CONTINUITY & DIFFERENTIABILITY Page # 3

A. DEFINITION OF CONTINUITY

Continuity at a Point : – A function f is continuous at c if the following three conditions are met.

(i) f(x) is defined. (ii) lim f(x) exists. (iii) lim f(x) = f(c).
x →c x →c

In other words function f(x) is said to be continuous at x = c , if Limit f(x) = f(c).

x →c

Symbolically f is continuous at x = c if Limit f(c - h) = Limit f(c+h) = f(c).

h→0 h→0

x=a x=a
(continuous) (Discontinuous)

One-sided Continuity :
A function f defined in some neighbourhood of a point c for c ≤ c is said to be continuous at c from the

left if xlim
→c −
f(x) = f(c).

A function f defined in some neigbourhood of a point c for x ≥ c is said to be continuous at c from the

right if xlim
→c +
f(x) = f(c).

One-sided continuity is a collective term for functions continuous from the left or from the right.

If the function f is continuous at c, then it is continuous at c from the left and from the right .

Conversely, if the function f is continuous at c from the left and from the right, then xlim
→c
f(x) exists &

lim f(x) = f(c).

x →c

The last equality means that f is continuous at c.

If one of the one-sided limits does not exist, then xlim

→c
f(x) does not exist either. In this case, the point

c is a discontinuity in the function, since the continuity condition is not met.

Continuity In An Interval :
(a) A function f is said to be continuous in an open interval (a , b) if f is continuous at each & every
point ∈ (a , b).

[
(b) A function f is said to be continuous in a closed interval a , b if : ]
(i) f is continuous in the open interval (a , b) &

(ii) f is right continuous at ‘a’ i.e. Limit

x →a +
f(x) = f(a) = a finite quantity..

(iii) f is left continuous at ‘b’ i.e. Limit

x →b −
f(x) = f(b) = a finite quantity..
 Page # 4 CONTINUITY & DIFFERENTIABILITY

A function f can be discontinuous due to any of the following three reasons :

(i) Limit f(x) does not exist i.e. Limit− f(x) ≠ Limit f (x)
x →c x →c x →c +

(ii) f(x) is not defined at x= c (iii) Limit f(x) ≠ f (c)

x →c

Geometrically, the graph of the function will exhibit a break at x= c.

Ex.1 Test the following functions for continuity

2x 5 − 8 x 2 + 11 3 sin3 x + cos 2 x + 1
(a) (b) f(x) =
x 4 + 4x3 + 8x 2 + 8x + 4 4 cos x − 2
Sol. (a) A function representing a ratio of two continous functions will be (polynomials in this case)
discontinuous only at points for which the denominator zero. But in this case
(x4 + 4x3 + 8x2 + 8x + 4) = (x2 + 2x + 2)2 = [(x + 1)2 + 1]2 > 0 (always greater than zero)
Hence f(x) is continuous throughout the entire real line.
(b) The function f(x) suffers discontinuities only at points for which the denominator is equal to zero
i.e. 4 cos x – 2 = 0 or cos x = 1/2 ⇒ x = xn = ± π/3 + 2nπ (n = 0, ±1, ±2...)
Thus the function f(x) is continuous everywhere, except at the point xn.

2 − (256 − 7 x )1/ 8
Ex.2 The function f(x) = , x ≠ 0 is continuous everywhere then find the value of f(0).
(5x + 32)1/ 5 − 2
Sol. Since f(x) is continuous

∴ f(0) = R.H.L. = xlim

→0 +
f(x) = hlim
→0
f(0 + h)

(256 − 7h)1/ 8 − (256 )1/ 8 (256 − 7h)1/ 8 − ( 256 )1/ 8

( 256 − 7h) − (256 )
1/ 8 1/ 8 .( −7h)
( 256 − 7h) − 256 7 ( 256 − 7h) − 256
= – hlim = – lim = lim
→0 (5h + 32)1/ 5 − (32)1/ 5 h →0 (5h + 32) − (32)
1/ 5 1/ 5
5 h → 0 (5h + 32)1/ 5 − (32)1/ 5
.(5h)
(5h + 32) − 32 (5h + 32) − 32

1
.( 256 )1/ 8−1 7 ( 2)−7 7 1 7 7
7 8 = . 3 =  x n − an 
= . = . −4 ⇒ f(0) = . ∵ xlim − nan−1 
5 1 8 ( 2) 8 2 64 64 →a x − a
.(32)1/ 5 −1  
5

 − 2 sin x if x ≤ −π / 2
 π
A sin x + B if − < x < π/2
Ex.3 Let f(x) =  2 Find A and B so as to make the function continuous.

 cos x if x ≥ π/2

Sol. At x = – π/2

L.H.L. = Lim ( −2 sin x ) R.H.L. = Lim A sin x + B

π− π+
x→ − x→ −
2 2
 CONTINUITY & DIFFERENTIABILITY Page # 5

π π
Replace x by – – h where h → 0 Replace x by – + h where h → 0.
2 2

 π   π 
Lim – 2 sin  − − h  = 2 = Lim A sin  − + h + B = B – A
h →0  2  h →0  2 
So B – A = 2 ...(i)
At x = π/2

L.H.L. = Lim− A sin x + B R.H.L. = Lim+ cos x

π π
x→ x→
2 2

π π
Replace x by –h Replace x by +h
2 2
where h → 0 where h → 0

π  π 
= Lim A sin  − h + B = A + B = Lim cos  + h = 0
h→ 0 2  h→0 2 
So A + B = 0 ...(ii)
Solving (i) & (ii), B = 1, A = –1

  1 1
2 − + 
( x + 1)  |x| x  , x ≠ 0
Ex.4 Test the continuity of f(x) at x = 0 if f(x) = 
 0 , x=0


Sol. For x < 0,

 1 1 
2 − + 
 |0 −h| ( 0 −h ) 
L.H.L. = xlim f(x) = hlim f(0 – h) = hlim (0 − h + 1) = hlim (1 – h)2 = (1 – 0)2 = 1
→0 − →0 →0 →0

 1 1 2
2 −  +  2−
f(0) = 0. & R.H.L = xlim
→0 +
f(x) = hlim
→0
f(0 + h) = lim (h + 1)  |h| h 
= lim (h + 1) h
= 1– ∞ = 1
h→0 h →0

∴ L.H.L. = R.H.L. ≠ f(0) Hence f(x) is discontinuous at x = 0.

Ex.5 If f(x) be continuous function for all real values of x and satisfies;

x2 + {f(x) – 2} x + 2 3 –3– 3 . f(x) = 0, ∀ x ∈ R. Then find the value of f( 3 ).

Sol. As f(x) is continuous for all x ∈ R.

lim f(x) = f ( 3 ) x 2 − 2x + 2 3 − 3
Thus, x→ 3
where f(x) = ,x≠ 3
3−x

( 2 − 3 − x )( 3 − x )
lim f(x) = lim x − 2x + 2 3 − 3
2
= xlim = 2( Ι − 3 )
x→ 3 x→ 3 3−x → 3 ( 3 − x)

f ( 3 ) = 2 (1 − 3 ) .
 Page # 6 CONTINUITY & DIFFERENTIABILITY

 π
(1+ | sin x | , − <x<0
a /(|sin x|)

 6
Ex.6 Let f(x) =  b , x = 0 Determine a and b such that f is continuous at x = 0.
 e tan 2 x / tan 3 x , x>0


Sol. For x < 0,

L.H.L. = xlim
→0 −
f(x) = hlim
→0
f(0 – h) = hlim
→0
(1 + |sin (0 – h)|)a/(|sin(0 – h)|)

lim (1+ sinh−1).a /(sinh)

= hlim
→0
(1 + sin h)a/(sin h) = e h →0 =ea

For x = 0, f(0) = b

tan 2h tan 2h / 2h 2
For x > 0,R.H.L. = xlim f(x) = hlim f(0 + h) = lim lim .
eh →0 tan 3h = eh→ 0 tan 3h / 3h 3 = e
(1/1) . (2/3)
→0 + →0
= e2/3

∵ f(x) is continuous at x = 0 ∴ L.H.L. = R.H.L. = f(0)

⇒ a
e =e 2/3
=b Hence a = 2/3 and b = e2/3.

Acosx +Bxsinx −5
Ex.7 If f (x) = (x ≠ 0) is continuous at x = 0, then find the value of A and B . Also find f
x4
(0).

Sol. For continuity Limit f (x) = f (0)

x→0

A cosx +Bxsinx −5
Now Limit
x→0 4
as x → 0 ; Numerator → A − 5 and Denominator → 0.
x

Hence A − 5 = 0 ⇒ A=5

5 sinx sin2 x
Bx sinx −5(1−cosx ) B. x − 1+cosx
Limit Limit x2
Hence x →0 = x →0
x4 x2

5 5
as x → 0 ; Numerator → B − and Denominator → 0 ⇒ B =
2 2

x cos x − 4sin 2 x
Limit xsinx −2(1−cosx )
5 5 2xsin 2
Hence l = = Limit 2 2
x →0 x →0
2 x4 2 x4

x
2sin 2 x −2cos x
xcos 2
5 Limit Limit 2
= x →0 x →0
Let x = 2 θ
2 x x3

5 Limit 2θcosθ−2sinθ 5 Limit ( θ−tanθ) 5 Limit θ−tanθ 5  1  5

= = 2 cos θ = = −  = −
16 θ →0
θ 3
16 θ→0
θ 3
8 θ →0
θ 3
8  3 24
 CONTINUITY & DIFFERENTIABILITY Page # 7

 a 2[ x ] + { x } − 1
 , x≠0
Ex.8 Discuss the continuity of the function f(x) =  2[ x ] + { x } (a ≠ 1)
 log a , x=0
 e

at x = 0, where [x] and {x} are the greatest integer function and fractional part of x respectively.
Sol. Value of function = f(0) = loge a

a 2[ 0 − h ] + { 0 − h } − 1 a 2[ 0 −h]+ { −1+(1−h )} − 1 a −1 − 1 1
L.H.L. = xlim f(x) = hlim f(0 – h) = hlim = hlim = = 1−
→0 − →0 →0 2[0 − h] + {0 − h} → 0 2[0 − h] + {−1 + (1 − h)} −1 a

2[ 0 +h]+ {0 +h} −1
lim f(0 + h) = lim a a 0 +h − 1 ah − 1
and R.H.L. = xlim f(x) = = lim = lim = loge a
→0 + h→ 0 h→0 2[0 + h] + {0 + h} h→0 0 + h h →0 h

∴ L.H.L. ≠ R.H.L. = f(0) Hence f(x) is discontinuous at x = 0.

1 + a cos 2x + b cos 4 x
 if x ≠ 0
 x 2 sin2 x
Ex.9 Let f (x) =  If f (x) is continuous at x = 0, then find the value of (b + c)3

c if x = 0

– 3a.

1 + a cos 2x + b cos 4 x
Sol. Lim as x → 0, Nr → 1 + a + b Dr → 0
x →0 x4

a cos 2x + b cos 4 x − (a + b)
for existence of limit a + b + 1 = 0 ∴ c = Lim .....(2)
x →0 x4

a(1 − cos 2x ) b(1 − cos 4 x )

+
= – Lim x2 x2
x →0 2
x

limit of Nr ⇒ 2a + 8b = 0 ⇒ a = – 4b

1 4
hence – 4b + b = – 1 ⇒ b= and a=–
3 3

4(1 − cos 2x ) − (1 − cos 4 x ) 8 sin2 x − 2 sin2 2 x 8 sin2 x − 8 sin2 x cos2 x

hence c = Lim = =
x →0 3x 2 3x 4 3x 4

2 2
8 8
· 2 · 2
 Page # 8 CONTINUITY & DIFFERENTIABILITY


 a(1− xsinx )+bcosx +5
 x <0
 x2
Ex.10 Let f(x) =  3 x =0 . If f is continuous at x = 0, then find the values of a,
  1

 1+ cx +dx  
3 x
x >0
   x2 
   

b, c & d .

a (1 − x sin x ) + b cos x + 5
Sol. f (0 −) = Limit
x →0
for existence of limit a + b + 5 = 0
x2

a (1 − x sin x ) − (a + 5) cos x + 5
a (1 − cos x ) + 5 (1 − cos x ) − a x sin x
= Limit x2 = Limit
x →0 x→0
x2
a 5
= + −a = 3 ⇒ a=−1 ⇒ b=−4
2 2

1/ x
 x (c + d x 2 ) 
f (0 ) = Limit
+
1+  for existence of limit c = 0
x →0
 x2 

Limit 1
x dx
Limit (1 + dx)1/x = e x → 0 = ed = 3 ⇒ d = ln 3
x →0

e 2x −1−x(e 2x +1)
Ex.11 The function , f (x) = is not defined at x = 0 . What should be the value of f (x) so
x3
that f (x) is continuous at x = 0 .

e 2x −1− x(e 2x +1)

Sol. l = Limit
x →0 Put x = 3t
x3

e 6t −1−3t( e 6t +1) ( e 2t −1)3 +3e 2t ( e 2t −1)−3t( e 6t +1)

= Limit
t →0
= Limit
t →0 3
27t 3 27t

= Limit
[ ] [
( e 2t −1)3 +3e 2t e 2t −1−t( e 2t +1) + 3t e 2t (e 2t +1)−e 6t −1 ]
t →0
27t 3

( e 2t −1)3 1 1
= Limit
t →0 3
+ Limit e2t × l − Limit
t →0 t →0
(e2t − 1) (e4t − 1)
27t 9 9t 2

8 8 8 Limit  e 2t −1   e 4t −1  1 2
⇒ = −   ×   ⇒ l = −1 = –
9 27 9 t →0     3 3
 2t   4t 
 CONTINUITY & DIFFERENTIABILITY Page # 9

max{ f ( t ) : x + 1 ≤ t ≤ x + 2,−3 ≤ x < 0}

Ex.12 Let f(x) = x3 – 3x2 + 6 ∀ x ∈ R and g(x) = 
1 − x, for x≥0

Test continuity of g (x) for x ∈ [–3, 1].

Sol. Since f(x) = x3 – 3x2 + 6 ⇒ f’(x) = 3x2 – 6x = 3x (x – 2)
for maxima and minima f′(x) = 0
∴ x = 0, 2
f″ (x) = 6x – 6
f″ (0) = –6 < 0 (local maxima at x = 0)
f″ (2) = 6 > 0 (local minima at x = 2)
x3 – 3x2 + 6 = 0 has maximum 2 positive and 1 negative real roots. Y
f(0) = 6. 6
Now graph of f(x) is :
Clearly f(x) is increasing in (– ∞, 0) ∪ (2, ∞) and decreasing in (0, 2)
y = f(x)
⇒ x+2<0⇒x<–2 ⇒ –3 ≤ x < – 2
⇒ –2 ≤ x + 1 < –1 and –1 ≤ x + 2 < 0
f(x) = 2
in both cases f(x) increases (maximum) of g(x) = f(x + 2) 2
∴ g(x) = f(x + 2); –3 ≤ x < – 2 ...(1)
and if x + 1 < 0 and 0 ≤ x + 2 < 2 X
–1 O 1 2 3
⇒ – 2 ≤ x < –1 then g(x) = f(0)
Now for x + 1 ≥ 0 and x + 2 < 2 ⇒ –1 ≤ x < 0, g(x) = f(x + 1)

 f ( x + 2) ; − 3 ≤ x < −2
 f (0) ; − 2 ≤ x < −1

Hence g(x) =  Hence g(x) is continuous in the interval [–3, 1].
 f ( x + 1) ; − 1 ≤ x < +0
 1 − x ; x≥0

Ex.13 Let y = f(x) be defined parametrically as y = t2 + t | t |, x = 2t – | t |, t ∈ R Then at x = 0, find f(x)

and discuss continuity.
Sol. As, y = t2 + t | t | and x = 2t – | t | Y
y = 2x2
Thus when t ≥ 0
⇒ x = 2t – t = t, y = t2 + t2 = 2t2 ∴ x = t and y = 2t2
⇒ y = 2x2 ∀ x ≥ 0
again when, t<0
X
y=0 0
⇒ x = 2t + t = 3t and y = t2 – t2 = 0 ⇒ y = 0 for all x < 0.
2x 2 , x ≥ 0
Hence, f(x) =  which is clearly continuous for all x as show graphically..
 0, x < 0
 Page # 10 CONTINUITY & DIFFERENTIABILITY

 1 1 1 
Ex.14 Given the function, f (x) = x  + + +......upto∞  .
 x(1+x ) (1+x )(1+2x ) (1+2x )(1+3x ) 
Find f (0) if f (x) is continuous at x = 0 .

1 (1+2x )−(1+ x ) (1+3x )−(1+2x ) (1+nx )−(1+n−1x )

Sol. f (x) = + + + ...... +
1+ x (1+ x )(1+2x ) (1+ 2x )(1+3x ) (1+n−1x )(1+nx )

 2 if x ≠ 0 and n → ∞
upto n terms when x ≠ 0. Hence f (x) =  1+ x
2 1
f (x) = −
1+ x 1 + nx  2 if x = 0 for continuity.

Ex.15 Let f : R → R be a function which satisfies f(x + y3) = f(x) + (f(y))3 ∀ x, y ∈ R. If f is continuous at x
= 0, prove that f is continuous every where.

Sol. To prove hlim

→0
f(x + h) = f(x).

Put x = y = 0 in the given relation f(0) = f(0) + (f(0))3 ⇒ f(0) = 0

Since f is continuous at x = 0, hlim

→0
f(h) = f(0) = 0.

Now, hlim
→0
f(x + h) = hlim
→0
f(x) + (f(h))3 = f(x) + hlim
→0
(f(h))3 = f(x) + 0 = f(x).

Hence f is continuous for all x ∈ R.

B. CLASSIFICATION OF DISCONTINUITY

Definition :– Let a function f be defined in the neighbourhood of a point c, except perhaps at c itself.

Also let both one–sided limits xlim

→c −
f(x) and xlim
→c +
f(x) exist, where xlim
→c +
f(x) ≠ xlim
→c −
f(x).

Then the point c is called a discontinuity of the first kind in the function f(x).

In more complicated case xlim

→c
f(x) may not exist because

one or both one-sided limits do not exist. Such condition

is called a discontinuity of the second kind.

 x 2 + 1 for x < 0,

x = 0,
y= 
5 for
The function
 −x for x > 0,


has a discontinuity of the first kind at x = 0

The function y = |x| /x is defined for all x ∈ R, x ≠ 0;
but at x = 0 it has a discontinuity of the first kind.

The left-hand limit is xlim

→0 −
y = –1, while the

right-hand limit is xlim

→0 +
y=1
CONTINUITY & DIFFERENTIABILITY Page # 11

1
The function y = has no limits (neither one-sided nor
( x − 2)( x − 3)

1
two-sided) at x = 2 and x = 3 since xlim = ∞. Therefore
→0 ( x − 2)( x − 3)
x = 2 and x = 3 are discontinuities of the second kind

The function y = ln |x| at the point x = 0 has the limits lim

x→0
ln |x| = –∞. Consequently, xlim
→0
f(x) (and

also the one-sided limits) do not exist; x = 0 is a discontinuity of the second kind.
It is not true that discontinuities of the second kind only

arise when xlim

→c
f(x) = ∞. The situation is more complicated.

Thus, the function y = sin (1/x), has no one-sided limits

– +
for x → 0 and x → 0 , and does not tend to infinity as x → 0
There is no limit as x → 0 since the values of the function
sin (1/x) do not approach a certain number, but repeat an
infinite number of times within the interval from –1 to 1 as x → 0.

Removable & Irremovable Discontinuity

(a) In case Limit f(x) exists but is not equal to f(c) then the function is said to have a removable
x →c

discontinuity. In this case we can redefine the function such that Limit f(x) = f(c) & make it
x →c

continuous at x = c.

Removable Type Of Discontinuity Can Be Further Classified As :

(i) Missing Point Discontinuity : where Limit f(x) exists finitely but f(a) is not defined . e.g. f(x)
x →a

(1 − x )(9 − x 2 )
= has a missing point discontinuity at x = 1 .
(1 − x )

(ii) Isolated Point Discontinuity : where

Limit f(x) exists & f(a) also exists but ; Limit ≠ f(a) .
x→ a x →a

e.g. f(x) =
x 2 − 16 , x ≠ 4 & f (4) = 9 has a break at x = 4.
x−4

(b) In case Limit f(x) does not exist then it is not possible to make the function continuous by
x →c

redefining it . Such discontinuities are known as non - removable discontinuity.

 Page # 12 CONTINUITY & DIFFERENTIABILITY

Irremovable Type Of Discontinuity Can Be Further Classified As :

(i) Finite discontinuity : e.g. f(x) = x − [x] at all integral x.

