .
2
On differentiating w.r.t. x, we get
dy
=m
dx
Again, differentiating w.r.t. x, we get
d2y
=0
dx 2

Q. 3 If dy = e -2 y and y = 0 when x = 5, then find the value of x when y = 3.

dx
dy dy
Sol. Given that, = e - 2y Þ = dx
dx e - 2y
2y
e
òe dy = ò dx Þ
2y
Þ =x+C ... (i)
2
 When x = 5 and y = 0, then substituting these values in Eq. (i), we get
e0
=5+C
2
1 1 9
Þ =5+C Þ C = -5=-
2 2 2
Eq. (i) becomes e 2y = 2x - 9
When y = 3, then e6 = 2x - 9 Þ 2x = e6 + 9
6
(e + 9)
\ x=
2

dy 1
Q. 4 Solve (x 2 - 1) + 2xy = 2 .
dx x -1
Sol. Given differential equation is
dy 1
(x 2 - 1) + 2 xy = 2
dx x -1
dy æ 2 x ö 1
Þ +ç ÷ y= 2
dx çè x 2 - 1 ÷ø (x - 1)2
which is a linear differential equation.
dy
On comparing it with + Py = Q, we get
dx
2x 1
P= 2 ,Q = 2
x -1 (x - 1)2
dx
æ 2x ö
ò çç x 2 - 1 ÷÷
IF = e ò
Pdx è ø
=e
2
Put x - 1 = t Þ 2xdx = dt
dt
ò
\ IF = e t = elog t = t = (x 2 - 1)
The complete solution is
y × IF =ò Q × IF + K
1
Þ y × (x 2
- 1) = ò 2 . (x 2 - 1) dx + K
(x - 1)2
dx
Þ y × (x 2 - 1) = ò (x 2 - 1) + K
1 æ x - 1ö
Þ y × (x 2 - 1) = log çç ÷÷ + K
2 è x + 1ø

Q. 5 Solve dy + 2xy = y.
dx
dy
Sol. Given that, + 2 xy = y
dx
dy
Þ + 2 xy - y = 0
dx
dy
Þ + (2 x - 1) y = 0
dx
which is a linear differential equation.
 dy
On comparing it with + Py = Q, we get
dx
P = (2 x - 1), Q = 0
IF = e ò = eò
Pdx ( 2 x - 1) dx

æ 2 x2 ö
ç - x÷
ç 2 ÷ 2
-x
= e è ø = ex
The complete solution is
2 2
-x -x
y × ex = ò Q × ex dx + C
2
-x
Þ y . ex =0+C
2
Þ y = C ex - x

Q. 6 Find the general solution of dy + ay = e mx .

dx
Sol. Given differential equation is
dy
+ ay = e mx
dx
which is a linear differential equation.
dy
On comparing it with + Py = Q, we get
dx
P = a, Q = e mx

IF = e ò = eò
Pdx adx
= e ax
The general solution is y × e ax = ò e mx × e ax dx + C
( m + a) x
Þ y × e ax = ò e dx + C
( m + a) x
e
Þ y × e ax = +C
( m + a)
e( m + a ) x (m + a) C
Þ ( m + a) y = +
e ax e ax
mx - ax
Þ ( m + a) y = e + K e [Q K = (m + a) C ]

Q. 7 Solve the differential equation dy + 1 = ex + y .

dx
dy
Sol. Given differential equation is + 1 = ex + y
... (i)
dx
On substituting x + y = t , we get
dy dt
1+
=
dx dx
dt
Eq. (i) becomes = et
dx
Þ e - t dt = dx
Þ - e- t = x + C
-1
Þ =x+C
ex + y
Þ - 1 = (x + C ) e x + y

Þ (x + C ) e x + y + 1 = 0
Q. 8 Solve ydx - xdy = x 2 ydx.
Sol. Given that, ydx - xdy = x 2 ydx
1 1 dy
Þ - . =1 [dividing throughout by x 2 ydx ]
x 2 xy dx
1 dy 1
Þ - . + 2 - 1= 0
xy dx x
dy xy
Þ - + xy = 0
dx x 2
dy y
Þ - + xy = 0
dx x
dy æ 1
Þ + ç x – ö÷ y = 0
dx è xø
which is a linear differential equation.
dy
On comparing it with + Py = Q, we get
dx
1
P = æç x – ö÷, Q = 0
è xø
IF = e ò
Pdx

æ 1ö
ò çè x - x ÷ø dx
=e
x2
- log x
= e2
x2
=e x , e - log x

x2
1
= e 2
x
The general solution is
1 x2/ 2 1 2
y× e = ò 0 × e x / 2 dx + C
x x
1 2
Þ y × ex / 2 = C
x
2
Þ y = C x e- x /2

dy
Q. 9 Solve the differential equation = 1 + x + y 2 + xy 2 , when
dx
y = 0 and x = 0.
dy
Sol. Given that, = 1 + x + y2 + xy2
dx
dy
Þ = (1 + x ) + y2 (1 + x )
dx
dy
Þ = (1 + y2 ) (1 + x )
dx
dy
Þ = (1 + x ) dx
1 + y2
On integrating both sides, we get
x2
tan-1 y = x + + K ... (i)
2
 When y = 0 and x = 0, then substituting these values in Eq. (i), we get
tan-1 (0) = 0 + 0 + K
Þ K=0
-1 x2
Þ tan y=x +
2
æ x2 ö
Þ y = tan çç x + ÷
è 2 ÷ø

Q. 10 Find the general solution of (x + 2 y 3 ) dy = y.

dx
dy
Sol. Given that, ( x + 2 y3 ) = y
dx
dx
Þ y× = x + 2 y3
dy
dx x
Þ = + 2 y2 [dividing throughout by y]
dy y
dx x
Þ - = 2 y2
dy y
which is a linear differential equation.
dx
On comparing it with + Px = Q, we get
dy
1
P = - , Q = 2 y2
y
1 1
ò - y dy -ò dy
IF = e =e y
1
\ = e - log y =
y
1 2 1
The general solution is x× = ò 2 y × dy + C
y y
x 2 y2
Þ = +C
y 2
x
Þ = y2 + C
y
Þ x = y3 + Cy

æ 2 + sin x ö dy
Q. 11 If y (x) is a solution of çç ÷÷ = - cos x and y (0) = 1, then find
è 1 + y ø dx
æpö
the value of y ç ÷.
è2ø
æ 2 + sin x ö dy
Sol. Given that, çç ÷÷ = - cos x
è 1 + y ø dx
dy cos x
Þ =- dx
1+ y 2 + sin x
On integrating both sides, we get
1 cos x
ò 1 + y dy = - ò 2 + sin x dx
Þ log (1 + y) = - log (2 + sin x ) + log C
 Þ log (1 + y) + log (2 + sin x ) = log C
Þ log (1 + y) (2 + sin x ) = log C
Þ (1 + y) (2 + sin x ) = C
C
Þ 1+ y =
2 + sin x
C
Þ y= -1 ... (i)
2 + sin x
When x = 0 and y = 1, then
C
1= - 1
2
Þ C=4
On putting C = 4 in Eq. (i), we get
4
y= -1
2 + sin x
p 4 4
\ y æç ö÷ = -1 = -1
è 2 ø 2 + sin p 2+1
2
4 1
= - 1=
3 3

dy
Q. 12 If y (t) is a solution of (1 + t) - ty = 1 and y (0) = - 1, then show
dt
1
that y (1) = - .
2
dy
Sol. Given that, (1 + t ) - ty = 1
dt
dy æ t ö 1
Þ -ç ÷ y=
dt çè 1 + t ÷ø 1+ t
which is a linear differential equation.
dy
On comparing it with + Py = Q, we get
dt
æ t ö 1
P = - çç ÷, Q =
è1 + t ÷ø 1+ t
t æ 1 ö
-ò dt - ò çç 1 - ÷ dt = e - [ t - log (1 + t )]
1+ t 1+ t ÷
IF = e = e è ø

= e - t × elog ( 1 + t )
= e - t (1 + t )
The general solution is
(1 + t ) (1 + t ) × e - t
y (t ) ×
et
= ò (1 + t )
dt + C

e- t et C et
Þ y (t ) = × + C ¢, where C ¢=
(-1) 1 + t 1+ t
1
Þ y (t ) = - + C¢
1+ t
When t = 0 and y = - 1, then
- 1 = - 1 + C¢Þ C¢ = 0
1 1
y (t ) = - Þ y (1) = -
1+ t 2
Q. 13 Form the differential equation having y = (sin - 1 x)2 + A cos - 1 x + B,
where A and B are arbitrary constants, as its general solution.
Sol. Given that, y = (sin- 1 x )2 + A cos - 1 x + B
On differentiating w.r.t. x, we get
dy 2 sin-1 x (- A)
= +
dx 1- x 2
1 - x2
dy
Þ 1 - x2 = 2 sin- 1 x - A
dx
Again, differentiating w.r.t. x, we get
d 2 y dy - 2x 2
1 - x2 + . =
dx 2 dx 2 1 + x 2 1 - x2
d2y x dy
Þ (1 - x 2 ) - . 1 - x2 =2
dx 2 1 - x2 dx
dy d2y
Þ (1 - x 2 )
-x =2
dx 2 dx
d2y dy
Þ (1 - x 2 ) 2 - x -2 = 0
dx dx
which is the required differential equation.

Q. 14 Form the differential equation of all circles which pass through origin
and whose centres lie on Y-axis.
Sol. It is given that, circles pass through origin and their centreslie on Y-axis. Let (0, k) be the
centre of the circle and radius is k.
So, the equation of circle is
(x - 0)2 + ( y - k )2 = k 2
Þ x 2 + ( y - k )2 = k 2
Þ x 2 + y2 - 2 ky = 0
x 2 + y2
Þ =k ... (i)
2y
On differentiating Eq. (i) w.r.t. x, we get
2 2dy
2 y æç 2 x + 2 y
dy ö 2
÷ - (x + y )
è dx ø dx
=0
4 y2
4 y æç x + y
dy ö 2 2 dy
Þ ÷ - 2 (x + y ) =0
è dx ø dx
dy dy
Þ 4xy + 4 y2 - 2 ( x 2 + y2 ) =0
dx dx
dy
Þ [4 y2 - 2 (x 2 + y2 )] + 4xy = 0
dx
dy
Þ ( 4 y2 - 2 x 2 - 2 y2 ) + 4xy = 0
dx
dy
Þ (2 y2 - 2 x 2 ) + 4xy = 0
dx
dy
Þ ( y2 - x 2 ) + 2 xy = 0
dx
dy
Þ ( x 2 - y2 ) - 2 xy = 0
dx
Q. 15 Find the equation of a curve passing through origin and satisfying the
dy
differential equation (1 + x 2 ) + 2xy = 4 x 2 .
dx
dy
Sol. Given that, (1 + x 2 ) + 2 xy = 4x 2
dx
dy 2x 4x 2
Þ + . y =
dx 1 + x 2 1 + x2
which is a linear differential equation.
dy
On comparing it with + Py = Q, we get
dx
2x 4x 2
P= 2
,Q =
1+ x 1 + x2
2x
ò dx
1 + x2
\ IF = e ò Pdx = e
Put 1 + x 2 = t Þ 2xdx = dt
dt
ò x2)
IF = 1 + x 2 = e t = elog t = elog ( 1 +
The general solution is
4x 2
y × (1 + x 2 ) = ò 1 + x 2 (1 + x
2
) dx + C

Þ y × (1 + x 2 ) = ò 4x 2 dx + C
x3
Þ y × (1 + x 2 ) = 4+C ... (i)
3
Since, the curve passes through origin, then substituting
x = 0 and y = 0 in Eq. (i), we get
C=0
The required equation of curve is
4x 3
y (1 + x 2 ) =
3
4x 3
Þ y=
3 (1 + x 2 )

dy
Q. 16 Solve x2 = x 2 + xy + y 2 .
dx
dy
Sol. Given that, x2 = x 2 + xy + y2
dx
dy y y2
Þ = 1+ + 2 ... (i)
dx x x
y y2
Let f (x, y) = 1 + + 2
x x
ly l2 y2
f (lx, ly) = 1 + + 2 2
lx l x
æ y y2 ö
f (lx, ly) = l0 çç 1 + + 2 ÷÷
è x x ø
0
= l f (x, y)
 which is homogeneous expression of degree 0.
dy dv
Put y = vx Þ =v+ x
dx dx
On substituting these values in Eq.(i), we get
æ v + x dv ö = 1 + v + v 2
ç ÷
è dx ø
dv
Þ x = 1 + v + v2 - v
dx
dv
Þ x = 1 + v2
dx
dv dx
Þ 2
=
1+ v x
On integrating both sides, we get
tan-1 v = log|x| + C

tan-1 æç ö÷ = log|x| + C
y
Þ
èx ø

Q. 17 Find the general solution of the differential equation

2 tan -1 y dy
(1 + y ) + (x - e ) = 0.
dx
Sol. Given, differential equation is
-1 dy
(1 + y2 ) + (x - e tan y
) =0
dx
-1 dy
Þ (1 + y2 ) = - (x - e tan y
)
dx
dx -1
(1 + y2 ) = - x + e tan y
dy
dx -1
Þ (1 + y2 ) + x = e tan y
dy
-1
dx x e tan y
Þ + 2
= [dividing throughout by (1 + y2 ) ]
dy 1 + y 1 + y2
which is a linear differential equation.
dx
On comparing it with + Px = Q, we get
dy
-1
1 e tan y
P= 2
,Q = 2
1+ y 1+ y
1
ò1+ dy
-1
IF = e ò y2
Pdy
=e = e tan y

-1
-1 e tan y -1
x × e tan ò1+ × e tan
y y
The general solution is = 2
dy + C
y
-1
-1 (e tan y 2
)
x × e tan ò
y
Þ = × dy + C
1 + y2
1
Put tan-1 y = t Þ dy = dt
1 + y2
-1
\ x × e tan y
= ò e 2t dt + C
 -1 1 2 tan -1 y
Þ x × e tan y
= e +C
2
-1 -1
Þ 2 x e tan y
= e 2 tan y
+2C
-1 -1
tan y 2 tan y
Þ 2x e =e + K [Q K = 2 C]

Q. 18 Find the general solution of y 2dx + (x 2 - x y + y 2 ) dy = 0.

Sol. Given, differential equation is
y2dx + (x 2 - xy + y2 ) dy = 0
Þ y2dx = - (x 2 - xy + y2 ) dy
dx
Þ y2 = - (x 2 - xy + y2 )
dy
dx æ x2 x ö
Þ = - çç 2 - + 1÷÷ ... (i)
dy è y y ø
which is a homogeneous differential equation.
x
Put = v or x = vy
y
dx dv
Þ =v+ y
dy dy
On substituting these values in Eq. (i), we get
dv
v+ y = - [v 2 - v + 1]
dy
dv
Þ y = - v2 + v - 1 - v
dy
dv dv dy
Þ y = - v2 - 1 Þ 2 =-
dy v +1 y
On integrating both sides, we get
tan-1 (v ) = - log y + C
æx ö é xù
Þ tan-1 ç ÷ + log y = C êQ v = y ú
è yø ë û

Q. 19 Solve (x + y) (dx - dy) = dx + dy.

Sol. Given differential equation is
(x + y) (dx - dy) = dx + dy
(x + y) æç 1 -
dy ö dy
Þ ÷ = 1+ …(i)
è dx ø dx
Put x + y= z
dy dz
Þ 1+ =
dx dx
On substituting these values in Eq. (i), we get
z æç 1 - + 1ö÷ =
dz dz
è dx ø dx
z æç 2 -
dz ö dz
Þ ÷=
è dx ø dx
dz dz
Þ 2z - z - =0
dx dx
 dz
Þ 2 z - (z + 1) =0
dx
dz 2z
Þ =
dx z + 1

Þ æz+ 1ö
ç ÷ dz = 2 dx
è z ø
On integrating both sides, we get
æ 1 + 1 ö dz = 2 dx
òç
è zø
÷ ò
Þ z + log z = 2 x - log C
Þ (x + y) + log (x + y) = 2 x - log C [Q z = x + y]
Þ 2x - x - y = log C + log (x + y)
Þ x - y = log|C (x + y)|
Þ e x - y = C (x + y)
1
Þ (x + y) = e x - y
C
Þ x + y = Ke x - y éQ K = 1 ù
êë C úû

dy
Q. 20 Solve 2 ( y + 3) - xy = 0, given that y (1) = - 2.
dx
dy
Sol. Given that, 2 ( y + 3) - xy =0
dx dy
Þ 2 ( y + 3) = xy
dx
dx æ y ö
Þ 2 = çç ÷÷ dy
x è y + 3ø
dx æ y + 3 - 3 ö
Þ 2× = çç ÷÷ dy
x è y+ 3 ø
dx æ 3 ö
Þ 2× = çç 1 - ÷ dy
x è y + 3 ÷ø
On integrating both sides, we get
2 log x = y - 3 log ( y + 3) + C ... (i)
When x = 1 and y = - 2, then
2 log 1 = - 2 - 3log (- 2 + 3) + C
Þ 2 × 0 = - 2 - 3× 0 + C
Þ C =2
On substituting the value of C in Eq. (i), we get
2 log x = y - 3 log ( y + 3) + 2
Þ 2 log x + 3 log ( y + 3) = y + 2
Þ log x 2 + log ( y + 3)3 = ( y + 2 )
Þ log x 2 ( y + 3)3 = y + 2
+ 2
Þ x 2 ( y + 3)3 = e y
Q. 21 Solve the differential equation dy = cos x (2 - y cosec x ) dx given that
p
y = 2, when x = .
2
Sol. Given differential equation,
dy = cos x (2 - y cosec x )dx
dy
Þ = cos x (2 - y cosec x )
dx
dy
Þ = 2 cos x - y cosec x × cos x
dx
dy
Þ = 2cos x - ycot x
dx
dy
Þ + ycot x = 2 cos x
dx
which is a linear differential equation.
dy
On comparing it with + Py = Q, we get
dx
P = cot x, Q = 2 cos x

IF = e ò = eò
Pdx cot x dx
= elog sin x = sin x
The general solution is
y × sin x = ò 2 cos x × sin x dx + C
Þ y × sin x = ò sin 2 x dx + C [Q sin 2 x = 2 sin x cos x ]
cos 2 x
Þ y × sin x = - +C ... (i)
2
p
When x = and y = 2, then
2
p
cos æç 2 ´ ö÷
p è 2ø
2 . sin = - +C
2 2
1
Þ 2 ×1= + + C
2
1 4-1
Þ 2 - =C Þ =C
2 2
3
Þ \ C=
2
On substituting the value of C in Eq. (i), we get
1 3
y sin x = - cos 2 x +
2 2

Q. 22 Form the differential equation by eliminating A and B in

Ax 2 + By 2 = 1.
Sol. Given differential equation is Ax 2 + By2 = 1
On differentiating both sides w.r.t. x, we get
dy
2 Ax + 2 By =0
dx
dy
Þ 2 By = - 2 Ax
dx
dy y dy A
Þ By = - Ax Þ . =-
dx x dx B
 Again, differentiating w.r.t. x, we get
æ x dy _ y ö
y d 2 y dy ç dx ÷
× + ×ç ÷=0
x dx 2 dx ç x 2 ÷
è ø
2
æ dy ö
x ç ÷ - yç ÷ æ dy ö
y d2y
× 2 + è ø 2 è ø =0
dx dx
Þ
x dx x
d2y dy 2
xy 2 + x æç ö÷ - y æç ö÷ = 0
dy
Þ
dx è dx ø è dx ø
Þ xy y¢¢ + x ( y¢)2 - y y¢ = 0

Q. 23 Solve the differential equation (1 + y 2 ) tan -1 x dx + 2 y (1 + x 2 ) dy = 0.

Sol. Given differential equation is
(1 + y2 ) tan-1 xdx + 2 y (1 + x 2 ) dy = 0
Þ (1 + y2 ) tan-1 xdx = - 2 y (1 + x 2 ) dy
tan-1 x dx 2 y
Þ =- dy
1 + x2 1 + y2
On integrating both sides, we get
tan-1 x 2y
ò 1 + x 2 dx = - ò 1 + y2
dy

Put tan-1 x = t in LHS, we get

1
dx = dt
1 + x2
and put 1 + y2 = u in RHS, we get
2 y dy = du
1 t2
Þ ò t dt = - ò u du Þ 2
= - log u + C
1
Þ (tan-1 x )2 = - log (1 + y2 ) + C
2
1
Þ (tan-1 x )2 + log (1 + y2 ) = C
2

Q. 24 Find the differential equation of system of concentric circles with

centre (1, 2).
Sol. The family of concentric circles with centre (1, 2) and radius a is given by
(x - 1)2 + ( y - 2 )2 = a2
Þ x 2 + 1 - 2 x + y2 + 4 - 4 y = a 2
Þ x 2 + y2 - 2 x - 4 y + 5 = a 2 ...(i)
On differentiating Eq. (i) w.r.t. x, we get
dy dy
2x + 2 y -2 - 4 =0
dx dx
dy
Þ (2 y - 4) + 2x - 2 = 0
dx
dy
Þ ( y - 2) + (x - 1) = 0
dx
Long Answer Type Questions
d
Q. 25 Solve y + (xy) = x (sin x + log x).
dx
Sol. Given differential equation is
d
y+ (xy) = x (sin x + log x )
dx
dy
Þ y+ x + y = x (sin x + log x )
dx
dy
Þ x + 2 y = x (sin x + log x )
dx
dy 2
Þ + y = sin x + log x
dx x
which is a linear differential equation.
dy
On comparing it with + Py = Q, we get
dx
2
P = , Q = sin x + log x
x
2
ò x dx
IF =e = e 2 log x
= x2
The general solution is
y × x 2 = ò (sin x + log x ) x 2dx + C
Þ y × x 2 = ò (x 2 sin x + x 2 log x ) dx + C
Þ y × x 2 = ò x 2 sin x dx + òx
2
log x dx + C
2
Þ y × x = I1 + I 2 + C ...(i)
Now, I1 = ò x sin x dx
2

= x 2 (- cos x ) + ò 2 x cos x dx
= - x 2 cos x + [2 x (sin x ) - ò 2 sin x dx ]
I1 = - x 2 cos x + 2 x sin x + 2 cos x ... (ii)
and I 2 = ò x log x dx
2

x3 1 x3
= log x × -ò × dx
3 x 3
x3 1
= log x × - ò x 2dx
3 3
x3 1 x3
= log x × - × ... (iii)
3 3 3
On substituting the value of I1 and I 2 in Eq. (i), we get
x3 1
y × x 2 = - x 2 cos x + 2 x sin x + 2 cos x + log x - x 3 + C
3 9
2 sin x 2 cos x x x
\ y = - cos x + + + log x - + Cx - 2
x x2 3 9
Q. 26 Find the general solution of (1 + tan y) (dx - dy) + 2x dy = 0.
Sol. Given differential equation is (1 + tan y) (dx - dy) + 2 x dy = 0
on dividing throughout by dy, we get
æ dx ö
(1 + tan y) ç - 1÷ + 2 x = 0
è dy ø
dx
Þ (1 + tan y) - (1 + tan y) + 2 x = 0
dy
dx
Þ (1 + tan y) + 2 x = (1 + tan y)
dy
dx 2x
Þ + =1
dy 1 + tan y
which is a linear differential equation.
dx
On comparing it with + Px = Q, we get
dy
2
P= ,Q = 1
1 + tan y
2 2 cos y
ò1+ tan y
dy ò cos y + sin y
dy
IF = e =e
cos y + sin y + cos y - sin y
ò cos y + sin y
dy
=e
æ cos y - sin y ö
ò çç 1 + cos y ÷ dy
+ sin y ÷ø + log (cos y + sin y )
=e è = ey
y
= e × (cos y + sin y) [Q elog x
= x]
The general solution is
x × e y (cos y + sin y) = ò 1× e y (cos y + sin y) dy + C
Þ x × e y (cos y + sin y) = ò e y (sin y + cos y) dy + C
Þ x × e y (cos y + sin y) = e y sin y + C [Q ò e x {f (x ) + f ¢ (x )} dx = e x f (x )]
Þ x (sin y + cos y) = sin y + Ce - y

dy
Q. 27 Solve = cos (x + y) + sin (x + y) .
dx
dy
Sol. Given, = cos (x + y) + sin (x + y) …(i)
dx
Put x + y= z
dy dz
Þ 1+ =
dx dx
On substituting these values in Eq. (i), we get
æ dz - 1ö = cos z + sin z
ç ÷
è dx ø
dz
Þ = (cos z + sin z + 1)
dx
dz
Þ = dx
cos z + sin z + 1
 On integrating both sides, we get
dz
ò cos z + sin z + 1 = ò 1d x
dz
Þ ò 1 - tan2 z /2 2 tan z /2
= ò dx
+ +1
1 + tan2 z /2 1 + tan2 z /2
dz
Þ ò 1 - tan2 z / 2 + 2 tan z / 2 + 1 + tan2 z/2
= ò dx

(1 + tan2 z / 2 )
(1 + tan2 z / 2 ) dz
Þ ò 2 + 2 tan2 z / 2
= ò dx

sec 2 z / 2 dz
Þ ò 2 (1 + tan z / 2 ) = ò dx
Put 1 + tan z / 2 = t Þ æ 1 sec 2 z / 2 ö d z = d t
ç ÷
è2 dt ø
Þ ò t = ò dx
Þ log|t| = x + C
Þ log|1 + tan z / 2| = x + C
(x + y)
Þ log 1 + tan =x+C
2

Q. 28 Find the general solution of dy - 3 y = sin 2x.

dx
dy
Sol. Given, - 3 y = sin 2 x
dx
which is a linear differential equation.
dy
On comparing it with + Py = Q, we get
dx
P = - 3, Q = sin2 x
IF = e -3 ò dx = e - 3 x
The general solution is
y . e - 3 x = ò sin2 x e - 3 x dx
I II

Let y × e - 3x = I … (i)
- 3x
\ I = òe sin2 x
II I
æ e - 3x ö æ e - 3x ö
Þ I = sin 2 x çç . ÷ - ò 2 cos 2 x ç ÷
÷ ç - 3 ÷ dx + C1
è -3 ø è ø
1 - 3x 2 - 3x
Þ I=- e sin 2 x + ò e cos 2 x dx + C1
3 3 II I

1 - 3x 2 æ e - 3x e - 3x ö
Þ I=- e sin 2 x + çç cos 2 x - ò (- 2 sin2 x ) dx ÷÷ + C1 + C 2
3 3è -3 -3 ø
1 - 3x 2 - 3x 4
Þ I=- e sin 2 x - cos 2 x e - I + C¢ [where, C ¢ = C1 + C 2 ]
3 9 9
4I 1 2
Þ I+ 2 = + e - 3 x æç - sin2 x - cos 2 x ö÷ + C ¢
9 è 3 9 ø
 13 1 2
Þ I = e - 3 x æç - sin2 x - cos 2 x ö÷ + C ¢
9 è 3 9 ø
Þ I=
9 -3x æ 1 2 ö é where C = 9C ¢ù
e ç - sin2 x - cos 2 x ÷ + C êë
13 è 3 9 ø 13 úû
3 - 3x æ 2 ö
Þ I= e ç - sin2 x - cos 2 x ÷ + C
13 è 3 ø
3 - 3 x (- 3 sin2 x - 2 cos 2 x )
Þ = e +C
13 3
e - 3x
Þ = (- 3 sin 2 x - 2 cos 2 x ) + C
13
e - 3x
Þ I=- (2 cos 2 x + 3sin 2 x ) + C
13
On substituting the value of I in Eq. (i), we get
e - 3x
y × e - 3x = - ( 2 cos 2 x + 3 sin 2 x ) + C
13
1
Þ y=- (2 cos 2 x + 3 sin 2 x ) + C e 3 x
13

Q. 29 Find the equation of a curve passing through (2, 1), if the slope of the
x2 + y 2
tangent to the curve at any point (x, y) is .
2x y
x 2 + y2
Sol. It is given that, the slope of tangent to the curve at point (x, y) is .
2 xy
æ dy ö x 2 + y2
\ ç ÷ =
è dx ø( x, y ) 2 xy
dy 1 æ x yö
Þ = ç + ÷ … (i)
dx 2 è y x ø
which is homogeneous differential equation.
Put y = vx
dy dv
Þ =v+ x
dx dx
On substituting these values in Eq. (i), we get
dv 1 æ 1
v+ x = ç + v ö÷
dx 2 è v ø
dv 1 æ 1 + v 2 ö
Þ v+ x = ç ÷
dx 2 çè v ÷ø
dv 1 + v 2
Þ x = -v
dx 2v
dv 1 + v 2 - 2 v 2
Þ x =
dx 2v
dv 1 - v 2
Þ x =
dx 2v
2v dx
Þ dv =
1 - v2 x
 On integrating both sides, we get
2v dx
ò 1 - v 2 dv = ò x
Put 1 - v 2 = t in LHS, we get
- 2 vdv = dt
dt dx
Þ -ò =ò
t x
Þ - log t = log x + log C
Þ - log (1 - v 2 ) = log x + log C
æ y2 ö
Þ - log çç 1 - 2 ÷÷ = log x + log C
è x ø
æ x 2 - y2 ö
Þ - log çç 2
÷ = log x + log C
÷
è x ø
æ x2 ö
Þ log çç 2 ÷ = log x + log C
2 ÷
èx - y ø
x2
Þ =C x …(ii)
x 2 - y2
Since, the curve passes through the point (2, 1).
(2 )2 2
\ = C (2 ) Þ C =
(2 ) - (1)2
2
3
So, the required solution is 2 (x 2 - y2 ) = 3x .

Q. 30 Find the equation of the curve through the point (1, 0), if the slope of
y -1
the tangent to the curve at any point (x, y) is .
x2 + x
y-1
Sol. It is given that, slope of tangent to the curve at any point (x, y) is 2 .
x + x
æ ö
dy y - 1
\ ç ÷ =
è dx ø( x, y ) x 2 + x
dy y-1
Þ =
dx x 2 + x
dy dx
Þ =
y - 1 x2 + x
On integrating both sides, we get
dy dx
ò y - 1 = ò x2 + x
dy dx
Þ ò y - 1 = ò x (x + 1)
dy æ1 1 ö
Þ ò y - 1 = ò ççè x -
x+
÷ dx
1 ÷ø
Þ log( y - 1) = log x - log(x + 1) + log C
æ xC ö
Þ log( y - 1) = log çç ÷÷
è x + 1ø
 Since, the given curve passes through point (1, 0).
1× C
\ 0 - 1= Þ C = -2
1+ 1
-2 x
The particular solution is y - 1=
x+1
Þ ( y - 1)(x + 1) = - 2 x
Þ ( y - 1)(x + 1) + 2 x = 0

Q. 31 Find the equation of a curve passing through origin, if the slope of the
tangent to the curve at any point (x, y) is equal to the square of the
difference of the abcissa and ordinate of the point.
dy
Sol. Slope of tangent to the curve =
dx
and difference of abscissa and ordinate = x - y
dy
According to the question, = (x - y)2 …(i)
dx
Put x - y= z
dy dz
Þ 1- =
dx dx
dy dz
Þ = 1-
dx dx
On substituting these values in Eq. (i), we get
dz
1- = z2
dx
dz
Þ 1 - z2 =
dx
dz
Þ dx =
1 - z2
On integrating both sides, we get
dz
ò dx = ò 1 - z2
1 1+ z
Þ x= log +C
2 1- z
1 1+ x - y
Þ tx = log +C ... (ii)
2 1- x + y
Since, the curve passes through the origin.
1 1+ 0 - 0
\ 0 = log +C
2 1- 0 + 0
Þ C=0
On substituting the value of C in Eq. (ii), we get
1 1+ x - y
x = log
2 1- x + y
1+ x - y
Þ 2 x = log
1- x + y
1+ x - y
Þ e 2x =
1- x + y
Þ (1 - x + y) e 2 x = 1 + x - y
Q. 32 Find the equation of a curve passing through the point (1, 1), if the
tangent drawn at any point P (x, y) on the curve meets the coordinate
axes at A and B such that P is the mid-point of AB.
Sol. The below figure obtained by the given information
(0, 2y)
B
P(
x,
y) A (2x, 0)

Let the coordinate of the point P is (x, y ). It is given that, P is mid-point of AB.
So, the coordinates of points A and B are (2 x, 0) and (0, 2 y), respectively.
0 - 2y y
\ Slope of AB = =-
2x - 0 x
Since, the segment AB is a tangent to the curve at P.
dy y
\ =-
dx x
dy dx
Þ =-
y x
On integrating both sides, we get
log y = - log x + log C
C
log y = log …(i)
x
Since, the given curve passes through (1, 1).
C
\ log 1 = log
1
Þ 0 = log C
Þ c =1
1
\ log y = log
x
1
Þ y=
x
Þ xy = 1

Q. 33 Solve x dy = y(log y - log x + 1)

dx
dy
Sol. Given, = y (log y - log x + 1)
x
dx
= ylog æç + 1ö÷
dy y
Þ x
dx èx ø
= ç log + 1ö÷
dy yæ y
Þ …(i)
dx x è x ø
which is a homogeneous equation.
y
Put = v or y = vx
x
dy dv
\ =v+ x
dx dx
 On substituting these values in Eq.(i), we get
dv
v+ x = v(log v + 1)
dx
dv
Þ x = v(log v + 1 - 1)
dx
dv
Þ x = v(log v )
dx
dv dx
Þ =
v log v x
On integrating both sides, we get
dv dx
ò v log v = ò x
On putting log v = u in LHS integral, we get
1
× dv = du
v
du dx
ò u =ò x
Þ log u = log x + log C
Þ log u = log C x
Þ u = Cx
Þ log v = Cx
log æç ö÷ = Cx
y
Þ
èx ø

Objective Type Questions

2 2
æ d2y ö æ dy ö æ dy ö
Q. 34 The degree of the differential equation çç 2 ÷÷ + ç ÷ = x sin ç ÷
è dx ø è dx ø è dx ø
is
(a) 1 (b) 2 (c) 3 (d) not defined
Sol. (d) The degree of above differential equation is not defined because when we expand
sin æç ö÷ we get an infinite series in the increasing powers of . Therefore its degree is
dy dy
è ø
dx dx
not defined.

3/2
é æ dy ö2 ù d2y
Q. 35 The degree of the differential equation ê1 + ç ÷ ú = is
êë è dx ø úû dx 2
3
(a) 4 (b) (c) not defined (d) 2
2
2 3/ 2
é dy ù d2y
Sol. (d) Given that ê1 + æç ö÷ ú =
ë è dx ø û dx 2
On squaring both sides, we get
3 2
é æ dy ö ù
2
æd2y ö
ê1 + ç ÷ ú = çç 2 ÷÷
ë è dx ø û è dx ø
So, the degree of differential equation is 2.
Q. 36 The order and degree of the differential equation
1/ 4
d2y æ dy ö
2
+ç ÷ + x 1/ 5 = 0 respectively, are
dx è ø
d x
(a) 2 and 4 (b) 2 and 2
(c) 2 and 3 (d) 3 and 3
1/ 4
d2y
+ æç ö÷
dy
Sol. (a) Given that, 2
= - x 1/ 5
dx è dx ø
1/ 4
d2y
+ æç ö÷
dy
Þ 2
= - x 1/ 5
dx è dx ø
1/ 4
æ dy ö æ d2y ö
Þ ç ÷ = - çç x 1/ 5 + ÷
è dx ø è dx 2 ÷ø
On squaring both sides, we get
1/ 2 2
æ
æ dy ö d2y ö
= çç x 1/ 5 +
ç ÷ ÷
è dx ø
è dx 2 ÷ø
Again, on squaring both sides, we have
4
dy æ 1/ 5 d 2 y ö
= çx + ÷
dx çè dx 2 ÷ø
order = 2, degree = 4

Q. 37 If y = e -x ( A cos x + B sin x), then y is a solution of

d 2y dy d2y dy
(a) 2 +2 =0 (b) -2 + 2y = 0
dx dx dx 2 dx
d 2y dy d 2y
(c) +2 + 2y = 0 (d) + 2y = 0
dx 2 dx dx 2

Sol. (c) Given that, y = e - x (A cos x + Bsin x )

On differentiating both sides w.r.t., x we get
dy
= - e - x (A cos x + Bsin x) + e - x (- A sin x + Bcos x )
dx
dy
= - y + e - x (- A sin x + Bcos x )
dx
Again, differentiating both sides w.r.t. x, we get
d 2 y - dy
= + e - x (- cos x - Bsin x ) - e - x (- A sin x + Bcos x )
dx 2 dx
d2y
- y-é + yù
dy dy
Þ =-
dx 2 dx ëê dx ûú
d2y dy dy
Þ 2
=- - y- - y
dx dx dx
d2y dy
Þ = -2 - 2y
dx 2 dx
d2y dy
Þ +2 + 2y = 0
dx 2 dx
Q. 38 The differential equation for y = A cos a x + B sin a x, where A and B
are arbitrary constants is
d 2y d 2y
(a) - a 2y = 0 (b) + a 2y = 0
dx 2 dx 2
d 2y d 2y
(c) + ay = 0 (d) - ay = 0
dx 2 dx 2

Sol. (b) Given, y = A cos a + Bsin a

dy
Þ = - a A sin a x + a Bcos a x
dx
Again, differentiating both sides w.r.t. x, we get
d2y
= - Aa 2 cos ax - a 2 Bsin ax
dx 2
d2y
Þ = - a 2 ( A cos ax + Bsin ax )
dx 2
d2y
Þ = - a2 y
dx 2
d2y
Þ + a2 y = 0
dx 2

Q. 39 The solution of differential equation xdy - ydx = 0 represents

(a) a rectangular hyperbola
(b) parabola whose vertex is at origin
(c) straight line passing through origin
(d) a circle whose centre is at origin
Sol. (c) Given that , xdy - ydx = 0
Þ xdy = ydx
dy dx
Þ =
y x
On integrating both sides, we get
log y = log x + log C
Þ log y = log Cx
Þ y = Cx
which is a straight line passing through origin.

Q. 40 The integrating factor of differential equation cos x dy + y sin x = 1 is

dx
(a) cos x (b) tan x (c) sec x (d) sin x
dy
Sol. (c) Given that, cos x + ysin x = 1
dx
dy
Þ + y tan x = sec x
dx
Here, P = tan x and Q = sec x
IF = e ò = eò
Pdx tan x dx
= elog sec x
\ = sec x
Q. 41 The solution of differential equation tan ysec 2 xdx + tan x sec 2 ydy = 0 is
(a) tan x + tan y = k (b) tan x - tan y = k
tan x
(c) =k (d) tan x × tan y = k
tan y
Sol. (d) Given that, tan ysec 2 x dx + tan x sec 2 ydy = 0
Þ tan sec 2 x dx = - tan x sec 2 ydy
sec 2 x -sec 2 y
Þ dx = dy …(i)
tan x tan y
On integrating both sides, we have
sec 2 x sec 2 y
ò tan x dx = - ò tan y dy
Put tanx = t in LHS integral, we get
sec 2 x dx = dt Þ sec 2 x dx = dt
and tan y = u in RHS integral, we get
sec 2 ydy = du
On substituting these values in Eq. (i), we get
dt du
ò t =-ò u
log t = - log u + log k
Þ log(t × u ) = log k
Þ log(tan x tan y) = log k
Þ tan x tan y = k

Q. 42 The family y = Ax + A 3 of curves is represented by differential

equation of degree
(a) 1 (b) 2 (c) 3 (d) 4
3
Sol. (a) Given that, y = Ax + A
dy
Þ = A
dx
[we can differential above equation only once because it has only one arbitrary constant]
\ Degree = 1

Q. 43 The integrating factor of xdy - y = x 4 - 3x is

dx
1
(a) x (b) log x (c) (d) - x
dy x
Sol. (c) Given that, x - y = x 4 - 3x
dx
dy y
Þ - = x3 - 3
dx x
1
Here, P = - , Q = x3 - 3
x
1
-ò dx
IF = e ò = e - log x
Pdx
\ =e x

1
=
x
Q. 44The solution of dy - y = 1, y(0) = 1 is given by
dx
(a) xy = - ex (b) xy = - e- x
(c) xy = - 1 (d) y = 2ex - 1
Sol. (b) Given that,
dy
- y=1
dx
dy
Þ = 1+ y
dx
dy
Þ = dx
1+ y
On integrating both sides, we get
log(1 + y) = x + C ...(i)
When x = 0 and y = 1, then
log 2 = 0 + c
Þ C = log 2
The required solution is
log (1 + y) = x + log 2
1+ yö
Þ logæç ÷=x
è 2 ø
1+ y
Þ = ex
2
Þ 1 + y = 2e x
Þ y = 2e x - 1

y +1
Q. 45 The number of solutions of dy = , when y(1) = 2 is
dx x -1
(a) none (b) one (c) two (d) infinite
dy y+1
Sol. (b) Given that, =
dx x -1
dy dx
Þ =
y+1 x -1
On integrating both sides, we get
log( y + 1) = log(x - 1) - log C
C( y + 1) = (x - 1)
x -1
Þ C=
y+1
When x = 1and y = 2, then C = 0
So, the required solution is x - 1 = 0.
Hence, only one solution exist.

Q. 46 Which of the following is a second order differential equation?

(a) (y ¢ ) 2 + x = y 2 (b) y ¢ y ¢¢ + y = sin x
(c) y ¢¢¢ + (y ¢¢ ) 2 + y = 0 (d) y ¢ = y 2
Sol. (b) The second order differential equation is y ¢ y ¢¢+ y = sin x .
 dy
Q. 47 The integrating factor of differential equation (1 - x 2 ) - xy = 1 is
dx
x 1
(a) - x (b) (c) 1 - x 2 (d) log(1 - x 2)
1+ x2 2
dy
Sol. (c) Given that, (1 - x 2 ) - xy = 1
dx
dy x 1
Þ - y=
dx 1 - x 2 1 - x2
which is a linear differential equation.
x
-ò dx
1 - x2
\ IF = e
dt
Put 1 - x 2 = t Þ - 2xdx = dt Þ xdx = -
2
1 dt 1 1
òt log t log( 1 - x 2 )
Now, IF = e 2 = e2 = e2 = 1 - x2

Q. 48 tan -1 x + tan -1 y = C is general solution of the differential equation

2 2
dy 1 + y dy 1 + x
(a) = (b) =
dx 1 + x 2 dx 1 + y 2

(c) (1 + x 2)dy + (1 + y 2)dx = 0 (d) (1 + x 2)dx + (1 + y 2)dy = 0

Sol. (c) Given that, tan-1 x + tan-1 y = C
On differentiating w.r.t. x, we get
1 1 dy
+ × =0
1 + x 2 1 + y2 dx
1 dy 1
Þ × =-
1 + y2 dx 1 + x2
Þ (1 + x 2 ) d y + (1 + y2 ) dx = 0

Q. 49 The differential equation y dy + x = C represents

dx
(a) family of hyperbolas (b) family of parabolas
(c) family of ellipses (d) family of circles
Sol. (d) dy
Given that, + x =C
y
dx
dy
Þ y =C - x
dx
Þ yd y = (C - x ) d x
On integrating both sides, we get
y2 x2
= Cx - + K
2 2
x2 y2
Þ + = Cx + K
2 2
x2 y2
Þ + - Cx = K
2 2
which represent family of circles.
Q. 50 The general solution of e x cos ydx - e x sin ydy = 0 is
(a) ex cos y = k (b) ex sin y = k
(c) ex = k cos y (d) ex = k sin y
Sol. (a) Given that, e x cos ydx - e x sin ydy = 0
Þ e x cos ydx = e x sin ydy
dx
Þ = tan y
dy
Þ dx = tan ydy
On integrating both sides, we get
x = log sec y + C
Þ x - C = log sec y
Þ sec y = e x - C
Þ sec y = e xe - C
1 ex
Þ = C
cos y e
Þ e x cos y = eC
Þ e x cos y = K [where, K = eC ]

2 3
æ dy ö
Q. 51 The degree of differential equation d 2
y
+ ç ÷ + 6 y 5 = 0 is
dx è dx ø
(a) 1 (b) 2 (c) 3 (d) 5
3
d2y
+ æç ö÷ + 6 y5 = 0
dy
Sol. (a)
dx 2 è dx ø
We know that, the degree of a differential equation is exponent heighest of order
derivative.
\ Degree = 1

Q. 52 The solution of dy + y = e - x , y(0) = 0 is

dx
(a) y = ex ( x - 1) (b) y = xe- x
(c) y = xe-x + 1 (d) y = ( x + 1) e-x
dy
Sol. (b) Given that, + y = e -x
dx
Here, P = 1, Q = e - x
IF = e ò =eò
Pd x d x
= ex
The general solution is
y × e x = ò e - x e xdx + C
Þ y × e x = ò dx + C
Þ y×ex = x + C ...(i)
When x = 0 and y = 0, then
0=0+C Þ C =0
Eq. (i) becomes y×ex = x
Þ y = x e- x
 dy
Q. 53The integrating factor of differential equation + y tan x - sec x = 0
dx
is
(a) cos x (b) sec x (c) ecos x (d) esec x
dy
Sol. (b) Given that, + y tan x - sec x = 0
dx
Here, P = tan x,Q = sec x
IF = e ò = eò
Pdx tan x dx

= e(log sec x )
= sec x

1 + y2
Q. 54 The solution of differential equation dy = is
dx 1 + x2
(a) y = tan -1 x (b) y - x = k(1 + xy)
(c) x = tan -1 y (d) tan( xy) = k
dy 1 + y2
Sol. (b) Given that, =
dx 1 + x 2
dy dx
Þ =
1 + y2 1 + x 2
On integrating both sides, we get
tan-1 y = tan-1 x + C
Þ tan-1 y - tan-1 x = C
æ y-x ö
Þ tan-1 çç ÷÷ = C
è 1 + xy ø
y-x
Þ = tan C
1 + xy
Þ y - x = tan c(1 + x y)
Þ y - x = K (1 + x y)
where, k = tan C

1+ y
Q. 55 The integrating factor of differential equation dy +y= is
dx x
x ex
(a) (b) (c) xex (d) ex
ex x
dy 1+ y
Sol. (b) Given that, + y=
dx x
dy 1 + y
Þ = - y
dx x
dy 1 + y - xy
Þ =
dx x
dy 1 y (1 - x )
Þ = +
dx x x
dy æ 1 - x ö 1
Þ -ç ÷ y=
dx è x ø x
 - (1 - x ) 1
Here, P= ,Q =
x x
1- x x -1
-ò dx ò dx
IF = e ò
Pdx
=e x =e x

æ 1ö
ò çè 1 - x ÷ø dx
=e
= eò
x - log x

æ1ö
log ç ÷
= ex ×e èxø

1
= ex ×
x

Q. 56 y = ae mx + be - mx satisfies which of the following differential

equation?
dy dy
(a) + my = 0 (b) - my = 0
dx dx
d 2y d 2y
(c) - m 2y = 0 (d) + m 2y = 0
dx 2 dx 2
Sol. (c) Given that, y = ae mx + be - mx
On differentiating both sides w.r.t. x, we get
dy
= mae mx - bme - mx
dx
Again, differentiating both sides w.r.t. x, we get
d2y
= m2 ae mx + bm2e - mx
dx 2
d2y
Þ = m2 (ae mn + be - mn )
dx 2
d2y
Þ = m2 y
dx 2
d2y
Þ - m2 y = 0
dx 2

Q. 57 The solution of differential equation cos x sin ydx + sin x cos ydy = 0 is
sin x
(a) =C (b) sin x sin y = C
sin y
(c) sin x + sin y = C (d) cos x cos y = C
Sol. (b) Given differential equation is
cos x sin ydx + sin x cos ydy = 0
Þ cos x sin ydx = - sin x cos ydy
cos x cos y
Þ dx = - dy
sin x sin y
Þ cot x dx = - cot ydy
On integrating both sides, we get
logsin x = - logsin y + log C
Þ logsin x sin y = log C
Þ sin x × sin y = C
 dy
Q. 58 The solution of x + y = e x is
dx
ex k ey k
(a) y = + (b) y = xex + Cx (c) y = xex + k (d) x = +
x x y y
dy
Sol. (a) Given that, x + y = ex
dx
dy y ex
Þ + =
dx x x
which is a linear differential equation.
1
ò x dx
\ IF = e = e(log x ) = x
æd x ö
The general solution is y × x = ò çç × x ÷÷ d x
è x ø
Þ y × x = ò e xdx
Þ y× x = ex + k
ex k
Þ y= +
x x

Q. 59 The differential equation of the family of curves x 2 + y 2 - 2ay = 0,

where a is arbitrary constant, is
dy dy
(a) ( x 2 - y 2 ) = 2xy (b) 2 ( x 2 + y 2 ) = xy
dx dx
dy dy
(c) 2 ( x 2 - y 2 ) = xy (d) ( x 2 + y 2 ) = 2xy
dx dx
Sol. (a) Given equation of curve is
x 2 + y2 - 2 ay = 0
x 2 + y2
Þ = 2a
y
On differentiating both sides w.r.t. x, we get
æ d yö dy
y çç 2 x + 2 y ÷÷ - (x 2 + y2 )
è d x ø dx
=0
y2
dy dy
Þ 2 xy + 2 y2 - ( x 2 + y2 ) =0
dx dx
dy
Þ (2 y2 - x 2 - y2 ) = - 2 xy
dx
dy
Þ ( y2 - x 2 ) = - 2 xy
dx
dy
Þ ( x 2 - y2 ) = 2 xy
dx

Q. 60 The family Y = Ax + A 3 of curves will correspond to a differential

equation of order
(a) 3 (b) 2
(c) 1 (d) not defined
Sol. (c) Given family of curves is y = Ax + A 3 ...(i)
dy
Þ = A
dx
dy
Replacing A by in Eq. (i), we get
dx
dy æ dy ö 3
y=x + ç ÷
dx è dx ø
\ Order = 1

Q. 61 The general solution of dy

2
-y
= 2x e x is
dx
2 2
(a) ex - y = C (b) e- y + ex = C
2 2
(c) ey = ex + C (d) ex + y = C
Sol. (c) dy 2 2
Given that, = 2 x e x - y = 2 x e x × e -y
dx
dy 2
Þ ey = 2x ex
dx
2
Þ e y dy = 2 x e x dx
On integrating both sides, we get
y 2
òe d y = 2 ò xe x d x
Put x 2 = t in RHS integral, we get
2x dx = dt
y t
òe dy = ò e dt
Þ e y = et + C
2
Þ ey = ex + C

Q. 62 The curve for which the slope of the tangent at any point is equal to
the ratio of the abcissa to the ordinate of the point is
(a) an ellipse (b) parabola
(c) circle (d) rectangular hyperbola
dy
Sol. (d) Slope of tangent to the curve =
dx
x
and ratio of abscissa to the ordinate =
y
dy x
According to the question, =
dx y
yd y = x d x
On integrating both sides, we get
y2 x 2
= +C
2 2
y2 x 2
Þ - = C Þ y2 - x 2 = 2C
2 2
which is an equation of rectangular hyperbola.
 x2
dy
Q. 63 The general solution of differential equation =e 2 + xy is
dx
2 2
(a) y = Ce-x / 2 (b) y = C ex / 2
2 2
(c) y = ( x + C) ex / 2 (d) y = (C - x) ex / 2
dy 2
Sol. (c) Given that, = e x / 2 + xy
dx
dy 2
Þ - xy = e x / 2
dx
2
/2
Here, P = - x, Q = e x
\ IF = e ò - x dx = e -x
2
/2

The general solution is

2 2 2
y × e -x /2
= ò e -x /2
- ex /2
dx + C
-x 2 / 2
Þ ye = ò 1d x + C
2
Þ y × e -x /2
=x+C
2 2
Þ y = x ex /2
+ C e+ x /2

2
/2
Þ y = (x + C ) e x

Q. 64 The solution of equation (2 y - 1) dx - (2x + 3) dy = 0 is

2x - 1 2y + 1
(a) =k (b) =k
2y + 3 2x - 3
2x + 3 2x - 1
(c) =k (d) =k
2y - 1 2y - 1

Sol. (c) Given that, (2 y -1) dx - (2 x + 3)d y = 0

Þ (2 y - 1) dx = (2 x + 3) dy
dx dy
Þ =
2x + 3 2 y - 1
On integrating both sides, we get
1 1
log (2 x + 3) = log (2 y - 1) + log C
2 2
1
Þ [log × (2 x + 3) - log (2 y - 1)] = log C
2
1 æ 2x + 3 ö
Þ log çç ÷÷ = log C
2 è 2y - 1 ø
1/ 2
æ 2x + 3 ö
Þ çç ÷÷ = C
è 2y - 1 ø
2x + 3
Þ = C2
2y - 1
2x + 3
Þ = k , where K = C 2
2y - 1
Q. 65 The differential equation for which y = a cos x + b sin x is a solution,
is
d 2y d 2y
(a) + y =0 (b) -y =0
dx 2 dx 2
d 2y d 2y
(c) 2 + ( a + b) y = 0 (d) 2 + ( a - b) y = 0
dx dx
Sol. (a) Given that, y = a cos x + b sin x
On differentiating both sides w.r.t. x, we get
dy
= - a sin x + b cos x
dx
Again, differentiating w.r.t. x, we get
d2y
= - a sin x + b cos x
dx 2
d2y
Þ =- y
dx 2
d2y
Þ + y=0
dx 2

dy
Q. 66 The solution of + y = e - x , y (0) = 0 is
dx
(a) y = e-x ( x -1) (b) y = xex
(c) y = xe-x + 1 (d) y = xe- x
dy
Sol. (d) Given that, + y = e -x
dx
which is a linear differential equation.
Here, P = 1and Q = e - x
IF = e ò = e x
dx

The general solution is

y × e x = ò e - x × e x dx + C
Þ y e x = ò dx + C
Þ y ex = x + C ...(i)
When x = 0 and y = 0 then,0 = 0 + C Þ C = 0
Eq. (i) becomes y × e x = x Þ y = x e - x

Q. 67 The order and degree of differential equation

2 4
æ d 3y ö d2y æ dy ö
ç ÷ -3 + 2 ç ÷ = y 4 are
ç dx 3 ÷ 2
è d x ø
è ø d x
(a) 1, 4 (b) 3, 4 (c) 2, 4 (d) 3, 2
2 4
æd3y ö d2y
ç 3 ÷ - 3 2 + 2 æç ö÷ = y4
dy
Sol. (d) Given that, ç dx ÷
è ø dx è dx ø
\ Order = 3
and degree = 2
 é 2ù
d2y
Q. 68 The order and degree of differential equation ê1 + æç dy ö÷ ú = are
êë è dx ø úû dx 2
3
(a) 2, (b) 2, 3 (c) 2, 1 (d) 3, 4
2
é æ dy ö ù d y
2 2
Sol. (c) Given that, ê1 + ç ÷ ú =
ë è dx ø û dx 2
\ Order = 2 and degree = 1

Q. 69 The differential equation of family of curves y 2 = 4a (x + a) is

dy æ dy ö dy
(a) y 2 = 4 çx + ÷ (b) 2y = 4a
dx è dx ø dx
2 2
d 2y æ dy ö dy æ dy ö
(c) y + ç ÷ =0 (d) 2x + y ç ÷ -y =0
dx 2 è dx ø dx è dx ø
Sol. (d) Given that, y 2 = 4 a ( x + a) .. (i)
On differentiating both sides w.r.t. x, we get
dy dy
2y = 4a Þ 2 y = 4a
dx dx
dy 1 dy
Þ y = 2a Þ a = y ...(ii)
dx 2 dx
On putting the value of a from Eq. (ii) in Eq. (i), we get
dy æ 1 dy ö
y2 = 2 y çx + y ÷
dx è 2 dx ø
2
+ y2 æç ö÷
dy dy
Þ y2 = 2 xy
dx è dx ø
2
+ y æç ö÷ - y = 0
dy dy
Þ 2x
dx è dx ø

Q. 70 Which of the following is the general solution of

d2y dy
2
-2 + y =0?
dx dx
(a) y = ( Ax + B ) e x (b) y = ( Ax + B ) e- x
(c) y = Aex + Be-x (d) y = A cos x + B sin x
d2y dy
Sol. (a) Given that, -2 + y=0
dx 2 dx
D2 y - 2 Dy + y = 0,
d
where D=
dx
(D2 - 2 D + 1) y = 0
The auxiliary equation is m2 - 2 m + 1 = 0
(m - 1)2 = 0 Þ m = 1, 1
Since, the roots are real and equal.
\ CF = ( A x + B )e x Þ y = ( Ax + B) e x
[since, if roots of Auxilliary equation are real and equal say (m) , then CF = (C1 x + C 2 ) e m x ]
Q. 71 The general solution of dy + y tan x = sec x is
dx
(a) y sec x = tan x + C (b) y tan x = sec x + C
(c) tan x = y tan x + C (d) x sec x = tan y + C
Sol. (a) Given differential equation is
dy
+ y tan x = sec x
dx
which is a linear differential equation
Here, P = tanx , Q = sec x,
IF = e ò
tan x d x
\ = elog | sec x|
= sec x
The general solution is
y × sec x = ò sec x × sec x + C
Þ y × sec x = ò sec 2 x dx + C
Þ y × sec x = tan x + C

Q. 72 The solution of differential equation dy +

y
= sin x is
dx x
(a) x (y + cos x) = sin x + C (b) x (y - cos x) = sin x + C
(c) xy cos x = sin x + C (d) x (y + cos x) = cos x + C
Sol. (a) Given differential equation is
dy 1
+ y = sin x
dx x
which is linear differential equation.
1
Here, P = and Q = sin x
x
1
ò x dx
\ IF = e = elog x
=x
The general solution is
y × x = ò x × sin x dx + C ...(i)
Take I = ò x sin x dx

- x cos x - ò - cos x dx
= - x cos x + sin x
Put the value of I in Eq. (i), we get
xy = - x cos x + sin x + C
Þ x ( y + cos x ) = sin x + C

Q. 73 The general solution of differential equation

(e x + 1) ydy = ( y + 1) e x dx is
(a) ( y + 1) = k ( ex + 1) (b) y + 1 = ex + 1 + k
ì ex + 1ü
(c) y = log {k (y + 1) ( ex + 1)} (d) y = log í ý+ k
î y +1þ
Sol. (c) Given differential equation
(e x + 1) ydy = ( y + 1) e x dx
dy e x (1 + y) dx (e x + 1) y
Þ = x Þ =
dx (e + 1)y dy e x (1 + y)
dx ex y y
Þ = x +
dy e (1 + y) e x (1 + y)
dx y y
Þ = +
dy 1 + y (1 + y) e x
dx y æ 1 ö
Þ = ç1 + x ÷
dy 1 + y è e ø
dx y æ ex + 1ö
Þ = çç x
÷÷
dy 1 + y è e ø
æ y ö æ ex ö
Þ çç ÷÷ dy = çç x ÷÷ dx
è 1+ y ø è e + 1ø
On integrating both sides, we get
y ex
ò 1 + y dy = ò 1 + e x dx
1+ y - 1 ex
Þ ò 1+ y
dy = ò
1 + ex
dx

1 ex
Þ ò 1 dy - ò 1 + y
dy = ò 1 + e x dx
Þ y - log|(1 + y) = log|(1 + e x ) + log k
Þ y = log (1 + y) + log (1 + e x ) + log (k )
Þ y = log {k (1 + y) (1 + e x )}

dy
Q. 74 The solution of differential equation = e x -y + x 2 e - y is
dx
x3
(a) y = ex - y - x 2 e- y + C (b) ey - ex = +C
3
x3 x3
(c) ex + ey = +C (d) ex - ey = +C
3 3
dy
Sol. (b) Given that, = e x - y + x 2 e -y
dx
dy
Þ = e xe - y + x 2 e - y
dx
dy e x + x 2
Þ =
dx ey
Þ e dy = (e + x 2 ) dx
y x

On integrating both sides, we get

y
òe dy = ò (e x + x 2 ) dx
x3
Þ ey = ex+ +C
3
x3
Þ ey - ex = +C
3
 dy 2xy 1
Q. 75 The solution of differential equation + = is
dx 1 +x 2 (1 + x 2 )2
y
(a) y (1 + x 2 ) = C + tan -1 x (b) = C + tan -1 x
1+ x2
(c) y log (1 + x 2) = C + tan -1 x (d) y (1 + x 2) = C + sin -1 x
dy 2 xy 1
Sol. (a) Given that, + =
dx 1 + x 2 (1 + x 2 )2
2x 1
Here, P= and Q =
1 + x2 (1 + x 2 )2
which is a linear differential equation.
2x
ò1+ x2
dx
\ IF = e
Put 1 + x 2 = t Þ 2x dx = dt
dt
ò x2)
\ IF = e t = elog t = elog ( 1 + = 1 + x2
The general solution is
1
y × (1 + x 2 ) = ò (1 + x 2 ) +C
(1 + x 2 )2
1
Þ y (1 + x 2 ) = ò 1 + x 2 dx + C
Þ y (1 + x 2 ) = tan-1 x + C

Fillers
d2 y
Q. 76 (i) The degree of the differential equation
2
+ e dy / dx = 0 is ....... .
dx
2
æ dy ö
(ii) The degree of the differential equation 1 + ç ÷ = x is ........ .
è dx ø
(iii) The number of arbitrary constants in the general solution of a
differential equation of order three is ....... .
dy y 1
(iv) + = is an equation of the type ........ .
dx x log x x
(v) General solution of the differential equation of the type
is given by ........ .
xdy
(vi) The solution of the differential equation + 2 y = x 2 is ........ .
dx
2 dy 2
(vii) The solution of (1 + x ) + 2xy - 4x = 0 is ........ .
dx
 (viii) The solution of the differential equation ydx + ( x + xy) dy = 0 is
...... .
dy
(ix) General solution of + y = sin x is ....... .
dx
(x) The solution of differential equation cot y dx = xdy is ....... .
dy 1+ y
(xi) The integrating factor of +y= is ......... .
dx x
Sol. (i) Given differential equation is
dy
d2 y
+ e dx = 0
dx 2
Degree of this equation is not defined.
2
(ii) Given differential equation is 1 + æç ö÷ = x
dy
è dx ø
So, degree of this equation is two.
(iii) There are three arbitrary constants.
dy y 1
(iv) Given differential equation is + =
dx x log x x
dy
The equation is of the type + Py = Q
dx
(v) Given differential equation is
dx
+ P1 x = Q1
dy
The general solution is
x × IF = ò Q ( IF ) dy + C i .e., x e ò = ò Q {e ò } d y + C
Pdy Pd y

(vi) Given differential equation is

dy dy 2 y
x + 2 y = x2 Þ + =x
dx dx x
dy
This equation of the form + Py = Q.
dx
2
ò x dx
\ IF = e = e 2 log x
= x2
The general solution is
yx 2 = ò x × x 2 dx + C

x4
Þ yx 2 = +C
4
x2
Þ y= + Cx -2
4
(vii) Given differential equation is
dy
(1 + x 2 ) + 2 xy - 4x 2 = 0
dx
dy 2 xy 4x 2
Þ + 2
- =0
dx 1 + x 1 + x2
 dy 2x 4x 2
Þ + 2
y=
dx 1 + x 1 + x2
2x
ò 1 + x2
dx
\ IF = e
Put 1 + x 2 = t Þ 2xdx = dt
dt
ò log ( 1 + x 2 )
\ IF = e t = elog t = e = 1 + x2
The general solution is
.4x 2
y × (1 + x 2 ) = ò (1 + x 2 ) dx + C
(1 + x 2 )
Þ (1 + x 2 ) y = ò 4x 2 dx + C

x3
Þ (1 + x 2 )y = 4 +C
3
4x 3
Þ y= + C (1 + x 2 )-1
3(1 + x 2 )
(viii) Given differential equation is
Þ ydx + (x + xy) dy = 0
Þ ydx + x (1 + y) d y = 0
dx æ1 + y ö
Þ = ç ÷ dy
-x è y ø
1 æ1 ö
Þ ò x dx = - ò çè y + 1÷ø dy [on integrating]

Þ log (x ) = - log ( y) - y + log A

log (x ) + log ( y) + y = log A
log (xy) + y = log A
Þ log xy + log e y = log A
Þ xy e y = A
Þ xy = Ae - y
(ix) Given differential equation is
dy
+ y = sin x
dx
IF = ò e1dx = e x
The general solution is
y × e x = ò e x sin x dx + C ...(i)
Let I = ò e sin x dx
x

I = sin x e x - ò cos x e x dx
= sin x e x - cos x e x + ò (- sin x ) e x dx
2I = e x (sin x - cos x )
1
I = e x (sin x - cos x )
2
 From Eq. (i),
x
y× ex = (sin x - cos x ) + C
2
1
Þ y = (sin x - cos x ) + C × e - x
2
(x) Given differential equation is
cot y dx = xdy
1
Þ dx = tan y dy
x
On integrating both sides, we get
1
Þ ò x dx = ò tan y d y
Þ log (x ) = log (sec y) + log C
æ x ö
Þ log çç ÷÷ = log C
è sec y ø
x
Þ =C
sec y
Þ x = C sec y
(xi) Given differential equation is
dy 1+ y
+ y=
dx x
dy 1 y
+ y= +
dx x x
1 1
+ y æç 1 - ö÷ =
dy
Þ
dx è x ø x
æ 1ö
ò çè 1 - x ÷ø dx
\ IF = e
= e x - log x
ex
= e x × e - log x
=
x

True/False
Q. 77 State True or False for the following
dx
(i) Integrating factor of the differential of the form + P1 x = Q 1 is
dy
given by e ò 1 .
P dy

dx
(ii) Solution of the differential equation of the type + P1 x = Q 1 is
dy
given by x × IF = ò (IF) ´ Q 1 dy.
 (iii) Correct substitution for the solution of the differential equation of
dy
the type = f (x, y), where f (x, y) is a homogeneous function of
dx
zero degree is y = vx.
(iv) Correct substitution for the solution of the differential equation of
dy
the type = g (x, y), where g (x, y) is a homogeneous function of
dx
the degree zero is x = vy.
(v) Number of arbitrary constants in the particular solution of a
differential equation of order two is two.
(vi) The differential equation representing the family of circles
x 2 + ( y - a)2 = a2 will be of order two.
1/ 3
dy æ y ö
(vii) The solution of =ç ÷ is y 2/ 3 - x 2 / 3 = c
dx è x ø
(viii) Differential equation representing the family of curves
d2 y dy
y = e x (A cos x + B sin x) is -2 + 2 y = 0.
2 dx
dx
dy x + 2 y
(ix) The solution of the differential equation = is x + y = kx 2 .
dx x
xdy y æ yö
(x) Solution of = y + x tan is sin ç ÷ = c x
dx x èxø
(xi) The differential equation of all non horizontal lines in a plane is
d2x
= 0.
dy 2
Sol. (i) True
Given differential equation,
dx
+ P1 x = Q1
dy
IF = e ò
p1dy
\
(ii) True
(iii) True
(iv) True
(v) False
There is no arbitrary constant in the particular solution of a differential equation.
(vi) False
We know that, order of the differential equation = number of arbitrary constant
Here, number of arbitrary constant = 1.
So order is one.
(vii) True
1/ 3
dy æ y ö
Given differential equation, =ç ÷
dx è x ø
dy y1/ 3
Þ = 1/ 3
dx x
Þ y- 1/ 3 dy = x -1/ 3 dx
On integrating both sides, we get
-1/ 3 -1/ 3
ò y dy = ò x dx
y-1/ 3 + 1 x -1/ 3 + 1
Þ = + C¢
-1 -1
+1 +1
3 3
3 2/ 3 3 2/ 3
Þ y = x + C¢
2 2
Þ y2 / 3 - x 2 / 3 = C ¢ é where, 2 C ¢ = C ù
êë 3 úû

(viii) True
Given that, y = e x ( A cos x + B sin x )
On differentiating w.r.t. x, we get
dy
= e x (- A sin x + Bcos x ) + e x ( A cos x + Bsin x )
dx
dy
Þ - y = e x (- A sin x + Bcos x )
dx
Again differentiating w.r.t. x, we get
d 2 y dy
- = e x (- A cos x - Bsin x ) + e x (- A sin x + Bcos x )
dx 2 dx
d 2 y dy dy
Þ - + y= -y
dx 2 dx dx
d2y dy
Þ -2 + 2y = 0
dx 2 dx
(ix) True
dy x + 2 y dy 2
Given that, = Þ = 1+ .y
dx x dx x
dy 2
Þ - y=1
dx x
-2
dx
IF = e x = e -2 log x
= x -2
The differential solution,
y × x -2 = ò x -2 × 1d x + k
y x -2 + 1
Þ = + k
x 2
-2 + 1
y -1
Þ = + k
x2 x
Þ y = - x + kx 2
Þ x + y = kx 2
(x) True
Given differential equation,
= y + x tan æç ö÷
xdy y
dx èx ø
= + tan ç ö÷. æ
dy y y
Þ …(i)
dx x èx ø
y
Put = v i .e., y = vx
x
dy xdv
Þ =v+
dx dx
On substituting these values in Eq. (i), we get
xdv
+ v = v + tan v
dx
dx dv
Þ =
x tan v
On integrating both sides, we get
1 1
ò x dx = ò tan v dx
Þ log (x ) = log (sin v ) + log C ¢
æ x ö
Þ log çç ÷÷ = log C ¢
è sin v ø
x
Þ = C¢
sin v

Þ sin v = Cx é where, C = 1 ù
êë C ¢úû
y
Þ sin = Cx
x
(xi) True
Let any non-horizontal line in a plane is given by
y = mx + c
On differentiating w.r.t. x, we get
dy
=m
dx
Again, differentiating w.r.t. x, we get
d2y
=0
dx 2
 10
Vector Algebra
Short Answer Type Questions
®
Q. 1 Find the unit vector in the direction of sum of vectors a = 2$i - $j + k$
®
and b = 2$j + k$ .
K Thinking Process
®
® a
We know that, unit vector in the direction of a vector a is . So, first we will find the
®
| a|
sum of vectors and then we will use this concept.
® ® ®
Sol. Let c denote the sum of a and b.
® ® ®
We have, c = a+ b
= 2 $i - $j + k$ + 2 $j + k$ = 2 $i + $j + 2 k$
®
® c 2 $i + $j + 2k$ 2 $i + $j + 2k$
\ Unit vector in the direction of c = = =
® 9
|c| 2 2 + 12 + 2 2
2 $i + $j + 2k$
c$ =
3
® ®
Q. 2 If a = $i + $j + 2k$ and b = 2$i + $j + 2k$ , then find the unit vector in the
direction of
® ® ®
(i) 6 b (ii) 2 a - b
® ®
Sol. Here, a = $i + $j + 2k$ and b = 2 $i + $j - 2k$
®
(i) Since, 6 b = 12 $i + 6$j - 12 k$
®
® 6b
\Unit vector in the direction of 6b =
®
|6b|
12 $i + 6$j - 12k$ 6(2 $i + $j - 2k$ )
= =
2
12 + 6 + 12 2 2 324
6(2 $i + $j - 2k$ ) 2 $i + $j - 2 k$
= =
18 3
 ® ®
(ii) Since, 2 a - b = 2($i + $j + 2k$ ) - (2 $i + $j - 2k$ )
= 2 $i + 2 $j + 4k$ - 2 $i - $j + 2k$ = $j + 6k$
® ®
® ® 2a - b $j + 6k$ 1 $
\Unit vector in the direction of 2 a - b = = = ( j + 6k$ )
®
|2 a - b|
® 1 + 36 37

®
Q. 3 Find a unit vector in the direction of PQ , where P and Q have
coordinates (5, 0, 8) and (3, 3, 2), respectively.
Sol. Since, the coordinates of P and Q are (5, 0, 8) and (3, 3, 2), respectively.
¾® ¾® ¾®
\ PQ = OQ - OP
= (3$i + 3$j + 2k$ ) - (5$i + 0$j + 8 k$ )
= - 2 $i + 3$j - 6k$
¾®
¾® PQ -2 $i + 3$j - 6 k$
\ Unit vector in the direction of PQ = ¾®
=
| PQ| 2 2 + 32 + 62
-2 i + 3 j - 6 k$ -2 $i + 3$j - 6 k$
$ $
= =
49 7