1 1
(ii) Infinite discontinuity : e.g. f(x) = or g(x) = at x = 4.
x−4 ( x − 4 )2

(iii) Oscillatory discontinuity : e.g. f(x) = sin

1 at x = 0 .
x
In all these cases the value of f(a) of the function at x= a (point of discontinuity) may or may not exist

but Limit does not exist.

x →a

Remark :
(i) In case of finite discontinuity the non-negative difference between the value of the RHL at x = c &
LHL at x = c is called THE JUMP OF DISCONTINUITY . A function having a finite number of jumps in a
given interval I is called a PIECE-WISE CONTINUOUS OR SECTIONALLY CONTINUOUS function in this interval.

(ii) All Polynomials, Trigonometrical functions, Exponential & Logarithmic functions are continuous in
their domains.

(iii) Point functions are to be treated as discontinuous . eg . f(x) = 1 − x + x − 1 is not continuous at

x = 1.

(iv)If f is continuous at x = c & g is continuous at x = f(c) then the composite g[f(x)] is

x sin x
continuous at x = c. eg . f(x) = & g(x) =  x  are continuous at x = 0 , hence the
x2 + 2

x sin x
composite gof (x) = will also be continuous at x = 0.
x2 + 2

C. THEOREMS OF CONTINUITY

THEOREM–1 If f & g are two functions that are continuous at x= c then the functions defined by
F1(x) = f(x) ± g(x) ; F2(x) = K f(x) , K any real number ; F3(x) = f(x).g(x) are also continuous

f ( x)
at x= c. Further, if g (c) is not zero, then F4(x) = is also continuous at x= c.
g( x )

THEOREM–2 If f(x) is continuous & g(x) is discontinuous at x = a then the product function

sin πx x ≠ 0
φ (x) = f(x) . g(x) is not necessarily discontinuous at x = a . e.g. f(x) = x & g(x) = 
 0 x=0

THEOREM–3 If f(x) and g(x) both are discontinuous at x = a then the product function

 1 x≥0
φ (x) = f(x) . g(x) is not necessarily discontinuous at x = a . e.g. f(x) = − g(x) = 
− 1 x < 0
 CONTINUITY & DIFFERENTIABILITY Page # 13

THEOREMS–4 : INTERMEDIATE VALUE THEOREM

If f is continuous on the closed interval [a, b] and k is any number between f(a) and f(b), then there is
at least one number c in [a, b] such that f(c) = k.
Note that the Intermediate Value Theorem tells that at least one c exists, but it does not give a
method for finding c. Such theorems are called existence theorems.
As a simple example of this theorem, consider a person’s height. Suppose that a girl is 5 feet tall on her
thirteenth birthday and 5 feet 7 inches tall on her fourteenth birthday. Then, for any height h between
5 feet and 7 inches, there must have been a time t when her height was exactly h. This seems
reasonable because human growth is continuous and a person’s height does not abruptly change from
one value to another.
The Intermediate Value Theorem guarantees the existence of at least one number c in the closed
interval [a, b]. There may, of course, be more than one number c such that f(c) = k, as shown in Figure
1. A function that is not continuous does not necessarily possess the intermediate value property. For
example, the graph of the function shown in Figure 2 jumps over the horizontal line given by y = k and
for this function there is no value of c in [a, b] such that f(c) = k.
The Intermediate Value Theorem often can be used to locate the zeroes of a function that is continuous
on a closed interval. Specifically, if f is continuous on [a, b] and f(a) and f(b) differ in sign, then the
intermediate Value Theorem guarantees the existence of at least one zero of f in the closed interval
[a, b].

f( x) = x 3 + 2 x − 1
y
y y

f(a) 2 (1, 2)
f(a)

k 1 (c, 0)
k

f(b) f(b) [ ] x
–1 1

[ ] x [ ] x –1 (0, –1)
c1 c2 c3 a
a b b
(Fig. 1) (Fig. 2) (Fig. 3)
f is continuous on [a, b]. (For k, f is not continuous on [a, b]. f is continuous on [0, 1] with
there exist 3 c’s.) (For k, there are no c’s.) f(0) < 0 and f(1) > 0.

3
Ex.16 Use the Intermediate Value Theorem to show that the polynomial function f(x) = x + 2x – 1 has a zero
in the interval [0, 1]
Sol. Note that f is continuous on the closed interval [0, 1]. Because
3 3
f(0) = 0 + 2(0) – 1 = –1 and f(1) = 1 + 2(1) – 1 = 2
it follows that f(0) < 0 and f(1) > 0. You can therefore apply the Intermediate Value Theorem to
conclude that there must be some c in [0, 1] such that f(c) = 0, as shown in Figure 3.
 Page # 14 CONTINUITY & DIFFERENTIABILITY

1
Ex.17 State intermediate value theorem and use it to prove that the equation x−5 = has at least
x+3
one real root.
1
Sol. Let f (x) = x −5 – first, f (x) is continuous on [5, 6]
x+3
1 1 1 8
Also f (5) = 0 – =– < 0, f (6) = 1 – = >0
5+3 8 9 9
Hence by intermediate value theorem ∃ at least one value of c ∈ (5, 6)
for which f (c) = 0

1 1
∴ c−5 – =0 ∴ c is root of the equation x −5 = and c ∈ (5, 6)
c+3 x +3

Ex.18 Let f : R → R satisfy f(x) – f(y) =ex – y – 1 ∀ x, y ∈ R. Prove that f is a continuous function. Also prove
that the function f(x) has atleast one zero if f(0) < 1.

Sol. lim f(x + h) – f(x) = lim (ex + h – x – 1) = 0. Hence f is continuous everywhere.

h→ 0 h→ 0

Now f(x) = f(0) + ex – 1. ⇒ xlim

→∞
f(x) = ∞ lim f(x) = f(0) – 1 < 0
x → −∞

Since f(x) is positive for large positive x and negative of large negative x, by Intermediate Value
Theorem f(x) = 0 has atleast one root.

Ex.19 If f(x) be a continuous function in [0, 2π] and f(0) = f(2π) then prove that there exists point
c ∈ (0, π) such that f(c) = f(c + π).
Sol. Let g(x) = f(x) – f(x + π) ....(i)
at x = π; g(π) = f(π) – f(2π) ....(ii)
at x = 0, g(0) = f(0) – f(π) ...(iii)
adding (ii) and (iii), g(0) + g(π) = f(0) – f(2π)
⇒ g(0) + g(π) = 0 [Given f(0) = f(2π) ⇒ g(0) = –g(π)
⇒ g(0) and g(π) are opposite in sign.
⇒ There exists a point c between 0 and π such g(c) = 0 as shown in graph;
From (i) putting x = c g(c) = f(c) – f(c + π) = 0 Hence, f(c) = f(c + π)

D. DIFFERENTIABILITY

Definition of Tangent : If f is defined on an open interval containing c, and if the limit

∆y f ( c + ∆x ) − f ( c )
lim = lim =m
∆x → 0 ∆x ∆x →0 ∆x
exists, then the line passing through (c, f(c)) with slope m is the tangent line to the graph of f at the
point (c, f(c)).
CONTINUITY & DIFFERENTIABILITY Page # 15

The slope of the tangent line to the graph of f at the point (c, f(c)) is also called the slope of the graph
of f at x = c.
The above definition of a tangent line to a curve does not cover the possibility of a vertical tangent
line. For vertical tangent lines, you can use the following definition. If f is continuous at c and

f (c + ∆x ) − f (c )
lim =∞
∆x →0 ∆x

then the vertical line, x = c, passing through (c, f(c)) is a vertical tangent line to the graph of f. For
example, the function shown in Figure has a vertical tangent line at (c, f(c)). If the domain of f is the
closed interval [a, b], then you can extend the definition of a vertical tangent line to include the
endpoints by considering continuity and limits from the right (for x = a) and from the left (for x = b).

f (a + ∆x ) − f (a) Vertical
lim + =∞ y
∆x →0 ∆x tangent
line.

f (b + ∆x ) − f (b)
lim − =∞
∆x → 0 ∆x (c, f(c))

In the preceding section we considered the x

derivative of a function f at a fixed number a : Figure
The graph of f has a vertical
f ( a + h ) − f (a ) tanent line at (c, f(c)).
f′(a) = lim ...(1)
h →0 h

Note that alternatively, we can define

f ¢(a) =
Limit f ( x )−f (a) , provided the limit exists.
x→ a x−a

Here we change our point of view and let the number a vary. If we replace a in Equation 1 by a variable

f ( x + h) − f ( x )
x, we obtain f′(x) = lim ...(2)
h →0 h

Given any number x for which this limit exists, we assign to x the number f′(x). So we can regard f′ as
a new function, called the derivative of f and defined by Equation 2. We know that the value of f′(x),
can be interpreted geometrically as the slope of the tangent line to the graph of f at the point (x, f(x)).
The function f′ is called the derivative of f because it has been “derived” from f by the limiting
operation in Equation 2. The domain of f′ is the set {x|f′(x) exists} and may be smaller than the
domain of f.
 Page # 16 CONTINUITY & DIFFERENTIABILITY

Average And Instantaneous Rate Of Change

Suppose y is a function of x, say y = f(x). Corresponding to a change from x to x + ∆x, the variable y
changes from f(x) to f(x + ∆x). The change in y is ∆y = f(x + ∆x) – f(x), and the average rate of
change of y with respect to x is

change in y ∆y f ( x + ∆x ) − f ( x )
Average rate of change = = =
change in x ∆x ∆x

As the interval over which we are averaging becomes shorter (that is, as ∆x → 0), the average rate of
change approaches what we would intuitively call the instantaneous rate of change of y with

dy
respect to x, and the difference quotient approaches the derivative . Thus, we have
dx

∆y f ( x + ∆x ) − f ( x )
Instantaneous Rate of Change = lim = lim = f′(x)
∆ x →0 ∆x ∆x →0 ∆x

To summarize :
Instantaneous Rate of Change
Suppopse f(x) is differentiable at x = x0. Then the instantaneous rate of cange of y = f(x) with
respect to x at x0 is the value of the derivative of f at x0. That is

dy
Instantaneous Rate of Change = f′(x0) = dx
x = x0

Ex.20 Find the rate at which the function y = x2 sin x is changing with respect to x when x = π.
For any x, the instantaneous rate of change in the derivative,

dy
Sol. = 2x sin x + x2 cos x
dx

dy
Thus, the rate when x = π is = 2π sin π + π2 cos π = 2π(0) + π2 (–1) = – π2
dx x=π

The negative sign indicates that when x = π, the function is decreasing at the rate of π2 ≈ 9.9 units of
y for each one-unit increase in x.
Let us consider an example comparing the average rate of change and the instantaneous rate of
change.

Ex.21 Let f(x) = x2 – 4x + 7.

(a) Find the instantaneous rate of change of f at x = 3.
(b) Find the average rate of change of f with respect to
x between x = 3 and 5.
 CONTINUITY & DIFFERENTIABILITY Page # 17

Sol. (a) The derivative of the function is f′(x) = 2x – 4

y
Thus, the instantaneous rate of change of f 
 Secant line
at x = 3 is f′(3) = 2(3) – 4 = 2 10  Change of 8

 units from y = 4
The tangent line at x = 3 has slope 2, as shown in the figure.  to y = 12

 f ( x 2 ) − f( x 1 )
(b) The (average) rate of change from x = 3 to x = 5 is found by m=
5  x 2 − x1

 8
dividing the change in f by the change in x. The change in f(x1) = =4
2
f from x = 3 to x = 5 is
f(5) – f(3) = [52 – 4(5) + 7] – [32 – 4(3) + 7] = 8 Tangent line
x
5 10
x1 x2
f (5 ) − f ( 3 ) 8 


Thus, the average rate of change is = =4 Change of 2 units
5.3 2 (from x1 = 3 to x2 = 5)
Figure
The slope of the secant line is 4, as shown in the figure.

E. RELATION BETWEEN CONTINUITY & DIFFERENTIABILITY

If a function f is derivable at x then f is continuous at x.

For : f ′(x) = Limit f( x + h)−f( x ) exists.

h→0 h

f( x + h)−f( x )
Also f( x + h)− f ( x )= .h[h ≠ 0]
h

Therefore Limit [f ( x + h)−f ( x )] = Limit f( x + h)−f( x ).h=f '( x ).0=0

h→0 h→0 h

Therefore Limit [ f ( x + h ) − f ( x) ] = 0 ⇒ Limit f (x+h) = f(x) ⇒ f is continuous at x.

h→0 h→0

If f(x) is derivable for every point of its domain, then it is continuous in that domain .

The converse of the above result is not true :

“If f is continuous at x, then f may or maynot be derivable at x”

1
The functions f(x) = x & g(x) = x sin ; x ≠ 0 & g(0) = 0 are continuous at x = 0 but not
x

derivable at x = 0.
 Page # 18 CONTINUITY & DIFFERENTIABILITY

Remark :
(a) Let f ′+(a) = p & f ′_(a) = q where p & q are finite then :

(i) p = q ⇒ f is derivable at x = a ⇒ f is continuous at x = a.

(ii) p ≠ q ⇒ f is not derivable at x = a but f is continuous at x = a

/ Discontinuous
Differentiable ⇒ Continuous ; Non-differentiable ⇒

But Discontinuous ⇒ Non-differentiable .

(b) If a function f is not differentiable but is continuous at x = a it geometrically implies a sharp

corner at x = a.

Ex.22 Given f (x) = x2 . sgn (x) examine the continuity and derivability at the origin.

 x 2 if x >0
 h 2 −0
Sol. f (x) =  0 if x =0 f ′ (0+) = Limit
h→0
= 0
h
 −x 2 if x<0


h2 − 0
f ′ (0 −) = Limit − = 0⇒ f is derivable at x = 0 ⇒ continuous at x = 0
h→0 −h

− 1 − x ; x ≤ −1
| x 2 − 1 | ; − 1< x ≤ 0


Ex.23 If f(x) = k( − x + 1) ; 0 < x ≤ 1 , then find the value of k so that f(x) becomes continuous at x = 0.
| x − 1 | ; x >1

Hence, find all the points where the functions is non-differentiable.

Sol. From the graph of f(x) it is clear that for the function
to be continuous only possible value of k is 1.
Points of non-differentiability are x = 0, ±1.

a1/ x −a −1/ x
Ex.24 Examine the function , f (x) = x . , x≠ 0 (a > 0) and f (0) = 0 for continuity and
a1/ x +a −1/ 2

existence of the derivative at the origin.

Sol. If a ∈ (0, 1) f ′ (0+) = − 1 ; f ′ (0−) = 1 ⇒ continuous but not derivable

a=1 ; f (x) = 0 which is constant ⇒ continuous and derivable

If a>1 f ′ (0−) = − 1 ; f ′ (0+) = 1 ⇒ continuous but not derivable

 CONTINUITY & DIFFERENTIABILITY Page # 19

Ex.25 If f(x) = |x + 1| {|x| + |x – 1|}, then draw the graph of f(x) in the interval [–2, 2] and discuss the
continuity and differentiablity in [–2, 2]
Y (2, 9)
Sol. Here, f(x) = |x + 1| { |x| + |x – 1| }
)
,2
 ( x + 1)(2x − 1); − 2 ≤ x < −1 (1
 +1
− ( x + 1)(2x − 1); − 1 ≤ x < 0 x
 1
f(x) =  ( x + 1); 0 ≤ x <1
 ( x + 1)(2x − 1); X
1≤ x ≤ 2 –2 –1 0 1 2

Thus the graph of f(x) is;

which is clearly, continuous for x ∈ R and, differentiablity for x ∈ R – {–1, 0, 1}

| 1 − 4x 2 | , 0 ≤ x < 1
Ex.26 If f(x) =  2 where [.] denotes the greatest integer function.
| x − 2x | , 1 ≤ x < 2
Discuss the continuity and differentiability of f(x) in [0, 2).
Sol. Since 1 ≤ x < 2 ⇒ 0 ≤ x – 1 < 1 then [x2 – 2x] = [(x – 1)2 – 1] = [(x – 1)2] – 1 = 0 – 1 = –1
Y

 1 3

1 − 4 x , 0 ≤ x < 2
2

 1
f(x) = 4 x − 1 , ≤ x <1
2
∴ ∴ Graph of f(x) : 1

 2
 −1 , 1≤ x < 2
O
X
 1/2 1 2

–1

1
It is clear from the graph that f(x) is discontinuous at x = 1 and not differentiable at x = and x = 1.
2

1 − 4 x 2 , 0 ≤ x < 1/ 2 − 8x, 0 ≤ x < 1/ 2

 2 
Further details are as follows : f(x) = 4 x − 1, 1/ 2 ≤ x < 1 ⇒ f′(x) =  8 x, 1 / 2 ≤ x < 1
 − 1, 1≤ x < 2  0, 1≤ x < 2
 

− 4 x < 1 / 2 8, x < 1

⇒ f′(x) =  and f′(x) = 
4 x > 1/ 2 0, x > 1
Hence, which shows f(x) is not differentiable at x = 1/2 (as RHD = 4 and LHD = –4)
and x = 1 (as RHD = 0 and LHD = 8). Therefore, f(x) is differentiable, ∀ x ∈ [0, 2) – {1/2, 1}

2 x 2 sin π x x≤1
Ex.27 Let f (x) = [ 3 be a differential function . Examine whether it is twice differentiable in R.
x + ax + b x > 1
2

f ′(1+ )=3+2a 2π + 3 2π + 3
Sol. differentiability of f gives : a + b + 1 = 0and −
]⇒a = − and b =
f ′(1 )=−2π 2 2
 Page # 20 CONTINUITY & DIFFERENTIABILITY

4xsinπx + 2πx 2 cosπx , x≤1 ( 4−2π 2 x 2 )sinπx +8πxcosπx , x<1

Thus f ′ (x) = [ & f ′′ (x) = [
3x 2 −(3+2π)x , x >1 6x −3−2π , x >1

f ′′ (1 −) = − 8 π & f ′′ (1 +) = 3 − 2 π ⇒ f ′′ (x) is discontinuous at x = 1

Hence f (x) is twice differentiable for x except at x = 1 ]

x3 if x < 1

Ex.28 Suppose f (x) =  . If f '' (1) exist then find the value of a2 + b2 + c2.
 ax 2 + bx + c if x ≥ 1

Sol. For continuity at x = 1 we leave f (1–) = 1 and f (1+) = a + b + c

∴ a+b+c=1 ....(1)

3x 2 if x < 1

f ' (x) =  for continuity of f ' (x) at x = 1 f ' (1–) = 3; f ' (1+) = 2a + b
 2ax + b if x ≥ 1

hence 2a + b = 3 ....(2)

 6 x if x < 1
f '' (x) =  f '' (1–) = 6; f '' (1+) = 2a for continuity of f '' (x) 2a = 6 ⇒ a = 3
 2a if x ≥ 1

from (2), b = – 3 ; c = 1. Hence a = 3, b = – 3 ; c = 1 ∴ ∑a2 = 19

Ex.29 Check the differentiability of the function f(x) = max {sin–1 |sin x|, cos–1 |sin x|}.

 π
 x , nπ ≤ x ≤ nπ +
Sol. sin–1 |sin x| is periodic with period π ⇒ sin–1 |sin x| =  2
π
 π − x , nπ + ≤ x ≤ nπ + π
 2
π
Also cos–1 |sin x| = – sin–1 |sin x|
2
π π
 2 − x, nπ ≤ x ≤ nπ +
4
 π π  π π
x, 2 − x , nπ ≤ x ≤ nπ +
2
x,

nπ + < nπ +
4 2
⇒ f(x) = max  π π ⇒ f(x) = 
π − x, x − π 3π
, nπ + ≤ x ≤ nπ + π π − x, nπ + < x ≤ nπ +
 2 2  2 4
π π 3π  π 3π
⇒ f(x) is not differentiable at x = 0, , , , π......... x − 2 , nπ +
4
< x ≤ nπ + π
4 2 4
nπ
⇒ f(x) is not differentiable at x = .
4
 CONTINUITY & DIFFERENTIABILITY Page # 21

 2x 
Ex.30 Let f(x) = cos–1  2 . Define f′(x) for every x ∈ R stating clearly the point where f(x) is not

 1+ x 
differentiable.

 2x  1 2(1 + x 2 ) − 4 x 2 2(1 − x 2 )
Sol. f(x) = cos–1   , x ∈ R ⇒ f′(x) = – · ; f′(x) = –
 1+ x 
2
 2x 
2 (1 + x 2 )2 | 1 − x 2 | (1 + x 2 )
1−  2 
 1+ x 

Since xlim
→1−
f′(x) and xlim
→1+
f′(x) have finite values which are unequal, f′(1) does not exist.