® ® ® ®
Q. 4 If a and b are the position vectors of A and B respectively, then find
® ® ® ®
the position vector of a point C in BA produced such that BC = 1 . 5 BA.
¾® ® ¾® ®
Sol. Since, OA = a and OB = b
¾® ¾® ¾® ® ®
\ BA = OA - OB = a - b
¾® ® ®
and 1.5 BA = 1.5(a - b )
¾® ¾® ® ®
Since, BC = 1.5 BA = 1.5(a - b )
¾® ¾® ® ®
OC - OB = 1.5 a - 1.5 b
¾® ® ® ® ¾® ®
OC = 15
. a - 15
. b+ b [Q OB = b]
® ®
= 1.5 a - 0.5 b
® ®
3a - b
=
2
Graphically, explanation of the above solution is given below
® ®
(a – b) A
B C
®

®
a

b ®
–b
3a 2
®
O
Q. 5 Using vectors, find the value of k, such that the points (k, - 10, 3),
(1, - 1, 3) and (3, 5, 3) are collinear.
K Thinking Process
® ® ®
Here, use the following stepwise approach first, get the values of | AB| , |BC and | AC|
® ® ®
and then use the concept that three points are collinear, if | AB | + |BC | = AC such that.
A B C

Sol. Let the points are A(k, - 10, 3) , B(1, - 1, 3) and C(3, 5, 3) .
® ® ®
So, AB = OB - OA
= ($i - $j + 3 k$ ) - (k $i - 10$j + 3 k$ )
= (1 - k )$i + (-1 + 10)$j + (3 - 3)k$
= (1 - k )$i + 9$j + 0 k$
®
\ |AB| = (1 - k )2 + (9)2 + 0 = (1 - k )2 + 81
® ® ®
Similarly, BC = OC - OB
= (3$i + 5$j + 3 k$ ) - ($i - $j + 3k$ )
= 2 $i + 6$j + 0 k$
®
\ |BC | = 2 2 + 62 + 0 = 2 10
® ® ®
and AC = OC - OA
= (3$i + 5$j + 3k$ ) - (k $i - 10$j + 3 k$ )
= (3 - k )$i + 15$j + 0 k$
®
\ |AC | = (3 - k )2 + 225
If A, B and C are collinear, then sum of modulus of any two vectors will be equal to the
modulus of third vectors
® ® ®
For|AB| + |BC | = |AC |,
(1 - k )2 + 81 + 2 10 = (3 - k )2 + 225

Þ (3 - k )2 + 225 - (1 - k )2 + 81 = 2 10

Þ 9 + k 2 - 6 k + 225 - 1 + k 2 - 2 k + 81 = 2 10

Þ k 2 - 6 k + 234 - 2 10 = k 2 - 2 k + 82

Þ k 2 - 6 k + 234 + 40 - 2 k 2 - 6k + 234 × 2 10 = k 2 - 2 k + 82

Þ k 2 - 6 k + 234 + 40 - k 2 + 2 k - 82 = 4 10 k 2 + 234 - 6 k

Þ -4 k + 192 = 4 10 k 2 + 234 - 6 k

Þ - k + 48 = 10 k 2 + 234 - 6 k
On squaring both sides, we get
48 ´ 48 + k 2 - 96 k = 10(k 2 + 234 - 6k )
Þ k 2 - 96 k - 10 k 2 + 60 k = - 48 ´ 48 + 2340
Þ -9k 2 - 36k = - 48 ´ 48 + 2340
 Þ (k 2 + 4k ) = + 16 ´ 16 - 260 [dividing by 9 in both sides]
Þ k2 + 4 k = - 4
k2 + 4 k + 4 = 0
Þ (k + 2 )2 = 0
\ k = -2
®
Q. 6 A vector r is inclined at equal angles to the three axes. If the
® ®
magnitude of r is 2 3 units, then find the value of r .
K Thinking Process
®
If a vector r is inclined at equal angles to the three axes, then direction cosines of vector,
® ® ® ®
r will be same and then use, r = r × | r | .
®
Sol. We have, | r|= 2 3
® ® ®
Since, r is equally inclined to the three axes, r so direction cosines of the unit vector r will
be same. i.e., l = m = n.
We know that,
l2 + m2 + n2 = 1
Þ l2 + l2 + l2 = 1
1
Þ l2 =
3
æ 1 ö
Þ l=±ç ÷
è 3ø
1 $ 1 $ 1 $
So, r$ = ± i ± j± k
3 3 3
®
® ® é ù
\ r = r$| r| êQ r$ = r ú
®
ê | r |úû
ë
é 1 $ 1 $ 1 $ù
= ê± i ± j± k 2 3 [Q |r |= 2 3 ]
ë 3 3 3 úû
= ± 2 $i ± 2 $j ± 2k$ = ± 2($i + $j + k$ )

Q. 7 If a vector ®r has magnitude 14 and direction ratios 2 , 3 and - 6. Then,

® ®
find the direction cosines and components of r , given that r makes an
acute angle with X-axis.
® ® ® ®
Sol. Here, | r | = 14, a = 2k, b = 3k and c = - 6 k
\Direction cosines l, m and n are
®
a 2k k
l= ®
= =
| r| 14 7
®
b 3k
m= ®
=
| r| 14
®
c - 6 k -3 k
and n= ®
= =
| r| 14 7
 Also, we know that
l 2 + m2 + n2 = 1
k2 9 k2 9 k2
Þ + + =1
49 196 49
4 k 2 + 9 k 2 + 36 k 2
Þ =1
196
196
Þ k2 = =4
49
Þ k =±2
2 3 -6
So, the direction cosines (l, m, n) are , and .
7 7 7
®
[since, r makes an acute angle with X-axis]
® ®
Q r = r$ ×| r|
® ®
\ r = (l $i + m $j + nk$ )| r |
æ +2 $ 3$ 6 $ ö
=ç i + j - k ÷ × 14
è 7 7 7 ø
= + 4 i + 6$j - 12k$
$

Q. 8 Find a vector of magnitude 6, which is perpendicular to both the vectors

2$i - $j + 2k$ and 4 $i - $j + 3k$ .
K Thinking Process
® ®
First, we will use this concept any vector perpendicular to both the vectors a and b is
i$ $j k$
® ®
given by a ´ b = a 1 a2 a3 and then we will find the vector with magnitude 6.
b1 b2 b3

® ®
Sol. Let a = 2 $i - $j + 2k$ and b = 4$i - $j + 3k$
® ®
So, any vector perpendicular to both the vectors a and b is given by
$i $j k$
® ®
a ´ b = 2 -1 2
4 -1 3
= i (-3 + 2 ) - $j(6 - 8) + k$ (-2 + 4)
$
®
= - $i + 2 $j + 2k$ = r [say]
r
A vector of magnitude 6 in the direction of r
®
r - $i + 2 $j + 2k$
= .6= .6
®
| r| 12 + 2 2 + 2 2
- 6 $ 12 $ 12 $
= i + j+ k
3 3 3
= - 2 i + 4j + 4 k
$ $ $
Q. 9 Find the angle between the vectors 2$i - $j + k$ and 3$i + 4 $j - k$ .
K Thinking Process
® ®
® ® a× b
If a and b are two vectors, making angle q with each other, then cos q = , using
® ®
| a || b |
this concept we will find q.
® ®
Sol. Let a = 2 $i - $j + k$ and b = 3$i + 4$j - k$
® ®
We know that, angle between two vectors a and b is given by
® ®
a× b
cos q =
® ®
| a||b|
(2 $i - $j + k$ )(3$i + 4$j - k$ )
=
4 + 1 + 1 9 + 16 + 1
6- 4-1 1
= =
6 26 2 39
æ 1 ö
\ q = cos -1 ç ÷
è 2 39 ø

® ® ® ® ® ® ® ® ®
Q. 10 If a + b + c = 0, then show that a ´ b = b ´ c = c ´ a . Interpret the
result geometrically.
® ® ®
Sol. Since, a + b+ c =0
® ® ®
Þ b=-c -a
® ® ® ® ®
Now, a ´ b = a ´ (- c - a )
® ® ® ® ® ®
= a ´ (- c ) + a ´ (- a ) = - a ´ c
® ® ® ®
Þ a ´b = c ´a …(i)
® ® ® ® ®
Also, b ´ c = (- c - a ) ´ c
® ® ® ® ® ®
= (- c ´ c ) + (- a ´ c ) = - a ´ c
® ® ® ®
Þ b ´c = c ´a …(ii)
® ® ® ® ® ®
From Eqs. (i) and (ii), a ´b = b ´c = c ´a
Geometrical interpretation of the result
D
E C

® ® ® ®
c b |b| sin q c

q
A ® B
a
® ® ® ®
If ABCD is a parallelogram such that AB = a and AD = b and these adjacent sides are
making angle q between each other, then we say that
® ® ® ®
Area of parallelogram ABCD = |a||b||sin q| = |a ´ b|
Since, parallelogram on the same base and between the same parallels are equal in area.
 ® ® ® ® ® ®
We can say that, |a ´ b| = |a ´ c| = |b ´ c|
® ® ® ® ® ®
This also implies that, a ´b =a ´c = b ´c
® ® ®
So, area of the parallelograms formed by taking any two sides represented by a , b and c as
adjacent are equal.

®
Q. 11 Find the sine of the angle between the vectors a = 3$i + $j + 2k$ and
®
b = 2$i - 2$j + 4 k$ .
K Thinking Process
® ®
We know that, if a and b are in their component form, then
a 1b1 + a2 b2 + a3 b3
cos q = . After getting cos q, we shall find the sine of the
a21 + a22 + a23 b21 + b22 + b23
angle.

Sol. Here, a1 = 3, a2 = 1, a3 = 2 and b1 = 2, b2 = - 2, b3 = 4

We know that,
a1b1 + a2 b2 + a3 b3
cos q =
a1 + a22 + a32 b12 + b22 + b32
2

3 ´ 2 + 1 ´ ( -2 ) + 2 ´ 4
=
3 + 12 + 2 2 2 2 + (-2 )2 + 42
2

6-2 + 8 12 6 6 3
= = = = =
14 24 2 14 6 84 2 21 21
\ sin q = 1 - cos 2 q
9 12 2 3 2
= 1- = = =
21 21 3 7 7

Q. 12 If A, B, C and D are the points with position vectors $i - $j + k$ ,

2$i - $j + 3k$ , 2$i - 3k$ and 3$i - 2$j + k$ respectively, then find the
® ®
projection of AB along CD.
K Thinking Process
® ®
® ® a× b
We shall use the concept that projection of a along b is r .
|b|
® ® ® ®
Sol. Here, OA = $i + $j - k$ , OB = 2 $i - $j + 3k$ , OC = 2 $i - 3 k$ and OD = 3$i - 2 $j + k$
® ® ®
\ AB = OB - OA = (2 - 1)$i + (-1 - 1)$j + (3 + 1)k$
= $i - 2 $j + 4 k$
® ® ®
and CD = OD - OC = (3 - 2 )$i + (-2 - 0)$j + (1 + 3) k$
= $i - 2 $j + 4 k$
 ®
® ® ® CD
So, the projection of AB along CD = AB × ®
| CD|
( i - 2 $j + 4 k$ ) × ($i - 2 $j + 4k$ )
$
=
12 + 2 2 + 42
1 + 4 + 16 21
= =
21 21
= 21 units

Q.13 Using vectors, find the area of the DABC with vertices A(1, 2 , 3),
B(2 , - 1, 4) and C(4, 5, - 1).
K Thinking Process
We know that,
1 ® ®
Area of DABC = | AB ´ AC | . So, here we shall use this concept.
2
®
Sol. Here, AB = (2 - 1)$i + (-1 - 2 )$j + (4 - 3)k$
= $i - 3$j + k$
®
and AC = (4 - 1)$i + (5 - 2 )$j + (- 1 - 3) k$
= 3$i + 3$j - 4 k$
C (4, 5, –1)

A B
(1, 2, 3) (2, –1, 4)
$i $j k$
® ®
\ AB ´ AC = 1 -3 1
3 3 -4
= $i (12 - 3) - $j(-4 - 3) + k$ (3 + 9)
= 9$i + 7 $j + 12k$
® ®
and |AB ´ AC| = 92 + 7 2 + 12 2
= 81 + 49 + 144
= 274
1 ® ®
\ Area of DABC = |AB ´ AC|
2
1
= 274 sq units
2
Q. 14 Using vectors, prove that the parallelogram on the same base and
between the same parallels are equal in area.
Sol. Let ABCD and ABFE are parallelograms on the same base AB and between the same
parallel lines AB and DF.
Here, AB||CD and AE || BF
D E C F

®
b

A B
®
a
® ® ® ®
Let AB = a and AD = b
® ®
\ Area of parallelogram ABCD = a ´ b
® ®
Now, area of parallelogram ABFF = AB ´ AE
® ® ®
= AB ´ (AD + DE )
® ® ® ® ®
= AB ´ (b + k a ) [let DE = k a , where k is a scalar]
® ® ®
= a ´ (b + k a )
® ® ® ®
= (a ´ b ) + (a ´ k a )
® ® ® ®
= (a ´ b ) + k(a ´ a )
® ® ® ®
= (a ´ b ) [Q a ´ a = 0 ]
= Area of parallelogram ABCD
Hence proved.

Long Answer Type Questions

b 2 + c 2 - a2
Q. 15 Prove that in any DABC, cos A = , where a, b and c are the
2bc
magnitudes of the sides opposite to the vertices A, B and C,
respectively.
Sol. Here, components of C are c cos A and c sin A is drawn.
B

a C
c sin A

C A
D C cos A

b
 ®
Since, CD = b - c cos A
In DBDC,
a2 = (b - c cos A)2 + (c sin A)2
Þ a2 = b 2 + c 2 cos 2 A - 2 bc cos A + c 2 sin2 A
Þ 2 bc cos A = b 2 - a2 + c 2 (cos 2 A + sin2 A)
b 2 + c 2 - a2
\ cos A =
2 bc

® ® ®
Q. 16 If a , b and c determine the vertices of a triangle, show that
1 ® ® ® ® ® ®
[ b ´ c + c ´ a + a ´ b] gives the vector area of the triangle. Hence,
2
® ® ®
deduce the condition that the three points a , b and c are collinear.
Also, find the unit vector normal to the plane of the triangle.
K Thinking Process
Here, we shall use the following two concepts.
® ® ®
(i) If a , b and c are collinear, then the area of the triangle formed by the vectors will be
zero.
® ® ® ®
(ii) We know that, a ´ b = | a || b | sin qn.
$
® ® ®
Sol. Since, a , b and c are the vertices of a DABC as shown.
C
a®

®
c–
c®–

®
b

A ® ® B
b–a
1 ® ®
\ Area of DABC = |AB ´ AC |
2
® ® ® ® ® ®
Now, AB = b - a and AC = c - a
1 ® ® ® ®
\ Area of DABC = |b - a ´ c - a |
2
1 ® ® ® ® ® ® ® ®
= |b ´ c - b ´ a - a ´ c + a ´ a |
2
1 ® ® ® ® ® ® ®
= |b ´ c + a ´ b + c ´ a + 0 |
2
1 ® ® ® ® ® ®
= |b ´ c + a ´ b + c ´ a | …(i)
2
For three points to be collinear, area of the D ABC should be equal to zero.
1 ® ® ® ® ® ®
Þ [b ´ c + c ´ a + a ´ b ] = 0
2
® ® ® ® ® ®
Þ b ´c +c ´a + a ´b = 0 …(ii)
® ® ®
This is the required condition for collinearity of three points a , b and c .
 Let n$ be the unit vector normal to the plane of the D ABC.
® ®
AB ´ AC
\ n$ = ® ®
|AB ´ AC |
® ® ® ® ® ®
a ´b + b ´c + c ´a
=
® ® ® ® ® ®
|a ´ b + b ´ c + c ´ a |

®
Q. 17 Show that area of the parallelogram whose diagonals are given by a
® ®
®|a´b|
and b is . Also, find the area of the parallelogram, whose
2
^ ^ ^ ^
diagonals are 2 i - j + k and $i + 3 j - k .
K Thinking Process
® ®
If p and q are adjacent sides of a parallelogram, then the area formed by parallelogram
® ®
= | p ´ q | and then we shall obtained the desired result.

Sol. Let ABCD be a parallelogram such that

D C
q ® ®
a b
A ® B
p

® ® ¾® ® ® ®
AB = p, AD = q Þ BC = q
By triangle law of addition, we get
® ® ® ®
AC = p + q = a [say] ...(i)
® ® ® ®
Similarly, BD = - p + q = b [say] ...(ii)
On adding Eqs. (i) and (ii), we get
® ® ® ® 1 ® ®
a + b = 2q Þ q = (a + b )
2
On subtracting Eq. (ii) from Eq. (i), we get
® ® ® ® 1 ® ®
a - b = 2p Þ p = (a - b )
2
® 1 ® ®
® ® ®
Now, p ´q =
(a - b ) ´ (a + b )
4
1 ® ® ® ® ® ® ® ®
= (a ´ a + a ´ b - b ´ a - b ´ b )
4
1 ® ® ® ®
= [a ´ b + a ´ b ]
4
1 ® ®
= (a ´ b )
2
® ® 1 ® ®
So, area of a parallelogram ABCD = |p ´ q| = |a ´ b|
2
 Now, area of a parallelogram, whose diagonals are 2 $i - $j + k$ and $i + 3$j - k$ .
1
= |(2 $i - $j + k$ ) ´ ($i + 3$j - k$ )|
2
$i $j k$
1
= 2 -1 1
2
1 3 -1
1
= |[$i (1 - 3) - $j(-2 - 1) + k$ (6 + 1)]|
2
1
= |-2 $i + 3$j + 7k$|
2
1
= 4 + 9 + 49
2
1
= 62 sq units
2

® ®
Q. 18 $ then find a vector ®
If a = $i - $j + k$ and b = $j - k,
® ® ®
c such that a ´ c = b
® ®
and a × c =3.
K Thinking Process
We know that, for any two vectors
$i $j k$
® ®
a ´ b = a1 a2 a3
b1 b2 b3
® ® ® ®
and a × b = a 1b1 + a2 b2 + a3 b3, where a = a 1$i + a2 $j + a3k$ and b = b1$i + b2 $j + b3 k$ .
So, we shall use this concept.
®
Sol. Let c = x $i + y $j + zk$
® ®
Also, a = $i + $j + k$ and b = $j - k$
® ® ®
For a ´ c = b,
½ $i $j k$ ½
½ 1 1 1½ = $j - k$
½ ½
½x y z½
Þ $i ( z - y) - $j( z - x ) + k$ ( y - x ) = $j - k$
\ z- y=0 ...(i)
x - z=1 …(ii)
x - y=1 …(iii)
® ®
Also, a× c = 3
( i + j + k ) × (x i + y j + zk$ ) = 3
$ $ $ $ $
Þ x + y+ z=3 …(iv)
On adding Eqs. (ii) and (iii), we get
2x - y - z = 2 …(v)
 On solving Eqs. (iv) and (v), we get
5
x=
3
5 2 2
\ y = - 1 = and z =
3 3 3
® 5 2 2
Now, c = $i + $j + k$
3 3 3
1 $
= (5 i + 2 j + 2k$ )
$
3

Objective Type Questions

Q. 19 The vector in the direction of the vector $i - 2$j + 2 k$ that has
magnitude 9 is
$i - 2$j + 2k$
(a) $i - 2$j + 2 k$ (b)
3
(c) 3 ( $i - 2$j + 2 k$ ) (d) 9 ( $i - 2$j + 2k$ )
®
Sol. (c) Let a = $i - 2 $j + 2 k$
®
® a
Any vector in the direction of a vector a is given by .
®
|a|
$i - 2 $j + 2k$ $i - 2 $j + 2 k$
= =
2
1 +2 +2 2 2 3

®
$i - 2 $j + 2 k$
\ Vector in the direction of a with magnitude 9 = 9 ×
3
= 3($i - 2 $j + 2 k$ )

Q. 20 The position vector of the point which divides the join of points
® ® ® ®
2 a - 3 b and a + b in the ratio 3 : 1, is
® ® ® ® ® ®
3a -2b 7a -8b 3a 5a
(a) (b) (c) (d)
2 4 4 4
® ® ® ®
Sol. (d) Let the position vector of the point R divides the join of points 2 a - 3 b and a + b.
® ® ® ®
3(a + b ) + 1 (2 a - 3 b )
\ Position vector R =
3+1
Since, the position vector of a point R dividing the line segment joining the points P and
® ®
® ® mq + n p
Q, whose position vectors are p and q in the ratio m : n internally, is given by .
m+ n
®
5a
\ R =
4
Q. 21 The vector having initial and terminal points as (2, 5, 0) and (-3, 7,
4), respectively is
(a) - $i + 12$j + 4k$ (b) 5$i + 2$j - 4k$
(c) -5$i + 2$j + 4k$ (d) $i + $j + k$

Sol. (c) Required vector = (-3 - 2 )$i + (7 - 5)$j + (4 - 0)k$

= - 5$i + 2 $j + 4k$
Similarly, we can say that for having initial and terminal points as
(i) (4, 1, 1) and (3, 13, 5), respectively.
(ii) (1, 1, 9) and (6, 3, 5), respectively.
(iii) (1, 2, 3) and (2, 3, 4), respectively, we shall get (a), (b) and (d) as its correct
options.

® ®
Q. 22 The angle between two vectors a and b with magnitudes 3 and 4,
® ®
respectively and a × b = 2 3 is
p p p 5p
(a) (b) (c) (d)
6 3 2 2
® ® ® ®
Sol. (b) Here, |a| = 3,|b| = 4 and a × b = 2 3 [given]
® ® ® ®
We know that, a × b =|a||b|cos q
Þ 2 3= 3 × 4 × cos q
2 3 1
Þ cos q = =
4 3 2
p
\ q=
3

®
Q. 23 Find the value of l such that the vectors a = 2$i + l$j + k$ and
®
b = $i + 2$j + 3k$ are orthogonal.
3 -5
(a) 0 (b) 1 (c) (d)
2 2

K Thinking Process
Two non-zero vectors are orthogonal, if their dot product is zero. So, by using this
concept, we shall get the value of l.
® ® ® ®
Sol. (d) Since, two non-zero vectors a and b are orthogonal i.e., a × b = 0.
\ (2 $i + l$j + k$ ) × ($i + 2 $j + 3k$ ) = 0
Þ 2 + 2l + 3 = 0
-5
\ l=
2
Q. 24 The value of l for which the vectors 3$i - 6$j + k$ and 2$i - 4 $j + lk$ are
parallel, is
2 3 5 2
(a) (b) (c) (d)
3 2 2 5
Sol. (a) Since, two vectors are parallel i.e., angle between them is zero.
\ (3$i - 6$j + k$ ) × (2 $i - 4$j + lk$ ) = |3$i - 6$j + k$| ×|2 $i - 4$j + lk$|
® ® ® ® ® ®
[Q a × b = |a||b|cos 0°Þ a × b =|a||b|]
Þ 6 + 24 + l = 9 + 36 + 1 4 + 16 + l2
Þ 30 + l = 46 20 + l2
Þ 900 + l2 + 60l = 46 (20 + l2 ) [on squaring both sides]
Þ l2 + 60l - 46l2 = 920 - 900
Þ - 45l2 + 60l - 20 = 0
2
Þ - 45l + 30l + 30l - 20 = 0
Þ - 15l(3l - 2 ) + 10 (3l - 2 ) = 0
Þ (10 - 15l)(3l - 2 ) = 0
2 2
\ l= ,
3 3
Alternate Method
® ®
Let a = 3$i - 6$j + k$ and b = 2 $i - 4$j + lk$
® ®
Since, a || b
3 -6 1 2
Þ = = Þl=
2 -4 l 3

®
Q. 25 The vectors from origin to the points A and B are a = 2$i - 3$j + 2k$ and
®
b = 2$i + 3$j + k$ respectively, then the area of DOAB is equal to
(a) 340 (b) 25
1
(c) 229 (d) 229
2
1 ® ®
Sol. (d) \ Area of DOAB = |OA ´ OB|
2
1
= |(2 $i - 3$j + 2 k$ ) ´ (2 $i + 3$j + k$ )|
2
½ $i $j k$ ½
1½
= 2 -3 2½
2½ ½
½2 3 1½
1
= | [$i (-3 - 6) - $j (2 - 4) + k$ (6 + 6)]|
2
1
= |- 9i + 2 $j + 12k$|
2
1 1
\ Area of DOAB = (81 + 4 + 144) = 229
2 2
 ® ® ® ®
Q. 26 For any vector a , the value of ( a ´ $i )2 + ( a ´ $j)2 + ( a ´ k$ )2 is
2 2
® ®
(a) a (b) 3 a
2 2
® ®
(c) 4 a (d) 2 a
®
Sol. (d) Let a = x $i + y$j + zk$
®2
\ a = x 2 + y2 + z2
½$i $j k$ ½
a ´ i = ½x y z½
®
\ $
½ ½
½1 0 0½
= $i [0] - $j[- z] + k$ [- y]
= z$j - yk$
®
\ (a ´ $i )2 = ( z$j - yk$ )( z$j - yk$ )
= y2 + z2
®
Similarly, (a ´ $j )2 = x 2 + z2
®
and (a ´ k$ )2 = x 2 + y2
® ® ®
\ (a ´ $i )2 + (a ´ $j )2 + (a ´ k$ )2 = y2 + z2 + x 2 + z2 + x 2 + y2
®2
= 2(x 2 + y2 + z2 ) = 2 a

® ® ® ® ® ®
Q. 27 If | a | = 10, | b | = 2 and a × b = 12 , then the value of | a ´ b | is
(a) 5 (b) 10 (c) 14 (d) 16
K Thinking Process
® ® ® ® ® ® ® ®
We know that, | a ´ b | = | a || b ||sin q| n$ and a × b = | a || b |cos q. So, we shall use these
® ®
formulae to get the value of | a ´ b | .
® ® ® ®
Sol. (d) Here, |a| = 10,|b| = 2 and a × b = 12 [given]
® ® ® ®
\ a × b =|a||b|cos q
12 = 10 ´ 2 cos q
12 3
Þ cos q = =
20 5
9
Þ sin q = 1 - cos 2 q = 1 -
25
4
sin q = ±
5
® ® ® ®
\ |a ´ b| = |a || b|| sin q|
4
= 10 ´ 2 ´
5
= 16
Q. 28 The vectors l$i + $j + 2 k$ , $i + l$j - k$ and 2$i - $j + l k$ are coplanar, if
(a) l = - 2 (b) l = 0
(c) l = 1 (d) l = -1
® ® ®
Sol. (a) Let a = l $i + $j + 2k$ , b = $i + l$j - k$ and c = 2 $i - $j + lk$
® ® ®
For a , b and c to be coplanar,
½l 1 2½
½ 1 l -1½ = 0
½ ½
½ 2 -1 l ½
Þ l(l2 - 1) - 1(l + 2 ) + 2(-1 - 2 l) = 0
Þ l3 - l - l - 2 - 2 - 4l = 0
Þ l3 - 6l - 4 = 0
Þ (l + 2 )(l2 - 2 l - 2 ) = 0
2± 12
Þ l = - 2 or l =
2
2±2 3
Þ l = - 2 or l = = 1± 3
2

® ® ® ® ® ® r
Q. 29 If a , b and c are unit vectors such that a + b + c = 0, then the value
® ® ® ® ® ®
of a × b + b× c + c × a is
(a) 1 (b) 3
3
(c) - (d) None of these
2
® ® ® ®2 ®2 ®2
Sol. (c) We have, a + b + c = 0 and a = 1, b = 1, c = 1
® ® ® ® ® ®
Q (a + b + c )(a + b + c ) = 0
®2 ®® ® ® ®® ®2 ® ® ®® ®® ®2
Þ a + a × b + a × c + b ×a + b + b × c + c × a + c × b + c = 0
®2 ®2 ®2 ® ® ® ® ® ®
Þ a + b + c + 2 (a × b + b× c + c × a ) = 0
® ® ® ® ® ® ® ® ® ® ® ®
[Q a × b = b× a , b× c = c × b and c × a = a × c ]
® ® ® ® ® ®
Þ 1 + 1 + 1 + 2(a × b + b× c + c × a ) = 0
® ® ® ® ® ® 3
Þ a × b + b× c + c × a = -
2

® ®
Q. 30 The projection vector of a on b is
æ ® ®ö ® ® ® ® æ ® ®ö
ç a× b ÷ ® a× b a× b ç a× b ÷ $
(a) ç ÷b (b) (c) (d) ç b
ç|® ÷ ® ®
ç| ® 2÷÷
è b| ø | b| | a| è a| ø
® æ ® ö
® ® ® b ® ç® b ÷ ®
Sol. (a) Projection vector of a on b is given by = a × b = ça× ÷÷ × b
® ç |®
|b| è b| ø
 ® ® ® ® ® ® ® ®
Q. 31 If a , b and c are three vectors such that a + b + c = 0 and | a | = 2,
® ® ® ® ® ® ® ®
| b | = 3 and | c | = 5, then the value of a × b + b× c + c × a is
(a) 0 (b) 1 (c) -19 (d) 38
® ® ®
® ® ® ® 2 2
Sol. (c) Here, a + b + c = 0 and a = 4, b = 9, c = 25
2

® ® ® ® ® ® ®
\ (a + b + c ) × (a + b + c ) = 0
® ®
® ® ® ® ® ® ®2 ® ® ® ® ® ® ®
Þ a 2 + a × b + a × c + b× a + b + b× c + c × a + c × b + c2 = 0
® ® ®
® ® ® ® ®® ® ® ® ®
Þ a 2 + b2 + c 2 + 2(a × b + b× c + c ×a ) = 0 [Q a × b = b× a ]
® ® ® ® ® ®
Þ 4 + 9 + 25 + 2(a × b + b× c + c × a ) = 0
® ® ® ® ® ® -38
Þ a × b + b× c + c × a = = - 19
2

® ®
Q. 32 If | a | = 4 and -3 £ l £ 2 , then the range of |l a | is
(a) [0, 8] (b) [-12, 8]
(c) [0, 12] (d) [8, 12]
®
Sol. (c) We have, | a | = 4 and -3 £ l £ 2
® ®
\ | la | = |l||a|= l|4|
®
Þ | la | = |-3 |4 = 12, at l = - 3
®
| la | = |0|4 = 0, at l = 0
®
and | la | = |2|4 = 8, at l = 2
®
So, the range of|la| is [0, 12].
Alternate Method
Since, -3 £ l £ 2
0 £|l| £ 3
Þ 0 £ 4| l| £ 12
®
|la|Î [0, 12 ]

Q. 33 The number of vectors of unit length perpendicular to the vectors

® ®
a = 2$i + $j + 2k$ and b = $j + k$ is
(a) one (b) two
(c) three (d) infinite
® ® ®
Sol. (b) The number of vectors of unit length perpendicular to the vectors a and b is c (say)
® ® ®
i.e., c = ± (a ´ b ) .
® ®
So, there will be two vectors of unit length perpendicular to the vectors a and b.
Fillers
® ®
Q. 34 ®
The vector
®
a + b bisects the angle between the non-collinear vectors
a and b, if…… .
® ®
Sol. If vector a + b bisects the angle between the non-collinear vectors, then
® ® ® ® ® ®
a × (a + b ) = |a||× a + b|cos q
® ® ®
a × (a + b ) = a a2 + b 2 cos q
® ® ®
a × (a + b )
Þ cos q = …(i)
a a2 + b 2
® ® ® ® ® ®
and b× (a + b ) = |b| ×|a + b|cos q
® ® ®
b× (a + b ) = b a2 + b 2 cos q [since, q should be same]
® ® ®
b× (a + b )
Þ cos q = …(ii)
b a2 + b 2
From Eqs. (i) and (ii),
® ® ® ® ® ® ® ®
a × (a + b ) b× (a + b ) a b
= Þ =
® ®
a a2 + b 2 b a2 + b 2 |a| |b|
® ®
\ a$ = b$ Þ a and b are equal vectors.

® ® ® ® ® ® ®
Q. 35 If r × a = 0, r × b = 0 and r × c = 0 for some non-zero vector r , then the
® ® ®
value of a × ( b ´ c ) is…… .
® ® ® ®
Sol. Since, r is a non-zero vector. So, we can say that a , b and c are in a same plane.
® ® ®
\ a × (b ´ c ) = 0
® ® ®
[since, angle between a , b and c are zero i.e., q = 0]

® ®
Q. 36 The vectors a = 3$i - 2$j + 2k$ and b = - $i - 2k$ are the adjacent sides of
a parallelogram. The angle between its diagonals is…… .
® ®
Sol. We have, a = 3$i - 2 $j + 2k$ and b = - $i - 2 k$
® ® ® ®
\ a + b = 2 $i - 2 $j and a - b = 4$i - 2 $j + 4k$
® ® ® ®
Now, let q is the acute angle between the diagonals a + b and a - b.
® ® ® ®
(a + b ) × (a - b )
\ cos q =
® ® ® ®
|a + b||a - b|
(2 $i - 2 $j ) × (4$i - 2 $j + 4k$ ) 8 + 4 1
= = =
8 16 + 4 + 16 2 2 ×6 2
p é p 1 ù
\ q= êQ cos =
4 ë 4 2 úû
 ® ® ® 1® ®
Q. 37 The values of k, for which | k a | < a | and k a + a is parallel to a
2
holds true are …… .
® ® ® 1® ®
Sol. We have,|k a| <|a| and ka + a is parallel to a .
2
® ® ® ®
\ |ka| <|a| Þ |k||a| <|a|
Þ |k| < 1 Þ -1 < k < 1
® 1® ® -1 ® 1 ®
Also, since ka + a is parallel to a , then we see that at k = , ka + a becomes a null
2 2 2
®
vector and then it will not be parallel to a .
® 1® ® -1
So, ka + a is parallel to a holds true when k Î] - 1, 1 [ k ¹ .
2 2

® ® ® ®
Q. 38 The value of the expression | a ´ b |2 + ( a × b)2 is …… .
® ® ® ® ® ® ® ®
Sol. |a ´ b|2 + (a × b )2 = |a|2|b|2 sin2 q + (a × b )2
® 2 ® 2 ® ®
= |a| |b| (1 - cos 2 q) + (a × b )2
® ® ® ® ® ®
= |a|2|b|2 - |a|2|b|2 cos 2 q + (a × b )2
® ® ® ® ® ®
= |a|2|b|2 - (a × b )2 + (a × b )2
® ® ® ® ® ®
|a ´ b|2 + (a × b )2 = |a|2|b|2

® ® ® ® ® ®
Q. 39 If | a ´ b |2 + | a × b |2 = 144 and | a | = 4, then | b | is equal to ……
K Thinking Process
® ® ® ® ® ®
We know that, | a ´ b | 2 + | a × a | 2 = | a | 2 | b | 2. So, we shall use this concept here to find
®
the value of | b | .
® ® ® ® ® ®
Sol. Q |a ´ b|2 + |a × b|2 = 144 = |a|2 ×| b|2
® ®
Þ |a| 2|b| 2 = 144
® 144 144
Þ |b| 2 = = =9
® 16
|a| 2
®
\ |b| = 3

® ® ® ®
Q. 40 If a is any non-zero vector, then ( a × $i ) × $i + ( a × $j) × $j + ( a × k$ ) k$ is equal
to ……
®
Sol. Let a = a1 $i + a2 $j + a3 k$
® ® ®
\ a × $i = a1, a × $j = a2 and a × k$ = a3
® ® ® ®
\ (a × $i ) $i + (a × $j )$j + (a × k$ )k$ = a1 $i + a2 $j + a3 k$ = a
True/False
® ® ® ®
Q. 41 If | a | = | b |, then necessarily it implies a = ± b.
Sol. True
® ® ® ®
If |a| = |b| Þ a = ± b
So, it is a true statement.

®
Q. 42 Position vector of a point P is a vector whose initial point is origin.
Sol. True
® ¾® ®
Since, P = OP = displacement of vector P from origin

® ® ® ® ® ®
Q. 43 If | a + b | = | a - b |, then the vectors a and b are orthogonal

Sol. True
® ® ® ®
Since, |a + b| = |a - b|
® ® ® ®
Þ |a + b|2 = |a - b|2
® ® ® ®
Þ 2|a||b| = - 2|a||b|
® ®
Þ 4|a||b| = 0
® ®
Þ |a || b | = 0
® ® ® ® ® ®
Hence, a and b are orthogonal. [Q a × b = |a|×|b|cos 90° = 0]

2 2
® ®2 ® ® ® ®
Q. 44 The formula ( a + b) = a + b + 2 a ´ b is valid for ®
®
b ®
a+ b
® ®
non-zero vectors a and b. ®
a
Sol. False
® ®2 ® ® ® ® ®
(a + b) = (a + b) × (a + b)
®

–b
a–
®
b

®2 ®2 ® ®
= a + b + 2a× b

® ®
Q. 45 If a and b are adjacent sides of a rhombus, then
® ®
a × b = 0.
Sol. False
® ® ® ® ® ®
If a × b = 0, then a × b = |a||b|cos 90°
® ®
Hence, angle between a and b is 90°, which is not possible in a rhombus.
Since, angle between adjacent sides in a rhombus is not equal to 90°.
 11
Three Dimensional
Geometry
Short Answer Type Questions
®
Q. 1 Find the position vector of a point A in space such that OA is inclined
®
at 60° to OX and at 45° to OY and|OA| = 10 units.
® ®
Sol. Since, OA is inclined at 60° to OX and at 45° to OY. Let OA makes angle a with OZ.
\ cos 2 60° + cos 2 45° + cos 2 a = 1
2 2
æ 1ö æ 1 ö
Þ ç ÷ + ç ÷ + cos 2 a = 1 [Q l2 + m2 + n2 = 1]
è2 ø è 2ø
1 1
Þ + + cos 2 a = 1
4 2
æ 1 1ö
Þ cos 2 a = 1 - ç + ÷
è2 4ø
æ 6ö
Þ cos 2 a = 1 - ç ÷
è 8ø
1
Þ cos 2 a =
4
1
Þ cos a = = cos 60°
2
\ a = 60°
® ® æ1 1 $ 1$ö
\ OA = |OA| ç $i + j + k÷
è2 2 2 ø
æ1 1 $ 1$ö ®
= 10 ç $i + j + k÷ [Q |OA| = 10]
è2 2 2 ø
= 5 $i + 5 2 $j + 5 k$
Q. 2 Find the vector equation of the line which is parallel to the vector
3i - 2$j + 6k$ and which passes through the point (1, - 2, 3).
$

K Thinking Process
® ® ® ®
Here, we use the formula r = b + l a , where r is the equation of the line which passes
® ®
through b and parallel to a .
® ®
Sol. Let a = 3$i - 2 $j + 6 k$ and b = $i - 2 $j + 3 k$
®
So, vector equation of the line, which is parallel to the vector a = 3$i - 2 $j + 6 k$ and passes
® ® ® ®
through the vector b = $i - 2 $j + 3 k$ is r = b + la .
®
\ r = ($i - 2 $j + 3 k$ ) + l(3$i - 2 $j + 6 k$ )
Þ (x i + y j + z k ) - ( i - 2 j + 3 k ) = l (3$i - 2 $j + 6 k$ )
$ $ $ $ $ $
Þ (x - 1)$i + ( y + 2 )$j + ( z - 3) k$ = l(3$i - 2 $j + 6 k)$

x -1 y -2 z -3 x - 4 y -1
Q. 3 Show that the lines = = and = = z intersect.
2 3 4 5 2
Also, find their point of intersection.
K Thinking Process
If shortest distance between the lines is zero, then they intersect.
Sol. We have, x1 = 1, y1 = 2, z1 = 3 and a1 = 2, b1 = 3, c1 = 4
Also, x 2 = 4, y2 = 1, z2 = 0 and a2 = 5, b2 = 2, c 2 = 1
If two lines intersect, then shortest distance between them should be zero.
\ Shortest distance between two given lines
x 2 - x1 y2 - y1 z2 - z1
a1 b1 c1
a2 b2 c2
=
(b1c 2 - b2c 1)2 + (c1a 2 - c 2 a1 )2 + (a1b 2 - a 2 b1 )2

4 - 1 1- 2 0 - 3
2 3 4
5 2 1
=
(3 × 1 - 2 × 4)2 + (4 × 5 - 1× 2 )2 + (2 × 2 - 5 × 3)2

3 -1 -3
2 3 4
5 2 1
=
25 + 324 + 121
3(3 - 8) + 1(2 - 20) - 3(4 - 15)
=
470
-15 - 18 + 33 0
= = =0
470 470
Therefore, the given two lines are intersecting.
 For finding their point of intersection for first line,
x -1 y-2 z- 3
= = =l
2 3 4
Þ x = 2 l + 1, y = 3l + 2 and z = 4l + 3
Since, the lines are intersecting. So, let us put these values in the equation of another line.
2 l + 1 - 4 3l + 2 - 1 4l + 3
Thus, = =
5 2 1
2 l - 3 3l + 1 4l + 3
Þ = =
5 2 1
2 l - 3 4l + 3
Þ =
5 1
Þ 2 l - 3 = 20l + 15
Þ 18l = - 18 = - 1
So, the required point of intersection is
x = 2 (-1) + 1 = - 1
y = 3 (-1) + 2 = - 1
z = 4 (-1) + 3 = - 1
Thus, the lines intersect at (-1, - 1, - 1.)

Q. 4 Find the angle between the lines

® ®
r = 3$i - 2$j + 6 k$ + l (2$i + $j + 2 k$ ) and r = (2 $j - 5 k$ ) + m(6 $i + 3$j + 2 k$ ).
K Thinking Process
® ®
| b 1× b 2| ® ®
We know that, cos q = ® ®
, where, q is the angle between the lines a 1 + l b 1
|b 1 | × | b 2|
® ®
and a2 + m b2.
®
Sol. We have, r = 3$i - 2 $j + 6 k$ + l(2 $i + $j + 2k$ )
®
and r = (2 $j - 5 k$ ) + m (6 $i + 3$j + 2k$ )
® ®
where, a 1 = 3 $i - 2 $j + 6 k$, b1 = 2 $i + $j + 2 k$
® ®
and a 2 = 2 $j - 5 k$, b2 = 6 $i + 3$j + 2 k$
If q is angle between the lines, then
® ®
|b1× b2|
cos q =
® ®
|b1| × |b2|
|(2 $i + $j + 2 k$ ) × (6 $i + 3$j + 2 k$ )|
=
|2 $i + $j + 2 k$||6 $i + 3$j + 2 k$|
|12 + 3 + 4| 19
= =
9 49 21
-1 19
\ q = cos
21
Q. 5 Prove that the line through A (0, - 1, - 1) and B (4, 5, 1) intersects the
line through C (3, 9, 4) and D (-4, 4, 4).
Sol. We know that,the cartesian equation of a line that passes through two points (x1, y1, z1 ) and
(x 2 , y2 , z2 ) is
x - x1 y - y1 z - z1
= =
x 2 - x1 y2 - y1 z2 - z1
Hence, the cartesian equation of line passes through A(0, - 1, - 1) and B(4, 5, 1) is
x -0 y+1 z+1
= =
4 - 0 5 + 1 1+ 1
x y+1 z+1
Þ = = ...(i)
4 6 2
and cartesian equation of the line passes through C(3, 9, 4) and D(-4, 4, 4) is
x-3 y-9 z-4
= =
-4 - 3 4 - 9 4 - 4
x -3 y-9 z-4
Þ = = ...(ii)
-7 -5 0
If the lines intersect, then shortest distance between both of them should be zero.
\ Shortest distance between the lines
x 2 - x1 y2 - y1 z2 - z1
a1 b1 c1
a2 b2 c2
=
(b1c 2 - b 2c 1 )2 + (c 1a 2 - c 2 a1 )2 + (a1b 2 - a 2 b1 )2
3-0 9+1 4+1
4 6 2
-7 -5 0
=
(6 × 0 + 10)2 + (-14 - 0)2 + (-20 + 42 )2
3 10 5
4 6 2
-7 -5 0
=
100 + 196 + 484
3(0 + 10) - 10 (14) + 5 (-20 + 42 )
=
780
30 - 140 + 110
= =0
780
So, the given lines intersect.

Q. 6 Prove that the lines x = py + q, z = ry + s and x = p¢ y + q¢, z = r ¢ y + s¢

are perpendicular, if pp¢ + rr ¢ + 1 = 0.
x -q
Sol. We have, x = py + q Þ y = ..(i)
P
z-s
and z = ry + s Þ y = ...(ii)
r
x -q y z-s
Þ = = [using Eqs. (i) and (ii)] ...(iii)
p 1 r
x - q ¢ y z - s¢
Similarly, = = ...(iv)
p¢ 1 r¢
 From Eqs. (iii) and (iv),
a1 = p, b1 = 1, c1 = r
and a2 = p¢, b2 = 1, c 2 = r ¢
If these given lines are perpendicular to each other, then
a1 a2 + b1b 2 + c1 c 2 = 0
Þ pp¢ + 1 + rr ¢ = 0
which is the required condition.

Q. 7 Find the equation of a plane which bisects perpendicularly the line

joining the points A(2, 3, 4) and B (4, 5, 8) at right angles.
Sol. Since, the equation of a plane is bisecting perpendicular the line joining the points A(2, 3, 4)
and B(4, 5, 8) at right angles.
æ2 + 4 3 + 5 4 + 8ö
So, mid-point of AB is ç , , ÷ i.e., (3, 4, 6) .
è 2 2 2 ø
®
Also, N =(4 - 2)$i +(5 - 3)$j +(8 - 4) k$ = 2 $i + 2 $j + 4k$
® ® ®
So, the required equation of the plane is ( r - a ) × N = 0.
®
Þ [(x - 3)$i + ( y - 4)$j + ( z - 6) k$ ]× (2 $i + 2 $j + 4 k$ ) = 0 [Q a = 3$i + 4$j + 6 k$ ]
Þ 2 x - 6 + 2 y - 8 + 4 z - 24 = 0
Þ 2 x + 2 y + 4 z = 38
\ x + y + 2 z = 19

Q. 8 Find the equation of a plane which is at a distance 3 3 units from

origin and the normal to which is equally inclined to coordinate axis.
Sol. Since, normal to the plane is equally inclined to the coordinate axis.
1
Therefore, cos a = cos b = cos g =
3
® 1 $ 1 $ 1 $
So, the normal is N = i + j+ k and plane is at a distance of 3 3 units from origin.
3 3 3
é ® ù
®
The equation of plane is r × N
$ =3 3 êQ N $ = Nú
ê |N|ú
ë û
®
[since, vector equation of the plane at a distance p from the origin is r × N = p]
$
æ 1 $ 1 $ 1 $ö
ç i + j+ k÷
Þ (x $i + y $j + z k$ ) × è 3 3 3 ø
= 3 3
1
x y z
Þ + + = 3 3
3 3 3
\ x + y+ z= 3 3× 3 = 9
So, the required equation of plane is x + y + z = 9.
Q. 9 If the line drawn from the point (-2, - 1, - 3) meets a plane at right
angle at the point (1, - 3, 3), then find the equation of the plane.
Sol. Since, the line drawn from the point (-2, - 1, - 3) meets a plane at right angle at the point
(1, - 3, 3). So, the plane passes through the point (1, - 3, 3) and normal to plane is
(-3$i + 2 $j - 6 k$ ).
®
Þ a = $i - 3$j + 3 k$
®
and N = - 3 $i + 2 $j - 6 k$
® ® ®
So, the equation of required plane is ( r - a ) × N = 0
Þ [(x $i + y$j + zk$ ) - ($i - 3$j + 3 k$ )]× (-3 $i + 2 $j - 6 k$ ) = 0
Þ [(x - 1) $i + ( y + 3) $j + ( z - 3 ) k$ ]× (-3 $i + 2 $j - 6 k)
$ =0
Þ -3x + 3 + 2 y + 6 - 6 z + 18 = 0
Þ - 3x + 2 y - 6 z = - 27
\ 3x - 2 y + 6 z - 27 = 0

Q. 10 Find the equation of the plane through the points (2, 1, 0), (3, - 2, - 2)
and (3, 1, 7).
K Thinking Process
Here, apply the equation of the plane passing through the points (x1, y1, z 1) , (x2 , y2 , z2)
x - x1 y - y1 z - z 1
and (x3 , y3 , z3) is given by x2 - x1 y2 - y1 z2 - z 1 = 0.
x3 - x1 y3 - y1 z3 - z 1

Sol. We know that, the equation of a plane passing through three non-collinear points (x1, y1, z1 ),
(x 2 , y2 , z2 ) and (x 3 , y3 , z3 ) is
x - x1 y - y1 z - z1
x 2 - x1 y2 - y1 z2 - z1 = 0
x 3 - x1 y3 - y1 z3 - z1
x-2 y-1 z-0
Þ 3 - 2 -2 - 1 -2 - 0 = 0
3- 2 1 - 1 7 - 0
x-2 y-1 z
Þ 1 -3 -2 = 0
1 0 7
Þ (x - 2 )(-21 + 0) - ( y - 1)(7 + 2 ) + z(3) = 0
Þ - 21x + 42 - 9 y + 9 + 3 z = 0
Þ - 21x - 9 y + 3 z = - 51
\ 7 x + 3 y - z = 17
So, the required equation of plane is 7 x + 3 y - z = 17.
Q. 11 Find the equations of the two lines through the origin which intersect
x -3 y -3 z p
the line = = at angles of each.
2 1 1 3
x -3 y-3 z
Sol. Given equation of the line is = = =l …(i)
2 1 1
Y¢ Q

(0, 0, 0)
O
X¢ X

Y (2l + 3, l + 3, l)
2 1 1
So, DR’s of the line are 2, 1, 1 and DC’s of the given line are , , .
6 6 16
p
Also, the required lines make angle with the given line.
3
From Eq. (i), x = (2 l + 3), y = (l + 3) and z = l
a1a2 + b1b 2 + c1c 2
Q cos q =
a1 + b12 + c12 a22 + b22 + c 22
2

p (4l + 6) + (l + 3) + (l)
\ cos =
3 6 (2 l + 3)2 + (l + 3)2 + l2
1 6l + 9
Þ =
2 6 (4l2 + 9 + 12 l + l2 + 9 + 6l + l2 )
6 6l + 9
Þ =
2 2
6l + 18l + 18
2
Þ 6 (l + 3l + 3) = 2 (6l + 9)
Þ 36 (l2 + 3l + 3) = 36(4l2 + 9 + 12 l)
Þ l2 + 3l + 3 = 4l2 + 9 + 12 l
Þ 3l2 + 9l + 6 = 0
Þ l2 + 3l + 2 = 0
Þ l(l + 2 ) + 1(l + 2 ) = 0
Þ (l + 1)(l + 2 ) = 0
\ l = - 1, -2
So, the DC’s are 1, 2, - 1 and -1, 1, - 2.
Also, both the required lines passes through origin.
x y z x y z
So, the equations of required lines are = = and = = .
1 2 -1 -1 1 -2
Q. 12 Find the angle between the lines whose direction cosines are given by
the equation l + m + n = 0 and l 2 + m2 - n2 = 0.
Sol. Eliminating n from both the equations, we have
l2 + m2 - (l - m)2 = 0
Þ l + m2 - l2 - m2 + 2 ml = 0
2
Þ 2 lm = 0
Þ lm = 0 Þ (- m - n) m = 0 [Q l = - m - n]
Þ (m + n) m = 0
Þ m= -n Þ m= 0
Þ l = 0, l = - n
Thus, Dr’s two lines are proportional to 0, -n, n and -n, 0, n i.e., 0, - 1, 1 and -1, 0, 1.
® ®
So, the vector parallel to these given lines are a = - $j + k$ and b = - $i + k$
®®
ab 1 1 1
Now, cos q = ® ®
= × Þ cos q =
|a||b| 2 2 2
p é p 1ù
\ q= êëQ cos 3 = 2 úû
3

Q. 13 If a variable line in two adjacent positions has direction cosines l, m, n

and l + dl, m + dm, n + dn, then show that the small angle dq between
the two positions is given by dq 2 = dl 2 + dm2 + dn2 .
Sol. We have l, m, n and l + dl, m + d m, n + d n as direction cosines of a variable line in two
different positions.
\ l2 + m2 + n2 = 1 ... (i)
and (l + d l)2 + (m + d m)2 + (n + d n)2 = 1 ...(ii)
Þ l + m + n + d l + d m2 + d n2 + 2 (l d l + m d m + n d n) = 1
2 2 2 2

Þ dl2 + dm2 + dn2 = - 2 (l d l + m d m + n d n) [Q l 2 + m 2 + n 2 = 1]

-1 2
Þ l d l + m d m + nd n = (d l + d m2 + d n2 ) ...(iii)
2
® ®
Now, a and b are unit vectors along a line with direction cosines l, m, n and
(l + dl), (m + dm), (n + dn), respectively.
® ®
\ a = l $i + m $j + nk$ and b = (l + d l) $i + (m + d m) $j + (n + d n) k$
® ®
a ×b ® ®
Þ cos dq = ® ®
= a ×b [Q |a$| = |b$| = 1]
|a||b|
Þ cos dq = l (l + dl) + m (m + dm) + n (n + dn)
= ( l 2 + m2 + n2 ) + ( l d l + m d m + n d n)
1
= 1 - (d l2 + d m2 + d n2 ) [using Eq. (iii)]
2
2 2 2
Þ 2(1 - cos d q) = ( d l + d m + d n )
dq é 2 qù
Þ 2 × 2 sin2 = d l2 + d m2 + d n2 êëQ 1 - cos q = 2 sin 2 úû
2
2
æ dq ö é dq dq dq ù
Þ 4 ç ÷ = d l2 + d m2 + d n2 êësince, 2 is small, then sin 2 = 2 úû
è2 ø
\ d q2 = d l2 + d m2 + d n2
Q. 14 If O is the origin and A is (a, b, c), then find the direction cosines of
the line OA and the equation of plane through A at right angle to OA.
a b c
Sol. Since, DC’s of line OA are
2 2 2
,
2 2 2
and .
a + b +c a + b +c 2
a + b2 + c 2
® ® ®
Also, n = OA = a = a $i + b$j + ck$
The equation of plane passes through (a, b, c) and perpendicular to OA is given by
® ® ®
[ r - a ]× n = 0
® ® ® ®
Þ r ×n =a ×n
Þ [(x $i + y$j + zk$ ) × (a $i + b$j + ck$ )] = (a $i + b$j + ck$ ) . (a $i + b$j + ck$ )
Þ ax + by + cz = a2 + b 2 + c 2

Q. 15 Two systems of rectangular axis have the same origin. If a plane cuts
them at distances a, b, c and a¢ , b¢ , c ¢ , respectively from the origin,
1 1 1 1 1 1
then prove that 2 + 2 + 2 = 2 + 2 + 2 .
a b c a¢ b¢ c¢
Sol. Consider OX, OY , OZ and ox, oy, oz are two system of rectangular axes.
Let their corresponding equation of plane be
x y z
+ + =1 ...(i)
a b c
x y z
and + + =1 ...(ii)
a¢ b ¢ c ¢
Also, the length of perpendicular from origin to Eqs. (i) and (ii) must be same.
0 0 0 0 0 0
+ + -1 + + -1
\ a b c = a ¢ b ¢ c ¢
1 1 1 1 1 1
2
+ 2 + 2 2
+ 2 + 2
a b c a¢ b¢ c¢
1 1 1 1 1 1
Þ + + = + 2 + 2
a¢2 b ¢2 c ¢2 a2 b c
1 1 1 1 1 1
Þ + 2 + 2 = 2 + 2 + 2
a2 b c a¢ b¢ c¢

Long Answer Type Questions

Q. 16 Find the foot of perpendicular from the point (2, 3, -8) to the line
4 - x y 1- z
= = . Also, find the perpendicular distance from the
2 6 3
given point to the line.
4-x y 1- z
Sol. We have, equation of line as = =
2 6 3
x - 4 y z- 1
Þ = = =l
-2 6 -3
Þ x = - 2 l + 4, y = 6 l and z = - 3 l + 1
 Let the coordinates of L be (4 - 2 l , 6l, 1 - 3l) and direction ratios of PL are proportional to
(4 - 2 l - 2, 6l - 3, 1 - 3l + 8) i.e., (2 - 2 l, 6l - 3, 9 - 3l) .
Also, direction ratios are proportional to -2, 6, - 3. Since, PL is perpendicular to give line.
\ -2 (2 - 2 l) + 6 (6l - 3) - 3 (9 - 3l) = 0
Þ -4 + 4l + 36l - 18 - 27 + 9l = 0
Þ 49l = 49 Þ l = 1
So, the coordinates of L are (4 - 2 l, 6l, 1 - 3l) i.e., (2, 6, - 2 ).
P (2, 3, – 8)

L
4–x= y =1–z
2 6 3

Also, length of PL = (2 - 2 )2 + (6 - 3)2 + (-2 + 8)2

= 0 + 9 + 36 = 3 5 units

Q. 17 Find the distance of a point (2, 4, - 1) from the line

x +5 y +3 z -6
= = .
1 4 -9
x + 5 y+ 3 z-6
Sol. We have, equation of the line as = = =l
1 4 -9
Þ x = l - 5, y = 4l - 3, z = 6 - 9l
Let the coordinates of L be (l - 5, 4l - 3, 6 - 9l), then Dr’s of PL are (l - 7, 4l - 7, 7 - 9l).
Also, the direction ratios of given line are proportional to 1, 4, -9.
Since, PL is perpendicular to the given line.
\ (l - 7 ) × 1 + (4l - 7 ) × 4 + (7 - 9l) × (-9) = 0
Þ l - 7 + 16l - 28 + 81l - 63 = 0
Þ 98l = 98 Þ l = 1
So, the coordinates of L are (-4, 1, - 3).
\ Required distance, PL = (-4 - 2 )2 + (1 - 4)2 + (-3 + 1)2
= 36 + 9 + 4 = 7 units

Q. 18 Find the length and the foot of perpendicular from the point æç 1, 3 , 2 ö÷
è 2 ø
to the plane 2x - 2 y + 4 z + 5 = 0.
Sol. Equation of the given plane is 2 x - 2 y + 4 z + 5 = 0 … (i)
®
Þ n = 2 $i - 2 $j + 4 k$
æ 3 ö ®
So, the equation of line through ç 1, , 2 ÷ and parallel to n is given by
è 2 ø
x -1 y - 3/2 z - 2
= = =l
2 -2 4
3
Þ x = 2 l + 1, y = - 2 l + and z = 4l + 2
2
 If this point lies on the given plane, then
æ 3 ö
2 (2 l + 1) - 2 ç -2 l + ÷ + 4 (4l + 2 ) + 5 = 0 [using Eq. (i)]
è 2 ø
Þ 4l + 2 + 4l - 3 + 16l + 8 + 5 = 0
-1
Þ 24l = -12 Þ l =
2
\ Required foot of perpendicular
é æ -1 ö æ -1 ö 3 æ -1 ö ù æ 5 ö
= ê2 ´ ç ÷ + 1, - 2 ´ ç ÷ + , 4 ´ ç ÷ + 2 ú i.e., ç 0, , 0 ÷
ë è2 ø è2 ø 2 è2 ø û è 2 ø
2
æ 3 5ö
\ Required length of perpendicular = (1 - 0)2 + ç - ÷ + (2 - 0)2
è2 2 ø
= 1+ 1+ 4 = 6 units

Q. 19 Find the equation of the line passing through the point (3, 0, 1) and
parallel to the planes x + 2 y = 0 and 3 y - z = 0.
Sol. Equation of the two planes are x + 2 y = 0 and 3 y - z = 0.
® ®
Let n1 and n 2 are the normals to the two planes, respectively.
® ®
\ n1 = $i + 2 $j and n 2 = 3$j - k$
Since, required line is parallel to the given two planes.
$i $j k$
® ® ®
Therefore, b = n1 ´ n 2 = 1 2 0
0 3 -1
= $i (-2 ) - $j (-1) + k$ (3)
= - 2 $i + $j + 3 k$
So, the equation of the lines through the point (3, 0, 1) and parallel to the given two planes are
(x - 3) $i + ( y - 0)$j + ( z - 1) k$ + l (-2 $i + $j + 3 k$ )
Þ (x - 3) $i + y $j + ( z - 1) k$ + l (-2 $i + $j + 3 k$ )

Q. 20 Find the equation of the plane through the points (2, 1, – 1), (–1, 3, 4)
and perpendicular to the plane x– 2 y + 4 z = 10.
Sol. The equation of the plane passing through (2, 1, – 1) is
a (x – 2 ) + b ( y – 1) + c ( z + 1) = 0 ...(i)
Sicne, this passes through (–1, 3, 4).
\ a (–1 – 2 ) + b (3 – 1) + c (4 + 1) = 0
Þ – 3 a + 2 b + 5c = 0 ... (ii)
Since, the plane (i) is perpendicular to the plane x – 2 y + 4 z = 10.
\ 1× a – 2 × b + 4 × c = 0
Þ a – 2 b + 4c = 0 ...(iii)
On solving Eqs. (ii) and (iii), we get
a –b c
= = =l
8 + 10 –17 4
Þ a = 18 l, b = 17 l, c = 4l
 From Eq. (i),
18 l (x - 2 ) + 17 l ( y – 1) + 4l ( z + 1) = 0
Þ 18x – 36 + 17 y – 17 + 4 z + 4 = 0
Þ 18x + 17 y + 4 z – 49 = 0
\ 18x + 17 y + 4 z = 49

Q. 21 Find the shortest distance between the lines gives by

®
r = (8 + 3l)$i - (9 + 16l)$j + (10 + 7l)k$
®
and r = 15$i + 29$j + 5k$ + m (3$i + 8 $j - 5k$ ).
®
Sol. We have, r = (8 + 3l) $i - (9 + 16l)$j + (10 + 7 l) k$ )
= 8$i - 9$j + 10 k$ + 3l $i – 16l$j + 7 lk$
= 8$i – 9$j + 10 k$ + l(3$i – 16$j + 7k$ )
® ®
Þ a 1 = 8$i - 9$j + 10 k$ and b1 = 3$i - 16 $j + 7k$ ...(i)
®
Also r = 15$i + 29 $j + 5 k$ + m (3$i + 8$j - 5k$ )
® ®
Þ a 2 = 15 $i + 29 $j + 5 k$ and b2 = 3 $i + 8 $j - 5 k$ ... (ii)
® ® ® ®
( b1 ´ b2 ) × (a 2 – a 1 )
Now, shortest distance betwen two lines is given by ® ®
|b1 ´ b2|
$i $j k$
® ®
\ b1 ´ b2 = 3 –16 7
3 8 –5
= i (80 – 56) – $j (– 15 – 21) + k$ (24 + 48)
$
= 24$i + 36$j + 72 k$
® ®
Now, |b1 ´ b2| = (24)2 + (36)2 + (72 )2
= 12 2 2 + 32 + 62 = 84
® ®
and (a 2 – a 1 ) = (15 – 8)$i + (29 + 9)$j + (5 – 10) k$
= 7 $i + 38$j - 5 k$
( 24$i + 36$j + 72 k$ ) × (7 $i + 38$j – 5 k$ )
\ Shortest distance =
84
168 + 1368 – 360 1176
= = = 14 units
84 84

Q. 22 Find the equation of the plane which is perpendicular to the plane

5x + 3 y + 6z + 8 = 0 and which contains the line of intersection of the
planes x + 2 y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
Sol. The equation of a plane through the line of intersection of the planes x + 2 y + 3 z – 4 = 0
and 2 x + y – z + 5 = 0 is
(x + 2 y + 3 z – 4) + l ( 2 x + y – z + 5) = 0
Þ x (1 + 2 l) + y (2 + l) + z (– l + 3) – 4 + 5l = 0 ...(i)
 Also, this is perpendicular to the plane 5x + 3 y + 6 z + 8 = 0.
\ 5(1 + 2 l) + 3 (2 + l) + 6 (3 – l) = 0 [Q a1 a2 + b1b2 + c1c 2 = 0 ]
Þ 5 + 10l + 6 + 3l + 18 - 6l = 0
\ l = – 29 / 7
From Eq. (i),
é æ –29 ö ù æ 29 ö æ 29 ö æ –29 ö
x ê1 + 2 ç ÷ + y ç2 – ÷+ zç + 3÷ – 4 + 5 ç ÷=0
ë è 7 ø úû è 7 ø è7 ø è 7 ø
Þ x (7 – 58) + y (14 – 29) + z (29 + 21) – 28 – 145 = 0
Þ –51x – 15 y + 50 z – 173 = 0
So, the required equation of plane is 51x + 15 y – 50 z + 173 = 0 .

Q. 23 If the plane ax + by = 0 is rotated about its line of intersection with

the plane z = 0 through an angle a, then prove that the equation of
the plane in its new position is ax + by ± ( a 2 + b 2 tan a ) z = 0.
Sol. Equation of the plane is ax + by = 0 ... (i)
\Equation of the plane after new position is
ax cos a by cos a
+ ± z sin a = 0
2 2
a + b b 2 + a2
ax by
Þ + ± z tan a = 0 [on dividing by cos a]
2 2
a + b b + a2
2

Þ ax + by ± z tana a 2 + b2 = 0 [on multiplying with a2 + b 2 ]

Alternate Method
Given,planes are ax + by = 0 …(i)
and z=0 …(ii)
Therefore, the equation of any plane passing through the line of intersection of planes
(i) and (ii) may be taken as ax + by + k = 0. …(iii)
a b
Then, direction cosines of a normal to the plane (iii) are , ,
a2 + b 2 + k 2 a2 + b 2 + k 2
c a b
and direction cosines of the normal to the plane (i) are , ,
2 2 2 2 2
a +b +k a +b a + b2
2

0.
Since, the angle between the planes (i) and (ii) is a,
a× a + b× b + k × 0
\ cos a =
a + b 2 + k 2 a2 + b 2
2

a2 + b 2
=
a + b2 + k 2
2

Þ k cos a = a (1 - cos 2 a ) + b 2 (1 - cos 2 a )

2 2 2

(a2 + b 2 ) sin2 a
Þ k2 =
cos 2 a
k=± a2 + b 2 tan a
On putting this value in plane (iii), we get the equation of the plane as
ax + by + z a2 + b 2 tan a = 0
Q. 24 Find the equation of the plane through the intersection of the planes
® ®
r × ($i + 3$j) – 6 = 0 and r × (3$i – $j – 4 k$ ) = 0, whose perpendicular
distance from origin is unity.
® ®
Sol. We have, n1 = ($i + 3$j ), d1 = 6 and n 2 = (3$i – $j – 4 k$ ), d 2 = 0
® ® ®
Using the relation, r × (n1 + ln 2 ) = d1 + d 2 l
®
Þ r × [($i + 3$j ) + l (3$i – $j – 4k$ )] = 6 + 0 × l
®
Þ r ×[(1 + 3l)$i + (3 – l)$j + k$ (– 4l)] = 6 ...(i)

On dviding both sides by (1 + 3 l)2 + (3 – l)2 + (– 4l)2 , we get

®
r × [(1 + 3l)$i + (3 – l)$j + k$ (– 4 l)] 6
=
(1 + 3l)2 + (3 – l)2 + (– 4 l)2 (1 + 3l)2 + (3 – l)2 + (– 4l)2
Since, the perpendicular distance from origin is unity.
6
\ =1
(1 + 3l) + (3 – l)2 + (– 4l)2
2

Þ (1 + 3l)2 + (3 – l)2 + (– 4l)2 = 36

Þ 1 + 9l2 + 6l + 9 + l2 – 6l + 16l2 = 36
Þ 26l2 + 10 = 36
Þ l2 = 1
\ l= ± 1
Using Eq. (i), the required equation of plane is
®
r × [(1 ± 3)$i + (3 m 1) $j + (m 4) k$ ] = 6
®
Þ r × [(1 + 3)$i + (3 – 1)$j + (– 4) k$ ] = 6
®
and r × [(1 – 3)$i + (3 + 1) $j + 4 k$ ] = 6
®
Þ r × (4$i + 2 $j – 4k$ ) = 6
®
and r . (– 2 $i + 4$j + 4 k$ ) = 6
Þ 4x + 2 y – 4 z – 6 = 0
and – 2x + 4y + 4z – 6 = 0

Q. 25 Show that the points ($i – $j + 3 k$ ) and 3 ($i + $j + k$ ) are equidistant from
®
the plane r × (5$i + 2$j – 7k$ ) + 9 = 0 and lies on opposite side of it.
Sol. To show that these given points ($i – $j + 3 k$ ) and 3($i + $j + k$ ) are equidistant from the plane
®
r × (5$i + 2 $j – 7k$ ) + 9 = 0, we first find out the mid- point of the points which is 2 $i + $j + 3 k$ .
®
On substituting r by the mid-point in plane, we get
LHS = (2 $i + $j + 3 k$ ) × (5$i + 2 $j - 7k$ ) + 9
= 10 + 2 – 21 + 9 = 0
= RHS
Hence, the two points lie on opposite sides of the plane are equidistant from the plane.
 ® ®
Q. 26 AB = 3 $i - $j + k$ and CD = - 3$i + 2$j + 4 k$ are two vectors. The position
vectors of the points A and C are 6$i + 7$j + 4 k$ and - 9$i + 2k, $
respectively. Find the position vector of a point P on the line AB and a
® ® ®
point Q on the line CD such that PQ is perpendicular to AB and CD
both.
® ®
Sol. We have, AB = 3 $i - $j + k$ and CD = - 3 $i + 2 $j + 4 k$
®
Also, the position vectors of A and C are 6$i + 7 $j + 4k$ and - 9$j + 2k$ , respectively. Since, PQ
® ®
is perpendicular to both AB and CD.
So, P and Q will be foot of perpendicular to both the lines through A and C.
®
Now, equation of the line through A and parallel to the vector AB is,
®
r = (6$i + 7 $j + 4 k$ ) + l (3$i - $j + k$ )
®
and the line through C and parallel to the vector CD is given by
®
r = – 9$j + 2 k$ + m (– 3 $i + 2 $j + 4 k$ ) .. .(i)
®
Let r = (6$i + 7 $j + 4 k$ ) + l (3$i - $j + k$ )
®
and r = – 9$j + 2 k$ + m (– 3$i + 2 $j + 4 k$ ) ...(ii)
Let P (6 + 3l, 7 - l, 4 + l) is any point on the first line and Q be any point on second line is
given by (– 3 m , - 9 + 2m , 2 + 4m ).
®
\ PQ = (– 3m – 6 – 3 l) $i + (-9 + 2 m - 7 + l)$j + (2 + 4 m - 4 - l) k$
= (– 3m – 6 – 3l) $i + (2 m + l – 16)$j + (4 m - l – 2 ) k$
®
If PQ is perpendicular to the first line, then
3 (– 3m – 6 – 3l) – (2 m + l - 16) + (4 m – l - 2 ) = 0
Þ – 9 m – 18 – 9 l –- 2m - l + 16 + 4m - l - 2 = 0
Þ - 7 m - 11l - 4 = 0 ...(iii)
®
If PQ is perpendicular to the second line, then
– 3 (– 3 m – 6 – 3l) + 2 (2m + l - 16) + 4 (4m – l - 2 ) = 0
Þ 9m + 18 + 9l + 4m + 2 l – 32 +16 m - 4l – 8 = 0
Þ 29 m + 7 l - 22 = 0 ...(iv)
On solving Eqs. (iii) and (iv), we get
- 49 m - 77 l - 28 = 0
Þ 319 m + 77 l - 242 = 0
Þ 270 m – 270 = 0
Þ m =1
Using m in Eq. (iii), we get
– 7 (1) – 11l – 4 = 0
Þ – 7 – 11l – 4 = 0
Þ – 11 – 11l = 0
Þ l= –1
®
\ PQ = [– 3 (1) – 6 – 3 (– 1)] $i + [2 (1) + (- 1) - 16] $j + [4(1) - (- 1) - 2 ] k$
= – 6$i - 15$j + 3 k$
Q. 27 Show that the straight lines whose direction cosines are given by
2l + 2m - n = 0 and mn + nl + lm = 0 are at right angles.
Sol. We have, 2l + 2m - n = 0 ...(i)
and mn + nl + lm = 0 ...(ii)
Eliminating m from the both equations, we get
n - 2l
m= [from Eq. (i)]
2
æ n - 2l ö æ n - 2l ö
Þ ç ÷ n + nl + l ç ÷=0
è 2 ø è 2 ø
2 2
n - 2 nl + 2 nl + nl - 2 l
Þ =0
2
2 2
Þ n + nl - 2 l = 0
Þ n2 + 2 nl - nl - 2 l2 = 0
Þ ( n + 2 l) ( n - l) = 0
Þ n = - 2l and n = l
-2 l - 2 l l - 2l
\ m= ,m =
2 2
-l
Þ m = - 2 l, m =
2
-l
Thus, the direction ratios of two lines are proportional to l, - 2 l, - 2 and l, , l.
2
-1
Þ 1, - 2, - 2 and 1, ,1
2
Þ 1, - 2, - 2 and 2, - 1, 2
® ®
Also, the vectors parallel to these lines are a = $i - 2 $j - 2 k$ and b = 2 $i - $j + 2 k$ ,
respectively.
® ®
a. b ($i - 2 $j - 2 k$ ) × (2 $i - $j + 2 k$ )
\ cos q = ® ® =
|a ||b| 3× 3
2 +2-4
= =0
9
p é p ù
\ q= êëQ cos 2 = 0úû
2

Q. 28 If l1 , m1 , n1 , l2 , m2 , n2 and l3 , m3 , n3 are the direction cosines of three

mutually perpendicular lines, then prove that the line whose direction
cosines are proportional to l1 + l2 + l3 , m1 + m2 + m3 and n 1 + n2 + n 3
makes equal angles with them.
®
Sol. Let a = l1$i + m1 $j + n1 k$
®
b = l2 $i + m2 $j + n 2 k$
®
c = l3 $i + m3 $j + n 3 k$
®
d = (l1 + l2 + l3 ) $i + (m1 + m2 + m2 ) $j + (n1 + n2 + n3 ) k$
® ® ® ® ® ®
Also, let a , b and g are the angles between a and d, b and d, c and d.
\ cos a = l1 (l1 + l2 + l3 ) + m1 (m1 + m2 + m3 ) + n1 (n1 + n2 + n3 )
= l12 + l1 l2 + l1 l3 + m12 + m1 m2 + m1 m3 + n12 + n1 n2 + n1 n3
 = (l12 + m12 + n12 ) + (l1 l2 + l1 l3 + m1 m2 + m1 m3 + n1 n2 + n1 n3 )
= 1+ 0 = 1
[Q l12 + m12 + n12 = 1 and l1 ^ l2 , l1 ^ l3 , m1 ^ m2 , m1 ^ m3 , n1 ^ n2 , n1 ^ n3 ]
Similarly, cosb = l2 (l1 + l2 + l3 ) + m2 (m1 + m2 + m3 ) + n2 (n1 + n2 + n3 )
= 1 + 0 and cos g = 1 + 0
Þ cos a = cos b = cos g
Þ a =b = g
So, the line whose direction cosines are proportional to l1 + l2 + l3, m1 + m2 + m3 ,
n1 + n2 + n3 makes equal angles with the three mutually perpendicular lines whose direction
cosines are l1, m1, n1, l2 , m2 , n2 and l3 , m3 , n3 respectively.