2
 − 1+ x2 if − 1 < x < 1

 non existent if x = ± 1
Similarly f′(–1) does not exist. hence f′(x) = 
2
 if x > 1 or x < −1
 1+ x
2

Ex.31 Find the interval of values of k for which the function f(x) = |x2 + (k – 1) |x| – k| is non differentiable
at five points.
Sol. f(x) = |x2 + (k – 1) |x| – k| = |(|x| – 1) (|x| + k)|
Also f(x) is an even function and f(x) is not differentiable at five points.
So |(x – 1) (x + k)| is non differentiable for two positive values of x.
⇒ Both the roots of (x – 1) (x + k) = 0 are positive.
⇒ k<0⇒ k ∈ (–∞, 0).

 1

 (1 + { x }) 1  { x }
 {x} 
 , x ≠ Integer
Ex.32 Let f(x) =  e  . Discuss the continuity and differentiability of f(x) at any
 

2
 , x = Integer
e

integeral point. (where { * } denotes the fractional part)

Page # 22 CONTINUITY & DIFFERENTIABILITY

 1

 (1 + {I − h}) {I0 −h}   (1 + 1)1 1 2
f (I0− ) = hlim f(I0 – h) = lim  0 =  =
→0 h → 0 e   e  e
 
 

1/ h
1/ h  (1+h )1 / h  1  (1+h )1 / h  1  (1+h )1/ h 
 (1 + h)1/ h  lim 
h→ 0 


lim 
h→ 0 h 
−1

lim 
h→0 h 
−1
 lim
(1+h )1/ h − e
f(I0 + h) = hlim (I + h) = lim  
 = e  e 
=e  e 
=e  e 
=e h →0 eh
→0 0 h →0 e
 

1  h2 h3   1 h h2 
1
ln(1+h ) h − + −....  −h − + −..... 
h  2 3  2 3 4 
Now (1 + h)1/h = e h =e 
= e.e  

  1 h h2  
 − h  2 − 3 + 4 −....    2 
e   −1  1 − h + h −....  ( −1)
  2 3 4 
   
 1 h h2   
−h  − + −....  lim
2 3 4  h→ 0  1 h h2 
  −1
e −h − + −....  −
1
lim 2 3 4 
⇒ f (I0+ ) = e h →0 h =e  
=e 2

Since f(I0 + 0) ≠ f(I0 – 0) ⇒ f(x) is discontinuous at any integeral point and hence non-differentiable.

Definition : A function f is differentiable at a if f′(a) exists. It is differentiable on an open interval

(a,b) [or (a, ∞) or (–∞, a) or (– ∞, ∞)] if it is differentiable at every number in the interval.

Derivability Over An Interval : f(x) is said to be derivable over an interval if it is derivable at

each & every point of the interval. f(x) is said to be derivable over the closed interval [a, b] if :
(i) for the points a and b, f ′(a+) & f ′(b −) exist &
(ii) for any point c such that a < c < b, f ′(c+) & f′(c −) exist & are equal .
How Can a Function Fail to Be Differentiable ?
We see that the function y = |x| is not differentiable at 0 and Figure
y
shows that its graph changes direction abruptly when x = 0. In general,
if the graph of a function f has a “corner” or “kink” in it, then the graph of
f has no tangent at this point and f is not differentiable there. [In trying
to compute f ′(a), we find that the left and right limits are different.]
x
There is another way for a function not to have a derivative. If f is 0

discontinuous at a, then f is not differentiable at a. So at any discontinuity y = f(x) = |x|

(for instance, a jump discontinuity), f fails to be differentiable. Figure
 CONTINUITY & DIFFERENTIABILITY Page # 23

A third possibility is that the curve has a vertical tangent line when at x = a, lim |f′(x)| = ∞
x →a

This means that the tangent lines become steeper and steeper as x → a. Figure (a, b, c) illustrates the
three posibilities that we have discussed.
y y y

0 x x x
a 0 a 0 a
(a) A corner (b) A discontinuity (c) A Vertical tangent

Right hand & Left hand Derivatives By definition : f ′(a) = Limit f(a + h)−f(a)
h→0 h
(i) The right hand derivative of f ′ at x = a denoted by f ′+(a) is defined by :

f(a + h)−f(a)
f ′+(a) = Limit
h→0 +
, provided the limit exists & is finite.
h
(ii) The left hand derivative of f at x = a denoted by f ′_(a) is defined by :

f(a − h)−f(a )
f ′ _(a) = Limit
h→0 +
, Provided the limit exists & is finite.
−h
We also write f ′+(a) = f ′(a+) & f ′_(a) = f ′(a-) .
f′(a) exists if and only if these one-sided derivatives exist and are equal.

 xe1/x
 ,x≠0
Ex.33 If a function f is defined by f(x) =  1 + e1/x show that f is continuous but not derivable at x = 0
 0 ,x=0

xe1/ x e1/ x x
Sol. We have f(0 + 0) = lim = lim = lim 1/ x =0
x →0 + 0 1 + e1 / x x →0 + 0 1 + e 1 / x x →0 + 0 e +1

xe1/ x
f(0 – 0) = lim =0
x →0 −0 1 + e1 / x

Alsof(0) = 0 ∴ f(0 + 0) = f(0 – 0) = f(0) ⇒ f is continuous at x = 0

xe1/ x
−0
f ( x ) − f (0 ) 1 + e1/ x e1/ x 1
Again f′(0 + 0) = lim = lim = lim = lim −1 / x =1
x →0 + 0 x−0 x →0 + 0 x x →0 + 0 1 + e1/ x x →0 + 0 e +1

xe1/ x
f ( x ) − f (0 ) −0 e1/ x
= lim 1 + e
1/ x
f′(0 – 0) = lim = lim =0
x →0 −0 x−0 x →0 −0 x x →0 −0 1 + e1/ x

Since f′(0 + 0) ≠ f′(0 – 0), the derivative of f(x) at x = 0 does not exist.
 Page # 24 CONTINUITY & DIFFERENTIABILITY

 π π π
Ex.34 A function f(x) is such that f  x + = – |x| ∀ x. Find f′   , if it exists.
 2 2 2

π   π
 π π + f + h − f  π − | h | − π
π 
Given that f  x +  = f′   2  2 = 2 2 = −1
 2  = hlim
Sol. – |x| ⇒
 2 2 →0 h h
 

π  π
f  − h  − f   π − | −h | − π π
 π− 
⇒ and f′   = lim  2  2 = 2 2 =1 ⇒ f′   doesn’t exist.
 2  h→0 − h − h 2
 

n
 f ( a + 1 n) 
Ex.35 Let f be differentiable at x = a and let f (a) ≠ 0. Evaluate Lim   .
n→ ∞
 f (a ) 
n
 f ( a + 1 n) 
Sol. l = Lim   (1∞ form)
n →∞
 f ( a) 
  f (a +1 n)− f (a )    f ( a + h )− f ( a ) 1 
 Lim n     Lim ·  f '( a )
 n→∞
  f (a )   h→0 h f (a )  f (a )
l= e = e = e (put n = 1/h)

 1 if |x|≥ 1
 |x| 2
Ex.36 A function f is defined as , f (x) =  . If f (x) is derivable at x =1/2 find the values
a+bx 2 if |x|< 1
 2
of 'a' and 'b'.

1 1
if x ≥
1 1 1 x 2
 |x| if x ≥ or x ≤ − 
2 2
 
  1 1
Sol. f (x) =  = − if x ≤ −
x 2
 1 1 
a + bx 2 if − < x < 
2 2 1 1
 a + bx 2 if − < x <
2 2
1
−2 1 
1   1  1  1 − 2 + h 
 1+  f + h − f    + h 
f '  =   2  = Lim  2   2  Lim − 2h
 2  Lim  2 = Lim
h →0 1 
= h →0
1 
=–4
  h →0 h h→0 h h + h  h + h 
2  2 

2
1   1 1   b 
 1−  f − h − f  a + b − h  − 2  a + − 2  − bh + h
2

f '   = Lim  2  2 =  2  = Lim  4 

2  Lim
  h →0 −h h→0 −h h→0 −h
 CONTINUITY & DIFFERENTIABILITY Page # 25

b  1− 
for existence of limit a + = 2 ....(1) ⇒  2  = b
f '
4  

 1−   1+ 
∴ f'   = b = f'   = – 4 ∴ a – 1 = 2⇒ a = 3
2  2  Hence a = 3 and b = – 4
   

Ex.37 Let f : R → R satisfying |f(x)| ≤ x2, ∀ x ∈ R then show f(x) is differentiable at x = 0.

Sol. Since, |f(x)| ≤ x2 , ∀ x ∈ R ∴ at x = 0, |f(0)| ≤ 0 ⇒ f(0) = 0 ...(i)

f (h) − f (0) f (h)

∴ f′(0) = lim lim ....(ii) {f(0) = 0 from (i)}
h→0 h h→0 h

(h) (h) (h)

Now, f ≤|h|⇒ –|h|≤f ≤ | h | ⇒ lim f → 0 ...(iii) {using Cauchy–Squeeze theo-
h h h→ 0 h
rem}
∴ from (ii) and (iii) , we get f′(0) = 0. i.e. f(x) is differentiable at x = 0.

F. OPERATION ON DIFFERENTIABLE FUNCTIONS

1. If f(x) & g(x) are derivable at x = a then the functions f(x) + g(x), f(x) − g(x), f(x). g(x) will also
be derivable at x = a & if g (a) ≠ 0 then the function f(x)/g(x) will also be derivable at x = a.
If f and g are differentiable functions, then prove that their product fg is differentiable.
Let a be a number in the domain of fg. By the definition of the product of two functions we have
(fg) (a) = f(a) g(a) (fg) (a + t) = f(a + t) g(a + t).

f ( g)(a + t ) − ( fg)(a) f (a + t )g(a + t ) − f (a)g(a)

Hence (fg)′ (a) = lim = lim
t →0 t t → 0 t
The following algebraic manipulation will enable us to put the above fraction into a form in which
we can see what the limit is:
f(a + t) g(a + t) – f(a) g(a) = f(a + t) g(a + t) – f(a) g (a + t) + f(a)g(a + t) – f(a) g(a)
= [f(a + t) – f(a)] g(a + t) + [g(a + t) – g(a)] f(a).

 f (a + t ) − f (a ) g(a + t ) − g(a) 
Thus (fg)′ (a) = lim g(a + t ) + f (a) .
t →0   t t 
The limit of a sum of products is the sum of the products of the limits. Moreover, f′(a) and g′(a)
exist by hypothesis. Finally, since g is differentiable at a, it is continuous there ; and so

lim g(a + t) = f(a). We conclude that

t →0

 f (a + t ) − f ( a )   g(a + t ) − g(a) 
(fg)′(a) = lim  lim g(a + t ) + lim  f (a )
 t →0 t  t →0  t →0 t 
= f′(a)g(a) + g′(a)f(a) = (f′g + g′f) (a).
 Page # 26 CONTINUITY & DIFFERENTIABILITY

2. If f(x) is differentiable at x = a & g(x) is not differentiable at x = a , then the product function
F(x) = f(x) . g(x) can still be differentiable at x = a e.g. f(x) = x and g(x) = x .

3. If f(x) & g(x) both are not differentiable at x = a then the product function ;
F(x) = f(x) . g(x) can still be differentiable at x = a e.g. f(x) = x & g(x) = x.

4. If f(x) & g(x) both are non-deri. at x = a then the sum function F(x) = f(x) + g(x) may be a
differentiable function . e.g. f(x) = x & g(x) = −x .

5. If f(x) is derivable at x = a ⇒ f ′(x) is continuous at x = a.

 x 2 sin 1
x if x ≠0
e.g. f(x) = 
 0 if x =0

G. FUNCTIONAL EQUATIONS

Ex.38 Let f(xy) = xf(y) + yf(x) for all x, y ∈ R+ and f(x) be differentiable in (0, ∞) then determine f(x).
Given f(xy)= xf(y) + yf(x)
Sol. Replacing x by 1 and y by x then we get x f(1) = 0 ∴ f(1) = 0, x ≠ 0 (∵ x, y, ∈ R+)

  h   h  h
f  x1 +   − f ( x ) xf 1 +  + 1 + f ( x ) − f ( x )
f ( x + h) − f ( x )  x 
= lim  = lim 
x  x
Now, f′(x) = lim
h →0 h h→0 h h →0 h

 h h  h
xf 1 +  + f ( x ) f 1 + 
 x x  x
+ lim
f (x) f ( x)
= lim = hlim
→ 0  h  h → 0 x = f′(x) +
h →0 h x
 
x

f (x) xf ′( x ) − f ( x ) f ′(1) d  f ( x )  = f ′(1)

⇒ f′(x) – = f′(1) ⇒ = ⇒  
x x2 x dx  x  x

f ( x ) f (1)
On integrating w.r.t.x and taking limit 1 to x then − = f′(1) (ln x – ln 1)
x 1

f(x)
⇒ – 0 = f′(1) ln x ∵ f(1) = 0) ∴ f(x) = f′(1) (x ln x)
x
Alternative Method :
Given f(xy) = xf(y) + yf(x)
Differentiating both sides w.r.t.x treating y as constant, f′(xy) . y = f(y) + yf′(x)
Putting y = x and x = 1, then
xf ′( x ) − f ( x ) f ′(1) d  f ( x )  = f ′(1)
f′(xy). x = f(x) + xf′(x) ⇒ = ⇒  
x2 x dx  x  x
Integrating both sides w.r.t.x taking limit 1 to x,

f(x) f (1) f (x)

– = f′ (1){ln x – ln 1} ⇒ – 0 = f′(1) ln x (∵ f(1) = 0)
x 1 x
Hence, f(x) =- f′(1)(x ln x).
 CONTINUITY & DIFFERENTIABILITY Page # 27

Ex.39 If e–xyf(xy) = e–xf(x) + e–yf(y) ∀ x, y ∈ R+, and f′(1) = e, determine f(x).

Sol. Given e–xy f(xy) = e–xf(x) + e–yf(y) ...(1)
Putting x = y = 1 in (1), we get f(1) = 0 ...(2)

  h 
f  x1 +   − f ( x.1)
f ( x + h) − f ( x )   x 
Now, f′(x) = lim = lim
h →0 h h →0 h

 h
−1  h  h
x +h−1 
e x +h .e − x f ( x ) + e x f 1 +  − 2 x (e − x f ( x ) + e −1f (1)) h
+ x f 1 + h  − f ( x ) − e x −1f (1)
  x  e f ( x ) e  
= lim = lim  x
h→0 h h →0 h

h
x f 1 + h
h−
e  
 e − 1
h
 x
= f(x) hlim   + e(x – 1) lim
→ 0  h→0 h (∵ f(1) = 0)
 h  x.
x

f ′(1) e x−1.e
= f(x) . 1 + ex –1
. = f(x) + (∵ f′(1) = e)
x x

ex 1 d 1
f′(x) = f(x) + ⇒ e–xf′(x) – e–x f(x) = ⇒ (e–x f(x)) =
x x dx x
On integrating we have e–xf(x) = ln x + c at x = 1, c = 0 ∴ f(x) = ex ln x

Ex.40 Find a function continuous and derivable for all x and satisfying the functional relation,
f (x + y) . f (x − y) = f2 (x) , where x & y are independent variables & f (0) ≠ 0 .
Sol. Put y = x and x = 0 to get f (x) . f (− x) = f2 (0) ..... (1)

f 2(0 )
f( x+h)−f( x ) f( x+h)−
Limit f( x+h).f ( −x )−f (0)
2
f ( −x ) 1
Now f ′ (x) = Limit = Limit = .
h→0
h h→0
h f( −x ) h →0
h

 h  h   h  h  h
1 f  + + x  .f  − + x  −f 2 (0) 1 f 2   − f 2 (0 )
= Limit  2  2    2  2   = Limit 2
f (− x) h→0 f( −x ) h→0
h h

 h   h 
f  2  + f (0) f  2  − f (0)
1      f( x ) f ′(0)
= Limit = . 2 f (0) . f ′ (0) = f(x)
2f( −x ) h →0 h 2 f(0)
2f ( x )
2

f ′( x ) f ′(0)
⇒ = =k ⇒ Result
f( x ) f (0 )
 Page # 28 CONTINUITY & DIFFERENTIABILITY

Ex.41 Let f be a function such that f(x + f(y)) = f(f(x)) + f(y) ∀ x, y ∈ R and f(h) = h for 0 < h < ε
where ε > 0, then determine f”(x) and f(x).
Sol. Given f(x + f(y)) = f(f(x) + f(y)) .....(1)
Put x = y = 0 in (1), then f(0 + f(0)) = f(f(0)) + f(0) ⇒ f(f(0)) = f(f(0)) + f(0)
∴ f(0) = 0 ....(2)

f ( x + h) − f ( x )
Now f’(x) = lim (for 0 < h < ε)
h→0 h

f ( f (h)) f (h) h
= lim {from (1)} = lim (∵ f(h) = h) = lim = 1.
h→0 h h→0 h h→0 h

Integrating both sides with limites 0 to x then f(x) = x ∴ f′(x) = 1.

Ex.42 Let f : R+ → R satisfies the functional equation f(xy) = exy – x – y {ey f(x) + ex f(y)} ∀ x, y ∈ R+.
If f′(1) = e, determine f(x).
Sol. Given that; f(xy) = exy – x – y {ey f(x) + ex f(y)} ∀ x, y ∈ R+ ...(i)
Putting x = y = 1, we get f(1) = e–1 {e1 f(1) + e1 f(1)} ⇒ f(1) = 0 ...(ii)

  h 
f  x1 +   − f ( x )
f ( x + h) − f ( x )   x 
Now, f′(x) = lim = lim
h →0 h h →0 h

 h  h
x  1+  − x − 1+   1+ h
 x  h  h
h −1− + x  h
e  x
e x
f ( x ) + e x f 1 +  − f ( x ) e f(x) +
h
e x f 1+   − f ( x)
  x   x
= lim = lim
h→0 h h →0 h
h
h −1− + x   h 
h −1
h
h −1− + x   h  e x f 1+
  − f (1)
f ( x )(e )+ e x f 1+   − f (1) f ( x )(eh − 1)  x  
=  x  = hlim + lim {∵ f(1) = 0}
lim →0 h h→0 h
h→0 h .x
x

  h 
f 1 +  − f (1)
e .f ′(1) 
x −1
 x e − 1 
h
ex
= f(x) + ∵ hlim = f ′(1) and lim = 1 = f(x) + . f′(1) {∵ f′(1) = e}
x  →0 h h →0 h  ex
 x 
x
e ex
∴ f′(x) = f(x) + ⇒ = f′(x) – f(x)
x x

1 e x f ′( x ) − f ( x ).e x  e x f ′( x ) − f ( x ).e x d  f ( x ) 
⇒ =  As by quotient rule we can write x 2
=  
x e2x  (e ) dx  e x 

1 d  f (x) 
∴ =  
x dx  e x 

f ( x)
Integrating both sides w.r.t. ‘x’, we get, log |x| + c = or f(x) = ex{log | x | + c}
ex
Since f(1) = 0 ⇒ c = 0 Thus f(x) = ex log | x |.
 CONTINUITY & DIFFERENTIABILITY Page # 29

EXERCISE – I SINGLE CORRECT (OBJECTIVE QUESTIONS)

1. A function f(x) is defined as below Sol.
cos(sin x ) − cos x
f(x) = , x ≠ 0 and f(0) = a, f(x) is
x2
continuous at x = 0 if a equals
(A) 0 (B) 4 (C) 5 (D) 6
Sol.

1 1
5. If y = where t = , then the number
t +t−2
2
x −1
of points of discontinuities of y = f(x), x ∈ R is
(A) 1 (B) 2 (C) 3 (D) infinite
Sol.

 (1 + px ) − (1 − px )
 ,−1 ≤ x < 0
 x
2. f(x ) = 2x + 1 is
 ,0 ≤ x ≤ 1
 x−2
continuous in the interval [–1, 1], then ‘p’ is equal to:
(A) –1 (B) – 1/2 (C) 1/2 (D) 1
Sol.
6. The equation 2 tan x + 5x – 2 = 0 has
(A) no solution
(B) at least one real solution in [0, π/4]
(C) two real solution in [0, π/4]
(D) None of these
Sol.

 1
3. Let f(x) =  x + [ x] when – 2 ≤ x ≤ 2. Then
 2
(where [ * ] represents greatest integer function) 7. If f(x) = x ( x − x + 1) , then indicate the correct
(A) f(x) is continuous at x = 2 alternative(s)
(B) f(x) is continuous at x = 1 (A) f(x) is continuous but not differentiable at x = 0
(C) f(x) is continuous at x = –1 (B) f(x) is differentiable at x = 0
(D) f(x) is discontinuous at x = 0 (C) f(x) is not differentiable at x = 0
Sol.
(D) None of these
Sol.