Objective Type Questions

Q. 29 Distance of the point (a , b, g) from Y-axis is
(a) b (b)|b|
(c)|b| + | g| (d) a 2 + g 2

Sol. (d) Required distance = (a - 0)2 + (b - b )2 + (g - 0)2 = a2 + g 2

Q. 30 If the direction cosines of a line are k, k and k, then

(a) > 0 (b) 0 < < 1
1 1
(c) = 1 (d) = or -
3 3
Sol. (d) Since, direction cosines of a line are k, k and k.
\ l = k, m = k and n = k
We know that, l2 + m2 + n2 = 1
Þ k2 + k2 + k2 = 1
1
Þ k2 =
3
1
\ k=±
3

® 2
plane r æç $i + $j - k$ ö÷ = 1 from the origin is
3 6
Q. 31 The distance of the
è 7 7 7 ø
(a) 1 (b) 7
1
(c) (d) None of these
7
® æ2 3 6 ö
Sol. (a) The distance of the plane r ç $i + $j - k$ ÷ = 1 from the origin is 1.
è7 7 7 ø
® ®
[since, r × n = d is the form of above equation, where d represents the distance of
plane from the origin i.e., d = 1]
 x -2 y -3 z -4
Q. 32 The sine of the angle between the straight line = =
3 4 5
and the plane 2x - 2 y + z = 5 is
10 4 2 3 2
(a) (b) (c) (d)
6 5 5 2 5 10
Sol. (d) We have, the equation of line as
x -2 y- 3 z- 4
= =
3 4 5
Now, the line passes through point (2, 3, 4) and having direction ratios (3, 4, 5).
Since, the line passes through point (2, 3, 4) and parallel to the vector (3$i + 4$j + 5 k$ ).
®
\ b = 3$i + 4$j + 5 k$
Also, the cartesian form of the given plane is 2 x - 2 y + z = 5.
Þ (x $i + y$j + zk$ ) (2 $i - 2 $j + k$ ) = 5
®
\ n = (2 $i - 2 $j + k$ )
® ®
|b . n| (3$i + 4$j + 5 k$ ) . (2 $i - 2 $j + k$ )
We know that, sin q = ® ®
=
|b| .|n| 32 + 42 + 52 × 4+ 4+1
|6 - 8 + 5| 3 1
= = =
50 × 3 15 2 5 2
2
sin q =
10

Q. 33 The reflection of the point (a, b, g) in the XY -plane is

(a) ( a , b , 0) (b) (0 , 0, g) (c) ( -a , - b , g) (d) ( a , b , - g)
Sol. (d) In XY-plane, the reflection of the point (a, b, g ) is (a, b, - g ).

Q. 34 The area of the quadrilateral ABCD where A (0,4,1),

B (2, 3, - 1), C (4, 5, 0), and D (2, 6, 2) is equal to
(a) 9 sq units (b) 18 sq units
(c) 27 sq units (d) 81 sq units
®
Sol. (a) We have, AB =(2 - 0)$i + (3 - 4)$j + (- 1 - 1) k$ = 2 $i - $j - 2 k$
®
BC = (4 - 2 )$i + (5 - 3)$j + (0 + 1) k$ = 2 $i + 2 $j + k$
®
CD = (2 - 4)$i + (6 - 5)$j + (2 - 0) k$ = - 2$i + $j + 2 k$
®
DA = (0 - 2 )$i + (4 - 6)$j + (1 - 2 ) k$ = - 2 $i - 2 $j - k$
$i $j k$
® ®
\ Area of quadrilateral ABCD = |AB ´ BC| = 2 - 1 - 2
2 2 1

= $i (- 1 + 4) - $j (2 + 4) + k$ (4 + 2 )

= 3$i - 6$j + 6 k$
= 9 + 36 + 36 = 9 sq units
Q. 35 The locus represented by xy + yz = 0 is
(a) a pair of perpendicular lines
(b) a pair of parallel lines
(c) a pair of parallel planes
(d) a pair of perpendicular planes
Sol. (d) We have, xy + yz = 0
Þ xy = - yz
So, a pair of perpendicular planes.

Q. 36 If the plane 2x - 3 y + 6z - 11 = 0 makes an angle sin -1 a with X-axis,

then the value of a is
3 2
(a) (b)
2 3
2 3
(c) (d)
7 7
Sol. (c) Since, 2 x - 3 y + 6 z - 11 = 0 makes an angle sin-1 a with X-axis.
® ®
b = (1$i + 0$j + 0 k$ ) and n = 2 $i - 3$j + 6 k$
® ®
|b × n|
We know that, sin q = ® ®
|b| ×|n|
|(1$i ) × (2 $i - 3$j + 6 k$ )| 2
= =
1 4 + 9 + 36 7

Fillers
Q. 37 If a plane passes through the points (2, 0,0) (0, 3, 0) and (0, 0, 4) the
equation of plane is ......... .
Sol. We know that, equation of a the plane that cut the coordinate axes at (a, 0, 0) (0, b, 0) and
x y z
(0, 0, c) is + + = 1.
a b c
Hence, the equation of plane passes through the points (2, 0, 0), (0, 3, 0) and (0, 0, 4) is
x y z
+ + = 1.
2 3 4

Q. 38 The direction cosines of the vector (2$i + 2$j - k$ ) are ......... .

2 2 -1 2 2 -1
Sol. Direction cosines of (2 $i + 2 $j - k$ ) are , , i.e., , , .
4+ 4+1 4 + 4+ 1 4+ 4+1 3 3 3
 x -5 y +4 z -6
Q. 39 The vector equation of the line = = is ......... .
3 7 2
® ®
Sol. We have, a = 5$i - 4$j + 6 k$ and b = 3$i + 7 $j + 2 k$
So, the vector equation will be
®
r = (5$i - 4$j + 6 k$ ) + l (3$i + 7 $j + 2 k$
Þ (x $i + y$j + z k$ ) - (5$i - 4$j + 6k$ ) = l (3$i + 7 $j + 2 k$ )
Þ (x - 5)$i + ( y + 4)$j + ( z - 6) k$ = l(3$i + 7 $j + 2 k$ )

Q. 40 The vector equation of the line through the points (3, 4, - 7) and (1, -1,
6) is ......... .
Sol. We know that, vector equation of a line passes through two points is represented by
® ® ® ®
r = a + l (b - a )
® ®
Here, r = x $i + y$j + 3k$ , a = 3$i + 4$j - 7k$
®
and b = $i - $j + 6k$
® ®
Þ ( b - a ) = - 2 $i - 5$j + 13k$
So, the required equation is
x $i + y$j + zk$ = 3$i + 4$j - 7k$ + l (-2 $i - 5$j + 13 k$ )
Þ (x - 3) $i + ( y - 4) $j + ( z + 7 ) k$ = l (-2 $i - 5$j + 13 k$ )

®
Q. 41 The cartesian equation of the plane r × ($i + $j - k$ ) = 2 is ......... .
®
Sol. We have, r × ($i + $j - k$ ) = 2

Þ (x $i + y$j + z k$ ) ×($i + $j - k$ ) = 2
Þ x + y- z=2
which is the required form

True/False
Q. 42 The unit vector normal to the plane x + 2 y + 3z - 6 = 0 is
1 $ 2 $ 3 $
i+ j+ k.
14 14 14
Sol. True
®
We have, n = $i + 2 $j + 3 k$
$i + 2 $j + 3 k$ $i 2 $j 3 k$
\ n$ = = + +
2
1 +2 + 32 2 14 14 14
Q. 43 The intercepts made by the plane 2x - 3 y + 5z + 4 = 0 on the
4 4
coordinate axis are - 2, and - .
3 5
Sol. True
We have, 2x - 3y + 5z + 4 = 0
Þ 2x - 3y + 5z = - 4
2x 3y 5z
Þ - + =1
-4 -4 -4
x y z
Þ + - =1
-2 4 4
3 5
x y z
Þ + + =1
-2 4 æ 4ö
ç- ÷
3 è 5ø
4 4
So, the intercepts are -2, and - .
3 5

®
Q. 44 The angle between the line r = (5$i - $j - 4 k$ ) + l (2$i - $j + k$ ) and the
®
æ 5 ö
plane r (3$i - 4 $j - k$ ) + 5 = 0 is sin -1 ç ÷.
è 2 91 ø
Sol. False
® ®
We have, b = 2 $i - $j + k$ and n = 3$i - 4$j - k$
Let q is the angle between line and plane.
® ®
|b × n| |(2 $i - $j + k$ ) × (3$i - 4$j - k$ )|
Then, sin q = ® ®
=
| b| ×|n| 6 × 26
|6 + 4 - 1| 9
= =
156 2 39
9
\ q = sin-1
2 39

® ®
Q. 45 The angle between the planes r (2$i - 3$j + k$ ) = 1 and r ($i - $j) = 4 is
æ -5 ö
cos -1 ç ÷.
è 58 ø
Sol. False
® ®
|n1 × n 2|
We know that, the angle between two planes is given by cos q = ® ®
|n1||n 2|
® ®
Here, n1 = (2 $i - 3$j + k$ ) and n 2 = ($i - $j )
|(2 $i - 3$j + k$ ) ($i - $j )|
\ cos q =
4 + 9 + 1 1+ 1
|2 + 3| 5
Þ cos q = =
14 × 2 2 7
æ 5 ö
\ q = cos -1 ç ÷
è2 7 ø
 ®
Q. 46 The line r = 2$i - 3$j - k$ + l ($i - $j + 2 k$ ) lies in the plane
®
r (3$i + $j - k$ ) + 2 = 0.
Sol. False
®
We have, r = 2 $i - 3$j - k$ + l ($i - $j + 2 k$ )
Þ (x $i + y$j + zk$ ) = $i (2 + l) + $j ( - 3 - l) + k$ (- 1 + 2 l)
Since, x = (2 + l), y = (- 3 - l) and z = (- 1 + 2l) are coordinates of general point which
should satisfy the equation of the given plane.
\ [(2 + l) $i + (- 3 - l) $j + (2 l - 1) k$ ] × [$i + $j - k$ ] = 2
Þ (2 + l) - 3 - l - 2 l + 1 = 2
Þ - 2l = 2
Þ l= -1
®
\ r = (2 - 1) $i + (- 3 + 1) $j + (- 2 - 1) k$
= $i - 2 $j - 3 k$
Again, from the equation of the plane
®
r × (3$i + $j - k$ ) + 2 = 0
Þ ($i - 2 $j - 3 k$ ) (3$i + $j - k$ ) + 2 = 0
Þ (3 - 2 + 3) + 2 = 0
Þ 6¹0
which is not true.
®
So, the line r = 2 $i - 3$j - k$ + l ($i - $j + 2 k$ ) does not lie in a plane.

x -5 y +4 z -6
Q. 47 The vector equation of the line = = is
3 7 2
®
r = 5$i - 4 $j + 6 k$ + l (3$i + 7$j + 2 k$ )
Sol. True
We have, x = 5, y = - 4, z = 6
and a = 3, b = 7, c = 2
®
\ r = (5$i - 4$j + 6 k$ ) + l (3$i + 7 $j + 2 k$ )

Q. 48 The equation of a line, which is parallel to 2 $i + $j + 3 k$ and which

x -5 y +2 z -4
passes through the point (5, - 2, 4) is = = .
2 -1 3
Sol. False
Here, x1 = 5, y1 = - 2, z1 = 4
and a = 2, b = 1, c = 3
x -5 y+2 z-4
Þ = =
2 1 3
Q. 49 If the foot of perpendicular drawn from the origin to a plane is
®
(5, - 3, - 2), then the equation of plane is r (5$i - 3$j - 2 k$ ) = 38.
Sol. True
®
Since, the required plane passes through the point P (5, - 3, - 2 ) and is perpendicular to OP.
®
\ a = 5 $i - 3$j - 2 k$
® ®
and n = OP = 5 $i - 3 $j - 2 k$
Now, the equation of the plane is
® ® ®
(r - a )× n = 0
® ® ® ®
Þ r ×n =a ×n
®
Þ r × (5 $i - 3$j - 2 k$ ) = (5 $i - 3 $j - 2 k$ ) × (5$i - 3$j - 2k$ )
®
Þ r . (5 $i - 3 $j - 2 k$ ) = 25 + 9 + 4
®
Þ r × (5 $i - 3 $j - 2 k$ ) = 38
 12
Linear Programming
Short Answer Type Questions
Q. 1 Determine the maximum value of Z = 11x + 7 y subject to the
constraints2x + y £ 6, x £ 2 , x ³ 0, y ³ 0.
K Thinking Process
Using constraints, get the corner points for the bounded region and then for each corner
point check the corresponding value of Z.
Sol. We have, maximise Z = 11x + 7 y ...(i)
Subject to the constraints
2x + y £ 6 …(ii)
x £2 …(iii)
x ³ 0, y ³ 0 …(iv)
We see that, the feasible region as shaded determined by the system of constraints (ii) to
(iv) is OABC and is bounded. So, now we shall use corner point method to determine the
maximum value of Z.
Y

(0, 6) C

B (2, 2)

(3, 0)
O
X¢ X
(0, 0) A (2, 0)
x=2
Y¢
2x + y = 6
 Corner points Corresponding value of Z
(0, 0) 0
(2, 0) 22
(2, 2) 36
(0, 6) 42 ¬ Maximum
Hence, the maximum value of Z is 42 at (0, 6).

Q. 2 Maximise Z = 3x + 4 y, subject to the constraints x + y £ 1, x ³ 0, y ³ 0.

Sol. Maximise Z = 3x + 4 y, Subject to the constraints
x + y £ 1, x ³ 0, y ³ 0.
The shaded region shown in the figure as OAB is bounded and the coordinates of corner
points O, A and B are (0, 0), (1, 0) and (0, 1), respectively.
Y

(0, 1) B
O A
X¢ X
(0, 0) (1, 0)
(x
+
y=
1)

Y¢

Corner points Corresponding value of Z

(0, 0) 0
(1, 0) 3
(0, 1) 4 ¬ Maximum

Hence, the maximum value of Z is 4 at (0, 1).

Q. 3 Maximise the function Z = 11x + 7 y, subject to the constraints x £ 3,

y £ 2, x ³ 0 and y ³ 0.
Sol. Maximise Z = 11x + 7 y, subject to the constraints x £ 3, y £ 2, x ³ 0, y ³ 0.
Y

C B
(0, 2) (3, 2) y = 2
A
X¢ X
O (0, 0) (3, 0)

x=3
Y¢
The shaded region as shown in the figure as OABC is bounded and the coordinates of
corner points are (0, 0), (3, 0), (3, 2) and (0, 2), respectively.
 Corner points Corresponding value of Z
(0, 0) 0
(3, 0) 33
(3, 2) 47 ¬ Maximum
(0, 2) 14

Hence, Z is maximum at (3, 2) and its maximum value is 47.

Q. 4 Minimise Z = 13x - 15 y subject to the constraints x + y £ 7,

2x - 3 y + 6 ³ 0, x ³ 0 and y ³ 0.
Sol. Minimise Z = 13x - 15 y subject to the constraints x + y £ 7, 2 x - 3 y + 6 ³ 0, x ³ 0, y ³ 0.
Y

(0, 7)
B
(3, 4)
C
(0, 2)

A
X¢ X
(–3, 0) (0, 0) (0, 7)
2x – 3y+6=0 x+y=7
Y¢

Shaded region shown as OABC is bounded and coordinates of its corner points are (0, 0),
(7, 0), (3, 4) and (0, 2), respectively.

Corner points Corresponding value of Z

(0, 0) 0
(7, 0) 91
(3, 4) -21
(0, 2) -30 ¬ Minimum

Hence, the minimum value of Z is (- 30) at (0, 2).

Q. 5 Determine the maximum value of Z = 3x + 4 y, if the feasible region
(shaded) for a LPP is shown in following figure.
Y

x+
2 y=
D 76
E
X
O A C
2x + y=104

Sol. As clear from the graph, corner points are O, A, E and D with coordinates (0, 0), (52, 0),
(144, 16) and (0, 38), respectively. Also, given region is bounded.
Here, Z = 3x + 4 y
Q 2 x + y = 104 and 2 x + 4 y = 152
Þ -3 y = - 48
Þ y = 16and x = 44

Corner points Corresponding value of Z

(0, 0) 0
(52, 0) 156
(44, 16) 196 ¬ Maximum
(0, 38) 152
Hence, Z is at (44, 16) is maximum and its maximum value is 196.

Q. 6 Feasible region (shaded) for a LPP is shown in following figure.

Maximise Z = 5x + 7 y.

B (3, 4)

C (0, 2)

O (0, 0) A (7, 0)

Sol. The shaded region is bounded and has coordinates of corner points as (0, 0), (7, 0), (3, 4)
and (0, 2). Also, Z = 5x + 7 y.
Corner points Corresponding value of Z
(0, 0) 0
(7, 0) 35
(3, 4) 43 ¬ Maximum
(0, 2) 14
Hence, the maximum value of Z is 43 at (3, 4).
Q. 7 The feasible region for a LPP is shown in following figure. Find the
minimum value of Z = 11x + 7 y.
Y

(0, 5)

(2, 3)
(3, 2)

x+ X
O
x+
3y =

y=
9

Sol. 5
From the figure, it is clear that feasible region is bounded with coordinates of corner points
as (0, 3), (3, 2) and (0, 5). Here, Z = 11x + 7 y.
Q x + 3 y = 9 and x + y = 5
Þ 2y = 4
\ y = 2 and x = ,3
So, intersection points of x + y = 5 and x + 3 y = 9 is (3, 2).

Corner points Corresponding value of Z

(0, 3) 21 ¬ Minimum
(3, 2) 47
(0, 5) 35
Hence, the minimum value of Z is 21 at (0, 3).

Q. 8 Refer to question 7 above. Find the maximum value of Z.

Sol. From question 7, above, it is clear that Z is maximum at (3, 2 ) and its maximum value is 47.

Q. 9 The feasible region for a LPP is shown in the following figure. Evaluate
Z = 4x + y at each of the corner points of this region. Find the
minimum value of Z, if it exists.
Y

x+2y=4

X
O
x+
y=
3

Sol. From the shaded region, it is clear that feasible region is unbounded with the corner points
A (4, 0), B (2, 1) and C (0, 3).
 Also, we have Z = 4 x + y.
[since, x + 2 y = 4 and x + y = 3Þ y = 1and x = 2]
Y

, 3)
C (0

B (2, 1)
X¢ X
O 3 ,0 A (4,
4 0)
x+ x+2
y= y=4
3
y¢
4 x+
y=3

Corner points Corresponding value of Z

(4, 0) 16
(2, 1) 9
(0, 3) 3 ¬ Minimum

Now, we see that 3 is the smallest value of Z at the corner point (0, 3). Note that here we
see that, the region is unbounded, therefore 3 may or may not be the minimum value of Z.
To decide this issue, we graph the inequality 4x + y < 3 and check whether the resulting
open half plan has no point in common with feasible region otherwise, Z has no minimum
value.
From the shown graph above, it is clear that there is no point in common with feasible
region and hence Z has minimum value 3 at (0, 3).

Q. 10 In following figure, the feasible region (shaded) for a LPP is shown.

Determine the maximum and minimum value of Z = x + 2 y.

3 , 15
Q
2 4
3 , 24
P 3
13 13 7, 4
R 2

18, 2
S
7 7

Sol. From the shaded bounded region, it is clear that the coordinates of corner points are
æ 3 24 ö æ 18 2 ö æ 7 3 ö æ 3 15 ö
ç , ÷, ç , ÷, ç , ÷ and ç , ÷.
è 13 13 ø è 7 7 ø è 2 4 ø è2 4 ø
 Also, we have to determine maximum and minimum value of Z = x + 2 y.

Corner points Corresponding value of Z

æ 3 24 ö 3
+
48 51
= =3
12
ç , ÷
è 13 13 ø 13 13 13 13
æ 18 2 ö 18 4 22
+ =
1
= 3 Minimum
ç , ÷
è 7 7ø 7 7 7 7
æ7 3 ö 7 6 20
+ = =5
ç , ÷
è2 4 ø 2 4 4
æ 3 15 ö 3 30 36
+ = = 9 Maximum
ç , ÷
è2 4 ø 2 4 4
1
Hence, the maximum and minimum values of Z are 9 and 3 , respectively.
7

Q. 11 A manufacturer of electronic circuits has a stock of 200 resistors, 120

transistors and 150 capacitors and is required to produce two types of
circuits A and B. Type A requires 20 resistors, 10 transistors and 10
capacitors. Type B requires 10 resistors, 20 transistors and 30
capacitors. If the profit on type A circuit is ` 50 and that on type B
circuit is ` 60, formulate this problem as a LPP, so that the
manufacturer can maximise his profit.
K Thinking Process
For maximising the profit, use resistor constraint, transistor constraint, capacitor
constraint and non-negative constraint.
Sol. Let the manufacturer produces x units of type A circuits and y units of type B circuits.
Form the given information, we have following corresponding constraint table.
Type A ( x) Type B (y) Maximum
stock
Resistors 20 10 200
Transistors 10 20 120
Capacitors 10 30 150
Profit ` 50 ` 60
Thus, we see that total profit Z = 50x + 60 y (in `).
Now, we have the following mathematical model for the given problem.
Maximise Z = 50x + 60 y ...(i)
Subject to the constraints.
20x + 10 y £ 200 [resistors constraint]
Þ 2 x + y £ 20 ...(ii)
and 10x + 20 y £ 120 [transistor constraint]
Þ x + 2 y £ 12 ...(iii)
and 10x + 30 y £ 150 [capacitor constraint]
Þ x + 3 y £ 15 ...(iv)
and x ³ 0, y ³ 0 [non-negative constraint] ...(v)
So, maximise Z = 50x + 60 y, subject to 2 x + y £ 20, x + 2 y £ 12, x + 3 y £ 15, x ³ 0, y ³ 0.
Q. 12 A firm has to transport 1200 packages using large vans which can carry
200 packages each and small vans which can take 80 packages each.
The cost for engaging each large van is ` 400 and each small van is `
200. Not more than ` 3000 is to be spent on the job and the number of
large vans cannot exceed the number of small vans. Formulate this
problem as a LPP given that the objective is to minimise cost.
Sol. Let the firm has x number of large vans and y number of small vans. From the given
information, we have following corresponding constraint table.

Large vans ( x) Small vans ( y ) Maximum /

Minimum
Packages 200 80 1200
Cost 400 200 3000
Thus, we see that objective function for minimum cost is Z = 400 x + 200 y.
Subject to constraints
200x + 80 y ³ 1200 [package constraint]
Þ 5x + 2 y ³ 30 ...(i)
and 400x + 200 y £ 3000 [cost constraint]
Þ 2 x + y £ 15 ...(ii)
and x£y [van constraint] ...(iii)
and x ³ 0, y ³ 0 [non-negative constraints] ...(iv)
Thus, required LPP to minimise cost is minimise Z = 400x + 200 y, subject to 5x + 2 y ³ 30.
2 x + y £ 15
x£y
x ³ 0, y ³ 0

Q. 13 A company manufactures two types of screws A and B. All the screws

have to pass through a threading machine and a slotting machine. A
box of type A screws requires 2 min on the threading machine and 3
min on the slotting machine. A box of type B screws requires 8 min on
the threading machine and 2 min on the slotting machine. In a week,
each machine is available for 60 h. On selling these screws, the
company gets a profit of ` 100 per box on type A screws and ` 170 per
box on type B screws.
Formulate this problem as a LPP given that the objective is to
maximise profit.
Sol. Let the company manufactures x boxes of type A screws and y boxes of type B screws.
From the given information, we have following corresponding constraint table

Type A ( x) Type B ( y ) Maximum time available

on each machine in a
week
Time required for screws on 2 8 60 ´ 60 (min)
threading machine
Time required for screws on 3 2 60 ´ 60 (min)
slotting machine
Profit ` 100 ` 170
 Thus, we see that objective function for maximum profit is Z = 100x + 170 y.
Subject to constraints
2 x + 8 y £ 60 ´ 60 [time constraint for threading machine]
Þ x + 4 y £ 1800 ...(i)
and 3x + 2 y £ 60 ´ 60 [time constraint for slotting machine]
Þ 3x + 2 y £ 3600 ... (ii)
Also, x ³ 0, y ³ 0 [non-negative constraints] ...(iii)
\ Required LPP is,
Maximise Z = 100x + 170 y
Subject to constraints x + 4 y £ 1800, 3x + 2 y £ 3600, x ³ 0, y ³ 0.

Q. 14 A company manufactures two types of sweaters type A and type B. It

costs ` 360 to make a type A sweater and ` 120 to make a type B
sweater. The company can make atmost 300 sweaters and spend
atmost ` 72000 a day. The number of sweaters of type B cannot exceed
the number of sweaters of type A by more than 100. The company
makes a profit of ` 200 for each sweater of type A and ` 120 for every
sweater of type B.
Formulate this problem as a LPP to maximise the profit to the
company.
Sol. Let the company manufactures x number of type A sweaters and y number of type B
sweaters.
From the given information we see that cost to make a type A sweater is `360 and cost to
make a type B sweater is ` 120.
Also, the company spend atmost ` 72000 a day.
\ 360x + 120 y £ 72000
Þ 3x + y £ 600 …(i)
Also, company can make atmost 300 sweaters.
\ x + y £ 300 …(ii)
Further, the number of sweaters of type B cannot exceed the number of sweaters of type A
by more than 100 i.e.,
x + 100 ³ y
Þ x - y ³ - 100 …(iii)
Also, we have non-negative constraints for x and y i.e., x ³ 0, y ³ 0 …(iv)
Hence, the company makes a profit of `200 for each sweater of type A and `120 for each
sweater of type B i.e.,
Profit (Z ) = 200x + 120 y
Thus, the required LPP to maximise the profit is
Maximise Z = 200x + 120 y is subject to constraints.
3x + y £ 600
x + y £ 300
x - y ³ - 100
x ³ 0, y ³ 0
Q. 15 A man rides his motorcycle at the speed of 50 km/h. He has to spend
` 2 per km on petrol. If he rides it at a faster speed of 80 km/h, the
petrol cost increases to ` 3 per km. He has atmost ` 120 to spend on
petrol and one hour’s time. He wishes to find the maximum distance that
he can travel. Express this problem as a linear programming problem.
Sol. Let the man rides to his motorcycle to a distance x km at the speed of 50 km/h and to a
distance y km at the speed of 80 km/h.
Therefore, cost on petrol is 2 x + 3 y.
Since, he has to spend ` 120 atmost on petrol.
\ 2 x + 3 y £ 120 ...(i)
Also, he has atmost one hour’s time.
x y
\ + £1
50 80
Þ 8x + 5 y £ 400 ...(ii)
Also, we have x ³ 0, y ³ 0 [non-negative constraints]
Thus, required LPP to travel maximum distance by him is
Maximise Z = x + y, subject to 2 x + 3 y £ 120, 8x + 5 y £ 400, x ³ 0, y ³ 0

Q. 16 Refer to question 11. How many of circuits of type A and of type B,

should be produced by the manufacturer, so as to maximise his profit?
Determine the maximum profit.
K Thinking Process
Using the constraints draw the graph to get the corner points and find the maximum
value of possible corner points (as asked).
Sol. Referring to solution 11, we have
Maximise Z = 50x + 60 y, subject to
2 x + y £ 20, x + 2 y £ 12, x + 3 y £ 15, x ³ 0, y ³ 0
From the shaded region it is clear that the feasible region determined by the system of
constraints is OABCD and is bounded and the coordinates of corner points are (0, 0),
æ 28 4 ö
(10, 0), ç , ÷, (6, 3) and (0, 5), respectively.
è 3 3ø
28 4
[since, x + 2 y = 12 and 2 x + y = 20 Þ x = , y = and x + 3 y = 15
3 3
and x + 2 y = 12 Þ y = 3 and x = 6]

(0, 20)

(0, 6)
D
C (6, 3)
(0, 5)
28, 4
B
3 3
(15, 0)
O (0, 0) A (10, 0) x+3y=15
(12, 0)
x+2y=12
2x+y=20
 Corner points Corresponding value of Z = 50 x + 60y
(0, 0) 0
(10, 0) 500
æ 28 4 ö 1400 240 1640
+ = = 546.66 ¬ Maximum
ç , ÷
è 3 3ø 3 3 3

(6 , 3 ) 480
(0 , 5 ) 300

Since, the manufacturer is required to produce two types of circuits A and B and it is clear
that parts of resistor, transistor and capacitor cannot be in fraction, so the required
maximum profit is 480 where circuits of type A is 6 and circuits of type B is 3.

Q. 17 Refer to question 12. What will be the minimum cost?

Sol. Referring to solution 12, we have minimise Z = 400x + 200 y, subject to 5x + 2 y ³ 30,
2 x + y £ 15, x £ y, x ³ 0, y ³ 0.
On solving x - y = 0 and 5x + 2 y = 30, we get
30 30
y= ,x =
7 7

(0, 15)
x)
=
5)

(y
,
(5

30, 30
7 7 5x 2x
+ +
2y y=
= 15
30

On solving x - y = 0 and 2 x + y = 15, we get x = 5, y = 5

So, from the shaded feasible region it is clear that coordinates of corner points are (0, 15),
æ 30 30 ö
(5, 5) and ç , ÷.
è7 7 ø

Corner points Corresponding value of Z = 400 x + 200y

(0, 15) 3000
(5, 5) 3000
æ 30 30 ö 400 ´
30 30 18000
+ 200 ´ =
ç , ÷
è7 7 ø 7 7 7
= 2571. 43 ¬ Minimum
Hence, the minimum cost is ` 2571.43.
Q. 18 Refer to question 13. Solve the linear programming problem and
determine the maximum profit to the manufacturer.
Sol. Referring to solution 13, we have
Maximise Z =100 x + 170 y subject to
3x + 2 y £ 3600, x + 4 y £ 1800, x ³ 0, y ³ 0
From the shaded feasible region it is clear that the coordinates of corner points are (0, 0),
(1200, 0), (1080, 180) and (0, 450).
On solving x + 4 y = 1800 and 3x + 2 y = 3600, we get x = 1080 and y = 180

(0, 1800)

(0, 450)
(1080, 180)

x+4y=1800
O (0, 0) (1200, 0) (1800, 0)
3x+2y=3600

Corner points Corresponding value of Z = 100 x + 170y

(0, 0) 0
(1200, 0) 1200 ´ 100 = 120000
(1080, 180) 100 ´ 1080 + 170 ´ 180 = 138600 ¬ Maximum
(0, 450) 0 + 170 ´ 450 = 76500
Hence, the maximum profit to the manufacturer is 138600.

Q. 19 Refer to question 14. How many sweaters of each type should the company
make in a day to get a maximum profit? What is the maximum profit?
Sol. Referring to solution 14, we have maximise Z =200x + 120 y
subject to x + y £ 300, 3x + y £ 600, x - y ³ - 100, x ³ 0, y ³ 0.
On solving x + y = 300 and 3x + y = 600, we get
x = 150, y = 150
On solving x - y = - 100 and x + y = 300, we get
x = 100, y = 200

(0, 600)

(0, 300)

(100, 200)
(150, 150)
(0, 100)

(–100, 0)
(0, 0) (200, 0) (300, 0)
x – y=–100 x+y=300
3x+y=600
 From the shaded feasible region it is clear that coordinates of corner points are (0, 0),
(200, 0), (150, 150), (100, 200) and (0, 100).

Corner points Corresponding value of Z = 200 x + 120y

(0, 0) 0
(200, 0) 40000
(150, 150) 150 ´ 200 + 120 ´ 150 = 48000 ¬ Maximum
(100, 200) 100 ´ 200 + 120 ´ 200 = 44000
(0, 100) 120 ´ 100 = 12000
Hence, 150 sweaters of each type made by company and maximum profit = ` 48000.

Q. 20 Refer to question 15. Determine the maximum distance that the man
can travel.
Sol. Referring to solution 15, we have

(0, 80)

(0, 40)
300, 80
7 7
(60, 0)
(0, 0) (50, 0)
2x+3y=120
8x+5y=400
Maximise Z = x + y, subject to
2 x + 3 y £ 120, 8x + 5 y £ 400, x ³ 0,y ³ 0
On solving, we get
8x + 5 y = 400 and 2 x + 3 y = 120, we get
300 80
x= , y=
7 7
From the shaded feasible region, it is clear that coordinates of corner points are (0, 0),
æ 300 80 ö
(50, 0), ç , ÷ and (0, 40).
è 7 7 ø

Corner points Corresponding value of Z = x + y

(0, 0) 0
(50, 0) 50
300 80 380 2
, = 54 km ¬ Maximum
7 7 7 7
(0, 40) 40
2
Hence, the maximum distance that the man can travel is 54 km.
7
Q. 21 Maximise Z = x + y subject to x + 4 y £ 8, 2x + 3 y £ 12, 3x + y £ 9,
x ³ 0 and y ³ 0.
Sol. Here, the given LPP is,

(0, 9)

(0, 4)
28 , 15
(0, 2) 11 11
(8, 0)
(0, 0) (3, 0) (6, 0) x+4y=8
2x+3y=12
3x+y=9
Maximise Z = x + y subject to,
x + 4 y £ 8, 2 x + 3 y £ 12, 3x + y £ 9, x ³ 0, y ³ 0.
On solving x + 4 y = 8 and 3x + y = 9, we get
28 15
x = , y= .
11 11
From the feasible region, it is clear that coordinates of corner points are (0, 0), (3, 0),
æ 28 15 ö
ç , ÷ and (0, 2).
è 11 11 ø

Corner points Value of Z = x + y

(0, 0) 0
(3, 0) 3
æ 28 15 ö 43 10
= 3 ¬ Maximum
ç , ÷
è 11 11 ø 11 11
(0, 2) 2
10
Hence, the maximum value is 3 .
11

Q. 22 A manufacturer produces two models of bikes-model X and model Y.

Model X takes a 6 man-hours to make per unit, while model Y takes
10 man hours per unit. There is a total of 450 man-hour available per
week. Handling and marketing costs are ` 2000 and ` 1000 per unit for
models X and Y, respectively. The total funds available for these
purposes are ` 80000 per week. Profits per unit for models X and Y are
` 1000 and ` 500, respectively. How many bikes of each model should
the manufacturer produce, so as to yield a maximum profit? Find the
maximum profit.
K Thinking Process
First check whether the drawn graph (by using constraints) gives bounded region or not
and then get the maximised profit on corresponding corner points.
Sol. Let the manufacturer produces x number of models X and y number of model Y bikes.
Model X takes a 6 man-hours to make per unit and model Y takes a 10 man-hours to make
per unit.
 There is total of 450 man-hour available per week.
\6x + 10 y £ 450
Þ 3x + 5 y £ 225 …(i)
For models X and Y, handling and marketing costs are ` 2000 and `1000, respectively, total
funds available for these purposes are ` 80000 per week.
\ 2000x + 1000 y £ 80000
Þ 2 x + y £ 80 …(ii)
Also, x ³ 0, y ³ 0
Hence, the profits per unit for models X and Y are ` 1000 and ` 500, respectively.
\Required LPP is
Maximise Z =1000x + 500 y
Subject to,3x + 5 y £ 225, 2 x + y £ 80, x ³ 0, y ³ 0
From the shaded feasible region, it is clear that coordinates of corner points are (0, 0),
(40, 0), (25, 30) and (0, 45).
On solving 3x + 5 y = 225 and 2 x + y = 80, we get
x = 25, y = 30

(0, 80)

(0, 45)
(25, 30)

(75, 0)
(40, 0)

2x+y=80 3x+5y=225

Corner points Value of Z = 1000 x + 500y

(0, 0) 0
(40, 0) 40000 ¬ Maximum
(25, 30) 25000 + 15000 = 40000 ¬ Maximum
(0, 45) 22500
So, the manufacturer should produce 25 bikes of model X and 30 bikes of model Y to get a
maximum profit of ` 40000.
Since, in question it is asked that each model bikes should be produced.

Q. 23 In order to supplement daily diet, a person wishes to take some X and

some wishes Y tablets. The contents of iron, calcium and vitamins in X
and Y (in mg/tablet) are given as below
Tablets Iron Calcium Vitamin
X 6 3 2
Y 2 3 4

The person needs atleast 18 mg of iron, 21 mg of calcium and 16 mg of

vitamins. The price of each tablet of X and Y is ` 2 and `1,
respectively. How many tablets of each should the person take in order
to satisfy the above requirement at the minimum cost?
Sol. Let the person takes x units of tablet X and y units of tablet Y.
So, from the given information, we have
6x + 2 y ³ 18 Þ 3x + y ³ 9 …(i)
3x + 3 y ³ 21Þ x + y ³ 7 …(ii)
and 2 x + 4 y ³ 16 Þ x + 2 y ³ 8 …(iii)
Also, we know that here, x ³ 0, y ³ 0 …(iv)
The price of each tablet of X and Y is ` 2 and ` 1, respectively.
So, the corresponding LPP is minimise Z = 2 x + y, subject to 3x + y ³ 9, x + y ³ 7,
x + 2 y ³ 8, x ³ 0, y ³ 0
From the shaded graph, we see that for the shown unbounded region, we have coordinates
of corner points A, B, C and D as (8, 0), (6, 1), (1, 6), and (0, 9), respectively.
[on solving x + 2 y = 8 and x + y = 7, we get x = 6, y= 1and on solving 3x + y = 9 and
x + y = 7, we get x = 1, y = 6 ]
(0, 9)

(0, 7)
(1, 6)
C
(0, 4)

(6, 1)B x+2y=8

(3, 0) (4, 0) (7, 0) A (8, 0)

3x+y=9 2x+y=8 x+y=7

Corner points Value of Z = 2 x + y

(8, 0) 16
(6, 1) 13
(1, 6) 8 ¬ Minimum
(0, 9) 9
Thus, we see that 8 is the minimum value of Z at the corner point (1, 6). Here, we see that
the feasible region is unbounded. Therefore, 8 may or may not be the minimum value of Z.
To decide this issue, we graph the inequality
2x + y < 8 …(v)
and check whether the resulting open half has points in common with feasible region or not.
If it has common point, then 8 will not be the minimum value of Z, otherwise 8 will be the
minimum value of Z.
Thus, from the graph it is clear that, it has no common point.
Therefore, Z =2 x + y has 8 as minimum value subject to the given constraints.
Hence, the person should take 1 unit of X tablet and 6 units of Y tablets to satisfy the given
requirements and at the minimum cost of ` 8.
Q. 24 A company makes 3 model of calculators; A, B and C at factory I and
factory II. The company has orders for atleast 6400 calculators of
model A, 4000 calculators of model B and 4800 calculators of model C.
At factory I, 50 calculators of model A, 50 of model B and 30 of model
C are made everyday; at factory II, 40 calculators of model A, 20 of
model B and 40 of model C are made everyday. It costs ` 12000 and `
15000 each day to operate factory I and II, respectively. Find the
number of days each factory should operate to minimise the operating
costs and still meet the demand.
Sol. Let the factory I operate for x days and the factory II operate for y days.
At factory I, 50 calculators of model A and at factory II, 40 calculators of model A are made
everyday. Also, company has ordered for atleast 6400 calculators of model A.
\ 50 x + 40 y ³ 6400 Þ 5x + 4 y ³ 640 …(i)
Also, at factory I, 50 calculators of model B and at factory II, 20 calculators of modal B are
made everyday.
Since, the company has ordered atleast 4000 calculators of model B.
\ 50x + 20 y ³ 4000 Þ 5x + 2 y ³ 400 …(ii)
Similarly, for model C, 30x + 40 y ³ 4800
Þ 3x + 4 y ³ 480 …(iii)
Also, x ³ 0, y ³ 0 …(iv)
[since, x and y are non-negative]
It costs ` 12000 and ` 15000 each day to operate factories I and II, respectively.
\Corresponding LPP is,
Minimise Z =12000x + 15000 y, subject to
5x + 4 y ³ 640
5x + 2 y ³ 400
3x + 4 y ³ 480
x ³ 0, y ³ 0
On solving 3x + 4 y = 480 and 5x + 4 y = 640, we get x = 80, y = 60.
On solving 5x + 4 y = 640 and 5x + 2 y = 400, we get x = 32, y = 120
Thus, from the graph, it is clear that feasible region is unbounded and the coordinates of
corner points A, B, C and D are (160, 0), (80, 60), (32, 120) and (0, 200), respectively.

(0, 200) D

(0, 160)
(0, 124)
(32, 120)
C
(0, 120)
(80, 60)
B

A
O (80, 0) (128, 0)(155, 0) 160
,0
3x+4y=480
5x+2y=400 5x+4y=640 4x+5y=620
 Corner points Value of Z = 12000 x + 15000y
(160, 0) 160 ´ 12000 = 1920000
(80, 60) (80 ´ 12 + 60 ´ 15 ) ´ 1000 = 1860000 ¬ Minimum
(32, 120) (32 ´ 12 + 120 ´ 15 ) ´ 1000 = 2184000
(0, 200) 0 + 200 ´ 15000 = 3000000
From the above table, it is clear that for given unbounded region the minimum value of Z
may or may not be 1860000.
Now, for deciding this, we graph the inequality
12000x + 15000 y < 1860000
Þ 4x + 5 y < 620
and check whether the resulting open half plane has points in common with feasible region
or not.
Thus, as shown in the figure, it has no common points so, Z = 12000x + 15000 y
has minimum value 1860000.
So, number of days factory I should be operated is 80 and number of days factory II should
be operated is 60 for the minimum cost and satisfying the given constraints.

Q. 25 Maximise and minimise Z = 3x - 4 y subject to x - 2 y £ 0, -3x + y £ 4,

x - y £ 6 and x, y ³ 0.
Sol. Given LPP is,
maximise and minimise Z = 3x - 4 y subject to x - 2 y £ 0, -3x + y £ 4, x - y £ 6, x, y ³ 0.
[on solving x - y = 6 and x - 2 y = 0, we get x = 12, y = 6]

(0, 4) (12, 6)
=4

(0, 0) (6, 0)
– 3x+y

x–2y=0
– 3x+4y=16

3x – 4y=12 (0, – 6)

x – y=6
From the shown graph, for the feasible region, we see that it is unbounded and coordinates
of corner points are (0, 0), (12, 6) and (0, 4).

Corner points Corresponding value of Z = 3 x - 4 y

(0, 0) 0
(0, 4) -16 ¬ Minimum
(12, 6) 12¬ Maximum
 For given unbounded region the minimum value of Z may or may not be -16. So, for
deciding this, we graph the inequality.
3x - 4 y < - 16
and check whether the resulting open half plane has common points with feasible region or
not.
Thus, from the figure it shows it has common points with feasible region, so it does not have
any minimum value.
Also, similarly for maximum value, we graph the inequality 3x - 4 y > 12
and see that resulting open half plane has no common points with the feasible region and
hence maximum value 12 exist for Z = 3x - 4 y.

Objective Type Questions

Q. 26 The corner points of the feasible region determined by the system of
linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The
objective function is Z = 4 x + 3 y. Compare the quantity in column A
and column B.
Column A Column B
Maximum of Z 325
(a) The quantity in column A is greater
(b) The quantity in column B is greater
(c) The two quantities are equal
(d) The relationship cannot be determined on the basis of the information supplied.
Sol. (b)
Corner points Corresponding value of
Z = 4 x + 3y
(0, 0) 0
(0, 40) 120
(20, 40) 200
(60, 20) 300 ¬ Maximum
(60, 0) 240
Hence, maximum value of Z = 300 < 325
So, the quantity in column B is greater.

Q. 27 The feasible solution for a LPP is shown in following figure. Let

Z = 3x - 4 y be the objective function. Minimum of Z occurs at
Y (4, 10)

(0, 8) (6, 8)

(6, 5)

(0, 0)
X
(5, 0)
(a) (0, 0) (b) (0, 8) (c) (5, 0) (d) (4, 10)
Sol. (b)
Corner points Corresponding value of Z = 3 x - 4 y
(0, 0) 0
(5, 0) 15¬ Maximum
(6, 5) -2
(6, 8) -14
(4, 10) -28
(0, 8) -32¬ Minimum
Hence, the minimum of Z occurs at (0, 8) and its minimum value is (-32).

Q. 28 Refer to question 27. Maximum of Z occurs at

(a) (5, 0) (b) (6, 5) (c) (6, 8) (d) (4, 10)
Sol. (a) Refer to solution 27, maximum of Z occurs at (5, 0).

Q. 29 Refer to question 7, maximum value of Z + minimum value of Z is

equal to
(a) 13 (b) 1 (c) -13 (d) -17
Sol. (d) Refer to solution 27, maximum value of Z + minimum value of Z
= 15 - 32 = - 17

Q. 30 The feasible region for an LPP is shown in the following figure. Let
F = 3x -4 y be the objective function. Maximum value of F is

(12, 6)

(0, 4)

(12, 0)
(a) 0 (b) 8 (c) 12 (d) -18
Sol. (c) The feasible region as shown in the figure, has objective function F = 3x - 4 y.

Corner points Corresponding value of F = 3 x - 4 y

(0, 0) 0
(12, 6) 12 ¬ Maximum
(0, 4) -16 ¬ Minimum
Hence, the maximum value of F is 12.

Q. 31 Refer to question 30. Minimum value of F is

(a) 0 (b) -16 (c) 12 (d) Does not exist
Sol. (b) Referring to solution 30, minimum value of F is -16 at (0, 4).
Q. 32 Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0),
(6, 8) and (0, 5). Let F = 4 x + 6 y be the objective function. The
minimum value of F occurs at
(a) Only (0, 2)
(b) Only (3, 0)
(c) the mid-point of the line segment joining the points (0, 2) and (3, 0)
(d) any point on the line segment joining the points (0, 2) and (3, 0)
Sol. (d)
Corner points Corresponding value of
F = 4 x + 6y
(0, 2) 12¬ Minimum
(3, 0) 12¬ Minimum
(6, 0) 24
(6, 8) 72¬ Maximum
(0, 5) 30
Hence, minimum value of F occurs at any points on the line segment joining the points (0, 2)
and (3, 0).

Q. 33 Refer to question 32, maximum of F - minimum of F is equal to

(a) 60 (b) 48 (c) 42 (d) 18
Sol. (a) Referring to the solution 32, maximum of F - minimum of F = 72 - 12 = 60

Q. 34 Corner points of the feasible region determined by the system of linear

constraints are (0, 3), (1, 1) and (3, 0). Let Z = px + qy, where p, q > 0.
Condition on p and q, so that the minimum of Z occurs at (3, 0) and
(1, 1) is
q
(a) p = 2q (b) p = (c) p = 3q (d) p = q
2
Sol. (b)
Corner points Corresponding value of Z = px + qy ; p, q > 0
(0, 3) 3q
(1, 1) p+ q
(3, 0) 3p
So, condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) is
p + q = 3p Þ 2 p = q
q
\ p=
2
Fillers
Q. 35 In a LPP, the linear inequalities or restrictions on the variables are
called… .
Sol. In a LPP, the linear inequalities or restrictions on the variables are called linear constraints.

Q. 36 In a LPP, the objective function is always… .

Sol. In a LPP, objective function is always linear.

Q. 37 In the feasible region for a LPP is ..., then the optimal value of the
objective function Z = ax + by may or may not exist.
Sol. If the feasible region for a LPP is unbounded, then the optimal value of the objective
function Z = ax + by may or may not exist.

Q. 38 In a LPP, if the objective function Z = ax + by has the same maximum

value on two corner points of the feasible region, then every point on
the line segment joining these two points give the same ... value.
Sol. In a LPP, if the objective function Z = ax + by has the same maximum value on two corner
points of the feasible region, then every point on the line segment joining these two points
give the same maximum value.

Q. 39 A feasible region of a system of linear inequalities is said to be ..., if it

can be enclosed within a circle.
Sol. A feasible region of a system of linear inequalities is said to be bounded, if it can be
enclosed within a circle.

Q. 40 A corner point of a feasible region is a point in the region which is the

... of two boundary lines.
Sol. A corner point of a feasible region is a point in the region which is the intersection of two
boundary lines.

Q. 41 The feasible region for an LPP is always a ... polygon.

Sol. The feasible region for an LPP is always a convex polygon.
True/False
Q. 42 If the feasible region for a LPP is unbounded, maximum or minimum of
the objective function Z = ax + by may or may not exist.
Sol. True

Q. 43 Maximum value of the objective function Z = ax + by in a LPP always

occurs at only one corner point of the feasible region.
Sol. False

Q. 44 In a LPP, the minimum value of the objective function Z = ax + by is

always 0, if origin is one of the corner point of the feasible region.
Sol. False
Q. 45 In a LPP, the maximum value of the objective function Z = ax + by is
always finite.
Sol. True
 13
Probability
Short Answer Type Questions
Q. 1 For a loaded die, the probabilities of outcomes are given as under
P (1) = P (2) = 0. 2, P (3) = P (5) = P (6) = 0.1 and P ( 4) = 0 . 3.
The die is thrown two times. Let A and B be the events, ‘same number
each time’ and ‘a total score is 10 or more’, respectively. Determine
whether or not A and B are independent.
Thinking Process
First, find P (A), P (B) and P (A Ç B) and then use the concept that two events A and B are
called independent events, if P (A Ç B) = P (A) × P (B).
Sol. For a loaded die, it is given that
P (1) = P (2 ) = 0.2,
P (3) = P (5) = P (6) = 01 . and P(4) = 0. 3
Also, die is thrown two times.
Here, A = Same number each time and B = Total score is 10 or more
\ A = {(1, 1), (2, 2 ), (3, 3), (4, 4), (5, 5), (6, 6)}
So, P ( A) = [P (1, 1) + P (2, 2) + P (3, 3) + P (4, 4) + P (5, 5) + P (6, 6)]
= [P (1) × P (1) + P (2 ) × P (2 ) + P (3) × P (3) + P (4) × P (4) + P (5) × P (5) + P (6) × P (6)]
= [0.2 ´ 0.2 + 0.2 ´ 0.2 + 01 . ´ 01. + 0. 3 ´ 0. 3 + 01 . ´ 01
. + 01 . ´ 01 .]
= 0.04 + 0.04 + 0.01 + 0.09 + 0.01 + 0.01 = 0.20
and B = {(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}
\ P (B) = P (4, 6) + P (6, 4) + P (5, 5) + P (5, 6) + P (6, 5) + P (6, 6)
= P (4) × P (6) + P (6) × P (4) + P (5) × P (5) + P (5) × P (6) + P (6) × P (5) + P (6) × P (6)
= 0. 3 ´ 01 . + 01 . ´ 0. 3 + 01 . ´ 01
. + 01 . ´ 01
. + 01 . ´ 01. + 01. ´ 01 .
= 0.03 + 0.03 + 0.01 + 0.01 + 0.01 + 0.01 = 010 .
Also, A Ç B = {(5, 5), (6, 6)}
\ P ( A Ç B) = P (5, 5) + P (6, 6) = P (5) × P (5) + P (6) × P (6)
= 01. ´ 01 . + 01 . ´ 01 . = 0.01 + 0.01 = 0.02
We know that, for two events A and B, if P ( A Ç B) = P ( A) × P (B), then both are independent
events.
Here, P ( A Ç B) = 0.02 and P ( A) × P (B) = 0.20 ´ 010 . = 0.02
Thus, P ( A Ç B) = P ( A) × P (B) = 0.02
Hence, A and B are independent events.
Q. 2 Refer to question 1 above. If the die were fair, determine whether or not
the events A and B are independent.
Thinking Process
In a fair die, we have equally likely outcomes. So, with the given events A and B, we first
find P (A), P (B) and P (A Ç B) and then check whether they are dependent or independent.
Sol. Referring to the above solution, we have
A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Þ n ( A) = 6 and n (S ) = 62 = 36 [where, S is sample space]
n( A) 6 1
\ P( A) = = =
n(S ) 36 6
and B = {(4, 6), (6, 4), (5, 5), (6, 5), (5, 6), (6, 6)}
Þ n(B) = 6 and n(S ) = 62 = 36
n(B) 6 1
\ P (B) = = =
n(S ) 36 6
Also, A Ç B = {(5, 5), (6, 6)}
Þ n ( A Ç B) = 2 and n(S ) = 36
2 1
\ P ( A Ç B) = =
36 18
1
Also, P ( A) × P (B) =
36
Thus, P ( A Ç B) ¹ P ( A) × P (B) éQ 1 ¹ 1 ù
ëê 18 36 ûú
So, we can say that both A and B are not independent events.

Q. 3 The probability that atleast one of the two events A and B occurs is 0.6.
If A and B occur simultaneously with probability 0.3, evaluate
P ( A) + P (B).
Sol. We know that, A È B denotes the occurrence of atleast one of A and B and A Ç B denotes
the occurrence of both A and B, simultaneously.
Thus, P ( A È B) = 0.6 and P ( A Ç B) = 0. 3
Also, P ( A È B) = P ( A) + P (B) - P ( A Ç B)
Þ 0. 6 = P ( A) + P (B) - 0. 3
Þ P ( A) + P (B) = 0. 9
Þ [1 - P ( A)] + [1 - P (B)] = 0. 9 [Q P ( A) = 1 - P ( A) and P (B) = 1 - P (B) ]
Þ P ( A) + P (B) = 2 - 0. 9 = 11
.

Q. 4 A bag contains 5 red marbles and 3 black marbles. Three marbles are
drawn one by one without replacement. What is the probability that
atleast one of the three marbles drawn be black, if the first marble is
red?
Sol. Let R = {5 red marbles} and B = {3 black marbles}
For atleast one of the three marbles drawn be black, if the first marble is red, then the
following three conditions will be followed
(i) Second ball is black and third is red (E1 ).
(ii) Second ball is black and third is also black (E 2 ).
(iii) Second ball is red and third is black (E3 ).
 5 3 4 60 5
\ P (E1 ) = P (R1 ) × P (B1 / R1 ) × P (R 2 / R1B1 ) = × × = =
8 7 6 336 28
5 3 2 30 5
P (E2 ) = P (R1 ) × P (B1 / R 1 ) × P (B 2 / R1B1 ) = × × = =
8 7 6 336 56
5 4 3 60 5
and P (E3 ) = P (R1 ) × P (R 2 / R1 ) × P (B1 / R1R 2 ) = × × = =
8 7 6 336 28
5 5 5
\ P (E ) = P (E1 ) + P (E2 ) + P (E3 ) = + +
28 56 28
10 + 5 + 10 25
= =
56 56

Q. 5 Two dice are thrown together and the total score is noted. The events E,
F and G are ‘a total of 4’, ‘a total of 9 or more’ and ‘a total divisible by 5’,
respectively. Calculate P (E ), P (F ) and P (G) and decide which pairs of
events, if any are independent.
Sol. Two dice are thrown together i.e., sample space (S ) = 36 Þ n(S ) = 36
E = A total of 4 = {(2, 2), (3, 1), (1, 3)}
Þ n (E ) = 3
F = A total of 9 or more
= {(3, 6), (6, 3), (4, 5), (4, 6), (5, 4), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}
Þ n(F ) = 10
G = a total divisible by 5 = {(1, 4), (4, 1), (2, 3), (3, 2), (4, 6), (6, 4), (5, 5)}
Þ n(G ) = 7
Here, (E Ç F ) = f and (E Ç G ) = f
Also, (F Ç G ) = {(4, 6), (6, 4), (5, 5)}
Þ n(F Ç G ) = 3 and (E Ç F Ç G ) = f
n(E ) 3 1
\ P(E ) = = =
n(S ) 36 12
n(F ) 10 5
P (F ) = = =
n(S ) 36 18
n(G ) 7
P (G ) = =
n(S ) 36
3 1
P (F Ç G ) = =
36 12
5 7 35
and P (F ) × P (G ) = × =
18 36 648
Here, we see that P (F Ç G ) ¹ P (F ) × P (G )
[since, only F and G have common events, so only F and G are used here]
Hence, there is no pair which is independent.

Q. 6 Explain why the experiment of tossing a coin three times is said to have
Binomial distribution.
Sol. We know that, a random variable X taking values 0, 1, 2, ..., n is said to have a binomial
distribution with parameters n and P, if its probability distribution is given by
P ( X = r ) = nC r pr q n - r
where, q = 1- p
and r = 0, 1, 2, ..., n
 Similarly, in an experiment of tossing a coin three times, we have n = 3 and random variable
1 1
X can take values r = 0, 1, 2 and 3 with p = and q =
2 2
X 0 1 2 3
3 3 3 2 3 2 3 3
P (X ) C0 q C1 Pq C2 P q C3 P

So, we see that in the experiment of tossing a coin three times, we have random variable X
1
which can take values 0, 1, 2 and 3 with parameters n = 3 and P = .
2
Therefore, it is said to have a Binomial distribution.

Q. 7 If A and B are two events such that

1 1 1
P ( A) = , P (B) = and P ( A Ç B) = , then find
2 3 4
(i) P (A / B). (ii) P (B / A).
(iii) P (A ¢ / B). (iv) P (A ¢ / B ¢).
1 1 1
Sol. Here, P ( A) = , P (B) = and P ( A Ç B) =
2 3 4
P ( A Ç B) 1 / 4 3
(i) P ( A / B) = = =
P (B) 1/ 3 4
P ( A Ç B) 1 / 4 1
(ii) P (B / A) = = =
P( A) 1/ 2 2
3 1
(iii) P ( A ¢ / B) = 1 - P ( A / B) = 1 - =
4 4
1 1 1
-
P ( A ¢ Ç B) P (B) - P ( A Ç B) 3 4 12 1
or P ( A ¢ / B) = = = = =
P (B) P (B) 1 1 4
3 3
P ( A ¢ Ç B¢) 1 - P ( A È B) 1 - [P ( A) + P (B) - P ( A Ç B)]
(iv) P ( A ¢ / B¢) = = =
P (B¢) 1 - P (B) 1 - P (B)
1 1 1 5 1
1 - é + - ù 1 - æç - ö÷
êë 2 3 4 úû è 6 4ø
= =
1 2
1-
3 3
1 - 14 / 24 10 / 24 30 5
= = = =
2/3 2/3 48 8

Q. 8 Three events A, B and C have probabilities 2 , 1 and 1, respectively. If,

5 3 2
1 1
P ( A Ç C ) = and P (B Ç C ) = , then find the values of P (C / B) and
5 4
P ( A ¢ Ç C ¢).
2 1 1 1 1
Sol. Here, P ( A) = , P (B) = , P (C ) = , P ( A Ç C ) = and P (B Ç C ) =
5 3 2 5 4
P (B Ç C ) 1 / 4 3
\ P (C / B) = = =
P (B) 1/ 3 4
and P ( A ¢ Ç C ¢) = 1 - P ( A È C ) = 1 - [P ( A) + P (C ) - P ( A Ç C )]
4 + 5 - 2ù
= 1- é + - ù = 1- é
2 1 1 7 3
= 1- =
êë 5 2 5 úû êë 10 úû 10 10
Q. 9 Let E 1 and E 2 be two independent events such that P (E 1 ) = P1 and
P (E 2 ) = P2 . Describe in words of the events whose probabilities are
(i) P1 P2 (ii) (1 - P1 )P2
(iii) 1 - (1 - P1 ) (1 - P2 ) (iv) P1 + P2 - 2P1 P2
Sol. P (E1 ) = P1 and P (E2 ) = P2
(i) P1 P2 Þ P (E1 ) × P (E2 ) = P (E1 Ç E2 )
So, E1 and E2 occur.
(ii) (1 - P1 ) P2 = P (E1 )¢ × P (E2 ) = P(E1 ¢ Ç E2 )
So, E1 does not occur but E2 occurs.
(iii) 1 - (1 - P1 ) (1 - P2 ) = 1 - P (E1 )¢P (E2 )¢ = 1 - P (E1 ¢ Ç E2 ¢)
= 1 - [1 - P (E1 È E2 )] = P (E1 È E2 )
So, either E1 or E2 or both E1 and E2 occurs.
(iv) P1 + P2 - 2 P1P2 = P (E1 ) + P (E2 ) - 2 P (E1 ) × P (E2 )
= P (E1 ) + P (E2 ) - 2 P (E1 Ç E2 )
= P (E1 È E2 ) - P (E1 Ç E2 )
So, either E1 or E2 occurs but not both.

Q. 10 A discrete random variable X has the probability distribution as given below

X 0.5 1 1.5 2
P(X) k k2 2k 2 k
(i) Find the value of k.
(ii) Determine the mean of the distribution.
Sol. We have,
X 0.5 1 1.5 2
P(X) k k2 2k 2 k
n
(i) We know that, S Pi = 1, where Pi ³ 0
i =1

Þ P1 + P2 + P3 + P4 = 1
Þ k + k2 + 2 k2 + k = 1
Þ 3 k2 + 2 k - 1 = 0
2
Þ 3 k + 3k - k - 1 = 0
Þ 3 k(k + 1) - 1(k + 1) = 0
Þ (3 k - 1) (k + 1) = 0
Þ k = 1/ 3 Þ k = - 1
Since, k is ³ 0 Þ k = 1 / 3
n
(ii) Mean of the distribution (m ) = E( X ) = S x i Pi
i = 1i

= 0. 5(k ) + 1 (k 2 ) + 1. 5(2 k 2 ) + 2 (k ) = 4k 2 + 2. 5k
1 1 éQ k = 1 ù
= 4 × + 2. 5 ×
9 3 êë 3 úû
4 + 7. 5 23
= =
9 18
Q. 11 Prove that
(i) P (A) = P (A Ç B) + P (A Ç B)
(ii) P (A È B) = P (A Ç B) + P (A Ç B) + P (A Ç B)
Sol. (i) Q P ( A) = P ( A Ç B) + P ( A Ç B)
\ RHS = P ( A Ç B) + P ( A Ç B)
= P ( A) × P (B) + P ( A) × P (B)
= P ( A)[P (B) + P (B)]
= P ( A)[P (B) + 1 - P (B)] [Q P (B) = 1 - P (B) ]
= P ( A) = LHS Hence proved.
(ii) Q P ( A È B) = P ( A Ç B) + P ( A Ç B) + P ( A Ç B)
\ RHS = P ( A) × P (B) + P ( A) × P (B) + P ( A) × P (B)
= P ( A) × P (B) + P ( A) × [1 - P(B)] + [1 - P ( A)] P (B)
= P ( A) × P (B) + P ( A) - P ( A) × P (B) + P (B) - P ( A) × P (B)
= P ( A) + P (B) - P ( A) × P (B)
= P ( A) + P (B) - P ( A Ç B)
= P ( A È B) = LHS Hence proved.

Q. 12 If X is the number of tails in three tosses of a coin, then determine

the standard deviation of X .
Thinking Process
First get the values of P (X) at x = 0, 1, 2 , 3 and then use the formula of standard
deviation of X = Var (X), where Var (X) = E (X2) - [E (X)]2 = S X2P (X) - [SXP (X)]2
Sol. Given that, random variable X is the number of tails in three tosses of a coin.
So, X = 0, 1, 2, 3.
Þ P ( X = x ) = nC x ( p)x q n - x ,
where n = 3, p = 1 / 2, q = 1 / 2 and x = 0, 1, 2, 3
X 0 1 2 3
P(X) 1 3 3 1
8 8 8 8
XP(X) 3 3 3
0
8 4 8
2 3 3 9
X P (X )
0
8 2 8
We know that, Var ( X ) = E ( X 2 ) - [E ( X )]2 ...(i)
n n
where, E( X 2 ) = S x i2 P(x i ) and E ( X ) = S x i P(x i )
i =1 i =1
n
3 3 9 24
\ E( X 2 ) = S x i2 P( X i ) = 0 + + + = = 3
i =1 8 2 8 8
2
3 3 3 2
2
é n ù 12 9
and [E ( X )]2 = ê S x i2 P (x i )ú = é 0 + + + ù = é ù =
ê
ë 8 4 8û ú ê
ë8û ú 4
ëi = 0 û
9 3
\ Var ( X ) = 3 - = [using Eq. (i)]
4 4
3 3
and standard deviation of X = Var( X ) = =
4 2
Q. 13 In a dice game, a player pays a stake of ` 1 for each throw of a die. She
receives ` 5, if the die shows a 3, ` 2, if the die shows a 1 or 6 and
nothing otherwise, then what is the player’s expected profit per throw
over a long series of throws?
Thinking Process
Take X as the random variable of profit per throw and at X = -1, 1 and 4 get the values
of P (X) and use the formula expected profit E (X) = S X P (X) to get the desired result.
Sol. Let X is the random variable of profit per throw.
X -1 1 4
1 1 1
P(X)
2 3 6
Since, she loss ` 1 on getting any of 2, 4 or 5.
1 1 1 3 1
So, at X = - 1, P( X ) = + + = =
6 6 6 6 2
1 1 1
Similarly, at X = 1, P( X ) = + = [ if die shows of either 1 or 6 ]
6 6 3
1
and at X = 4, P( X ) = [ if die shows a 3]
6
\ Player’s expected profit = E ( X ) = SX P ( X )
1 1 1
= -1´ + 1´ + 4 ´
2 3 6
-3 + 2 + 4 3 1
= = = = ` 0.50
6 6 2

Q. 14 Three dice are thrown at the same time. Find the probability of getting
three two’s, if it is known that the sum of the numbers on the dice
was six.
Sol. On a throw of three dice, we have sample space [n(S )] = 63 = 216
Let E1 is the event when the sum of numbers on the dice was six and E2 is the event when
three two’s occurs.
Þ E1 = {(1, 1, 4), (1, 2, 3), (1, 3, 2), (1, 4, 1), (2, 1, 3), (2, 2, 2), (2, 3, 1), (3, 1, 2),
(3, 2, 1), (4, 1, 1)}
Þ n(E1 ) = 10 and E2 = {2, 2, 2}
Þ n ( E2 ) = 1
Also, (E1 Ç E2 ) = 1
P × (E1 Ç E2 ) 1 / 216 1
\ P (E2 / E1 ) = = =
P(E1 ) 10 / 216 10

Q. 15 Suppose 10000 tickets are sold in a lottery each for ` 1. First prize is
of ` 3000 and the second prize is of ` 2000. There are three third
prizes of ` 500 each. If you buy one ticket, then what is your
expectation?
Thinking Process
Take X is the random variable for the prize, so at X = 0, 500, 2000 and 3000, get P (X) for
each X and then use the formula of E (X) = SX P (X) to get the answer.
Sol. Let X is the random variable for the prize.
X 0 500 2000 3000
9995 3 1 1
P(X)
10000 10000 10000 10000
Since, E ( X ) = SX P ( X )
9995 1500 2000 3000
\ E( X ) = 0 ´ + + +
10000 10000 10000 10000
1500 + 2000 + 3000
=
10000
6500 13
= = = ` 0.65
10000 20

Q. 16 A bag contains 4 white and 5 black balls. Another bag contains 9

white and 7 black balls. A ball is transferred from the first bag to the
second and then a ball is drawn at random from the second bag. Find
the probability that the ball drawn is white.
Sol. Here, W1 = {4 white balls} and B1 = {5 black balls}
and W2 = {9 white balls} and B 2 = {7 black balls}
Let E1 is the event that ball transferred from the first bag is white and E2 is the event that the
ball transferred from the first bag is black.
Also, E is the event that the ball drawn from the second bag is white.
10 9
\ P (E / E1 ) = , P (E/E2 ) =
17 17
4 5
and P (E1 ) = and P (E2 ) =
9 9
\ P (E ) = P (E1 ) × P (E / E1 ) + P(E2 ) × P (E / E2 )
4 10 5 9
= × + ×
9 17 9 17
40 + 45 85 5
= = =
153 153 9

Q. 17 Bag I contains 3 black and 2 white balls, bag II contains 2 black and 4
white balls. A bag and a ball is selected at random. Determine the
probability of selecting a black ball.
Sol. Bag I = {3B, 2W }, Bag II = {2 B, 4W }
Let E1 = Event that bag I is selected
E2 = Event that bag II is selected
and E = Event that a black ball is selected
1 3 2 1
Þ P (E1 ) = 1 / 2, P (E2 ) = , P (E / E1 ) = , P (E / E2 ) = =
2 5 6 3
\ P (E ) = P (E1 ) × P (E / E1 ) + P (E2 ) × P (E / E2 )
1 3 1 2 3 2
= × + × = +
2 5 2 6 10 12
18 + 10 28 7
= = =
60 60 15
Q. 18 A box has 5 blue and 4 red balls. One ball is drawn at random and not
replaced. Its colour is also not noted. Then, another ball is drawn at
random. What is the probability of second ball being blue?
Sol. A box = {5 blue, 4 red}
Let E1 is the event that first ball drawn is blue, E2 is the event that first ball drawn is red and E
is the event that second ball drawn is blue.
\ P (E ) = P (E1 ) × P (E / E1 ) + P (E2 ) × P (E/E2 )
5 4 4 5 20 20 40 5
= × + × = + = =
9 8 9 8 72 72 72 9