2
4. Let f(x) = sgn (x) and g(x) = x (x – 5x + 6).
The function f(g(x)) is discontinuous at
(A) infinitely many points (B) exactly one point
(C) exactly three points (D) no point
 Page # 30 CONTINUITY & DIFFERENTIABILITY

Sol.
 x(3e1/ x + 4)
 , x≠0
8. If f(x) =  2 − e1/ x then f(x) is
 0 , x=0

(A) continuous as well differentiable at x = 0
(B) continuous but not differentiable at x = 0
(C) neither differentiable at x = 0 not continuous at x = 0
(D) none of these
Sol.
12. Let f(x) be defined in [–2, 2] by

max ( 4 − x 2 , 1 + x 2 ) ,−2 ≤ x ≤ 0
f(x) =  then f(x)
 min ( 4 − x 2 , 1 + x 2 ) , 0 < x ≤ 2

(A) is continuous at all points

(B) is not continuous at more than one point
x (C) is not differentiable only at one point
9. If f(x) = be a real valued function then
x +1− x (D) is not differentiable at more than one point.
(A) f(x) is continuous, but f′(0) does not exist Sol.
(B) f(x) is differentiable at x = 0
(C) f(x) is not continuous at x = 0
(D) f(x) is not differentiable at x = 0
Sol.

–1
13. The number of points at which the function
10. The function f(x) = sin (cos x) is f(x) = max. {a – x, a + x, b} – ∞ < x < ∞, 0 < a < b
(A) discontinuous at x = 0 (B) continuous at x = 0 cannot be differentiable is
(C) differentiable at x = 0 (D) none of these (A) 1 (B) 2
Sol. (C) 3 (D) none of these
Sol.

 x2 − 1
 , 0<x≤2
 x2 + 1
 1 3
11. If f(x) =  (x − x2 ) , 2 < x ≤ 3 , then
 4
 9 (| x − 4 | + | 2 − x |) , 3 < x < 4
4 14. If f(x) is differentiable everywhere, then

(A) f(x) is differentiable at x = 2 & x = 3 (A) | f | is differentiable everywhere
2
(B) f (x) is non–differentiable at x = 2 & x = 3 (B) | f | is differentiable everywhere
(C) f(x) is differentiable at x = 3 but not at x = 2 (C) f | f | is not differentiable at some point
(D) f(x) is differentiable at x = 2 but not at x = 3. (D) f + | f | is differentiable everywhere
 CONTINUITY & DIFFERENTIABILITY Page # 31

Sol. Sol.

15. Let f(x + y) = f(x) f(y) all x and y. Suppose that

f(3) = 3 and f′(0) = 11 then f′(3) is given by
(A) 22 (B) 44 (C) 28 (D) 33 19. The function f(x) is defined by
Sol.
 3
log( 4 x −3 ) ( x 2 − 2x + 5) if < x < 1 & x > 1
f(x) =  4
 4 if x = 1
(A) is continuous at x = 1
+
(B) is discontinuous at x = 1 since f(1 ) does not exist
–
though f(1 ) exists
–
(C) is discontinuous at x = 1 since f(1 ) does not
+
exist though f(1 ) exists
16. If f : R → R be a differentiable function, such that – +
(D) is discontinuous since neither f(1 ) nor f(1 ) exists.
f(x + 2y) = f(x) + f(2y) + 4xy ∀ x, y ∈ R, then Sol.
(A) f′ (1) = f′(0) + 1 (B) f′(1) = f′(0) – 1
(C) f′(0) = f′(1) + 2 (D) f′(0) = f′(1) – 2
Sol.

x 2 if x is irrational
20. Let f(x) =  then
2 max f ( t ),0 ≤ t ≤ x,0 ≤ x ≤ 1 1 if x is rational
17. Let f(x) = x – x and g(x) =  . (A) f(x) is discontinuous for all x
sin πx, x > 1
Then in the interval [0, ∞) (B) discontinuous for all x except at x = 0
(A) g(x) is everywhere continuous except at two points (C) discontinuous for all x except at x = 1 or – 1
(B) g(x) is everywhere differentiable except at two points (D) none of these
(C) g(x) is everywhere differentiable except at x = 1 Sol.
(D) none of these
Sol.

21. A point where function f(x) is not continuous where

f(x) = [sin [x]] in (0, 2π) ; is
([ * ] denotes greatest integer ≤ x)
(A) (3, 0) (B) (2, 0) (C) (1, 0) (D) (4, – 1)
Sol.
18. Let [x] denote the integral part of x ∈ R and
g(x) = x – [x]. Let f(x) be any continuous function
with f(0) = f(1) then the function h(x) = f(g(x))
(A) has finitely many discontinuities
(B) is continuous on R
(C) is discontinuous at some x = c
(D) is a constant function.
 Page # 32 CONTINUITY & DIFFERENTIABILITY

 (1 + sin πx )t − 1 25. Let f(x) = [n + p sin x], x ∈ (0, π), n ∈ Ι and p is

22. The function f defined by f(x)= tlim 
→ ∞ (1 + sin πx ) t + 1
 is a prime number. Then number of points where f(x) is
 
(A) everywhere continuous not differentiable is
(B) discontinuous at all integer values of x (where [ * ] denotes greatest integer function)
(C) continuous at x = 0 (D) none of these (A) p – 1 (B) p + 1 (C) 2p + 1 (D) 2p – 1
Sol. Sol.

3 2
26. Let f(x) = x – x + x + 1 and
max{f(t)} for 0 ≤ t ≤ x for 0 ≤ x ≤ 1
  1 g(x) =  then
 3 − x + x2 for 1 < x ≤ 2
 x 1 + sin  , x>0
  x (A) g(x) is continuous & derivable at x = 1
  1 (B) g(x) is continuous but not derivable at x = 1
23. If f(x) = − − x  1 + sin  , x < 0 , then f(x) is
  x (C) g(x) is neither continuous nor derivable at x = 1
 0 , x=0 (D) g(x) is derivable but not continuous at x = 1
 Sol.

(A) continuous as well diff. at x = 0
(B) continuous at x = 0, but not diff. at = 0
(C) neither continuous at x=0 nor diff. at x=0
(D) none of these
Sol.

27. Let f″(x) be continuous at x = 0 and f″(0) = 4 then

2f ( x ) − 3 f ( 2 x ) + f ( 4 x )
value of lim is
x →0 x2
(A) 11 (B) 2 (C) 12 (D) none of these
2 2
24. The functions defined by f(x) = max {x , (x – 1) , Sol.
2x (1 – x)}, 0 ≤ x ≤ 1
(A) is differentiable for all x
(B) is differentiable for all x except at one point
(C) is differentiable for all x except at two points
(D) is not differentiable at more than two points
Sol.

28. Let f : R → R be a function such that

 x + y  f ( x) + f ( y)
f  = , f(0) = 0 and f′(0) = 3, then
 3  3
f ( x)
(A) is differentiable in R
x
(B) f(x) is continuous but not differentiable in R
(C) f(x) is continuous in R (D) f(x) is bounded in R
 CONTINUITY & DIFFERENTIABILITY Page # 33

Sol. 32. The function f : R /{0} → R given by

1 2
f(x) = − can be made continuous at x = 0 by
x e2x − 1
defining f(0) as
(A) 2 (B) –1 (C) 0 (D) 1
Sol.

29. Suppose that f is a differentiable function with

1
the property that f(x + y) = f(x) + f(y) + xy and lim
h→0 h
f(h) = 3 then
2
(A) f is a linear function (B) f(x) = 3x + x 33. If f is a real-valued differentiable function satisfying
x 2 |f(x) – f(y)| ≤ (x – y)2, x, y ∈ R and f(0) = 0, then
(C) f(x) = 3x + (D) none of these f(1) equals
2
Sol. (A) 1 (B) 2 (C) 0 (D) –1
Sol.

34. Function f(x) = (|x – 1| + |x – 2| + cos x) where

30. If a differentiable function f satisfies x ∈ [0, 4] is not continuous at number of points
(A) 3 (B) 2 (C) 1 (D) 0
 x + y  4 − 2( f ( x ) + f ( y ))
f  = ∀ x, y ∈ R, find f(x)
Sol.
 3  3
(A) 1/7 (B) 2/7 (C) 8/7 (D) 4/7
Sol.

31. Let f : R → R be a function defined by f(x) = Min 1− | x |

 , x ≠ −1
{x + 1, |x| + 1}. Then which of the following is true ? 35. If f(x) =  1 + x , then f([2x]) is
(A) f(x) ≥ 1 for all x ∈ R
 1 , x = −1
(where [ * ] represent greatest integer function)
(B) f(x) is not differentiable at x = 1
(A) continuous at x = –1 (B) continuous at x=0
(C) f(x) is differentiable everywhere
(C) discontinuous at x = 1/2 (D) all to these
(D) f(x) is not differentiable at x = 0
Sol.
Sol.
 Page # 34 CONTINUITY & DIFFERENTIABILITY

36. Let f(x + y) = f(x) f(y) for all x, y, where f(0) ≠ 0. Sol.
If f′(0) = 2, then f(x) is equal to
(A) Aex (B) e2x (C) 2x (D) None of these
Sol.

37. A function f : R → R satisfies the equation 40. Let f(x) = [cos x + sin x], 0 < x < 2π where [x]
f(x + y) = f(x) . f(y) for all x, y ∈ R, f(x) ≠ 0. Suppose denotes the greatest integer less than or equal to x.
that the function is differentiable at x = 0 and f′(0) = 2 the number of points of discontinuity of f(x) is
then f′(x) = (A) 6 (B) 5 (C) 4 (D) 3
(A) f(x) (B) 2 f(x) (C) – f(x) (D) – 2 f(x) Sol.
Sol.

38. The value of f(0), so that the function,

(a 2 − ax + x 2 ) − (a 2 + ax + x 2 )  2 1 
x   ; x ≠ 0
f(x)= (a > 0) becomes 41. The function f(x) =   x2  , is [ x ]
( a + x ) − (a − x )  0 ; x=0
continuous for all x, is given by
represents the greatest integer less than or equal to x
(A) a a (B) a (C) – a (D) –a a (A) continuous at x = 1 (B) continuous at x = –1
Sol. (C) continuous at x = 0 (D) continuous at x = 2
Sol.

sin(ln | x |) x ≠ 0
 sin{cos x} π 42. The function f(x) = 
 , x≠  1 x=0
π 2
 x− (A) is continuous at x = 0
39. If f(x) =  2 , then f(x) is (B) has removable discontinuity at x = 0
 π (C) has jump discontinuity at x = 0
 1 , x=
 2 (D) has discontinuity of IInd type at x = 0
Sol.
(where { * } represents the fractional part function)
π
(A) continuous at x =
2
π
(B) Lim f(x) exists, but f is not continuous at x =
x→
π 2
2
(C) Lim f(x) does not exist (D) Lim f(x) = 1
π π
x→ x→
2 2
 CONTINUITY & DIFFERENTIABILITY Page # 35

|x −3| 1 Sol.
43. The set of all point for which f(x) = +
| x − 2 | [1 + x]
is continuous is
(where [ * ] represents greatest integer function)
(A) R (B) R – [–1, 0]
(C) R – ({2} ∪ [–1, 0]) (D) R – {(–1, 0) ∪ n, n ∈ Ι}
Sol.

47. Let f(x) = sin x, g(x) = [x + 1] and g(f(x)) = h(x)

π
then h′   is (where [*] is the greatest integer
2
function)
(A) nonexistent (B) 1
(C) –1 (D) None of these
Sol.

44. Let f(x) be a continuous function defined for

1 ≤ x ≤ 3. If f(x) takes rational values of for all x and
f(2) = 10 then the value of f(1.5) is
(A) 7.5 (B) 10 (C) 8 (D) None of these
Sol.

48. If f(x) = [tan2 x] then

(where [ * ] denotes the greatest integer function)

(A) Lim
x →0
f(x) does not exist
45. If [x] and {x} represents integral and fractional (B) f(x) is continuous at x = 0
a 2[ x ]+ { x } − 1 (C) f(x) is non-differentiable at x = 0 (D) f(0) = 1
parts of a real number x, and f(x) = , x ≠ 0, Sol.
2[ x ] + { x}
f(0) = loge a, where a > 0, a ≠ 1, then
(A) f(x) is continuous at x = 0
(B) f(x) has a removable discontinuity at x = 0

(C) xlim
→0
f(x) does not exist (D) None of these
Sol.

49. If f(x) = [x]2 + { x } 2 , then

(where, [ * ] and { * } denote the greatest integer
and fractional part functions respectively)
(A) f(x) is continuous at all integral points
(B) f(x) is continuous and differentiable at x = 0
(C) f(x) is discontinuous ∀ x ∈ Ι – {1}
(D) f(x) is differentiable ∀ x ∈ Ι.
Sol.

46. If f(x) = p |sin x| + q . e|x| + r|x|3 and f(x) is

differentiable at x = 0, then
(A) p = q = r = 0 (B) p = 0, q = 0, r ∈ R
(C) q = 0, r = 0, p ∈ R (D) p + q = 0, r ∈ R
 Page # 36 CONTINUITY & DIFFERENTIABILITY

f (h) − f (0) Sol.

50. If f is an even function such that Lim+
h→0 h
has some finite non-zero value, then
(A) f is continuous and derivable at x = 0
(B) f is continuous but not derivable at x = 0
(C) f may be discontinuous at x = 0
(D) None of these
Sol.

x n − sin x n
54. Consider f(x) = Limit for x > 0, x ≠ 1,
n→∞ x n + sin x n
f(1) = 0 then
51. f is a continuous function on the real line. Given (A) f is continuous at x = 1
2 (B) f has a finite discontinuity at x = 1
that x + (f(x) – 2) x – 3 . f(x) + 2 3 – 3 = 0. then (C) f has an infinite or oscillatory discontinuity at x = 1.
the value of f( 3 ) (D) f has a removable type of discontinuity at x = 1.
(A) can not be determined (B) is 2 (1 – Sol.
3)
2( 3 − 2)
(C) is zero (D) is
3
Sol.

2
52. If f(x) = sgn (cos 2x – 2 sin x + 3) then f(x)  [{| x |}]ex {[x + {x}]}
 for x ≠ 0
(where sgn ( ) is the signum function) 2
55. Given f(x)=  (e1/ x − 1)sgn(sin x) then, f(x)
(A) is continuous over its domain 
 0 for x = 0
(B) has a missing point discontinuity
(C) has isolated point discontinuity (where {x} is the fractional part function; [x] is the
(D) has irremovable discontinuity. step up function and sgn(x) is the signum function of x)
Sol. (A) is continuous at x = 0
(B) is discontinuous at x = 0
(C) has a removable discontinuity at x = 0
(D) has an irremovable discontinuity at x = 0
Sol.

–1 –1
[ x]
53. Let g(x) = tan |x| – cot |x|, f(x) = {x},
[ x + 1]
h(x) = |g (f (x) ) | then which of the following holds
good ?
(where { * } denotes fractional part and [ * ] denotes
the integral part)
(A) h is continuous at x = 0
(B) h is discontinuous at x = 0
– +
(C) h(0 ) = π/2 (D) h(0 ) = –π/2
 CONTINUITY & DIFFERENTIABILITY Page # 37

58. Consider the function defined on [0, 1] → R,

 x[ x] 2 log(1+ x ) 2 for − 1 < x < 0
 sin x − x cos x
2
56. Consider f(x) =  ln( e x + 2 { x} ) the f(x) = if x ≠ 0 and f(0) = 0, then the
for 0 < x < 1 x2
 function f(x)
 tan x
(where [ * ] & { * } are the greatest integer function (A) has a removable discontinuity at x = 0
& fractional part function respectively) (B) has a non removable finite discontinuity at x = 0
(A) f(0) = ln 2 ⇒ f is continuous at x = 0 (C) has a non removable infinite discontinuity at x = 0
(B) f(0) = 2 ⇒ f is continuous at x = 0 (D) is continuous at x = 0
2 Sol.
(C) f(0) = e ⇒ f is continuous at x = 0
(D) f has an irremovable discontinuity at x = 0
Sol.

|x|
59. Let f(x) = for x ≠ 0 & f(0) = 1 then ,
sin x
1+ x − 1− x (A) f(x) is conti. & diff. at x = 0
57. Consider f(x) = , x ≠ 0 ; g(x) = cos
{x} (B) f(x) is continuous & not derivable at x = 0
(C) f(x) is discont. & not diff. at x = 0
 1
 f (g( x )) for x<0 (D) None of these
π  2 Sol.
2x, – < x< 0, h(x) =  1 for x=0
4
 f ( x) for x>0


then, which of the following holds good
(where { * } denotes fractional part function)
(A) ‘h’ is continuous at x = 0
(B) ‘h’ is discontinuous at x = 0
(C) f(g(x)) is an even function
(D) f(x) is an even function
Sol. 60. Given

  2
−5 
   [ x ]+[ − x ]  
 
   |x|  
x a
loga (a | [ x ] + [ − x ] |)   for | x |≠ 0 ; a > 1
f(x)=   1
 then
 3+a
|x |
 
 
  
 0 for x=0
(where [*] represent the integral part function)
(A) f is continuous but not differentiable at x = 0
(B) f is cont. & diff. at x = 0
(C) the differentiability of ‘f’ at x = 0 depends on the
value of a
(D) f is cont. & diff. at x = 0 and for a = e only.
 Page # 38 CONTINUITY & DIFFERENTIABILITY

Sol. 63. The function f(x) is defined as follows

 −x if x<0
 x2 if 0 ≤ x ≤1
f(x) =  then f(x) is
 x − x + 1 if
3
x >1
(A) derivable & cont. at x = 0
(B) derivable at x = 1 but not cont. at x = 1
(C) neither derivable nor cont. at x = 1
(D) not derivable at x = 0 but cont. at x = 1
Sol.

61. For what triplets of real number (a, b, c) with

 x x ≤1
a ≠ 0 the function f(x) =  2 is
ax + bx + c otherwise
differentiable for all real x ?
(A) {(a, 1 – 2a, a)| a ε R, a ≠ 0}
(B) {(a, 1 – 2a, c)| a, c ε R, a ≠ 0}
(C) {(a, b) | a, b, c ε R, a + b + c = 1}
(D) {(a, 1 – 2a, 0)| a ε R, a ≠ 0}
Sol.

 x + { x } + x sin{ x } for x≠0

64. If f(x) =  then
 0 for x=0
(where { * } denotes the fractional part function)
(A) ‘f’ is cont. & diff. at x = 0
(B) ‘f’ is cont. but not diff. at x = 0
(C) ‘f’ is cont. & diff. at x = 2 (D) None of these
62. A function f defined as f(x) = x[x] for –1 ≤ x ≤ 3 Sol.
where [x] defines the greatest integer ≤ x is
(A) conti. at all points in the domain of but non-
derivable at a finite number of points
(B) discontinuous at all points & hence non-derivable
at all points in the domain of f
(C) discont. at a finite number of points but not
derivable at all points i the domain of f
(D) discont. & also non-derivable at a finite number of
points of f.
Sol.
 CONTINUITY & DIFFERENTIABILITY Page # 39

EXERCISE – II MULTIPLE CORRECT (OBJECTIVE QUESTIONS)

1. Which of the following function(x) not defined at x = 0 Sol.
has/have non–removable discontinuity at the origin ?

1
1 –1 1
(A) f(x) = (B) f(x) = tan
1+ 2x x
1
ex −1 1
(C) f(x) = 1 (D) f(x) =
ln | x |
ex +1
Sol. 4. The points at which the function,
f(x) = |x – 0.5| + |x – 1| + tan x does not have a
derivative in the interval (0, 2) are
(A) 1 (B) π/2 (C) 3 (D) 1/2
Sol.

2. Which of the following function(s) defined below

has / have single point continuity. 5. Let f(x) and g(x) be defined by f(x) = [x] and
0 , x∈Ι
1 if x ∈ Q  x if x ∈ Q g(x) =  then
x , x ∈ R − Ι
2
(A) f(x) =  (B) g(x) = 
0 if x ∉ Q 1 − x if x ∉ Q (where [ * ] denotes the greatest integer function)

 x if x ∈ Q  x if x ∈ Q (A) lim
x →1
g(x) exists, but g is not continuous at x = 1
(C) h(x) =  (D) k(x) = 
0 if x ∉ Q − x if x ∉ Q
(B) lim
x →1
f(x) does not exist and f is not continuous
Sol.
at x = 1.
(C) gof is continuous for all x
(D) fog is continuous for all x
Sol.

 | x −3| ,x ≥ 1
 x 2   3 x   13 
3. The function f(x) =   −   +   , x < 1 is
 4   2   4 

(A) continuous at x = 1 (B) differentiable at x = 1

(C) continuous at x = 3 (D) differentiable at x = 3
 Page # 40 CONTINUITY & DIFFERENTIABILITY

Sol.
   2x − 3 3
3 − cot −1 2
 for x > 0

6. Given f(x) =    x  then
 { x 2 } cos( e1/ x ) for x < 0


which of the following statement does not hold good.

(where { * } & [ * ] denotes the fractional part and
the integral part function respectively)
– 9. Let [x] denotes the greatest integer less than or
(A) f(0 ) = 0
+ equal to x. If f(x) = [x sin πx], then f(x) is
(B) f(0 ) = 3
(C) f(0) = 0 ⇒ continuity of f at x = 0 (A) continuous at x = 0
(D) irremovable discontinuity of f at x = 0 (B) continuous in (–1, 0)
Sol. (C) differentiable at x=1
(D) differentiable in (–1, 1)
Sol.

7. Let f(x) = [x] + x − [ x ] . Then 10. Let f(x) = nlim (sin x)2n, then f is
→∞
(where [ * ] denotes the greatest integer function)
+ (A) continuous at x = π/2
(A) f(x) is continuous on R
(B) discontinuous at x = π/2
(B) f(x) is continuous on R
(C) discontinuous at x = 0
(C) f(x) is continuous on R – Ι
(D) discontinuous at x = 1 (D) discontinuous at an infinite number of points
Sol. Sol.

n 1
8. If f(x) = ∑a
k =0
k | x |k , where a ’s are real constants,
i
11. Let f(x) =
[sin x]
then

then f(x) is (where [ * ] denotes the greatest integer function)

(A) continuous at x = 0 for all ai (A) domain of f(x) is (2n π + π, 2n π + 2π) ∪ {2n π + π/2}
(B) differentiable at x = 0 for all ai ∈ R (B) f(x) is continuous when x ∈ (2n π + π, 2n π + 2π)
(C) differentiable at x = 0 for all a2k + 1 = 0 (C) f(x) is continuous at x = 2nπ + π/2
(D) none of these (D) f(x) has the period 2π
 CONTINUITY & DIFFERENTIABILITY Page # 41

Sol. x
15. Indicate all correct alternatives if, f(x) = – 1,
2
then on the interval [0, π]
1
(A) tan (f(x)) & are both continuous
f ( x)
1
(B) tan (f(x)) & are both discontinuous
f ( x)
–1
(C) tan (f(x)) & f (x) are both continuous
12. The function f(x) = 1− 1− x 2 1
(D) tan (f(x)) is continuous but is not
(A) has its domain –1 ≤ x ≤ 1 f ( x)
(B) both f′(0–) and f′(0) are finite Sol.
(C) is continuous and differentiable at x = 0
(D) is continuous but not differentiable at x = 0
Sol.

1 − xn
13. Let f(x) = lim . Then 16. f(x) = |[x] x| in – 1 ≤ x ≤ 2, then f(x) is
n→∞ 1 + x n
(where [ * ] denotes greatest integer ≤ x)
(A) f(x) is a constant in 0 < x < 1
(A) cont. at x = 0 (B) discont. x = 0
(B) f(x) is continuous at x = 1
(C) f(x) is not differentiable at x = 1 (C) not diff. at x = 2 (D) diff. at x = 2
Sol.
(D) None of these
Sol.

14. Let ‘f’ be a continuous function on R.

−n 2 n2 17. f(x) = 1 + x . [cos x] in 0 < x ≤ π/2, then f(x) is

n n
If f(1/4 ) = (sin e ) e + then f(0) is
n2 + 1 (where [ * ] denotes greatest integer ≤ x)
(A) It is continuous in 0 < x < π/2
(A) not unique (B) 1
(B) It is differentiable in 0 < x < π/2
(C) data sufficient to find f(0)
(C) Its maximum value is 2
(D) data insufficient of find f(0)
(D) Its not differentiable in 0 < x < π/2
Sol.
Sol.
 Page # 42 CONTINUITY & DIFFERENTIABILITY

18. f(x) = (sin–1 x)2 . cos (1/x) if x ≠ 0 ; f(0) = 0, f(x) is Sol.

(A) cont. no where in –1 ≤ x ≤ 1
(B) cont. every where in –1 ≤ x ≤ 1
(C) differentiable no where in –1 ≤ x ≤ 1
(D) differentiable everywhere in –1 < x< 1
Sol.

22. If f(x) = x2 . sin (1/x), x ≠ 0 and f(0) = 0 then,

(A) f(x) is continuous at x = 0
(B) f(x) is derivable at x = 0
(C) f′(x) is continuous at x = 0
(D) f′(x) is not derivable at x = 0
Sol.
 π π
19. f(x) = |x| + |sin x| in  − ,  . It is
 2 2
(A) conti. no where
(B) conti. every where
(C) differentiable no where
(D) Differentiable every where except at x = 0
23. A function which is continuous & not differentiable
Sol.
at x = 0 is
(A) f(x) = x for x < 0 & f(x) = x2 for x ≥ 0
(B) g(x) = x for x < 0 & g(x) = 2x for x ≥ 0
(C) h(x) = x |x| x ∈ R (D) K(x) = 1 + |x|, x ∈ R
Sol.

20. If f(x) = 3(2x + 3)2/3 + 2x + 3 then

(A) f(x) is cont. but not diff. at x = –3/2
24. If sin–1 x + |y| = 2y then y as a function of x is
(B) f(x) is diff. at x = 0
(A) defined for –1 ≤ x ≤ 1
(C) f(x) is cont. at x = 0
(B) continuous at x = 0
(D) f(x) is diff. but not cont. at x = –3/2
(C) differentiable for all x
Sol.
dy 1
(D) such that dx = for –1 < x < 0
3 1− x2
Sol.

21. If f(x) = 2 + |sin–1 x|, it is

(A) continuous no where
(B) continuous every where in its domain
(C) differentiable no where in its domain
(D) Not differentiable at x = 0
 CONTINUITY & DIFFERENTIABILITY Page # 43

EXERCISE – III SUBJECTIVE QUESTIONS

3 x 2 + ax + a + 3
1. If the function f(x) = is continuous 4. Suppose that f(x) = x3 – 3x2 – 4x + 12 and
x2 + x − 2
at x = –2. Find f(–2).
 f(x)
Sol.
 , x≠3
h(x) =  x − 3 then
 K , x=3

(a) find all zeros of f(x)

(b) find the value of K that makes h continuous at x = 3
(c) using the value of K found in (b), determine whether
h is an even function.
Sol.
2. Find all possible values of a and b so that f(x) is
continuous for all x ∈ R if

 | ax + 3 | if x ≤ −1
 | 3x + a | if − 1 < x ≤ 0
f(x) =  bsin2x
− 2b if 0 < x < π
 x
 2
cos x − 3 if x ≥ π
Sol.

x2 x2
5. Let y n (x) = x2 + + +..........+
1 + x2 (1 + x 2 )2

x2
and y(x) = nLim
→ ∞ yn(x)
(1 + x 2 )n−1

Discuss the continuity of yn(x) (n ∈ N) and y(x) at x = 0

ln cos x Sol.
if x > 0
 4
1+ x2 − 1
3. Let f(x) =  Is it possible to
 esin 4 x − 1
if x < 0
ln (1 + tan 2x )
define f(0) to make the function continuous at x = 0.
If yes what is the value of f(0), if not then indicate
the nature of discontinuity.
Sol.

6. Draw the graph of the function

f(x) = x – |x – x2|, –1 ≤ x ≤ 1 & discuss the continuity
or discontinuity of f in the interval –1 ≤ x ≤ 1.
 Page # 44 CONTINUITY & DIFFERENTIABILITY

Sol.
1 + x , 0 ≤ x ≤ 2
9. Let f(x) = 3 − x , 2 < x ≤ 3 . Determine the form of

g(x) = f[f(x)] & hence find the point of discontinuity
of g, if any.
Sol.

1 − sin πx 1
 1 + cos 2πx , x<
2
 1
7. Let f(x) =  p, x = . Determine the
 2
 2x − 1 1
, x>
4 + 2x − 1 − 2 2
value of p, if possible, so that the function is continuous
10. Let [x] denote the greatest integer function &
at x = 1/2
f(x) be defined in a neighbourhood of 2 by
Sol.

 [x +1]
 (exp{(x + 2)n4}) 4 − 16
 , x<2
f(x) =  4x − 16
 1 − cos(x − 2)
A , x>2
 (x − 2)tan(x − 2)

Find the values of A & f(2) in order that f(x) may be

continuous at x = 2.
Sol.
8. Given the function g(x) = 6 − 2x and
2
h(x) = 2x – 3x + a. Then
(a) evaluate h (g(2))
Sol.

(b) If f(x)= [
g( x ), x ≤ 1
h( x ), x > 1
, find ‘a’ so that f is continuous.
Sol. tan 6 x
 6  tan 5 x π
  if 0<x<
 5 2
 π
11. The function f(x) =  b + 2 if x=
2
  a |tan x| 
   π
(1+ | cos x |) b  if <x<π
2
Determine the values of ‘a’ & ‘b’, if f is continuous
at x = π/2.
 CONTINUITY & DIFFERENTIABILITY Page # 45

Sol. sin 3 x + A sin 2x + B sin x

14. If f(x) = (x ≠ 0) is cont. at
x5
x = 0. Find A & B. Also find f(0).
Sol.

π
12. Determine a & b so that f is continuous at x =
2
 1 − sin3 x
 if x<
π  π

−1
( )
2 

−1
  2 − sin 1 − {x}  sin (1 − {x})
 3 cos 2 x 2  for x ≠ 0

where f(x) =  a if x=
π 15. Let f(x) = 
 (
2 {x} − {x}3 )
2  π
 b(1 − sin x ) π for x = 0
 
if x> 2
 ( π − 2x )2 2
Sol.  where {x} is the fractional part of x.
Consider another function g(x); such that
 f ( x ) for x ≥ 0
g(x) = 
2 2 f ( x ) for x < 0
Discuss the continuity of the functions f(x) & g(x)
at x = 0.
Sol.

13. Determine the values of a, b & c for which the

sin(a + 1)x + sinx

 for x < 0
 x
function f(x) =  c for x = 0
(x + bx2 )1/2 − x1/2 16. Discuss the continuity of f in [0, 2] where
 for x > 0
bx3/2 | 4 x − 5 | [ x ] for x > 1
f(x) = [cos πx ] for x ≤ 1 ; where [x] is the greatest
is continuous at x = 0. 
integer not greater than x. Also draw the graph.
Sol.
Sol.
 Page # 46 CONTINUITY & DIFFERENTIABILITY

17. If f(x) = x + {–x} + [x], where [x] is the integral Sol.

part & {x} is the fractional part of x. Discuss the
continuity of f in [–2, 2].
Sol.

 2 + cos x 3 
21. The function f(x) =  3 −  is not defined
 x sin x x 4 
at x = 0. How should the function be defined at x = 0
to make it continuous at x = 0.
Sol.
18. Find the locus of (a, b) for which the function

 ax − b for x ≤1
f(x) =  3x for 1< x < 2
 2
 bx − a for x≥2

is continuous at x = 1 but discontinuous at x = 2.

Sol.
 a sin x − a tan x
 for x > 0
tan x − sin x
22. f(x) =  2 2 , if f
 n (1 + x + x ) + n(1 − x + x ) for x < 0
 sec x − cos x
is continuous at x = 0, find ‘a’

 x
Now if g(x) = n  2 −  cot (x – a) for x ≠ a, a ≠ 0, a > 0.
 a 
19. A function f : R → R is defined as –1
If g is continuous at x = a then show that g(e ) = – e.
ax 2 + bx + c + enx Sol.
f(x) = nLim
→∞ where f is continuous on
1 + c.enx
R. Find the values of a, b and c.
Sol.

23. Let f(x + y) = f(x) + f(y) for all x, y & if the

function f(x) is continuous at x = 0, then show that
f(x) is continuous at all x.
Sol.
xn f ( x ) + h( x ) + 1
20. Let g(x) = nLim
→∞ , x ≠ 1 and
2x n + 3 x + 3

sin2 ( π.2 x )
g(1) = Lim
x →1 be a continuous function
ln (sec( π.2 x ))
at x = 1, find the value of 4 g (1) + 2 f(1) – h(1).
Assume that f(x) and h(x) are continuous at x = 1.
 CONTINUITY & DIFFERENTIABILITY Page # 47

26. (a) If g : [a, b] onto [a, b] is continuous show

n
 x  x 
24. Given f(x) = ∑
tan  r  sec  r −1  ; r, n ∈ N
2  2 
that there is some c ∈ [a, b] such that g (c) = c.
r =1 Sol.

n
 x   x    x 
n  f ( x ) + tan n  −  f ( x ) + tan n  . sin tan  
 2   2    2 
g(x) = Limit
n→∞ n =k
 x 
1 +  f ( x ) + tan n 
 2 

π
for x = and the domain of g(x) is (0, π/2).
4
(where [ * ] denotes the greatest integer function) (b) Let f be continuous on the interval [0, 1] to R
Find the value of k, if possible, so that g(x) is
such that f(0) = f(1). Prove that there exists a point
continuous at x = π/4. Also state the points of
discontinuity of g(x) in (0, π/4), if any.  1  1
Sol. c in 0,  such that f(c) = f  c + 
 2   2

Sol.

27. Consider the function

3 2 f ( x)
25. Let f(x) = x – x – 3x – 1 and h(x) = where
g( x )
h is a rational function such that  1 − a x + xa x na
 for x < 0
(a) it is continuous every where except when x = –1,  ax x2
 x x
g(x) = 2 a − x n 2 − x na − 1 where a > 0.
1  for x > 0
(b) xLim Lim
→ ∞ h(x) = ∞ and (c) x → −1 h(x) = 2 .  x2

Find xLim find the value of ‘a’ & ‘g(0)’ so that the function g(x)
→ 0 (3h(x) + f(x) – 2g(x))
is continuous at x = 0.
Sol.
Sol.
 Page # 48 CONTINUITY & DIFFERENTIABILITY

EXERCISE – IV ADVANCED SUBJECTIVE QUESTIONS

1. Discuss the continuity & differentiability of the 4. A function f is defined as follows :
function f(x) = sin x + sin | x |, x ∈ R. Draw a rough
sketch of the graph of f(x).

Sol.
 1 for −∞ < x < 0

π
f(x) =  1 + | sin x | for 0≤x<
 2
 2
  π π
 2 +  x −  for ≤ x < +∞
  2 2

Discuss the continuity & differentiability at x = 0

& x = π/2.
Sol.

2. Examine the continuity and differentiability of

f(x) = | x | + | x – 1 | + | x – 2 | x ∈ R.
Also draw the graph of f(x).
Sol.

5. Examine the origin for continuity & derivability in

the case of the function f defined by
–1
f(x) = x tan (1/x), x ≠ 0 and f(0) = 0.
Sol.
3. If the function f(x) defined as
x2
− for x ≤ 0
 2
f(x) =  is continuous but not
 xn sin 1 for x > 0
x
derivable at x = 0. Then find the range of n.
Sol.

6. Let f(0) = 0 and f ’(0) = 1. For a positive integer k,

show that

Lim 1  f(x) + f  x  + ....f  x   = 1 + 1 + 1 +.....+ 1

x →0  2  k  3
x      2 k
Sol.
 CONTINUITY & DIFFERENTIABILITY Page # 49

 1 1 Sol.
− + 
| x | x 
7. Let f(x) = xe  ; x ≠ 0, f(0) = 0, test the
continuity & differentiability at x = 0.
Sol.

  2[ x ]  
11. Given f(x) = cos–1  sgn    Discuss the

  3x − [ x]  
continuity & differentiability of f(x) at x = ± 1.
(where sgn ( * ) denotes the signum function & [ * ]
denotes the greatest integer function)
Sol.
+
8. If f(x) = |x – 1|. ([x] – [–x]), then find f ’(1 ) &
–
f ’(1 ) (where [ * ] denotes greatest integer function)
Sol.

12. Examine for continuity & differentiabilty the points

x = 1 & x = 2, the function f defined by
 x[ x ] , 0≤x<2
f(x) = ( x − 1) [ x ] , 2 ≤ x ≤ 3
 ax 2 − b if | x | < 1 
 where [x] = greatest integer less than or equal to x.
9. If f(x) =  1 is derivable at x = 1.
− if | x | ≥ 1 Sol.
 | x |
Find the values of a & b.
Sol.

10. Let f(x) be defined in the interval [–2, 2] such  e[ x ] + |x| − 2 

13. f(x) = x.   , x ≠ 0 & f(0) = – 1
−1 , −2 ≤ x ≤ 0  [ x] + | x | 
that f(x) = x − 1 ,  
 0 < x ≤ 2 & g(x) = f(| x |) + | f(x) |. where [x] denotes greatest integer less than or equal to x.
Test the differentiability of g(x) in (–2, 2). Test the differentiability of f(x) at x = 0.
 Page # 50 CONTINUITY & DIFFERENTIABILITY

Sol. Sol.

14. Discuss the continuity & the derivability in [0, 2]

| 2x − 3 | [ x ] for x ≥ 1 a1/ x − a −1/ x

of f(x) =  πx 17. Examine the function, f(x) = x. ,x≠0
sin for x < 1 a1/ x + a −1/ x
 2
(a > 0) and f(0) = 0 for continuity and existence of
(where [ * ] denotes greatest integer function) the derivative at the origin.
Sol. Sol.

15. Let f(x) = [3 + 4 sin x]. If sum of all the values of

‘x’ in [π, 2π] where f(x) fails to be differentiable, is 18. Discuss the continuity on 0 ≤ x ≤ 1 & differentiability
kπ at x = 0 for the function.
, then find the value of k.
2
1 1
(where [ * ] denotes the greatest integer function) f(x) = x.sin sin where x ≠ 0, x ≠ 1/ rπ &
x 1
x. sin
Sol. x

f(0) = f(1 /rπ) = 0, r = 1, 2, 3, ..........

Sol.

ax( x − 1) + b when x < 1


16. The function f(x) =  x − 1 when 1 ≤ x ≤ 3 .
2
px + qx + 2 when x > 3

Find the values of the constants a, b, p, q so that  1 − x , (0 ≤ x ≤ 1)


(i) f(x) is continuous for all x 19. f(x) =  x + 2 , (1 < x < 2) Discuss the continuity &
 4 − x , (2 ≤ x ≤ 4)
(ii) f ’(1) does not exist
(iii) f ’(x) is continuous at x = 3 differentiability of y = f[f(x)] for 0 ≤ x ≤ 4.
 CONTINUITY & DIFFERENTIABILITY Page # 51

Sol. (b) Show that f ’(1/3) does not exist.

Sol.

20. Let f be a function that is differentiable every (c) For what values of x, f ’(x) fails to exist.
where and that has the following properties Sol.
(i) f(x +h) = f(x) . f(h) (ii) f(x) > 0 for all real x.
(iii) f ‘(0) = – 1
Use the definition of derivative to find f ’(x) in terms
of f(x).
Sol.

23. Let f(x) be a real valued function not identically

zero satisfies the equation, f(x + yn) = f(x) + (f(y))n
for all real x & y and f ‘(0) ≥ 0 where n (> 1) is an odd
natural number. Find f(10).
Sol.

21. Let f(x) be a function defined on (–a, a) with

a > 0. Assume that f(x) is continuous at x = 0 and

Lim f(x) − f(k x) = α, where k ∈ (0, 1) then compute

x →0 x
f’(0+) and f’(0–), and comment upon the differentiability
of f at x = 0.
24. A derivable function f : R+ → R satisfies the
Sol.
condition f(x) – f(y) ≥ n (x/y) + x – y for every x,
y ∈ R+. If g denotes the derivative of f then compute
100
 1
the value of the sum ∑ g  n  .
n =1
Sol.
π
 x 2 cos if x ≠ 0
22. Consider the function, f(x) =  2x
 0 if x = 0
(a) Show that f ’(0) exists and find its value
Sol.
 Page # 52 CONTINUITY & DIFFERENTIABILITY

EXERCISE – V JEE PROBLEMS

2 2
1. The function f(x) = [x] – [x ] (where [y] is 4. Let f : R → R be any function. Define g : R → R by
the greatest integer less than or equal to y), is g(x) = | f(x) | for all x. Then g is [JEE 2000(Scr.), 1]
discontinuous at [JEE 99,2] (A) onto if f is onto (B) one one if f is one one
(A) all integers (B) all integers except 0 & 1 (C) continuous if f is continuous
(C) all integers except 0 (D) all integers except 1 (D) differentiable if f is differentiable.
Sol. Sol.

5. Discuss the continuity and differentiability of the

 x
2. Determine the constants a, b & c for which the 1 + | x | , | x | ≥ 1
function, f(x) =  x . [REE 2000, 3]
 ,| x | <1
  1 − | x |
 (1 + ax )1/ x for x < 0 Sol.

function f(x) =  b for x = 0 is continuous
 1/ 3
 ( x + c ) − 1
for x > 0
 ( x + 1)1/ 2 − 1
at x = 0. [REE 99,6]
Sol.

6. (a) Let f : R → R be a function defined by,

3
f(x) = max [x, x ]. The set of all points where f(x) is NOT
differentiable is [JEE 2001 (Scr.)]
(A) {–1, 1} (B) {–1, 0} (C) {0, 1} (D) {–1, 0, 1}
3. Discuss the continuity of the function Sol.

 e1/(x −1) − 2
 , x ≠1
f(x) =  e1/(x−1) + 2 at x=1. [REE 2001 (Mains), 3]
1, x =1
Sol.