Q. 19 Four cards are successively drawn without replacement from a deck of

52 playing cards. What is the probability that all the four cards are
king?
Sol. Let E1, E2 , E3 and E4 are the events that the first, second, third and fourth card is king,
respectively.
\ P (E1 Ç E2 Ç E3 Ç E4 ) = P (E1 ) × P (E2 / E1 ) × P (E3 / E1 Ç E2 ) × P [E4 / (E1 Ç E2 Ç E3 Ç E4 )]
4 3 2 1 24
= × × × =
52 51 50 49 52 × 51 × 50 × 49
1 1
= =
13 × 17 × 25 × 49 270725

Q. 20 If a die is thrown 5 times, then find the probability that an odd

number will come up exactly three times.
1 1 1 1 1 1
Sol. Here, n = 5, p = æç + + ö÷ = and q = 1 - p = 1 - =
è 6 6 6ø 2 2 2
Also, r=3
1 3 1 5-3
\ P ( X = r ) = n C r ( p)r (q )n - r = 5C 3 æç ö÷ æç ö÷
è2 ø è2 ø
5! 1 1 10 5
= × × = =
3!2 ! 8 4 32 16

Q. 21 If ten coins are tossed, then what is the probability of getting atleast
8 heads?
Thinking Process
For getting atleast 8 heads, take random variable X for getting head on tossing a coin.
So, get sum of P (8), P (9) and P (10) to get the answer.
Sol. In this case, we have to find out the probability of getting atleast 8 heads. Let X is the
random variable for getting a head.
Here, n = 10 , r ³ 8,
1 1
i.e., r = 8, 9, 10, p = , q =
2 2
We know that, P ( X = r ) = nC r pr q n - r
 \ P ( X = r ) = P (r = 8) + P (r = 9) + P (r = 10)
8 10 - 8 9 10 - 9 10 10 - 10
1 1 1 1 1 1
= C 8 æç ö÷ æç ö÷
10
+ 10C 9 æç ö÷ æç ö÷ + C10 æç ö÷ æç ö÷
10
è2 ø è2 ø è2 ø è2 ø è2 ø è2 ø
10! æ 1 ö10 10! æ 1 ö10
10
10! æ 1 ö
= ç ÷ + ç ÷ + ç ÷
8!2 ! è 2 ø 9!1! è 2 ø 0!10! è 2 ø
10
10 ´ 9
= æç ö÷ é
1
+ 10 + 1ù
è 2 ø êë 2 úû
1 10 1 7
= æç ö÷ × 56 = 7 3 × 56 =
è2 ø 2 ×2 128

Q. 22 The probability of a man hitting a target is 0.25. If he shoots 7 times,

then what is the probability of his hitting atleast twice?
Thinking Process
Using Binomial distribution P(X = r) to get the answer. Here,
P (X = r) = 1 - [P (r = 0) + P(r = 1)]
1 1 3
Sol. Here, n = 7 p = 0.25 = , q = 1 - = r ³ 2,
4 4 4
where, P ( X ) = nC r ( p)r (q )n - r
In this case for easy approach we shall first find out the probability of his hitting atmost once
(i.e., r = 0, 1) and then subtract this probability from 1 to get the desired probability.
\ P ( X = r ) = 1 - [P (r = 0) + P(r = 1)]
é 1 0 3 7 - 0 7 æ 1 ö1 æ 3 ö 7 - 1 ù
= 1 - ê 7C 0 æç ö÷ æç ö÷ + C1 ç ÷ ç ÷ ú
ë è 4ø è 4ø è 4ø è 4ø û
é 7 ! æ 3 ö7 7! æ 1 ö æ 3 ö ù
6
= 1- ê ç ÷ + ç ÷ç ÷ ú
ë 0!7 ! è 4 ø 1! 6! è 4 ø è 4 ø û
é 3 6 3 1 ù
= 1 - ê æç ö÷ æç × 1 + × 7 ö÷ ú
ëè 4 ø è 4 4 øû
é 36 æ 10 ö ù é 36 ´ 10 ù é 27 × 27 × 10 ù
= 1 - ê 6 ç ÷ú = 1 - ê ú = 1- ê ú
ë 64 × 256 û
7
ë 4 è 4 øû ë 4 û
7290 ù 3645 4547
= 1- é = 1- =
ëê 16384 ûú 8192 8192

Q. 23 A lot of 100 watches is known to have 10 defective watches. If

8 watches are selected (one by one with replacement) at random, then
what is the probability that there will be atleast one defective watch?
Sol. Probability of defective watch from a lot of 100 watches = 10 = 1
100 10
9
\ p = 1 / 10, q = , n = 8 and r ³1
10
1 0 9 8-0
\ P (r ³ 1) = 1 - P (r = 0) = 1 - 8C 0 æç ö÷ æç ö÷
è 10 ø è 10 ø
8! æ 9 ö 8 9
8
= 1- × ç ÷ = 1 - æç ö÷
0! 8! è 10 ø è 10 ø
Q. 24 Consider the probability distribution of a random variable X.
X 0 1 2 3 4
P(X) 0.1 0.25 0.3 0.2 0.15
Calculate
æX ö
(i) V ç ÷. (ii) Variance of X.
è2 ø
Sol. We have,
X 0 1 2 3 4
P(X) 0.1 0.25 0.3 0.2 0.15
XP(X) 0 0.25 0.6 0.6 0.60
2
X P (X ) 0 0.25 1.2 1.8 2.40
2 2
Var ( X ) = E ( X ) - [E ( X )]
n
where, E ( X ) = m = S x i Pi(xi )
i =1
n
and E ( X 2 ) = S x i2 P (x i )
i =1

\ E ( X ) = 0 + 0.25 + 0.6 + 0.6 + 0. 60 = 2.05

E ( X 2 ) = 0 + 0.25 + 12
. + 1. 8 + 2.40 = 5. 65
æ Xö 1 1 2
(i) V ç ÷ = V( X ) = [5. 65 - ( 2. 05) ]
è2 ø 4 4
1 1
= [5. 65 - 4.2025] = ´ 14475 . = 0. 361875
4 4
(ii) V( X ) = 14475
.

Q. 25 The probability distribution of a random variable X is given below

X 0 1 2 3
k k k
P(X) k
2 4 8
(i) Determine the value of k.
(ii) Determine P ( X £ 2) and P ( X > 2).
(iii) Find P ( X £ 2) + P ( X > 2).
Sol. We have,
X 0 1 2 3
k k k
P(X) k
2 4 8
n
(i) Since, S Pi = 1, i = 1, 2 ,..., n and Pi ³ 0
i =1
k k k
\ k+ + + = 1
2 4 8
Þ 8k + 4k + 2 k + k = 8
8
\ k=
15
 k k
(ii) P ( X £ 2 ) = P (0) + P (1) + P (2 ) = k + +
2 4
(4k + 2 k + k ) 7 k 7 8 14
= = = × =
4 4 4 15 15
k 1 8 1
and P ( X > 2 ) = P (3) = = × =
8 8 15 15
14 1
(iii) P ( X £ 2 ) + P ( X > 2 ) = + =1
15 15

Q. 26 For the following probability distribution determine standard

deviation of the random variable X .
X 2 3 4
P(X) 0.2 0.5 0.3
Sol. We have,
X 2 3 4
P(X) 0.2 0.5 0.3
XP(X) 0.4 1.5 1.2
X 2 P (X ) 0.8 4.5 4.8
We know that, standard deviation of X = Var X
where, Var X = E ( X 2 ) - [E ( X )]2
2
n é n ù
= S x i2 P(x i ) - ê S x i Pi ú
i =1 ë i = 1 û
\ . ]2
Var X = [0. 8 + 4. 5 + 4. 8] - [0.4 + 1. 5 + 12
= 101 . )2 = 101
. - (31 . - 9. 61 = 0.49
\ Standard deviation of X = Var X = 0.49 = 07
.

1
Q. 27 A biased die is such that P (4) = and other scores being equally
10
likely. The die is tossed twice. If X is the ‘number of fours seen’, then
find the variance of the random variable X .
Sol. Since, X = Number of fours seen
On tossing two die, X = 0, 1, 2.
1 9
Also, P( 4 ) = and P( not 4) =
10 10
9 9 81
So, P ( X = 0) = P(not 4) × P(not 4) = × =
10 10 100
9 1 1 9 18
P ( X = 1) = P(not 4) × P(4) + P(4) × P(not 4) = × + × =
10 10 10 10 100
P ( X = 2 ) = P( 4 ) × P( 4 ) = 1 × 1 = 1
10 10 100
Thus, we get following table
X 0 1 2
P(X) 81 18 1
100 100 100
XP(X) 0 18/100 2/100
X 2 P (X ) 0 18/100 4/100
 \ Var ( X ) = E ( X 2 ) - [E ( X )]2 = SX 2 P ( X ) - [S XP ( X )]2
18 4 ù é 18 2 ù2
= é0 + + - 0+ +
ëê 100 100 ûú êë 100 100 ûú
22 20 ö 2 11 1
= - æç ÷ = -
100 è 100 ø 50 25
11 - 2 9 18
= = = = 018
.
50 50 100

Q. 28 A die is thrown three times. Let X be the ‘number of twos seen’, find
the expectation of X.
Sol. We have, X = number of twos seen
So, on throwing a die three times, we will have X = 0, 1, 2, 3.
5 5 5 125
\ P ( X = 0) = P(not 2) × P(not 2) × P(not 2) = × × =
6 6 6 216
P ( X = 1) = P( (not 2) × P(not 2) × P( 2 ) + P(not 2) × P( 2 ) × P(not 2) + P( 2 ) × P(not 2) × P(not 2)
5 5 1 5 1 5 1 5 5 25 3 25
= × + × × + × × = × =
6 6 6 6 6 6 6 6 6 36 6 72
P ( X = 2 ) = P(not 2) × P( 2 ) × P( 2 ) + P( 2 ) × P( 2 ) × P(not 2) + P( 2 ) × P(not 2) + P( 2 )
5 1 1 1 1 5 1 5 1
=
× × + × × + × ×
6 6 6 6 6 6 6 6 6
1 é 15 ù 15
= × =
36 êë 6 úû 216
1 1 1 1
P ( X = 3) = P( 2 ) × P( 2 ) × P( 2 ) = × × =
6 6 6 216
125 25 15 1
We know that, E( X ) = SX P( X ) = 0 × + 1× +2× + 3×
216 72 216 216
75 + 30 + 3 108 1
= = =
216 216 2

1
Q. 29 Two biased dice are thrown together. For the first die P(6) = , the
2
2
other scores being equally likely while for the second die P(1) = and
5
the other scores are equally likely. Find the probability distribution of
‘the number of one’s seen’.
1 1
Sol. For first die, P(6) = and P(6¢) =
2 2
1
Þ P(1) + P(2 ) + P(3) + P(4) + P(5) =
2
1 9
Þ P(1) = and P(1¢) = [Q P(1) = P(2 ) = P(3) = P(4) = P(5)]
10 10
2 2 3
For second die, P(1) = and P(1¢) = 1 - =
5 5 5
 Let X = Number of one’s seen
9 3 27
For X = 0, P ( X = 0) = P(1¢) × P(1¢) = × = = 0.54
10 5 50
9 2 1 3
P ( X = 1) = P(1¢) × P(1¢) + P(1¢) × P(1¢) = × + ×
10 5 10 5
18 3 21
= + = = 0.42
50 50 50
1 2 2
P ( X = 2 ) = P(1) × P(1) = × = = 0. 04
10 5 50
Hence, the required probability distribution is as below
X 0 1 2
P (X) 0.54 0.42 0.04

Q. 30 Two probability distributions of the discrete random variables X and Y

are given below.
X 0 1 2 3
1 2 1 1
P(X)
5 5 5 5

Y 0 1 2 3
1 3 2 1
P (Y)
5 10 5 10

Prove that E (Y 2 ) = 2E ( X ).
Sol. X 0 1 2 3
1 2 1 1
P (X)
5 5 5 5

Y 0 1 2 3
1 3 2 1
P (Y) 5 10 5 10
Since, we have to prove that, E (Y 2 ) = 2 E ( X )
\ E( X ) = S X P ( X )
1 2 1 1 7
= 0 × + 1× + 2 × + 3 × =
5 5 5 5 5
14
Þ 2 E( X ) = ...(i)
5
E (Y )2 = SY 2 P (Y )
1 3 2 1
= 0×+ 1× + 4× + 9×
5 10 5 10
3 8 9 28 14
= + + = =
10 5 10 10 5
2 14
Þ E (Y ) = ...(ii)
5
From Eqs. (i) and (ii),
E(Y 2 ) = 2 E( X ) Hence proved.
Q. 31 A factory produces bulbs. The probability that any one bulb is
1
defective is and they are packed in 10 boxes. From a single box,
50
find the probability that
(i) none of the bulbs is defective.
(ii) exactly two bulbs are defective.
(iii) more than 8 bulbs work properly
Sol. Let X is the random variable which denotes that a bulb is defective.
1 49
Also, n = 10, p = and q = and P ( X = r ) = nC r pr q n - r
50 50
(i) None of the bulbs is defective i.e., r = 0
1 0 49 10 - 0 æ 49 ö10
\ p( X = r ) = P( 0 ) = 10C 0 æç ö÷ æç ö÷ =ç ÷
è 50 ø è 50 ø è 50 ø
(ii) Exactly two bulbs are defective i.e., r = 2
1 2 49 8
\ P ( X = r ) = P( 2 ) = 10
C 2 æç ö÷ æç ö÷
è 50 ø è 50 ø
2 8 10
10! æ 1 ö æ 49 ö æ 1 ö 8
= ç ÷ × ç ÷ = 45 ´ ç ÷ ´ (49)
8!2 ! è 50 ø è 50 ø è 50 ø
(iii) More than 8 bulbs work properly i.e., there is less than 2 bulbs which are defective.
So, r < 2 Þ r = 0, 1
\ P ( X = r ) = P (r < 2 ) = P (0) + P (1)
0 10 - 0 10 -1
1 49 1 1 49
= C 0 æç ö÷ æç ö÷
10
+ C1 æç ö÷ æç ö÷
10
è 50 ø è 50 ø è 50 ø è 50 ø
10
49 10! 1 æ 49 ö 9
= æç ö÷ + × ×ç ÷
è ø
50 1! 9! 50 è 50 ø
10 9 9
49 1 æ 49 ö 49 49 1 ö
= æç ö÷ + × ç ÷ = æç ö÷ æç + ÷
è 50 ø 5 è 50 ø è 50 ø è 50 5 ø
49 9 59 59 (49)9
= æç ö÷ æç ö÷ =
è 50 ø è 50 ø (50)10

Q. 32 Suppose you have two coins which appear identical in your pocket.
You know that, one is fair and one is 2 headed. If you take one out,
toss it and get a head, what is the probability that it was a fair coin?
Sol. Let E1 = Event that fair coin is drawn
E2 = Event that 2 headed coin is drawn
E = Event that tossed coin get a head
\ P (E1 ) = 1 / 2, P (E2 ) = 1 / 2, P (E / E1 ) = 1 / 2 and P (E / E2 ) = 1
P (E1 ) × P(E / E1 )
Now, using Baye’s theorem P (E1 / E ) =
P (E1 ) × P (E / E1 ) + P (E2 ) × P (E / E2 )
1 1 1 1
×
2 2 4 1
= = = 4 =
1 1 1 1 1 3 3
× + ×1 +
2 2 2 4 2 4
Q. 33 Suppose that 6% of the people with blood group O are left handed and
10% of those with other blood groups are left handed, 30% of the
people have blood group O. If a left handed person is selected at
random, what is the probability that he/she will have blood group O?
Sol. Other than
Blood group ‘O’ blood group ‘O’
I. Number of people 30 % 70 %
Percentage of left
II. 6% 10 %
handed people
E1 = Event that the person selected is of blood group O
E2 = Event that the person selected is of other than blood group O
(E3 ) = Event that selected person is left handed
\ P (E1 ) = 0. 30, P (E2 ) = 070
.
P (E3 / E1 ) = 0.06 and P (E3 / E2 ) = 010 .
P (E1 ) × P (E3 / E1 )
By using Baye’s theorem, P (E1 / E3 ) =
P (E1 ) × P (E3 / E1 ) + P (E2 ) × P (E3 / E2 )
0. 30 ´ 0.06
=
0. 30 × 0.06 + 070
. × 010
.
0.0180
=
0. 0180 + 0. 0700
0. 0180 180 9
= = =
0.0880 880 44

Q. 34 If two natural numbers r and s are drawn one at a time, without

replacement from the set S = { 1, 2, 3, ... n}, then find P (r £ p / s £ p),
where p Î S .
Sol. Q Set S = {1, 2, 3, ..., n}
P( p Ç S )
\ P (r < p/s < p) =
P (S )
p-1 n p-1
= ´ =
n n-1 n - 1

Q. 35 Find the probability distribution of the maximum of the two scores

obtained when a die is thrown twice. Determine also the mean of the
distribution.
Sol. Let X is the random variable score obtained when a die is thrown twice.
\ X = 1, 2, 3, 4, 5, 6
Here, S = {(1, 1), (1, 2 ), (2, 1), (2, 2 ),(1, 3), (2, 3), (3, 1), (3, 2 ), (3, 3), ...,(6, 6)}
1 1 1
\ P ( X = 1) = × =
6 6 36
1 1 1 1 1 1 3
P( X = 2 ) = × + × + × =
6 6 6 6 6 6 36
1 1 1 1 1 1 1 1 1 1 5
P ( X = 3) = × + × + × + × + × =
6 6 6 6 6 6 6 6 6 6 36
 7
Similarly, P ( X = 4) =
36
9
P ( X = 5) =
36
11
P ( X = 6) =
36
So, the required distribution is,
X 1 2 3 4 5 6
P(x) 1/36 3/36 5/36 7/36 9/36 11/36
Also, we know that, Mean {E ( X )} = S X P ( X )
1 6 15 28 45 66 161
= + + + + + =
36 36 36 36 36 36 36

Q. 36 The random variable X can take only the values 0, 1, 2. If

P ( X = 0) = P ( X = 1) = p and E ( X 2 ) = E[ X ],
then find the value of p.
Sol. Since, X = 0, 1, 2 and P (X) at X = 0 and 1 is p, let at X = 2, P (X) is x.
Þ p+ p+ x =1
Þ x = 1- 2p
We get, the following distribution.
X 0 1 2
P (X) p p 1- 2 p
\ E[ X ] = S X P ( X )
= 0 × p + 1 × p + 2 (1 - 2 p)
= p + 2 - 4p = 2 - 3p
and E ( X 2 ) = S X 2 P( X )
= 0 × p + 1 × p + 4 × (1 - 2 p)
= p + 4 - 8p = 4 - 7 p
Also, given that E ( X 2 ) = E [ X ]
Þ 4 - 7 p = 2 - 3p
1
Þ 4p = 2 Þ p =
2

Q. 37 Find the variance of the following distribution.

X 0 1 2 3 4 5
1 5 2 1 1 1
P (X)
6 18 9 6 9 18
Sol. We have,
X 0 1 2 3 4 5
P (X) 1 5 2 1 1 1
6 18 9 6 9 18
X P (X) 5 4 1 4 5
0
18 9 2 9 18
5 8 3 16 25
X 2 P (X ) 0
18 9 2 9 18
 \ Variance = E ( X 2 ) - [E ( X )]2 = SX 2 P ( X ) - [S X P ( X )]2
2
5 8 3 16 25 ù é 5 4 1 4 5ù
= é0 + + + + + - 0+ + + + +
êë 18 9 2 9 18 úû êë 18 9 2 9 18 úû
5 + 16 + 27 + 32 + 25 ù é 5 + 8 + 9 + 8 + 5 ù 2
=é -
êë 18 úû êë 18 úû
105 35 × 35 18 × 105 - 35 × 35
= - =
18 18 × 18 18 × 18
35 19 × 35 665
= [54 - 35] = =
18 × 18 324 324

Q. 38 A and B throw a pair of dice alternately. A wins the game, if he gets a

total of 6 and B wins, if she gets a total of 7. If A starts the game, then
find the probability of winning the game by A in third throw of the
pair of dice.
Sol. Let A1 = A total of 6 = {(2, 4), (1, 5), (5, 1), (4, 2), (3, 3)}
and B1 = A total of 7 = {(2, 5), (1, 6), (6, 1), (5, 2), (3, 4), (4, 3)}
5
Let P (A) is the probability, if A wins in a throw Þ P ( A) =
36
1
and P (B) is the probability, if B wins in a throw Þ P (B) =
6
31 5 5 775 775
\ Required probability = P ( A ) × P (B ) × P ( A) = × × = =
36 6 36 216 × 36 7776

Q. 39 Two dice are tossed. Find whether the following two events A and B
are independent A = {(x, y) : x + y = 11} and B = {(x, y): x ¹ 5}, where
(x, y) denotes a typical sample point.
Sol. We have, A = {(x, y) : x + y = 11} and B = {(x, y): x ¹ 5}
\ A = {(5, 6), (6, 5 )}, B = {(1,1), (1,2), (1,3), (1,4), (1,5) (1,6), (2,1), (2,2), (2,3), (2,4),
(2,5) (2,6), (3,1), (3,2), (3,3), (3,4), (3,5) (3,6), (4,1), (4,2), (4,3), (4,4), (4,5) (4,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Þ n( A) = 2, n(B) = 30 and n( A Ç B) = 1
2 1 30 5
\ P( A) = = and P (B) = =
36 18 36 6
5 1
Þ P ( A) × P (B) = and P ( A Ç B) = ¹ P ( A) × P (B)
108 36
So, A and B are not independent.

Q. 40 An urn contains m white and n black balls. A ball is drawn at random

and is put back into the urn along with k additional balls of the same
colour as that of the ball drawn. A ball is again drawn at random.
Show that the probability of drawing a white ball now does not
depend on k.
Sol. Let U = {m white, n black balls}
E1 = {First ball drawn of white colour}
E2 = {First ball drawn of black colour}
 and E3 = {Second ball drawn of white colour}
m n
\ P (E1 ) = and P (E2 ) =
m+ n m+ n
m+ k m
Also, P (E3 / E1 ) = and P (E3 / E2 ) =
m+ n+ k m+ n+ k
\ P (E3 ) = P (E1 ) × P (E3 / E1 ) + P (E2 ) × P (E3 / E2 )
m m+ k n m
= × + ×
m+ n m+ n+ k m+ n m+ n+ k
m (m + k ) + nm m2 + mk + nm
= =
(m + n + k ) (m + n) (m + n + k ) (m + n)
m (m + k + n) m
= =
(m + n + k ) (m + n) m + n
Hence, the probability of drawing a white ball does not depend on k.

Long Answer Type Questions

Q. 41 Three bags contain a number of red and white balls as follows Bag I : 3
red balls, Bag II : 2 red balls and 1 white ball and Bag III : 3 white
balls. The probability that bag i will be chosen and a ball is selected
i
from it is , where i =1, 2, 3. What is the probability that
6
(i) a red ball will be selected? (ii) a white ball is selected?
Sol. Bag I : 3 red balls and 0 white ball.
Bag II : 2 red balls and 1 white ball.
Beg III : 0 red ball and 3 white balls.
Let E1, E2 and E3 be the events that bag I , bag II and bag III is selected and a ball is chosen
from it.
1 2 3
P (E1 ) = , P (E2 ) = and P (E3 ) =
6 6 6
(i) Let E be the event that a red ball is selected. Then, probability that red ball will be
selected
P (E ) = P (E1 ) × P (E / E1 ) + P (E2 ) × P (E / E2 ) + P (E3 ) × P (E / E3 )
1 3 2 2 3
= × + × + ×0
6 3 6 3 6
1 2
= + + 0
6 9
3+ 4 7
= =
18 18
(ii) Let F be the event that a white ball is selected.
\ P (F ) = P (E1 ) × P (F / E1 ) + P (E2 ) × P (F / E2 ) + P (E3 ) × P (F / E3 )
1 2 1 3 1 3 11
= × 0 + × + ×1= + =
6 6 3 6 9 6 18
7 11
Note P (F) = 1 - P (E) = 1 - = [since, we know that P (E) + P (F) =1]
18 18
Q. 42 Refer to question 41 above. If a white ball is selected, what is the
probability that it came from
(i) Bag II? (ii) Bag III?
Sol. Referring to the previous solution, using Bay’s theorem, we have
P ( E2 ) × P ( F / E2 )
(i) P (E2 / F ) =
P (E1 ) × P (F / E1 ) + P (E2 ) × P (F / E2 ) + P (E3 ) × P (F / E3 )
2 1 2
×
6 3 18
= =
1 2 1 3 2 3
× 0+ × + ×1 +
6 6 3 6 18 6
2 / 18 2
= =
2 + 9 11
18
P ( E3 ) × P ( F / E3 )
(ii) P (E3 / F ) =
P (E1 ) × P (F / E1 ) + P (E2 ) × P (F / E2 ) + P (E3 ) × P (F / E3 )
3
×1
= 6
1 2 1 3
× 0+ × + ×1
6 6 3 6
3
6 3/ 6 9
= = =
2 3 2 9 11
+ +
18 6 18 18

Q. 43 A shopkeeper sells three types of flower seeds A 1 , A2 and A 3 . They are

sold as a mixture, where the proportions are 4 : 4 : 2, respectively. The
germination rates of the three types of seeds are 45%, 60% and 35%.
Calculate the probability
(i) of a randomly chosen seed to germinate.
(ii) that it will not germinate given that the seed is of type A 3 .
(iii) that it is of the type A2 given that a randomly chosen seed does not
germinate.
Sol. We have, A1 : A2 : A3 = 4 : 4 : 2
4 4 2
P ( A1 ) = , P ( A2 ) = and P( A3 ) =
10 10 10
where A1, A2 and A3 denote the three types of flower seeds.
Let E be the event that a seed germinates and E be the event that a seed does not
germinate.
45 60 35
\ P (E / A1 ) = , P (E / A2 ) = and P (E / A3 ) =
100 100 100
55 40 65
and P (E / A1 ) = , P (E / A2 ) = and P (E / A3 ) =
100 100 100
(i) \ P (E ) = P ( A1 ) × P (E / A1 ) + P ( A2 ) × P (E / A2 ) + P ( A3 )× P (E / A3 )
4 45 4 60 2 35
= × + × + ×
10 100 10 100 10 100
180 240 70 490
= + + = = 0.49
1000 1000 1000 1000
 35 65
(ii) P (E / A3 ) = 1 - P (E / A3 ) = 1 - = [as given above]
100 100
P ( A2 ) × P (E / A2 )
(iii) P ( A2 / E ) =
P ( A1 ) × P ( E / A1 ) + P ( A2 ) × P (E / A2 ) + P ( A3 ) × P (E / A3 )
4 40 160
×
10 100 1000
= =
4 55 4 40 2 65 220 160 130
× + × + × + +
10 100 10 100 10 100 1000 1000 1000
160 / 1000 16
= = = 0. 313725 = 0.314
510 / 1000 51

Q. 44 A letter is known to have come either from ‘TATA NAGAR’ or from

‘CALCUTTA’. On the envelope, just two consecutive letters TA are visible.
What is the probability that the letter came from ‘TATA NAGAR’?
Sol. Let E1 be the event that letter is from TATA NAGAR and E2 be the event that letter is from
CALCUTTA.
Also, let E3 be the event that on the letter, two consecutive letters TA are visible.
1 1
\ P (E1 ) = and P (E2 ) =
2 2
2 1
and P (E3 / E1 ) = and P (E3 / E2 ) =
8 7
[since, if letter is from TATA NAGAR, we see that the events of two consecutive letters visible
2
are {TA, AT, TA, AN, NA, AG, GA , AR }. So, P (E3 / E1 ) = and if letter is from CALCUTTA,
8
we see that the events of two consecutive letters to visible are {CA, AL, LC, CU, UT, TT, TA}.
1
So, P (E3 / E2 ) = ]
7
P (E1 ) × P (E3 / E1 )
\ P (E1 / E3 ) =
P (E1 ) × P (E3 / E1 ) + P (E2 ) × P (E3 / E2 )
1 2 1 1
×
2 8 8 1/ 8 7
= = = = 8 =
1 2 1 1 1 1 22 11 11
× + × +
2 8 2 7 8 14 8 ´ 14 56

Q. 45 There are two bags, one of which contains 3 black and 4 white balls
while the other contains 4 black and 3 white balls. A die is thrown. If
it shows up 1 or 3, a ball is taken from the Ist bag but it shows up any
other number, a ball is chosen from the II bag. Find the probability of
choosing a black ball.
Sol. Since, Bag I = {3 black, 4 white balls}, Bag II = {4 black, 3 white balls}
Let E1 be the event that bag I is selected and E2 be the event that bag II is selected.
Let E3 be the event that black ball is chosen.
1 1 1 1 2
\ P (E1 ) = + = and P (E2 ) = 1 - =
6 6 3 3 3
3 4
and P (E3 / E1 ) = and P (E3 / E2 ) =
7 7
\ P (E3 ) = P (E1 ) ×P (E3 / E1 ) + P (E2 ) × P (E3 / E2 )
1 3 2 4 11
= × + × =
3 7 3 7 21
Q. 46 There are three urns containing 2 white and 3 black balls, 3 white and
2 black balls and 4 white and 1 black balls, respectively. There is an
equal probability of each urn being chosen. A ball is drawn at random
from the chosen urn and it is found to be white. Find the probability
that the ball drawn was from the second urn.
Sol. Let U1 = {2 white, 3 black balls }
U 2 = {3 white, 2 black balls}
and U 3 = {4 white, 1 black balls}
1
\ P (U1 ) = P (U 2 ) = P (U 3 ) =
3
Let E1 be the event that a ball is chosen from urn U1, E 2 be the event that a ball is chosen
from urn U 2 and E3 be the event that a ball is chosen from urn U 3 .
Also, P (E1 ) = P (E2 ) = P (E3 ) = 1 / 3
Now, let E be the event that white ball is drawn.
2 3 4
\ P (E / E1 ) = , P (E / E2 ) = , P (E / E3 ) =
5 5 5
P ( E2 ) × P ( E / E2 )
Now, P ( E2 / E ) =
P (E1 ) × P (E / E1 ) + P (E2 ) × P (E / E2 ) + P (E3 ) × P (E / E3 )
1 3
×
= 3 5
1 2 1 3 1 4
× + × + ×
3 5 3 5 3 5
3
15 3 1
= = =
2 3 4 9 3
+ +
15 15 15

Q. 47 By examining the chest X-ray, the probability that TB is detected

when a person is actually suffering is 0.99. The probability of an
healthy person diagnosed to have TB is 0.001. In a certain city, 1 in
1000 people suffers from TB. A person is selected at random and is
diagnosed to have TB. What is the probability that he actually has TB?
Sol. Let E1 = Event that person has TB
E2 = Event that person does not have TB
E = Event that the person is diagnosed to have TB
1 999
\ P (E1 ) = = 0.001, P (E2 ) = = 0. 999
1000 1000
and P (E / E1 ) = 0. 99 and P (E / E2 ) = 0. 001
P (E1 ) × P (E / E1 )
\ P (E1 / E ) =
P (E1 ) × P (E / E1 ) + P (E2 ) × P (E / E2 )
0.001 ´ 0. 99
=
0.001 ´ 0. 99 + 0. 999 ´ 0.001
0.000990
=
0.000990 + 0.000999
990 110
= =
1989 221
Q. 48 An item is manufactured by three machines A, B and C. Out of the
total number of items manufactured during a specified period, 50% are
manufactured on A, 30% on B and 20% on C. 2% of the items produced
on A and 2% of items produced on B are defective and 3% of these
produced on C are defective. All the items are stored at one godown.
One item is drawn at random and is found to be defective. What is the
probability that it was manufactured on machine A?
Sol. Let E1 = Event that item is manufactured on A,
E2 = Event that an item is manufactured on B,
E3 = Event that an item is manufactured on C,
Let E be the event that an item is defective.
50 1 30 3 20 1
\ P (E1 ) = = , P ( E2 ) = = and P (E3 ) = =
100 2 100 10 100 5
æEö 2 1 æ E ö 2 1 æ E ö 3
P çç ÷÷ = = , P çç ÷= ç ÷
÷ 100 = 50 and P ç E ÷ = 100
E
è 1ø 100 50 è E2 ø è 3ø
æEö
P (E1 ) × P çç ÷÷
\
æE ö
Pç 1 ÷ = è E1 ø
èEø æEö æ E ö æ E ö
P (E1 ) × P çç ÷÷ + P (E2 ) × P çç ÷÷ + P (E3 ) × P çç ÷
÷
è E1 ø è E2 ø è E3 ø
1 1
×
2 50
=
1 1 3 1 1 3
× + × + ×
2 50 10 50 5 100
1 1
100 100 5
= = =
1 3 3 5 + 3 + 3 11
+ +
100 500 500 500

Q. 49 Let X be a discrete random variable whose probability distribution is

defined as follows.
ì k(x + 1), for x = 1, 2, 3, 4
ï
P ( X = x) = í2kx, for x = 5, 6, 7
ï0 , otherwise
î
where, k is a constant. Calculate
(i) the value of k. (ii) E ( X ).
(iii) standard deviation of X.
ìk(x + 1), for x = 1, 2 , 3, 4
ï
Sol. P ( X = x ) = í2 kx, for x = 5, 6, 7
ï0, otherwise
î
Thus, we have following table
X 1 2 3 4 5 6 7 Otherwise
P (X ) 2k 3k 4k 5k 10k 12k 14k 0
XP (X ) 2k 6k 12k 20k 50k 72k 98k 0
X 2 P (X ) 2k 12k 36k 80k 250k 432k 686k 0
 (i) Since, SPi = 1
1
Þ k(2 + 3 + 4 + 5 + 10 + 12 + 14) = 1 Þ k =
50
(ii) Q E ( X ) = S XP ( X )
\ E ( X ) = 2 k + 6k + 12 k + 20 k + 50 k + 72 k + 98 k + 0 = 260 k
1 26 éQ k = 1 ù ...(i)
= 260 ´ = = 5.2
50 5 êë 50 úû
(iii) We know that,
Var( X ) = [E ( X 2 )] - [E ( X )]2 = SX 2 P ( X ) - [S{ XP ( X )}]2
= [2 k + 12 k + 36 k + 80 k + 250 k + 432 k + 686 k + 0] - [5.2 ] 2 [using Eq. (i)]
1 éQ k = 1 ù
= [1498k ] - 27.04 = é1498 ´ ù - 27.04
êë 50 úû êë 50 úû
= 29. 96 - 27.04 = 2. 92
We know that, standard deviation of X = Var( X ) = 2. 92 = 17088
. = 17
. (approx)

Q. 50 The probability distribution of a discrete random variable X is given as

under
X 1 2 4 2A 3A 5A
1 1 3 1 1 1
P (X )
2 5 25 10 25 25

Calculate
(i) the value of A, if E ( X ) = 2. 94.
(ii) variance of X.
1 2 12 2 A 3 A 5 A
Sol. (i) We have, S XP ( X ) = + + + + +
2 5 25 10 25 25
25 + 20 + 24 + 10 A + 6 A + 10 A 69 + 26 A
= =
50 50
Since, E ( X ) = S XP ( X )
69 + 26 A
Þ 2. 94 =
50
Þ 26 A = 50 ´ 2. 94 - 69
147 - 69 78
Þ A= = =3
26 26
(ii) We know that,
Var( X ) = E ( X 2 ) - [E ( X )]2
= S X 2 P ( X ) - [ SXP ( X )]2
1 4 48 4 A 2 9 A 2 25 A 2
= + + + + + - [E ( X )]2
2 5 25 10 25 25
25 + 40 + 96 + 20 A 2 + 18 A 2 + 50 A 2
= - [E ( X )]2
50
161 + 88 A 2 161 + 88 ´ (3)2
= - [E ( X )]2 = - [E ( X )]2 [Q A = 3]
50 50
953
= - [2. 94]2 [Q E ( X ) = 2. 94]
50
= 19. 0600 - 8. 6436 = 10.4164
Q. 51 The probability distribution of a random variable x is given as under
ì kx 2 , x = 1, 2, 3
ï
P(X = x) = í2kx, x = 4 , 5, 6
ï0 , otherwise
î
where, k is a constant. Calculate
(i) E ( X ) (ii) E (3 X 2 ) (iii) P ( X ³ 4)
Sol. X 1 2 3 4 5 6 Otherwise
P (X ) k 4k 9k 8k 10k 12k 0
We know that, S Pi = 1
1
Þ 44k = 1 Þ k =
44
\ SXP ( X ) = k + 8k + 27 k + 32 k + 50k + 72 k + 0
1 95
= 190k = 190 ´ =
44 22
95
(i) So, E ( X ) = SXP ( X ) = = 4. 32
22
2 2
(ii) Also, E ( X ) = SX P ( X ) = k + 16 k + 81k + 128k + 250 k + 432 k
1 éQ k = 1 ù
= 908 k = 908 ´
44 êë 44 úû
= 20. 636 = 20. 64 (approx)
\ E (3 X 2 ) = 3 E ( X 2 ) = 3 ´ 20. 64 = 61. 92 = 61. 9
(iii) P ( X ³ 4) = P ( X = 4) + P ( X = 5) + P ( X = 6)
1 15
= 8k + 10k + 12 k = 30k = 30 × =
44 22

Q. 52 A bag contains (2n + 1) coins. It is known that n of these coins have a

head on both sides whereas the rest of the coins are fair. A coin is
picked up at random from the bag and is tossed. If the probability that
the toss results in a head is 31, then determine the value of n.
42
Sol. Given, n coins have head on both sides and (n + 1) coins are fair coins.
Let E1 = Event that an unfair coin is selected
E2 = Event that a fair coin is selected
E = Event that the toss results in a head
n n+1
\ P (E1 ) = and P (E2 ) =
2n + 1 2n + 1
æEö æ E ö 1
Also, P çç ÷÷ = 1 and P çç ÷÷ =
E
è 1ø è E2 ø 2
æEö æ E ö n n+1 1
\ P (E ) = P (E1 ) × P çç ÷÷ + P (E2 ) × P çç ÷÷ = ×1+ ×
E
è 1ø E
è 2ø 2 n + 1 2 n+1 2
31 2 n + n + 1 31 3 n + 1
Þ = Þ =
42 2 ( 2 n + 1) 42 4 n + 2
Þ 124 n + 62 = 126 n + 42
Þ 2 n = 20 Þ n = 10
Q. 53 Two cards are drawn successively without replacement from a well
shuffled deck of cards. Find the mean and standard variation of the
random variable X , where X is the number of aces.
Sol. Let X denotes a random variable of number of aces.
\ X = 0, 1, 2
48 47 2256
Now, P ( X = 0) = × =
52 51 2652
48 4 4 48 384
P ( X = 1) = × + × =
52 51 52 51 2652
4 3 12
P( X = 2 ) = × =
52 51 2652
X 0 1 2
P (X ) 2256 384 12
2652 2652 2652
XP (X ) 384 24
0
2652 2652
384 48
X 2 P (X ) 0
2652 2652
We know that, Mean (m) = E ( X ) = SXP( X )
384 24
=0+ +
2652 2652
408 2
= =
2652 13

Also, Var ( X ) = E ( X 2 ) - [E ( X )]2 = SX 2 P ( X ) - [E ( X )]2

2
384 48 ù æ 2 ö éQ E ( X ) = 2 ù
= é0 + + -ç ÷
ëê 2652 2652 ûú è 13 ø ëê 13 úû
432 4
= - = 01628
. - 0.0236 = 01391
.
2652 169
\ Standard deviation = Var( X ) = 0139. = 0. 373 (approx)

Q. 54 A die is tossed twice. If a ‘success’ is getting an even number on a toss,

then find the variance of the number of successes.
Sol. Let X be the random variable for a ‘success’ for getting an even number on a toss.
3 1 1
\ X = 0, 1, 2, n = 2, p = = and q =
6 2 2
1 0 1 2-0 1
At X = 0, P ( X = 0) = 2C 0 æç ö÷ æç ö÷ =
è2 ø è2 ø 4
1 1 1 2 -1 1 1 1
At X = 1, P ( X = 1) = 2C1 æç ö÷ æç ö÷ = 2 × × =
è2 ø è2 ø 2 2 2
1 2 1 2-2 1
At X = 2, P ( X = 2 ) = 2C 2 æç ö÷ æç ö÷ =
è2 ø è2 ø 4
 Thus, X 0 1 2
P (X ) 1 1 1
4 2 4
XP (X ) 1 1
0
2 2
1
X 2 P (X ) 0 1
2
1 1
\ SXP ( X ) = 0 + + =1 ...(i)
2 2
1 3
and SX 2 P ( X ) = 0 + + 1 = ...(ii)
2 2
Q Var( X ) = E ( X 2 ) - [E ( X )]2
3 1
= SX 2 P ( X ) - [SXP ( X )]2 = - (1)2 = [using Eqs. (i) and (ii)]
2 2

Q. 55 There are 5 cards numbered 1 to 5, one number on one card. Two cards
are drawn at random without replacement. Let X denotes the sum of
the numbers on two cards drawn. Find the mean and variance of X .
Sol. Here, S = {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2), (1, 4), (4, 1), (1, 5), (5, 1), (2, 4), (4, 2),
(2, 5), (5, 2), (3, 4), (4, 3), (3, 5), (5, 3), (5, 4), (4, 5)}.
Þ n(S ) = 20
Let random variable be X which denotes the sum of the numbers on two cards drawn.
\ X = 3, 4, 5, 6, 7, 8, 9
2 1
At X = 3, P ( X ) = =
20 10
2 1
At X = 4, P ( X ) = =
20 10
4 1
At X = 5, P ( X ) = =
20 5
4 1
At X = 6, P ( X ) = =
20 5
4 1
At X = 7, P ( X ) = =
20 5
2 1
At X = 8, P ( X ) = =
20 10
2 1
At X = 9, P ( X ) = =
20 10
3 4 5 6 7 8 9
\ Mean, E ( X ) = SX P ( X ) = + + + + + +
10 10 5 5 5 10 10
3 + 4 + 10 + 12 + 14 + 8 + 9
= =6
10
9 16 25 36 49 64 81
Also, SX 2 P ( X ) = + + + + + +
10 10 5 5 5 10 10
9 + 16 + 50 + 72 + 98 + 64 + 81
= = 39
10
\ Var( X ) = SX 2 P ( X ) - [SXP ( X )]2
= 39 - (6)2 = 39 - 36 = 3
Objective Type Questions
Q. 56 If P (A)= 4 and P ( A Ç B) = 7
, then P (B / A) is equal to
5 10
1 1 7 17
(a) (b) (c) (d)
10 8 8 20
4 7
Sol. (c) Q P ( A) = , P ( A Ç B) =
5 10
P ( A Ç B) 7 / 10 7
\ P (B / A) = = =
P ( A) 4/ 5 8

7 17
Q. 57 If P ( A Ç B) = and P (B) = , then P ( A / B) equals to
10 20
14 17 7 1
(a) (b) (c) (d)
17 20 8 8
7 17
Sol. (a) Here, P ( A Ç B) = and P (B) =
10 20
P ( A Ç B) 7 / 10 14
\ P ( A / B) = = =
P (B) 17 / 20 17

3 2 3
Q. 58 If P ( A) = , P (B) = and P ( A È B) = , then P (B / A ) + P ( A / B)
10 5 5
equals to
1 1 5 7
(a) (b) (c) (d)
4 3 12 12
3 2 3
Sol. (d) Here, P( A) = , P(B) and P ( A È B) =
10 5 5
P (B Ç A) P ( A Ç B)
P (B / A) + P ( A / B) = +
P ( A) P (B)
P ( A) + P (B) - P ( A È B) P ( A) + P (B) - P ( A È B)
= +
P( A) P (B)
éQ P ( A È B) = P ( A) + P (B) - P ( A Ç B)ù
ê i .e., P( A Ç B) = P( A) + P(B) - P( A È B) ú
ë û
3 2 3 3 2 3
+ - + -
= 10 5 5 + 10 5 5
3 2
10 5
1 1
10 10 1 1 7
= + = + =
3 2 3 4 12
10 5
 2 3 1
Q. 59 If P ( A) = , P (B) = and P ( A Ç B) = , then P ( A¢ / B¢ ) × P (B¢ / A¢ ) is
5 10 5
equal to
5 5 25
(a) (b) (c) (d) 1
6 7 42
2 3 1
Sol. (c) Here, P( A) = , P (B) = and P( A Ç B) =
5 10 5
P ( A¢ Ç B¢) 1 - P ( A È B)
P ( A¢/ B¢) = =
P (B¢) 1 - P (B)
1 - [P ( A) + P (B) - P ( A Ç B)]
=
1 - P (B)
2 3 1
1 - æç + - ö÷
= è 5 10 5 ø
3
1-
10
4+ 3-2ö
1 - æç
1
÷ 1-
è 10 ø 2 5
= = =
7 7 7
10 10
P (B¢ Ç A¢) 1 - P ( A È B)
and P (B¢/ A¢) = =
P ( A ¢) 1 - P( A)
1
1-
= 2 = 1/ 2 = 5 éQ P ( A È B) = 1 ù
2 3/ 5 6 êë 2 úû
1-
5
5 5 25
\ P ( A¢/ B¢) × P (B¢/ A¢) = × =
7 6 42

1 1
Q. 60 If A and B are two events such that P ( A) = , P (B) = and
2 3
1
P ( A / B) = , then P ( A¢ Ç B¢ ) equals to
4
1 3
(a) (b)
12 4
1 3
(c) (d)
4 16
1 1 1
Sol. (c) Here, P( A) = , P(B) = and P( A / B) =
2 3 4
P( A Ç B)
Q P ( A / B) =
P (B)
1 1 1
Þ P ( A Ç B) = P ( A / B) × P (B) = × =
4 3 12
Now, P ( A¢ Ç B¢) = 1 - P ( A È B)
= 1 - [P ( A) + P (B) - P ( A Ç B)]
1 1 1ù 6 + 4 - 1ù
= 1- é + - = 1- é
êë 2 3 12 úû êë 12 úû
9 3 1
= 1- = =
12 12 4
Q. 61 If P ( A) = 0.4, P (B) = 0. 8 and P (B / A) = 0.6, then P ( A È B ) is equal to
(a) 0.24 (b) 0.3
(c) 0.48 (d) 0.96
Sol. (d) Here, P( A) = 0.4, P(B) = 0.8 and P( A / B) = 0.6
P (B Ç A)
Q P (B / A) =
P( A)
Þ P (B Ç A) = P (B / A) × P ( A)
= 0. 6 ´ 0. 4 = 0.24
Q P ( A È B ) = P ( A) + P (B) - P ( A Ç B)
= 0.4 + 0. 8 - 0.24
= 12
. - 0.24 = 0. 96

Q. 62 If A and B are two events and A ¹ f, B ¹ f, then

P ( A Ç B)
(a) P ( A / B ) = P ( A) × P (B) (b) P ( A / B ) =
P (B)
(c) P ( A / B ) × P (B / A) = 1 (d) P ( A / B ) = P ( A) / P (B)
Sol. (b) P ( A Ç B)
If A ¹ f and B ¹ f, then P ( A / B) =
P (B)

Q. 63 If A and B are events such that P ( A) = 0.4, P (B) = 0. 3 and

P ( A È B) = 0. 5, then P (B¢ Ç A) equals to
2 1
(a) (b)
3 2
3 1
(c) (d)
10 5
Sol. (d) Here, P ( A) = 0.4, P (B) = 0. 3 and P ( A È B) = 0. 5
Q P ( A È B) = P ( A) + P (B) - P ( A Ç B)
Þ P ( A Ç B) = 0.4 + 0. 3 - 0. 5 = 0.2
\ P (B¢ Ç A) = P ( A) - P ( A Ç B)
1
= 0.4 - 0.2 = 0.2 =
5

3 1
Q. 64 If A and B are two events such that P (B) = , P ( A / B) = and
5 2
4
P ( A È B) = , then P ( A) equals to
5
3 1 1 3
(a) (b) (c) (d)
10 5 2 5
3 1 4
Sol. (c) Here, P(B) = , P( A / B) = and P( A È B) =
5 2 5
P ( A Ç B)
Q P ( A / B) =
P (B)
1 P ( A Ç B)
Þ =
2 3/ 5
 3 1 3
Þ P ( A Ç B) = ´ =
5 2 10
and P ( A È B) = P ( A) + P (B) - P ( A Ç B)
4 3 3
Þ = P( A) + -
5 5 10
4 3 3 8-6+ 3 1
\ P( A) = - + = =
5 5 10 10 2

Q. 65 In question 64 (above), P (B / A¢ ) is equal to

1 3 1 3
(a) (b) (c) (d)
5 10 2 5
P (B Ç A¢) P (B) - P (B Ç A)
Sol. (d) P ( B / A ¢) = =
P ( A ¢) 1 - P( A)
3 3 6- 3
-
6 3
= 5 10 = 10 = =
1 1 10 5
1-
2 2

3 1 4
Q. 66 If P (B) = , P ( A / B) = and P ( A È B) = , then
5 2 5
P ( A È B)¢ + P ( A¢ È B) is equal to
1 4 1
(a) (b) (c) (d) 1
5 5 2
3 1
Sol. (d) Here, P (B) = , P ( A / B) =
5 2
4
and P ( A È B) =
5
P ( A Ç B)
Since, P ( A / B) =
P (B)
Þ P ( A Ç B) = P ( A / B) × P (B)
1 3 3
= ´ =
2 5 10
Also, P ( A È B) = P ( A) + P (B) - P ( A Ç B)
4 3 3 1
Þ P( A) = - + =
5 5 10 2
4 1
\ P ( A È B)¢ = 1 - P ( A È B) = 1 - =
5 5
and P ( A ¢ È B) = 1 - P ( A - B) = 1 - P ( A Ç B¢)
= 1 - P ( A) × P (B¢)
1 2 4
= 1- × =
2 5 5
1 4 5
Þ P ( A È B)¢ + P ( A ¢ È B) = + = = 1
5 5 5
 7 9 4
Q. 67 If P ( A) = , P (B) = and P ( A Ç B) = , then P ( A ¢ / B) is equal to
13 13 13
6 4 4 5
(a) (b) (c) (d)
13 13 9 9
7 9 4
Sol. (d) Here, P( A) = , P(B) = and P( A Ç B) =
13 13 13
P ( A ¢ Ç B) P (B) - P ( A Ç B)
\ P ( A ¢ / B) = =
P (B) P (B)
9 4 5
-
13 5
= 13 13 = =
9 9 9
13 13

Q. 68 If A and B are such events that P ( A) > 0 and P (B) ¹ 1, then P ( A ¢ / B ¢)

equals to
1 - P ( A È B)
(a) 1 - P ( A / B) (b) 1 - P ( A¢ / B) (c) (d) P ( A¢) /P (B ¢ )
P (B ¢ )
Sol. (c) Q P ( A) > 0 and P (B) ¹ 1
P ( A ¢ Ç B¢) 1 - P ( A È B)
P ( A ¢ / B¢) = =
P (B¢) P (B¢)

3 4
Q. 69 If A and B are two independent events with P ( A) = and P (B) = ,
5 9
then P ( A ¢ Ç B ¢) equals to
4 8 1 2
(a) (b) (c) (d)
15 45 3 9
Sol. (d) P ( A ¢ Ç B¢) = 1 - P ( A È B)
= 1 - [P ( A) + P (B) - P ( A Ç B )]
3 4 3 4
= 1- é + - ´ ù [QP ( A Ç B) = P ( A) × P (B)]
êë 5 9 5 9 úû
27 + 20 - 12 ù
= 1- é
35 10 2
= 1- = =
êë 45 úû 45 45 9

Q. 70 If two events are independent, then

(a) they must be mutually exclusive
(b) the sum of their probabilities must be equal to 1
(c) Both (a) and (b) are correct
(d) None of the above is correct
Sol. (d) If two events A and B are independent ,then we know that
P ( A Ç B) = P ( A) × P (B), P ( A) ¹ 0, P (B) ¹ 0
Since, A and B have a common outcome.
Further, mutually exclusive events never have a common outcome.
In other words, two independent events having non-zero probabilities of occurrence
cannot be mutually exclusive and conversely, i .e., two mutually exclusive events having
non-zero probabilities of outcome cannot be independent.
 3 5
Q. 71 If A and B be two events such that P ( A) = , P (B) = and
8 8
3
P ( A È B) = , then P ( A / B) × P ( A ¢ / B) is equal to
4
2 3 3 6
(a) (b) (c) (d)
5 8 20 25
3 5 3
Sol. (d) Here, P( A) = , P (B) = and P ( A È B) =
8 8 4
Q P ( A È B) = P ( A) + P (B) - P ( A Ç B)
3 5 3 3+ 5-6 2 1
Þ P ( A Ç B) = + - = = =
8 8 4 8 8 4
P ( A Ç B) 1 / 4 8 2
Q P ( A / B) = = = =
P (B) 5 / 8 20 5
P ( A ¢ Ç B) P (B) - P ( A Ç B)
and P ( A ¢ / B) = =
P (B) P (B)
5 1 5-2
-
3
= 8 4 = 8 =
5 5 5
8 8
2 3 6
\ P ( A / B) × P ( A ¢ / B) = × =
5 5 25

Q. 72 If the events A and B are independent, then P ( A Ç B) is equal to

(a) P ( A) + P (B) (b) P ( A) - P (B)
(c) P ( A) × P (B) (d) P ( A) / P (B)
Sol. (c) If A and B are independent, then P ( A Ç B) = P ( A) × P (B)

Q. 73 Two events E and F are independent. If P (E ) = 0. 3 and P (E È F ) = 0. 5,

then P (E / F ) - P (F / E ) equals to
2 3 1 1
(a) (b) (c) (d)
7 35 70 7
Sol. (c) Here, P(E ) = 0.3 and P(E È F ) = 0.5
Let P (F ) = x
Q P (E È F ) = P (E ) + P (F ) - P (E Ç F )
= P (E ) + P (F ) - P (E ) × P(F )
Þ 0. 5 = 0. 3 + x - 0. 3x
0.5 - 0.3 2
Þ x= = = P(F )
07. 7
P (E Ç F ) P (F Ç E )
\ P (E / F ) - P (F / E ) = -
P (F ) P (E )
P (E Ç F ) × P(E ) - P (F Ç E ) × P(F )
=
P (E ) × P (F )
P (E Ç F ) [P (E ) - P (F )]
= = P (E ) - P (F )
P (E Ç F )
3 2 21 - 20 1
= - = =
10 7 70 70
Q. 74 A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random
without replacement, them the probability of getting exactly one red
ball is
45 135 15 15
(a) (b) (c) (d)
196 392 56 29
Sol. (c) Probability of getting exactly one red (R ) ball = PR × PR × PR + PR × PR × PR + PR × PR × PR
5 3 2 3 5 2 3 2 5
= × + × × + × ×
8 7 6 8 7 6 8 7 6
15 15 15
= + +
4×7 × 6 4×7 × 6 4×7 × 6
5 5 5 15
= + + =
56 56 56 56

Q. 75 Refer to question 74 above. If the probability that exactly two of the

three balls were red, then the first ball being red, is
1 4 15 5
(a) (b) (c) (d)
3 7 28 28
Sol. (b) Let E1 = Event that first ball being red
and E2 = Event that exactly two of the three balls being red
\ P(E1 ) = PR × PR × PR + PR × PR × PR + PR × PR × PR + PR × PR × PR
5 4 3 5 4 3 5 3 4 5 3 2
= × × + × × + × × + × ×
8 7 6 8 7 6 8 7 6 8 7 6
60 + 60 + 60 + 30 210
= =
336 336
P (E1 Ç E2 ) = PR × PR × PR + PR × PR × PR
5 3 4 5 4 3 120
= × × + × × =
8 7 6 8 7 6 336
P (E1 Ç E2 ) 120 / 336 4
\ P (E2 / E1 ) = = =
P (E1 ) 210 / 336 7

Q. 76 Three persons A, B and C, fire at a target in turn, starting with A. Their

probability of hitting the target are 0.4, 0.3 and 0.2, respectively. The
probability of two hits is
(a) 0.024 (b) 0.188 (c) 0.336 (d) 0.452
Sol. (b) Here, P( A) = 0.4,P ( A ) = 0. 6,P (B) = 0. 3, P (B ) = 07,
.
P (C ) = 0.2 and P (C ) = 0. 8
\ Probability of two hits = PA × PB × PC + PA × PB × PC + PA × PB × PC
= 0.4 ´ 0. 3 ´ 0. 8 + 0.4 ´ 07
. ´ 0.2 + 0. 6 ´ 0. 3 ´ 0.2
= 0.096 + 0.056 + 0.036 = 0188
.

Q. 77 Assume that in a family, each child is equally likely to be a boy or a

girl. A family with three children is chosen at random. The probability
that the eldest child is a girl given that the family has atleast one girl is
1 1 2 4
(a) (b) (c) (d)
2 3 3 7
Sol. (d) Here, S = {(B, B, B), (G, G, G), (B, G, G), (G, B, G), (G, G, B), (G, B, B), (B, G, B), (B, B, G)}
E1 = Event that a family has atleast one girl, then
E1 ={(G, B, B), (B, G, B), (B, B, G), (G, G, B), (B, G, G), (G, B, G), (G, G, G)}
E2 = Event that the eldest child is a girl, then
E2 = {(G, B, B), (G, G, B), (G, B, G), (G, G, G)}
\ E1 Ç E2 = {(G, B, B), (G, G, B), (G, B, G), (G, G, G)}
P (E1 Ç E2 ) 4 / 8 4
\ P (E2 / E1 ) = = =
P (E1 ) 7/8 7

Q. 78 If a die is thrown and a card is selected at random from a deck of 52

playing cards, then the probability of getting an even number on the
die and a spade card is
1 1 1 3
(a) (b) (c) (d)
2 4 8 4
Sol. (c) Let E1 = Event for getting an even number on the die
and E2 = Event that a spade card is selected
3 1 13 1
\ P (E1 ) = = and P (E2 ) = =
6 2 52 4
1 1 1
Then, P (E1 Ç E2 ) = P (E1 ) × P (E2 ) = × =
2 4 8

Q. 79 A box contains 3 orange balls, 3 green balls and 2 blue balls. Three
balls are drawn at random from the box without replacement. The
probability of drawing 2 green balls and one blue ball is
3 2 1 167
(a) (b) (c) (d)
28 21 28 168
Sol. (a) Probability of drawing 2 green balls and one blue ball
= PG × PG × PB + PB × PG × PG + PG × PB × PG
3 2 2 2 3 2 3 2 2
= × × + × × + × ×
8 7 6 8 7 6 8 7 6
1 1 1 3
= + + =
28 28 28 28

Q. 80 A flashlight has 8 batteries out of which 3 are dead. If two batteries

are selected without replacement and tested, then probability that
both are dead is
33 9 1 3
(a) (b) (c) (d)
56 64 14 28
3 2 3
Sol. (d) Required probability = PD × PD = × =
8 7 28

Q. 81 If eight coins are tossed together, then the probability of getting

exactly 3 heads is
1 7 5 3
(a) (b) (c) (d)
256 32 32 32
Sol. (b) We know that, probability distribution P ( X = r ) = nC r ( p)r q n - r
1 1
Here, n = 8, r = 3, p = and q =
2 2
1 3 1 8-3 8! æ 1 ö
8
\ Required probability = 8C 3 æç ö÷ æç ö÷ = ç ÷
è2 ø è2 ø 5! 3! è 2 ø
8×7 × 6 1 7
= × =
3 × 2 16 × 16 32

Q. 82 Two dice are thrown. If it is known that the sum of numbers on the
dice was less than 6, the probability of getting a sum 3, is
1 5 1 2
(a) (b) (c) (d)
18 18 5 5
Sol. (c) Let E1 = Event that the sum of numbers on the dice was less than 6
and E2 = Event that the sum of numbers on the dice is 3
\ E1 = {(1, 4), (4, 1), (2, 3), (3, 2 ), (2, 2 ), (1, 3), (3, 1), (1, 2 ), (2, 1), (1, 1}
)
Þ n(E1 ) = 10
and E2 = {(1, 2 ), (2, 1)} Þ n(E2 ) = 2
2 1
\ Required probability = =
10 5

Q. 83 Which one is not a requirement of a Binomial distribution?

(a) There are 2 outcomes for each trial
(b) There is a fixed number of trials
(c) The outcomes must be dependent on each other
(d) The probability of success must be the same for all the trials
Sol. (c) We know that, in a Binomial distribution,
(i) There are 2 outcomes for each trial.
(ii) There is a fixed number of trials.
(iii) The probability of success must be the same for all the trials.

Q. 84 If two cards are drawn from a well shuffled deck of 52 playing cards
with replacement, then the probability that both cards are queens, is
1 1 1 1 1 1 1 4
(a) × (b) + (c) × (d) ×
13 13 13 13 13 17 13 51
4 4 1 1
Sol. (a) Required probability = × = ´ [with replacement]
52 52 13 13

Q. 85 The probability of guessing correctly atleast 8 out of 10 answers on a

true false type examination is
7 7 45 7
(a) (b) (c) (d)
64 128 1024 41
Sol. (b) We know that, P ( X = r ) = nC r ( p)r (q )n - r
1 1
Here, n = 10, p = , q =
2 2
and r ³ 8 i .e., r = 8, 9, 10
 Þ P ( X = r ) = P(r = 8) + P(r = 9) + P(r = 10)
8 10 - 8 9 10 0
1 1 1 1 1 1
= 10C 8 æç ö÷ æç ö÷ + 10C 9 æç ö÷ æç ö÷ + C10 æç ö÷ . æç ö÷
10
è2 ø è2 ø è2 ø è2 ø è2 ø è2 ø
10 10 10
10! æ 1 ö 10! æ 1 ö æ 1ö
= ç ÷ + ç ÷ + ç ÷
8!2 ! è 2 ø 9!1! è 2 ø è2 ø
1 10 1
10
= æç ö÷ × [45 + 10 + 1] = æç ö÷ × 56
è2 ø è2 ø
1 7
= × 56 =
16 × 64 128

Q. 86 If the probability that a person is not a swimmer is 0.3, then the

probability that out of 5 persons 4 are swimmers is
(a) 5C 4(0.7) 4 (0. 3) (b) 5C1 (0.7) (0. 3) 4
(c) 5C 4 (0.7) (0. 3) 4 (d) (0.7) 4 (0. 3)
Sol. (a) Here, p = 0. 3 Þ p = 07
. and q = 0. 3, n = 5 and r = 4
\ Required probability = 5C 4 (07
. )4 (0. 3)

Q. 87 The probability distribution of a discrete random variable X is given

below
X 2 3 4 5
5 7 9 11
P(X)
k k k k

The value of k is
(a) 8 (b) 16 (c) 32 (d) 48
Sol. (c) We know that, SP ( X ) = 1
5 7 9 11
Þ + + + =1
k k k k
32
Þ =1
k
\ k = 32

Q. 88 For the following probability distribution.

X -4 -3 -2 -1 0
P(X) 0.1 0.2 0.3 0.2 0.2

E ( X ) is equal to
(a) 0 (b) -1 (c) -2 (d) -1.8
Sol. (d) E ( X ) = SX P ( X )
= - 4 ´ (01. ) + (-3 ´ 0.2 ) + (-2 ´ 0. 3) + (-1 ´ 0.2 ) + (0 ´ 0.2 )
= - 0.4 - 0.6 - 0.6 - 0.2 = - 1. 8
Q. 89 For the following probability distribution.
X 1 2 3 4
1 1 3 2
P(X)
10 5 10 5

E ( X 2 ) is equal to
(a) 3 (b) 5 (c) 7 (d) 10
2 2 1 1 3 2
Sol. (d) E ( X ) = SX P ( X ) = 1 × + 4 × + 9× + 16 ×
10 5 10 5
1 4 27 32
= + + +
10 5 10 5
1 + 8 + 27 + 64
= = 10
10

Q. 90 Suppose a random variable X follows the Binomial distribution with

parameters n and p, where 0 < p < 1. If P (x = r ) / P (x = n - r ) is
independent of n and r, then p equals to
1 1 1 1
(a) (b) (c) (d)
2 3 5 7
n!
Sol. (a) Q P ( X = r ) = nC r ( p)r (q )n - r = ( p)r (1 - p)n - r [Q q = 1 - p] ...(i)
(n - r )! r !
P ( X = 0) = (1 - p)n
and P ( X = n - r ) = nC n - r ( p)n - r (q )n -( n - r )
n!
= ( p)n - r (1 - p)+ r [Q q = 1 - p] [Q n C r = nC n - r ] ...(ii)
(n - r )! r !
n!
pr (1 - p)n - r
P (x = r ) (n - r )! r !
Now, = [using Eqs. (i) and (ii)]
P (x = n - r ) n!
pn - r (1 - p)+ r
(n - r )! r !
n-r
æ 1- pö 1
=ç ÷ ´ r
è p ø æ 1- pö
ç ÷
è p ø
1- p 1 1
Above expression is independent of n and r, if = 1Þ = 2 Þ p =
p p 2

Q. 91 In a college, 30% students fail in Physics, 25% fail in Mathematics and

10% fail in both. One student is chosen at random. The probability
that she fails in Physics, if she has failed in Mathematics is
1 2 9 1
(a) (b) (c) (d)
10 5 20 3
30 3 25 1
Sol. (b) Here, P( Ph ) = = ,P = =
100 10 ( M ) 100 4
10 1
and P( M Ç Ph ) = =
100 10
P (Ph Ç M ) 1 / 10 2
P æç ö÷
Ph
\ = = =
èMø P(M ) 1/ 4 5
Q. 92 A and B are two students. Their chances of solving a problem correctly
1 1
are and , respectively. If the probability of their making a common
3 4
1
error is, and they obtain the same answer, then the probability of
20
their answer to be correct is
1 1 13 10
(a) (b) (c) (d)
12 40 120 13
Sol. (d) Let E1 = Event that both A and B solve the problem
1 1 1
\ P(E1 ) = ´ = .
3 4 12
Let E2 = Event that both A and B got incorrect solution of the problem
2 3 1
\ P ( E2 ) = ´ =
3 4 2
Let E = Event that they got same answer

1
Here, P (E / E1 ) = 1, P (E / E2 ) =
20
P (E1 Ç E ) P (E1 ) × P (E / E1 )
\ P (E1 / E ) = =
P (E ) P (E1 ) × P (E / E1 ) + P (E2 ) P (E / E2 )
1
´1
12 1 / 12 120 10
= = = =
1 1 1 10 + 3 12 ´ 13 13
´1+ ´
12 2 20 120

Q. 93 If a box has 100 pens of which 10 are defective, then what is the
probability that out of a sample of 5 pens drawn one by one with
replacement atmost one is defective?
5 4
æ9ö 1æ 9 ö
(a) ç ÷ (b) ç ÷
è10 ø 2 è10 ø
4 5 4
1æ 9 ö æ9ö 1æ 9 ö
(c) ç ÷ (d) ç ÷ + ç ÷
2 è10 ø è10 ø 2 è10 ø
10 1 9
Sol. (d) Here, n = 5, p = = and q =
100 10 10
r £1
Þ r = 0, 1
Also, P ( X = r ) = nC r pr q n - r
\ P ( X = r ) = P (r = 0) + P (r = 1)
1 0 9 5
1 4
1 9
= 5C 0 æç ö÷ æç ö÷ + 5C1 æç ö÷ æç ö÷
è 10 ø è 10 ø è 10 ø è 10 ø
9 5 1 æ 9 ö4
= æç ö÷ + 5 × ×ç ÷
è 10 ø 10 è 10 ø
5 4
9 1 9
= æç ö÷ + æç ö÷
è 10 ø 2 è 10 ø
True/False
Q. 94 If P (A) > 0 and P (B) > 0. Then, A and B can be both mutually
exclusive and independent.
Sol. False

Q. 95 If A and B are independent events, then A ¢ and B ¢ are also

independent.
Sol. True

Q. 96 If A and B are mutually exclusive events, then they will be

independent also.
Sol. False
Q. 97 Two independent events are always mutually exclusive.
Sol. False

Q. 98 If A and B are two independent events, then P (A and B) = P ( A)× P (B).

Sol. True

Q. 99 Another name for the mean of a probability distribution is expected

value.
Sol. True
E ( X ) = SX P ( X ) = m

Q. 100 If A and B¢ are independent events, then P (A ¢ È B) = 1 - P (A) P (B¢).

Sol. True
P ( A ¢ È B) = 1 - P ( A Ç B¢) = 1 - P( A) P(B¢)

A B

(A¢ È B)

Q. 101 If A and B are independent, then P (exactly one of A, B occurs)

= P (A) P (B¢) + P (B) P (A ¢).
Sol. True
Q. 102 If A and B are two events such that P (A) > 0 and P (A) + P (B) > 1,
P (B¢ )
then P (B/ A) ³ 1 -
P (A)
Sol. False
P ( A Ç B)
Q P(B / A) =
P ( A)
P( A) + P(B) - P( A È B) 1 - P( A È B)
= >
P( A) P( A)

Q. 103 If A, B and C are three independent events such that

P (A) = P (B) = P (C) = p,
then P (atleast two of A, B and C occur) = 3p 2 - 2p 3 .
Sol. True
P (atleast two of A, B and C occur)
= p ´ p ´ (1 - p) + (1 - p) × p × p + p(1 - p) × p + p × p × p
= p2 [1 - p + 1 - p + 1 - p + p]
= p2 (3 - 3 p) + p3
= 3 p2 - 3 p3 + p3 = 3 p2 - 2 p3

Fillers
1
Q. 104 If A and B are two events such that P (A / B) = p, P (A) = p, P (B) =
3
5
and P (A È B) = , then p is equal to ......... .
9
1 5
Sol. Here, P ( A) = p, P (B) = and P ( A È B) =
3 9
P ( A Ç B) p
Q P ( A / B) = = p Þ P ( A Ç B) =
P (B) 3
and P ( A È B) = P ( A) + P (B) - P ( A Ç B)
5 1 p 5 1 2p
Þ = p+ - Þ - =
9 3 3 9 3 3
5 - 3 2p 2 3 1
Þ = Þ p= ´ =
9 3 9 2 3
2 5
Q. 105 If A and B are such that P (A¢ È B¢ ) = and P (A È B) = , then
3 9
P (A ¢) + P (B¢) is equal to ......... .
2 5
Sol. Here, P ( A ¢ È B¢) = and P ( A È B) =
3 9
P ( A ¢ È B¢) = 1 - P ( A Ç B)
2
Þ = 1 - P ( A Ç B)
3
2 1
Þ P ( A Ç B) = 1 - =
3 3
 Q P ( A ¢) + P (B¢) = 1 - P ( A) + 1 - P (B)
U
= 2 - [P ( A) + P (B)]
= 2 - [P ( A È B) + P ( A Ç B)]
A B
5 + 3ö
= 2 - æç + ö÷ = 2 - æç
5 1
÷
è 9 3ø è 9 ø
18 - 8 10
= =
9 9

Q. 106 If X follows Binomial distribution with parameters n = 5, p and

P (X = 2) = 9P (X = 3), then p is equal to ......... .
Sol. Q P ( X = 2 ) = 9 × P ( X = 3) (where, n = 5 and q = 1 - p)
5
Þ C 2 p2 (1 - p)3 = 9 × 5C 3 p3 (1 - p)2
5! 2 5! 3
Þ p (1 - p)3 = 9 × p (1 - p)2
2 ! 3! 3!2 !
p2 (1 - p)3
Þ =9
p3 (1 - p)2
(1 - p)
Þ = 9 Þ 9p + p = 1
p
1
\ p=
10

Q. 107 If X be a random variable taking values x 1 , x 2 , x 3 , K , xn with

probabilities P1 , P2 , P3 , ..., Pn , respectively. Then,Var (x) is equal to
......... .
Sol. Var ( X ) = E ( X )2 - [E ( X )]2
n n
= S X 2 P( X ) - [ S X P( X ) ]2
i =1 i =1
2 2
= S Px
i i - ( SPi x i )

Q. 108 Let A and B be two events. If P (A / B) = P (A), then A is ... of B.

P ( A Ç B)
Sol. Q P ( A / B) =
P (B)
P ( A Ç B)
Þ P ( A) =
P (B)
Þ P ( A) × P (B) = P ( A Ç B)
So, A is independent of B.