(b) The left hand derivative of, f(x) = [x] sin (π x) at

x = k, k an integer is
(where [ * ] denotes the greatest function)
k k–1
(A) (–1) (k – 1)π (B) (–1) (k – 1)π
k k–1
(C) (–1) k π (D) (–1) kπ
 CONTINUITY & DIFFERENTIABILITY Page # 53

Sol. 9. Let f : R → R be such that f(1) = 3 and f’(1) = 6.

1/ x
 f (1 + x ) 
The Limit 
x → 0  f (1) 
 equals [JEE 2002 (Scr.), 3]
 
1/2 2 3
(A) 1 (B) e (C) e (D) e
Sol.

(c) Which of the following functions is differentiable

at x = 0 ?
(A) cos (| x |) + | x | (B) cos (| x |) – | x |
(C) sin (| x |) + | x | (D) sin (| x |) – | x |
Sol.

{x + a if x < 0
10. f(x) = | x − 1| if x ≥ 0 and g(x) = 
x + 1
2
 ( x − 1) + b
if x < 0
if x ≥ 0

Where a and b are non negative real numbers. Determine

the composite function gof. If (gof) (x) is continuous for
all real x, determine the values of a and b. Further, for
these values of a and b, is gof differentiable at x = 0 ?
7. Let α ∈ R. Prove that a function f : R → R is
Justify your answer. [JEE 2002, 5]
differentiable at α if and only if there is a function
Sol.
g : R → R which is continuous at α and satisfies
f(x) – f(α) = g(x) (x – α) for all x ∈ R.
[JEE 2001 (mains), 5]
Sol.

11. If a function f : [–2a, 2a] → R is an odd function

8. The domain of the derivative of the function such that f(x) = f(2a – x) for x ∈ [a, 2a] and the left
hand derivative at x = a is 0 then find the left hand
 tan x−1 derivative at x = – a. [JEE 2003 (Mains), 2]
 if | x | ≤ 1
f(x) =  1 is [JEE 2002 (Scr.), 3] Sol.
 2 (| x | −1) if | x | > 1

(A) R – {0} (B) R – {1}

(C) R – {–1} (D) R – {–1, 1}
Sol.

12. (a) The function given by y = ||x| – 1| is

differentiable for all real numbers except the points
[JEE 2005 (Scr.), 3]
(A) {0, 1, –1} (B) ± 1 (C) 1 (D) –1
 Page # 54 CONTINUITY & DIFFERENTIABILITY

Sol. Sol.

(b) If |f(x1)–f(x2)| ≤ (x1–x2) , for all x1, x2 ∈ R. Find

2

the equation of tangent to the curve y = f(x) at the

point (1,2). [JEE 2005 (Mains), 2] b−x
16. Let f : (0, 1) → R be defined by f(x) =
Sol. 1 − bx
where b is a constant such that 0 < b < 1. Then
(A) f is not invertible on (0, 1) [JEE 2011, 4]
1
(B) f ≠ f–1 on (0, 1) and f'(b) =
2 3 f' (0)
13. If f(x) = min. (1, x , x ), then [JEE 2006, 5]
1
(A) f(x) is continuous ∀ x ∈ R (C) f = f–1 on (0, 1) and f'(b) =
f' (0)
(B) f’(x) > 0, ∀ x > 1
(D) f–1 is differentiable on (0, 1)
(C) f(x) is not differentiable but continuous ∀ x ∈ R Sol.
(D) f(x) is not differentiable for two values of x
Sol.

 π π
( x − 1)n  −x − 2, x≤−
2
14. Let g(x) = ; 0 < x < 2, m and n are  π
ln cosm ( x − 1) 
17. If f(x) =  − cos x, − < x ≤ 0 then
 2
integers m ≠ 0, n > 0 and let p be the left hand
 x − 1, 0< x ≤1
Lim
derivative of |x – 1| at x = 1. If x →1+ g(x) = p, then  ln x, x >1
π
[JEE 2008, 3] (A) f(x) is continuous at x = – [JEE 2011, 4]
2
(A) n = 1, m = 1 (B) n = 1, m = –1 (B) f(x) is not differentiable at x = 0
(C) n = 2, m = 2 (D) n > 2, m = n (C) f(x) is differentiable at x = 1
Sol. (D) f(x) is differentiable at x = – 3/2
Sol.

15. Let f : R → R be a function such that

f(x + y) = f(x) + f(y), ∀x, y ∈ R.
If f(x) is differentiable at x = 0, then [JEE 2011, 4]
(A) f(x) is differentiable only in a finite interval
containing zero
(B) f(x) is continuous ∀ x ∈ R
(C) f'(x) is constant ∀ x ∈ R
(D) f(x) is differentiable except at finitely many points
 CONTINUITY & DIFFERENTIABILITY Page # 55

Answer Ex–I SINGLE CORRECT (OBJECTIVE QUESTIONS)

1. A 2. B 3. D 4. C 5. C 6. B 7. B 8. B
9. B 10. B 11. B 12. D 13. B 14. B 15. D 16. D
17. C 18. B 19. D 20. C 21. D 22. B 23. B 24. C
25. D 26. C 27. C 28. C 29. C 30. D 31. C 32. D
33. C 34. D 35. D 36. B 37. B 38. C 39. C 40. B
41. D 42. D 43. D 44. B 45. C 46. D 47. A 48. B
49. C 50. B 51. B 52. C 53. A 54. B 55. A 56. D
57. A 58. D 59. C 60. B 61. A 62. D 63. D 64. D

Answer Ex–II MULTIPLE CORRECT (OBJECTIVE QUESTIONS)

1. ABC 2. BCD 3. ABC 4. ABD 5. ABC 6. AC 7. ABC

8. AC 9. ABD 10. BD 11. ABD 12. ABD 13. AC 14. BC
15. CD 16. AC 17. AB 18. BD 19. BD 20. ABC 21. BD
22. ABD 23. ABD 24. ABD

Answer Ex–III SUBJECTIVE QUESTIONS

+ –
1. –1 2. a = 0, b = 1 3. f(0 ) = – 2 ; f(0 ) = 2 hence f(0) not possible to define
4. (a) –2, 2, 3 (b) K = 5 (c) even 5. yn (x) is continuous at x = 0 for all n and y(x) is dicontinuous at x = 0

6. f is cont. in –1 ≤ x ≤1 7. P not possible. 8. (a) 4 – 3 2 + a, (b) a = 3

9. g(x) = 2 + x for 0 ≤ x ≤ 1, 2 – x for 1 < x ≤ 2, 4 – x for 2 < x ≤ 3, g is discontinuous at x = 1 & x = 2
10. A = 1 ; f(2) = 1/2 11. a = 0 ; b = –1 12. a = 1/2, b = 4 13. a = – 3/2, b ≠ 0, c = 1/2

+ π – π
14. A = –4, B = 5, f(0) = 1 15. f(0 ) = ; f(0 ) = ⇒ f is discont. at x = 0 ;
2 4 2
+ –
g(0 ) = g(0 ) = g(0) = π/2 ⇒ g is cont. at x = 0

1
16. the function f is continuous everywhere in [0, 2] except for x = 0, ,1&2
2

17. discontinuous at all integral values in [–2, 2]

18. locus (a, b) → x, y is y = x – 3 excluding the points where y = 3 intersects it.

19. c = 1, a, b ∈ R 20. 5 21.

 Page # 56 CONTINUITY & DIFFERENTIABILITY

 π
  n (tan x ) if 0 < x < 4
24. k = 0 ; g(x) =  π π . Hence g(x) is continuous everywhere.
0 if ≤x<
 4 2

39 1 ( n 2)2
25. g(x) = 4(x + 1) and limit = – 27. a = , g(0) =
4 2 8

Answer Ex–IV ADVANCED SUBJECTIVE QUESTIONS

1. f(x) is conti. but not derivable at x = 0 2. conti. ∀ x ∈ R, not diff. at x = 0, 1 & 2

3. 0 < n ≤ 1 4. conti. but not diff. at x = 0; diff. & conti. at x = π/2
5. conti. but not diff. at x = 0 7. f is cont. but not diff. at x = 0
+ –
8. f ’(1 ) = 3, f ’(1 ) = –1 9. a = 1/2, b = 3/2 10. not derivable at x = 0 & x = 1
11. f is cont. & derivable at x = –1 but f is neither cont. nor derivable at x = 1
12. discontinuous & not derivable at x = 1, continuous but not derivable at x = 2 13. not derivable at x = 0
14. f is conti. at x = 1, 3/2 & disconti. at x = 2, f is not diff. at x = 1, 3/2, 2 15. 24

1 + –
16. a ≠ 1, b = 0, p = and q = – 1 17. If a ∈ (0, 1) f ’(0 ) = – 1; f ’(0 ) = 1 ⇒ continuous but not derivable
3

If a = 1 ; f(x) = 0 which is constant ⇒ continuous but not derivable

– +
If a > 1 f ’(0 ) = – 1 ; f ’(0 ) = 1 ⇒ continuous but not derivable
18. conti. in 0 ≤ x ≤ 1 & not diff. at x = 0
19. f is conti. but not diff. at x = 1, disconti. at x = 2 & x = 3. cont. & diff. at all other points

 1−   1+ 
α   π   π 1
20. f ’(x) = – f(x) 21. f ’(0) = 22. (a) f ’(0) = 0, (b) f ’  3  = – and f ’  3  = , (c) x = n∈I
1− k   2   2 2n +1

23. f(x) = x ⇒ f(10) = 10 24. 5150

Answer Ex–V JEE PROBLEMS

2 2 + –
1. D 2. a = ln ;b= ; c=1 3. Discontinuous at x = 1; f(1 ) =1 and f(1 ) = –1
3 3

4. C 5. Discont. Hence hot deri. at x = 1 & –1. Cont. & deri. at x = 0

6. (a) D, (b) A, (c) D 8. D 9. C 10. a = 1 ; b = 0 (gof)’(0) = 0
–
11. f’(a ) = 0 12. (a) A, (b) y – 2 = 0 13. A, C 14. C
15. B,C 16. C,D 17. A,B,C,D
MATHS FOR JEE MAINS & ADVANCED

SOLVED EXAMPLES
Ex. 1 How many numbers between 10 and 10,000 can be formed by using the digits 1, 2, 3, 4, 5 if
(i) No digit is repeated in any number. (ii) Digits can be repeated.
Sol.
(i) Number of two digit numbers = 5 × 4 = 20
Number of three digit numbers = 5 × 4 × 3 = 60
Number of four digit numbers = 5 × 4 × 3 × 2 = 120
Total = 200

(ii) Number of two digit numbers = 5 × 5 = 25

Number of three digit numbers = 5 × 5 × 5 = 125
Number of four digit numbers = 5 × 5 × 5 × 5 = 625
Total = 775

Ex. 2 If 7all the letters of the word 'RAPID' are arranged in all possible manner as they are in a dictionary, then find the rank
of the word 'RAPID'.
Sol. First of all, arrange all letters of given word alphabetically : 'ADIPR'
Total number of words starting with A _ _ _ _ = 4! = 24
Total number of words starting with D _ _ _ _ = 4! = 24
Total number of words starting with I_ _ _ _ = 4! = 24
Total number of words starting with P _ _ _ _ = 4! = 24
Total number of words starting with RAD _ _ = 2! = 2
Total number of words starting with RAI _ _ = 2! = 2
Total number of words starting with RAPD _ =1
Total number of words starting with RAPI _ =1
 Rank of the word RAPID = 24 + 24 + 24 + 24 + 2 + 2 + 1 + 1 = 102

Ex.3 A delegation of four students is to be selected from a total of 12 students. In how many ways can the delegation
be selected, if-
(A) all the students are equally willing ?
(B) two particular students have to be included in the delegation ?
(C) two particular students do not wish to be together in the delegation ?
(D) two particular students wish to be included together only ?
(E) two particular students refuse to be together and two other particular students wish to be together only in
the delegation ?
Sol. (A) Formation of delegation means selection of 4 out of 12.
12
Hence the number of ways = C4 = 495.
(B) If two particular students are already selected. Here we need to select only 2 out of the remaining 10. Hence
10
the number of ways = C2 = 45.
(C) The number of ways in which both are selected = 45. Hence the number of ways in which the two are not
included together = 495 – 45 = 450
(D) There are two possible cases
(i) Either both are selected. In this case, the number of ways in which the selection can be made = 45.
(ii) Or both are not selected. In this case all the four students are selected from the remaining ten students.
10
This can be done in C4 = 210 ways.
Hence the total number of ways of selection = 45 + 210 = 255

16
 PERMUTATION AND COMBINATION

(e) We assume that students A and B wish to be selected together and students C and D do not wish to be
together. Now there are following 6 cases.
(i) (A, B, C) selected, (D) not selected
(ii) (A, B, D) selected, (C) not selected
(iii) (A, B) selected, (C, D) not selected
(iv) (C) selected, (A, B, D) not selected
(v) (D) selected, (A, B, C) not selected
(vi) A, B, C, D not selected
8
For (i) the number of ways of selection = C1 = 8
8
For (ii) the number of ways of selection = C1 = 8
8
For (iii) the number of ways of selection = C2 = 28
8
For (iv) the number of ways of selection = C3 = 56
8
For (v) the number of ways of selection = C3 = 56
8
For (vi) the number of ways of selection = C4 = 70
Hence total number of ways = 8 + 8 + 28 + 56 + 56 + 70 = 226.

Ex. 4 A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be drawn so that there are
atleast two balls of each color ?
Sol. The selections of 6 balls, consisting of atleast two balls of each color from 5 red and 6 white balls, can be made in the
following ways

Red balls (5) White balls (6) Number of ways

5 6
2 3 C2× C4 = 150
5 6
3 3 C3× C3 = 200
5 6
4 2 C4× C2 = 75

Therefore total number of ways = 425

Ex. 5 How many functions can be defined from a set A containing 5 elements to a set B having 3 elements ? How
many of these are surjective functions ?
Sol. Image of each element of A can be taken in 3 ways.
 Number of functions from A to B = 3 5 = 243.
Number of into functions from A to B = 25 + 25 + 2 5 – 3 = 93.
 Number of onto functions = 150.

Ex. 6 In how many ways the letters of the word "ARRANGE" can be arranged without altering the relative position of
vowels & consonants.
4!
Sol. The consonants in their positions can be arranged in = 12 ways.
2!
3!
The vowels in their positions can be arranged in = 3 ways
2!
 Total number of arrangements = 12 × 3 = 36

17
MATHS FOR JEE MAINS & ADVANCED

Ex. 7 (A) How many permutations can be made by using all the letters of the word HINDUSTAN ?
(B) How many of these permutations begin and end with a vowel ?
(C) In how many of these permutations, all the vowels come together ?
(D) In how many of these permutations, none of the vowels come together ?
(E) In how many of these permutations, do the vowels and the consonants occupy the same relative positions as in
HINDUSTAN ?
Sol.

9!
(A) The total number of permutations = Arrangements of nine letters taken all at a time = = 181440.
2!
(B) We have 3 vowels and 6 consonants, in which 2 consonants are alike. The first place can be filled in 3 ways a n d

7!
the last in 2 ways. The rest of the places can be filled in ways.
2!
7!
Hence the total number of permutations = 3 × 2 × = 15120.
2!
7!
(C) Assume the vowels (I, U, A) as a single letter. The letters (IUA), H, D, S, T, N, N can be arranged in ways.
2!
Also IUA can be arranged among themselves in 3! = 6 ways.
7!
Hence the total number of permutations = × 6 = 15120.
2!
6!
(D) Let us divide the task into two parts. In the first, we arrange the 6 consonants as shown below in ways.
2!
× C × C × C × C × C × C × (Here C stands for a consonant and × stands for a gap between two consonants)
7
Now 3 vowels can be placed in 7 places (gaps between the consonants) in C3.3! = 210 ways.

6!
Hence the total number of permutations = × 210 = 75600.
2!
(E) In this case, the vowels can be arranged among themselves in 3! = 6 ways.
6!
Also, the consonants can be arranged among themselves in ways.
2!
6!
Hence the total number of permutations = × 6 = 2160.
2!

Ex. 8 In how many ways can 8 different books be distributed among 3 students if each receives at least 2 books ?
Sol. If each receives at least two books, then the division trees would be as shown below :

8 8

2 2 4 3 3 2
(i) (ii)

 8! 
The number of ways of division for tree in figure (i) is  2 .
 (2!) 4!2! 

18
 PERMUTATION AND COMBINATION

 8! 
The number of ways of division for tree in figure (ii) is  .
 (3!) 2
2!2! 

 8! 8! 
The total number of ways of distribution of these groups among 3 students is  2
 2   3!
 (2!) 4!2! (3!) 2!2! 

Ex. 9 In how many ways 10 persons can be divided into 5 pairs ?

Sol. We have each group having 2 persons and the qualitative characteristic are same (Since there is no purpose
mentioned or names for each pair).
10 !
Thus the number of ways = (2 !)5 5! = 945.

Ex. 10 Find the number of all 6 digit numbers such that all the digits of each number are selected from the set {1,2,3,4,5}
and any digit that appears in the number appears at least twice.

No. of ways No. of ways of Total

Cases
of selection arrangements
5 5
All alike C1 C1 × 1 5
5
C2 × 2 ! 5 6!
4 alike + 2 other alike C2  2  300
2!4!
5
C2 5 6!
Sol. 3 alike + 3 other alike C2  200
3!3!
2 alike + 2 other alike + 5
C3 5 6!
C3  900
2 other alike 2!2!2!
Total 1405
Ex. 11 A student is allowed to select at most n books from a collection of (2n + 1) books. If the total number of ways in
which he can select books is 63, find the value of n.
Sol. Given student selects at most n books from a collection of (2n + 1) books. It means that he selects one book or two
books or three books or ............ or n books. Hence, by the given condition-
2n+1C + 2n+1C + 2n+1C +.........+ 2n+1C = 63 .....(i)
1 2 3 n
But we know that
2n+1C + 2n+1C + 2n+1C + 2n+1C + ....... + 2n+1C 2n+1
0 1 2 3 2n + 1 = 2 .....(ii)
Since 2n+1C = 2n+1C
0 2n + 1 = 1, equation (ii) can also be written as
2 + (2n+1C1 + 2n+1C2 + 2n+1 C3 + ....... + 2n+1Cn) +
(2n+1Cn+1 + 2n+1Cn+2 + 2n+1 Cn + 3 + ....... + 2n+1C2n–1 + 2n+1C2n) = 22n + 1

 2 + (2n+1C1 + 2n+1C2 + 2n+1C3 + ......... + 2n+1Cn) + (2n+1Cn + 2n+1Cn–1 + ........ + 2n+1C2 + 2n+1C1) = 22n+1
( 2n+1Cr = 2n+1C2n + 1 – r)
 2 + 2 (2n+1C1 + 2n+1C2 + 2n+1C3 + ....... + 2n+1Cn) = 22n + 1 [from (i)]
 2 + 2.63 = 22n+1  1+ 63 = 22n
 64 = 22n  26 = 22n  2n = 6
Hence, n = 3.

19
MATHS FOR JEE MAINS & ADVANCED

Ex. 12 Find the number of ways in which 20 different pearls of two colours can be set alternately on a necklace, there being
10 pearls of each color.
1
Sol. Ten pearls of one color can be arranged in . 10  1 ! ways. The number of arrangements of 10 pearls of the other
2
color in 10 places between the pearls of the first color = 10!
1 2
 The required number of ways   9! 10! = 5 (9!)
2
Ex. 13 Find the number of positive integral solutions of the inequation x + y + z 150, where 0 < x  60, 0 < y  60, 0 < z  60.
Sol. Let x = 60 – t1, y = 60 – t2, z = 60 – t3 (where 0  t1  59, 0  t2  59, 0  t3  59)
Given x + y + z  150
or x + y + z – w = 150 (where 0 w  147) .....(i)
Putting values of x, y, z in equation (i)
60 – t1 + 60 – t2 + 60 – t3 – w = 150
30 = t1 + t2 + t3 + w
Total solutions = 33C3

Ex. 14 Find sum of all numbers formed using the digits 2,4,6,8 taken all at a time and no digit being repeated.
Sol. All possible numbers = 4! = 24
If 2 occupies the unit's place then total numbers = 6
Hence, 2 comes at unit's place 6 times.
Sum of all the digits occuring at unit's place
= 6 × (2 + 4 + 6 + 8)
Same summation will occur for ten's, hundred's & thousand's place. Hence required sum
= 6 × (2 + 4 + 6 + 8) × (1 + 10 + 100 + 1000) = 133320

Ex.15 Four slip of papers with the numbers 1, 2, 3, 4 written on them are put in a box. They are drawn one by one
(without replacement) at random. In how many ways it can happen that the ordinal number of atleast one slip
coincide with its own number ?
Sol. Total number of ways = 4 ! = 24.
The number of ways in which ordinal number of any slip does not coincide with its own number is the number
1 1 1
of dearrangements of 4 objects = 4 !     = 9
 2 ! 3! 4 ! 
Thus the required number of ways. = 24 – 9 = 15

Ex. 16 Find the total number of proper factors of the number 35700. Also find
(i) sum of all these factors, (ii) sum of the odd proper divisors,
(iii) the number of proper divisors divisible by 10 and the sum of these divisors.
Sol. 35700 = 52 × 22 × 31 × 71 × 171
The total number of factors is equal to the total number of selections from (5,5), (2,2), (3), (7) and (17), which is given
by 3 × 3 × 2 × 2 × 2 = 72.
These include 1 and 35700. Therefore, the number of proper divisors (excluding 1 and 35700) is 72 – 2 = 70

(i) Sum of all these factors (proper) is :

(5° + 51 + 52) (2° + 21 + 22) (3° + 31) (7° + 71) (17° + 171) –1 –35700
= 31 × 7 × 4 × 8 × 18 – 1 – 35700 = 89291

20
 PERMUTATION AND COMBINATION

(ii) The sum of odd proper divisors is :

(5° + 51 + 52) (3° + 31) (7° + 71) (17° + 171) – 1
= 31 × 4 × 8 × 18 – 1 = 17856 – 1 = 17855

(iii) The number of proper divisors divisible by 10 is equal to number of selections from (5,5), (2,2), (3), (7), (17)
consisting of at least one 5 and at least one 2 and 35700 is to be excluded and is given by 2 × 2 × 2 × 2 × 2 – 1 = 31.
Sum of these divisors is :
(51 + 52) (21 + 22) (3° + 31) (7° + 71) (17° + 171) – 35700
= 30 × 6 × 4 × 8 × 18 – 35700 = 67980
Ex. 17 Find the number of solutions of the equation xyz = 360 when (i) x,y,z  N (ii) x,y,z  I
Sol. (i) xyz = 360 = 23 × 32 × 5 (x,y,z  N)
x = 2a1 3a2 5a3 (where 0  a1  3, 0  a2  2, 0  a3  1)
y = 2b1 3b2 5 b3 (where 0  b1  3, 0  b2  2, 0  b3  1)
z = 2c1 3c2 5 c3 (where 0  c1  3, 0  c2  2, 0  c3  1)
 2a1 3a2 5a3 .2b1 3b2 5 b3 .2c1 3c2 5c3  23  32  51

 2a1  b1  c1 .3a2  b2  c2 .5a3  b3  c3  23  33  51

 a1 + b1 + c1 = 3 5C2 10
a2 + b2 + c2 = 2 4C2  6
a3 + b3 + c3 = 1 3C2  3
Total solutions = 10 × 6 × 3 = 180.
(ii) If x,y,z  I then, (A) all positive (B) 1 positive and 2 negative.
Total number of ways = 180 + 3C2 × 180 = 720

Ex. 18 Find the number of positive integral solutions of x + y + z = 20, if x  y  z.

Sol. x1
y = x + t1 t1  1, z = y + t2 t2  1, x + x + t1 + x + t1 + t2 = 20
3x + 2t1 + t2 = 20
(i) x=1 2t1 + t2 = 17
t1 = 1,2 ......... 8  8 ways
(ii) x=2 2t1 + t2 = 14
t1 = 1,2 ......... 6  6 ways
(iii) x=3 2t1 + t2 = 11
t1 = 1,2 ......... 5  5 ways
(vi) x=4 2t1 + t2 = 8
t1 = 1,2,3  3 ways
(v) x=5 2t1 + t2 = 5
t1 = 1, 2  2 ways
Total = 8 + 6 + 5 + 3 + 2 = 24
But each solution can be arranged by 3! ways.
So total solutions = 24 × 3! = 144.

21
MATHS FOR JEE MAINS & ADVANCED

Exercise # 1 [Single Correct Choice Type Questions]

1. A 5 digit number divisible by 3 is to be formed using the numerals 0, 1, 2,3,4 & 5 without repetition. The total number
of ways this can be done is -
(A) 3125 (B) 600 (C) 240 (D) 216

2. The total number of words which can be formed using all the letters of the word "AKSHI" if each word begins with
vowel or terminates with vowel -
(A) 84 (B) 12 (C) 48 (D) 60

3. The number of signals that can be made with 3 flags each of different colour by hoisting 1 or 2 or 3 above the
other, is:
(A) 3 (B) 7 (C) 15 (D) 16

4. Number of words that can be made with the letters of the word "GENIUS" if each word neither begins with G
nor ends in S, is:
(A) 24 (B) 240 (C) 480 (D) 504

5. Number of ways in which 9 different prizes can be given to 5 students, if one particular student receives 4 prizes and
the rest of the students can get any numbers of prizes is -
(A) 9C4 . 210 (B) 9C5 . 54 (C) 4 . 45 (D) none of these

6. Boxes numbered 1, 2, 3, 4 and 5 are kept in a row and they are necessarily to be filled with either a red or a blue ball
such that no two adjacent boxes can be filled with blue balls. How many different arrangements are possible, given
that the balls of a given colour are exactly identical in all respects ?
(A) 8 (B) 10 (C) 13 (D) 22

7. In a conference 10 speakers are present. If S1 wants to speak before S2 & S2 wants to speak after S3, then
the number of ways all the 10 speakers can give their speeches with the above restriction if the remaining
seven speakers have no objection to speak at any number is :

10 !
(A) 10C3 (B) 10P8 (C) 10P3 (D)
3
8. If all the letters of the word “QUEUE” are arranged in all possible manner as they are in a dictionary, then the rank
of the word QUEUE is -
(A) 15th (B) 16th (C) 17th (D) 18th

9. Number of ways in which 9 different toys can be distributed among 4 children belonging to different age groups in such
a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more is -

(5!) 2 9! 9!
(A) (B) (C) (D) none of these
8 2 3!(2!)3
10. The number of ways of arranging the letters AAAAA, BBB, CCC, D, EE & F in a row if no two 'C's are
together :

12! 13! 14! 13!

(A) 13C3 . (B) (C) (D) 11.
5!3!2! 5!3!3!2! 5!3!2! 6!

22
 PERMUTATION AND COMBINATION

11. Passengers are to travel by a double decked bus which can accommodate 13 in the upper deck and 7 in the
lower deck. The number of ways that they can be divided if 5 refuse to sit in the upper deck and 8 refuse to
sit in the lower deck, is
(A) 25 (B) 21 (C) 18 (D) 15

12. How many nine digit numbers can be formed using the digits 2, 2, 3, 3, 5, 5, 8, 8, 8 so that the odd digits occupy
even positions?
(A) 7560 (B) 180 (C) 16 (D) 60

13. A box contains 2 white balls, 3 black balls & 4 red balls. In how many ways can three balls be drawn from the
box if atleast one black ball is to be included in draw (the balls of the same color are different).
(A) 60 (B) 64 (C) 56 (D) none

14. Number of ways in which 25 identical pens can be distributed among Keshav, Madhav, Mukund and Radhika such
that at least 1, 2, 3 and 4 pens are given to Keshav, Madhav, Mukund and Radhika respectively, is -
(A) 18C4 (B) 28C3 (C) 24C3 (D) 18C3

15. The number of proper divisors of apbqcrds where a, b, c, d are primes & p, q, r, s  N is -
(A) pqrs (B) (p + 1) (q + 1) (r + 1) (s + 1) – 4
(C) pqrs – 2 (D) (p + 1) (q + 1) (r + 1) (s + 1) – 2

16. The number of way in which 10 identical apples can be distributed among 6 children so that each child receives
atleast one apple is -
(A) 126 (B) 252 (C) 378 (D) none of these

17. How many divisors of 21600 are divisible by 10 but not by 15?
(A) 10 (B) 30 (C) 40 (D) none

18. If chocolates of a particular brand are all identical then the number of ways in which we can choose
6 chocolates out of 8 different brands available in the market, is:.
(A) 13C6 (B) 13C8 (C) 86 (D) none

19. The sum of all numbers greater than 1000 formed by using the digits 1, 3, 5, 7 such that no digit is being repeated in
any number is -
(A) 72215 (B) 83911 (C) 106656 (D) 114712

20. A set contains (2n + 1) elements. The number of subset of the set which contain at most n elements is : -
(A) 2n (B) 2n+1 (C) 2n – 1 (D) 2 2n

21. The number of ways in which we can arrange n ladies & n gentlemen at a round table so that 2 ladies or 2 gentlemen
may not sit next to one another is -
(A) (n – 1)! (n – 2)! (B) (n)! (n – !)! (C) (n + 1)! (n)! (D) none of these

22. If  = mC2, then aC2 is equal to

(A) m+1 C4 (B) m–1C4 (C) 3 m+2C4 (D) 3 m+1C4

23. The number of ways in which the number 27720 can be split into two factors which are co-primes, is:
(A) 15 (B) 16 (C) 25 (D) 49

23
MATHS FOR JEE MAINS & ADVANCED

24. Number of numbers greater than a million and divisible by 5 which can be formed by using only the digits
1, 2, 1, 2, 0, 5 & 2 is -
(A) 120 (B) 110 (C) 90 (D) none of these

25. The number of ways in which 6 red roses and 3 white roses (all roses different) can form a garland so that all
the white roses come together, is
(A) 2170 (B) 2165 (C) 2160 (D) 2155

26. Ten different letters of alphabet are given. Words with four letters are formed from these letters, then the number of
words which have at least one letter repeated is -
(A) 104 (B) 10P4 (C) 10C4 (D) 4960

27. Number of positive integral solutions of x1 . x2 . x3 = 30, is

(A) 25 (B) 26 (C) 27 (D) 28

28. The number of ways in which 5 different books can be distributed among 10 people if each person can get at most
one book is -
(A) 252 (B) 105 (C) 510 (D) 10C5 . 5!

29. A shopkeeper has 10 copies of each of nine different books, then number of ways in which atleast one book
can be selected is
(A) 911 – 1 (B) 1010 – 1 (C) 119 – 1 (D) 109

30. Out of seven consonants and four vowels, the number of words of six letters, formed by taking four consonants and
two vowels is (Assume that each ordered group of letter is a word) -
(A) 210 (B) 462 (C) 151200 (D) 332640

24
 PERMUTATION AND COMBINATION

Exercise # 2 Part # I [Multiple Correct Choice Type Questions]

1. N = 22. 33.54.7, then -

(A) Number of proper divisors of N(excluding 1 & N) is 118
(B) Number of proper divisors of N(excluding 1 & N) is 120
(C) Number of positive integral solutions of xy = N is 60
(D) Number of positive integral solutions of xy = N is 120

2. There are (p + q) different books on different topics in Mathematics. (p  q)

If L = the number of ways in which these books are distributed between two students X and Y such that X
get p books and Y gets q books.
M = The number of ways in which these books are distributed between two students X and Y such that one
of them gets p books and another gets q books.
N = The number of ways in which these books are divided into two groups of p books and q books then -
(A) L = N (B) L = 2M = 2N (C) 2L = M (D) L = M

3. A student has to answer 10 out of 13 questions in an examination. The number of ways in which he can
answer if he must answer atleast 3 of the first five questions is:
(A) 276 (B) 267 (C) 13C10 – 5C3 (D) 5C3 . 8C7 + 5C4 . 8C6 + 8C5

4. The number of ways in which 10 students can be divided into three teams, one containing 4 and others 3
each, is
10 ! 10 ! 1
(A) (B) 2100 (C) 10C4 . 5C3 (D) .
4 !3!3! 6 !3!3! 2
5. Number of dissimilar terms in the expansion of (x1 + x2 + ...... + xn)3 is -

n 2 (n  1)2 n(n  1)( n  2) n3  3n 2

(A) (B) (C) n+1C2 + n+1C3 (D)
4 6 4

6. The number of ways in which 200 different things can be divided into groups of 100 pairs, is:

200 !  101   102   103   200 

(A) (B)     ....  
2100  2  2  2   2 

200 !
(C) 100 (D) (1. 3. 5...... 199)
2 (100) !

50
7. C36 is divisible by
(A) 19 (B) 52 (C) 192 (D) 53

8. The number of five digit numbers that can be formed using all the digits 0, 1, 3, 6, 8 which are -
(A) divisible by 4 is 30
(B) greater than 30,000 and divisible by 11 is 12
(C) smaller than 60,000 when digit 8 always appears at ten's place is 6
(D) between 30,000 and 60,000 and divisible by 6 is 18.

9. The number of non-negative integral solutions of x1 + x2 + x3 + x4  n (where n is a positive integer) is

(A) n+3C3 (B) n+4C4 (C) n+5C5 (D) n+4Cn

25
MATHS FOR JEE MAINS & ADVANCED

10. Which of the following statement(s) is/are true :-

(A) 100C50 is not divisible by 10
(B) n(n – 1)(n – 2) .........(n – r + 1) is always divisible by r! (n  N and 0  r  n)
(C) Morse telegraph has 5 arms and each arm moves on 6 different positions including the position of rest.
Number of different signals that can be transmitted is 56 – 1.
(D) There are 5 different books each having 5 copies. Number of different selections is 65 –1.

11. The number of ways of arranging the letters AAAAA, BBB, CCC, D, EE & F in a row if the letter C are
separated from one another is:
12! 13 ! 14 ! 13 !
(A) 13C3. 5! 3! 2! (B) 5 ! 3 ! 3 ! 2 ! (C) 3 ! 3 ! 2 ! (D) 11. 6 !

12. Number of ways in which 3 numbers in A.P. can be selected from 1, 2, 3,...... n is:
(n  2)(n  4) n 2  4n  5
(A) if n is even (B) if n is odd
4 2
2
 n 1 n  n  2
(C) if n is odd (D) if n is even
4 4

13. All the 7 digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once and not divisible by 5 are arranged
in the increasing order. Then -
(A) 1800th number in the list is 3124567 (B) 1897th number in the list is 4213567
th
(C) 1994 number in the list is 4312567 (D) 2001th number in the list is 4315726

14. The kindergarten teacher has 25 kids in her class. She takes 5 of them at a time, to zoological garden as often
as she can, without taking the same 5 kids more than once. Then the number of visits, the teacher makes to the
garden exceeds that of a kid by:
(A) 25C5  24C4 (B) 24C5 (C) 25C5  24C5 (D) 24C4

15. A persons wants to invite one or more of his friend for a dinner party. In how many ways can he do so if he has
eight friends : -
(A) 28 (B) 28 – 1 (C) 82 (D) 8C1 + 8C2 + .....+ 8C8

16. You are given 8 balls of different color (black, white,...). The number of ways in which these balls can be
arranged in a row so that the two balls of particular colour (say red & white) may never come together is:
(A) 8 !  2.7 ! (B) 6. 7 ! (C) 2. 6 !. 7C2 (D) none

17. In an examination, a candidate is required to pass in all the four subjects he is studying. The number of ways
in which he can fail is
(A) 4P1 + 4P2 + 4P3 + 4P4 (B) 44 – 1 (C) 2 4 – 1 (D) 4C1 + 4C2 + 4C3 + 4C4

18. If P(n, n) denotes the number of permutations of n different things taken all at a time then P(n, n) is also identical
to
(A) n.P(n – 1, n – 1) (B) P(n, n – 1) (C) r! . P(n, n – r) (D) (n – r) . P(n, r)
(where 0  r  n)

19. There are 12 points in a plane of which 5 are collinear. The number of distinct quadrilaterals which can be
formed with vertices at these points is:
(A) 2. 7P3 (B) 7P3 (C) 10. 7C3 (D) 420
20. There are 10 seats in the first row of a theatre of which 4 are to be occupied. The number of ways of arranging
4 persons so that no two persons sit side by side is:
(A) 7C4 (B) 4. 7P3 (C) 7C3. 4 ! (D) 840

26
 PERMUTATION AND COMBINATION

Part # II [Assertion & Reason Type Questions]

These questions contain, Statement I (assertion) and Statement II (reason).

(A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
(B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I.
(C) Statement-I is true, Statement-II is false.
(D) Statement-I is false, Statement-II is true.

1. Statement-I : If a, b, c are positive integers such that a + b + c  8, then the number of possible values of
the ordered triplets (a, b, c) is 56.
Statement-II : The number of ways in which n distinct things can be distributed among r girls such that each
get at least one is n–1Cr–1.

2. Statement-I : Number of terms in the expansion of (x1 + x2 + x3 + .... + x11)6 = 16C6.

Statement-II : Number of ways of distributing n identical things among r persons when each person get zero
or more things = n+r–1Cn

3. Statement -I : The maximum number of points of intersection of 8 unequal circles is 56.

Statement -II : The maximum number of points into which 4 unequal circles and 4 non coincident straight lines
intersect, is 50.

4. Statement-I : Number of ways in which 400 different things can be distributed between Ramu & Shamu so that each
receives 200 things > Number of ways in which 400 different things can be distributed between Sita
& Geeta. So that Sita receives 238 things & Geeta receives 162 things.
Statement-II : Number of ways in which (m + n) different things can be distributed between two receivers such that
one receives m and other receives n is equal to m+nCm, for any two non-negative integers m and n.

5. Statement-I : The number of positive integral solutions of the equation xyzw = 770 is 28.
Statement-II : The number of ways of selection of atleast one thing from n things of which 'p' are alike of one kind,
q are alike of 2nd kind and rest of the things are different is (p + 1)(q + 1) 2n–(p+q) – 1.

6. Statement-I : If a polygon has 45 diagonals, then its number of sides is 10.

Statement-II : Number of ways of selecting 2 points from n non collinear points is nC2.

7. Statement -I : If there are six letters L1, L2 , L3, L4, L5, L6 and their corresponding six envelopes E1, E2, E3, E4, E5, E6.
Letters having odd value can be put into odd value envelopes and even value letters can be put into
even value envelopes, so that no letter go into the right envelopes,the number of arrangement will
be equal to 4.
Statement -II : If Pn number of ways in which n letter can be put in ‘n’ corresponding envelopes such that
 1 1 (1) n 
no letter goes to correct envelope, then Pn = n!  1   – ....  
 1! 2! n! 

27
MATHS FOR JEE MAINS & ADVANCED

Exercise # 3 Part # I [Matrix Match Type Questions]

Following question contains statements given in two columns, which have to be matched. The statements in
Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given
statement in Column-I can have correct matching with one or more statement(s) in Column-II.
1. Column-I Column-II
(A) 24C + 23C + 22C + 21C + 20C + 20C is equal to (p) 102
2 2 2 2 2 3
(B) In the adjoining figure number of progressive
4 (4,4)

3
(2,2)
2

1
(q) 2300

0
1 2 3 4
ways to reach from (0,0) to (4, 4) passing
through point (2, 2) are
(particle can move on horizontal or vertical line)
(C) The number of 4 digit numbers that can be made with the digits (r) 82
1, 2, 3, 4, 3, 2

 500! 
(D) If  k   0, then the maximum natural value of k is equal to (s) 36
 14 
(where {.} is fractional part function)

2. Column – I Column – II
(A) The total number of selections of fruits which can be made (p) Greater than 50
from, 3 bananas, 4 apples and 2 oranges is, it is given that
fruits of one kind are identical
(B) If 7 points out of 12 are in the same straight line, then (q) Greater than 100
the number of triangles formed is
(C) The number of ways of selecting 10 balls from unlimited (r) Greater than 150
number of red, black, white and green balls is, it is given
that balls of same colours are identical
(D) The total number of proper divisors of 38808 is (s) Greater than 200

3. 5 balls are to be placed in 3 boxes. Each box can hold all the 5 balls. Number of ways in which the balls can be placed
so that no box remains empty, if :

Column-I Column-II
(A) balls are identical but boxes are different (p) 2
(B) balls are different but boxes are identical (q) 25
(C) balls as well as boxes are identical (r) 50
(D) balls as well as boxes are identical but boxes are kept in a row (s) 6

28
 PERMUTATION AND COMBINATION

4. Consider the word "HONOLULU".

Column – I Column – II
(A) Number of words that can be formed using (p) 26
the letters of the given word in which consonants
& vowels are alternate is
(B) Number of words that can be formed without (q) 144
changing the order of vowels is
(C) Number of ways in which 4 letters can be (r) 840
selected from the letters of the given word is
(D) Number of words in which two O's are together (s) 900
but U's are separated is

5. Consider all the different words that can be formed using the letters of the word HAVANA, taken 4 at a time.
Column-I Column-II
(A) Number of such words in which all the 4 letters are different (p) 36
(B) Number of such words in which there are 2 alike letters & (q) 42
2 different letters.
(C) Number of such words in which A's never appear together (r) 37
(D) If all such 4 letters words are written, by the rule of dictionary then (s) 24
the rank of the word HANA

Part # II [Comprehension Type Questions]

Comprehension # 1

Let p be a prime number and n be a positive integer, then exponent of p is n! is denoted by Ep (n!) and is given
by
n n n n
Ep(n!) =   +  2  +  3  + ..... +  k 
p  p  p  p 
where pk < n < pk+1
and [x] denotes the integral part of x.
If we isolate the power of each prime contained in any number N, then N can be written as
   
N = 2 1 · 3 2 · 5 3 · 7 4 ....
where i are whole numbers.

100
1. The exponent of 7 in C50 is -
(A) 0 (B) 1 (C) 2 (D) 3

2. The number of zeros at the end of 108! is -

(A) 10 (B) 13 (C) 25 (D) 26

3. The exponent of 12 in 100! is -

(A) 32 (B) 48 (C) 97 (D) none of these

29
MATHS FOR JEE MAINS & ADVANCED

Comprehension # 2

There are 8 official and 4 non-official members, out of these 12 members a committee of 5 members is to be formed,
then answer the following questions.

1. Number of committees consisting of 3 official and 2 non-official members, are

(A) 363 (B) 336 (C) 236 (D) 326

2. Number of committees consisting of at least two non-official members, are

(A) 456 (B) 546 (C) 654 (D) 466

3. Number of committees in which a particular official member is never included, are

(A) 264 (B) 642 (C) 266 (D) 462

Comprehension # 3

S = {0, 2, 4, 6, 8}. A natural number is said to be divisible by 2 if the digit at the unit place is an even number. The
number is divisible by 5, if the number at the unit place is 0 or 5. If four numbers are selected from S and a four digit
number ABCD is formed.

1. The number of such numbers which are even (all digits are different) is
(A) 60 (B) 96 (C) 120 (D) 204
2. The number of such numbers which are even (all digits are not different) is
(A) 404 (B) 500 (C) 380 (D) none of these
3. The number of such numbers which are divisible by two and five (all digits are not different) is
(A) 125 (B) 76 (C) 65 (D) 100

Comprehension # 4

Consider the letters of the word MATHEMATICS.

1. Possible number of words taking all letters at a time such that at least one repeating letter is at odd position in each
word is
11! 9! 9! 9! 11!
(A)  (B) (C) (D)
2!2!2! 2!2! 2!2!2! 2!2! 2!2!2!

2. Possible number of words taking all letters at a time such that in each word both M’s are together and both T’s are
together but both A’s are not together is
11! 10! 6!4! 9!
(A)  (B) 7! 8C2 (C) (D)
2!2!2! 2!2! 2!2! 2!2!2!

3. Possible number of words in which no two vowels are together is

4! 7! 8 4! 7! 8 4! 7! 8 4!
(A) 7! 8C4 (B) C (C) C (D) C
2! 2! 4 2! 2!2! 4 2! 2!2!2! 4 2!

30
 PERMUTATION AND COMBINATION

Comprehension # 5
We have to choose 11 players for cricket team from 8 batsmen. 6 bowlers, 4 allrounders and 2 wicketkeeper, in the
following conditions.

1. The number of selections when at most 1 allrounder and 1 wicketkeeper will play -

(A) 4C1 .14C10 + 2C1 .14C10 + 4C1 .2C1 .14C9 + 14C11 (B) 4C1 . 15C11 + 15C11

(C) 4C1 . 15C10 + 15C11 (D) none of these

2. Number of selections when 2 particular batsmen don't want to play, if a particular bowler will play -

(A) 17C10 + 19C11 (B) 17C10 + 19C11 + 17C11

(C) 17C10 + 20C11 (D) 19C10 + 19C11

3. Number of selections when a particular batsman and a particular wicketkeeper don't want to play together -

(A) 2 18C10 (B) 19C11 + 18C

10 (C) 19C10 + 19C11 (D) none of these

31
MATHS FOR JEE MAINS & ADVANCED

Exercise # 4 [Subjective Type Questions]

1. In how many ways can a team of 6 horses be selected out of a stud of 16, so that there shall always be 3 out of
ABC A'B'C', but never AA', BB' or CC' together ?

2. How many different permutations are possible using all the letters of the word MISSISSIPPI, if no two 's are
together ?

3. There are n straight lines in a plane, no 2 of which are parallel & no 3 pass through the same point. Their points of
n(n  1)(n  2)(n  3)
intersection are joined. Show that the number of fresh lines introduced is .
8
4. A family consists of a grandfather, m sons and daughters and 2n grand children. They are to be seated in a
row for dinner. The grand children wish to occupy the n seats at each end and the grandfather refuses to have
a grand children on either side of him. In how many ways can the family be made to sit.

5. There are 2 women participating in a chess tournament. Every participant played 2 games with the other partici-
pants. The number of games that the men played between themselves exceeded by 66 as compared to the number
of games that the men played with the women. Find the number of participants & the total number of games played
in the tournament.
200!
6. Prove that : is an integer
(10!) 20 19!

7. A party of 10 consists of 2 Americans, 2 Britishmen, 2 Chinese & 4 men of other nationalities (all different). Find the
number of ways in which they can stand in a row so that no two men of the same nationality are next to one another.
Find also the number of ways in which they can sit at a round table.

8. Find the number of words those can be formed by using all letters of the word ‘DAUGHTER’. If
(i) Vowels occurs in first and last place.
(ii) Start with letter G and end with letters H.
(iii) Letters G,H,T always occurs together.
(iv) No two letters of G,H,T are consecutive
(v) No vowel occurs together
(vi) Vowels always occupy even place.
(vii) Order of vowels remains same.
(viii) Relative order of vowels and consonants remains same.
(ix) Number of words are possible by selecting 2 vowels and 3 consonants.
9. How many different ways can 15 candy bars be distributed between Ram, Shyam, Ghanshyam and Balram, if Ram can
not have more than 5 candy bars and Shyam must have at least two ? Assume all candy bars to be alike.

10. In how many other ways can the letters of the word MULTIPLE be arranged ;
(i) without changing the order of the vowels ?
(ii) keeping the position of each vowel fixed ?
(iii) without changing the relative order/position of vowels & consonants ?
11. Find the number of ways to invite one of the three friends for dinner on 6 successive nights such that no friend is
invited more than 3 times.

32
 PERMUTATION AND COMBINATION

12. Let N = 24500, then find

(i) The number of ways by which N can be resolved into two factors.
(ii) The number of ways by which 5N can be resolved into two factors.
(iii) The number of ways by which N can be resolved into two coprime factors.
13. A man has 7 relatives, 4 of them are ladies & 3 gentlemen; his wife has also 7 relatives, 3 of them are ladies & 4
gentlemen. In how many ways can they invite a dinner party of 3 ladies & 3 gentlemen so that there are 3 of the man's
relatives & 3 of the wife's relatives ?
14. Words are formed by arranging the letters of the word "STRANGE" in all possible manner. Let m be the
number of words in which vowels do not come together and 'n' be the number of words in which vowels come
together. Then find the ratio of m: n.
15. A firm of Chartered Accountants in Bombay has to send 10 clerks to 5 different companies, two clerks in each. Two of
the companies are in Bombay and the others are outside. Two of the clerks prefer to work in Bombay while three others
prefer to work outside. In how many ways can the assignment be made if the preferences are to be satisfied ?
16. One hundred management students who read at least one of the three business magazines are surveyed to study
the readership pattern. It is found that 80 read Business India, 50 read Business world, and 30 read Business Today.
Five students read all the three magazines. Find how many read exactly two magazines?
17. (A) Prove that : nPr = n–1Pr + r. n–1Pr–1
(B) If 20Cr+2 = 20C2r–3 find 12Cr
(C) Find r if 15C3r = 15Cr+3
(D) Find the ratio 20Cr to 25Cr when each of them has the greatest value possible.
18. Find number of ways in which five vowels of English alphabets and ten decimal digits can be placed in a row
such that between any two vowels odd number of digits are placed and both end places are occupied by vowels
19. Prove by combinatorial argument that :
(A) n+1C = nC + nC
r r r–1
n+mC = nC . mC + nC . mC n m + ..... + nCr . mC0
(B) r 0 r 1 r–1 + C2 . Cr–2

20. A number lock has 4 dials, each dial has the digits 0, 1, 2, ........, 9. What is the maximum unsuccessful attempts
to open the lock ?
21. 5 boys & 4 girls sit in a straight line. Find the number of ways in which they can be seated if 2 girls are together &
the other 2 are also together but separated from the first 2.
22. X = {1, 2, 3, 4, ...... n} and A  X ; B  X ; A  B  X here P  Q denotes that P is subset of
Q(P  Q).Find number of ways of selecting unordered pair of sets A and B such that A  B  X.
23. In how many ways 11 players can be selected from 15 players, if only 6 of these players can bowl and the 11
players must include atleast 4 bowlers ?
24. There are 20 books on Algebra & Calculus in our library. Prove that the greatest number of selections each of which
consists of 5 books on each topic is possible only when there are 10 books on each topic in the library.
25. Find number of divisiors of 1980.
(i) How many of them are multiple of 11 ? find their sum
(ii) How many of them are divisible by 4 but not by 15.

33
MATHS FOR JEE MAINS & ADVANCED

Exercise # 5 Part # I [Previous Year Questions] [AIEEE/JEE-MAIN]

1. Numbers greater than 1000 but not greater than 4000 which can be formed with the digits 0, 1, 2, 3, 4 (repetition of
digits is allowed), are
(A) 350 (B) 375 (C) 450 (D) 576

2. A five digit number divisible by 3 has to formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total
number of ways in which this can be done is
(A) 216 (B) 240 (C) 600 (D) 3125

3. Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 are
(A) 192 (B) 375 (C) 400 (D) 720

4. The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together
is given by
(A) 6! × 5! (B) 30 (C) 5! × 4! (D) 7! × 5!

5. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first
five question. The number of choices available to him is
(A) 140 (B) 196 (C) 280 (D) 346
6. If nCr denotes the number of combinations of n things taken r at a time, then the expression
n
Cr 1  n Cr 1  2  n Cr equals
(A) n+2Cr (B) n+2Cr+1 (C) n+1Cr (D) n+1Cr+1

7. How many ways are there to arrange the letters in the word 'GARDEN' with the vowels in alphabetical order ?

(A) 120 (B) 240 (C) 360 (D) 480

8. The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

(A) 5 (B) 21 (C) 38 (C) 8C3

9. If the letters of the word 'SACHIN' are arranged in all possible ways and these words are written out as in dictionary,
then the word 'SACHIN' appears at serial number
(A) 602 (B) 603 (C) 600 (D) 601
6
50
10. The value of C4   56  r C3 is
r 1
56 56 55 55
(A) C4 (B) C3 (C) C3 (D) C4

11. At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There
are 10 candidated and 4 are to be elected. If a voter votes for at least one candidates, then the number of ways
in which he can vote is
(A) 385 (B) 1110 (C) 5040 (D) 6210

12. The set S = {1, 2, 3,.....,12} is to partitioned into three sets A, B, C of equal size. Thus,A B  C = S,
A B = B C = C A = , then number of ways to partition S are-
12! 12! 12! 12!
(A) 4 (B) 3 (C) 4 (D)
3!(3!) (4!) (3!) 3!(4!)3

34
 PERMUTATION AND COMBINATION

13. In a shop there are five types of ice–creams available. A child buys six ice–creams.
Statement –I : The number of different ways the child can buy the six ice–creams is 10C5.
Statement –2 : The number of different ways the child can buy the six ice–creams is equal to the number of different
ways of arranging 6 A’s and 4 B’s in a row.
(A) Statement –1 is false, Statement –2 is true
(B) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1
(C) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for Statement–1
(D) Statement–1 is true, Statement–2 is false

14. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in
a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangements is :-

(A) At least 750 but less than 1000 (B) At least 1000
(C) Less than 500 (D) At least 500 but less than 750
15. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are
taken out at random and then transferred to the other. The number of ways in which this can be done
is :-
(A) 3 (B) 36 (C) 66 (D) 108
16. Statement - I : The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty
is 9C3.
Statement - II : The number of ways of choosing any 3 places from 9 different places is 9C3.
(A) Statement-1 is true, Statement-2 is false.
(B) Statement-1 is false, Statement-2 is true
(C) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(D) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
17. There are 10 points in a plane, out of these 6 are collinear. If N is the number of triangles formed by joining these
points, then :
(A) N > 190 (B) N < 100 (C) 100 < N < 140 (D) 140 < N < 190
18. Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls
can be selected from 10 white, 9 green and 7 black balls is :-
(A) 879 (B) 880 (C) 629 (D) 630
19. Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having
3 or more elements is
(A) 256 (B) 220 (C) 219 (D) 211
20. Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If
Tn+1 – Tn = 10, then the value of n is :
(A) 7 (B) 5 (C) 10 (D) 8

21. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0),
(0, 41) and (41, 0), is
(A) 820 (B) 780 (C) 901 (D) 861

22. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is :
(A) 120 (B) 72 (C) 216 (D) 192
23. Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set
A × B, each having at least three elements is :
(A) 275 (B) 510 (C) 219 (D) 256

35
MATHS FOR JEE MAINS & ADVANCED

24. Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the
event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the
following statements is NOT true ?
(A) E2 and E3 are independent (B) E1 and E3 are independent
(C) E1, E2 and E3 are independent (D) E1 and E2 are independent

Part # II [Previous Year Questions][IIT-JEE ADVANCED]

1. How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that
the odd digits occupy even positions ?
(A) 16 (B) 36 (C) 60 (D) 180

2. (A) Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of ‘n’ sides. If
Tn+1 – Tn = 21, then ‘n’ equals -
(A) 5 (B) 7 (C) 6 (D) 4

(B) Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is -
(A) 14 (B) 16 (C) 12 (D) 8

3. The number of arrangements of the letters of the word BANANA in which two 'N's do not appear adjacently
is -
(A) 40 (B) 60 (C) 80 (D) 100

4. Number of points with integral co-ordinates that lie inside a triangle whose co-ordinates are (0, 0), (0, 21) and (21, 0)

(A) 210 (B) 190 (C) 220 (D) none of these

2
(n )!
5. Using permutation or otherwise prove that is an integer, where n is a positive integer..
(n!)n
6. A rectangle with sides 2m – 1 and 2n – 1 is divided into squares of unit length by drawing

parallel lines as shown then number of rectangles possible with odd side lengths is -

(A) (m + n + 1)2 (B) 4m + n – 1 (C) m2 n2 (D) mn (m + 1) (n + 1)

7. If r, s, t are the prime numbers and p, q are the positive integers such that the LCM of p & q is r2t4s2, then the number
of ordered pair (p, q) is :
(A) 252 (B) 254 (C) 225 (D) 224

8. The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an
English dictionary. The number of words that appear before the word COCHIN is -
(A) 360 (B) 192 (C) 96 (D) 48

36
 PERMUTATION AND COMBINATION

9. Consider all possible permutations of the letters of the word ENDEANOEL.

Match the Statements / Expressions in Column I with the Statements / Expressions in Column II.
Column I Column II
(A) The number of permutations containing the word ENDEA is (p) 5!
(B) The number of permutations in which the letter E occurs in the first (q) 2 × 5!
and the last positions is
(C) The number of permutations in which none of the letters D, L, N (r) 7 × 5!
occurs in the last five positions is
(D) The number of permutations in which the letters A, E, O occur only in (s) 21 × 5!
odd positions is

10. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is -

(A) 55 (B) 66 (C) 77 (D) 88

11. Let S = {1,2,3,4}. The total number of unordered pairs of disjoint subsets of S is equal to -

(A) 25 (B) 34 (C) 42 (D) 41

12. The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that
each person gets at least one ball is -
(A) 75 (B) 150 (C) 210 (D) 243

Paragraph for Question 13 and 14

Let an denotes the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive
digits in them are 0. Let bn = the number of such n-digit integers ending with digit 1 and cn = the number of such n-
digit integers ending with digit 0.

13. The value of b6 is

(A) 7 (B) 8 (C) 9 (D) 11

14. Which of the following is correct ?

(A) a17 = a16 + a15 (B) c17  c16 + c15 (C) b17  b16 + c16 (D) a17 = c17 + b16

15. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each
envelope contains exactly one card and no card is placed in the envelope bearing the same number and
moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be
done is
(A) 264 (B) 265 (C) 53 (D) 67

16. Let n  2 be an integer. Take n distinct points on a circle and join each pair of points by a line segment. Colour
the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and
blue line segment are equal, then the value of n is
17. Let n1 < n2 < n3 < n4 < n5 be positive integers such that n1 + n2 + n3 + n4 + n5 = 20. Then the number of such
distinct arrangements (n1, n2, n3, n4, n5) is
18. A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the
selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then
the number of ways of selecting the team is
(A) 380 (B) 320 (C) 260 (D) 95

37
MATHS FOR JEE MAINS & ADVANCED

MOCK TEST
SECTION - I : STRAIGHT OBJECTIVE TYPE
1. A shopkeeper has 10 copies of each of nine different books, then number of ways in which atleast one book
can be selected is
(A) 911 – 1 (B) 1010 – 1 (C) 119 – 1 (D) 109

2. No. of different squares of any size (side of square be natural no.) which can be made from a rectangle of size 15 × 8, is
(A) 456 (B) 120 (C) 228 (D) None of these

3. The number of different ways in which five ‘alike dashes’ and eight ‘alike dots’ can be arranged, using only seven
of these ‘dashes’ & ‘dots’ is
(A) 1287 (B) 119 (C) 120 (D) 1235520

4. In a hockey series between team X and Y, they decide to play till a team win ‘m’ match. Then the no. of ways in which
team X wins -
(A) 2m (B) 2mPm (C) 2mCm (D) None of these

5. There are three coplanar parallel lines. If any p points are taken on each the lines, the maximum number of
triangles with vertices at these points is
(A) 3p 2 ( p – 1) + 1 (B) 3p2 (p – 1) (C) p 2 (4p – 3) (D) none of these

6. A gentleman invites a party of m + n (m  n) friends to a dinner & places m at one table and n at another, the
table being round. If the clockwise & anticlockwise arrangements are not to be distinguished and assuming
sufficient space on both tables, then the number of ways in which he can arrange the guest is
(m  n)! 1 (m  n)! (m  n)!
(A) (B) (C) 2 (D) none
4 mn 2 4 mn 4 mn

7. Number of ways in which 6 different toys can be distributed among two brothers in ratio 1 : 2, is
(A) 30 (B) 60 (C) 20 (D) 40

8. There are m apples and n oranges to be placed in a line such that the two extreme fruits being both oranges.
Let P denotes the number of arrangements if the fruits of the same species are different and Q the corresponding
figure when the fruits of the same species are alike, then the ratio P/Q has the value equal to :
(A) nP 2. mP m. (n  2) ! (B) mP 2. nP n. (n  2) ! (C) nP2. nPn. (m  2) ! (D) none

9. S1 : For some natural N, the number of positive integral ‘x’ satisfying the equation,
1 ! + 2 ! + 3 ! + ..... + (x !) = (N)2 is 3
S2 : A women has 11 close friends then the number of ways in which she can invite 5 of them to
dinner, if two particular of them are not on speaking terms & will not attend together is 378
S3 : An old man while dialing a 7 digit telephone number remembers that the first four digits consists
of one 1’s, one 2’s and two 3’s. He also remembers that the fifth digit is either a 4 or 5 while has
no memorizing of the sixth digit, he remembers that the seventh digit is 9 minus the sixth digit.
Maximum number of distinct trials he has to try to make sure that he dials the correct telephone
number, is 240
S4 : The number of times the digit 3 will be written when listing the integers from 1 to 1000 is 300
(A) TTTF (B) FTTF (C) FTTT (D) FTFT

38
 PERMUTATION AND COMBINATION

10. A five letter word is to be formed such that the letters appearing in the odd numbered positions are taken from the
letters which appear without repetition in the word "MATHEMATICS". Further the letters appearing in the even
numbered positions are taken from the letters which appear with repetition in the same word "MATHEMATICS".
The number of ways in which the five letter word can be formed is:
(A) 720 (B) 540 (C) 360 (D) none

SECTION - II : MULTIPLE CORRECT ANSWER TYPE

11. Identify the correct statement(s).
(A) Number of zeroes standing at the end of 125 ! is 30.
(B) A telegraph has 10 arms and each arm is capable of 9 distinct positions excluding the position of rest.
The number of signals that can be transmitted is 1010  1.
(C) Numb er of number s greater than 4 lacs which can be for med by using o nly the d igits
0, 2, 2, 4, 4 and 5 is 90.
(D) In a table tennis tournament , every player plays with every other player. If the number of
games played is 5050 then the number of players in the tournament is 100.

12. Six cards are drawn one by one from a set of unlimited number of cards, each card is marked with numbers 
1, 0 or 1. Number of different ways in which they can be drawn if the sum of the numbers shown by them
vanishes, is:
(A) 111 (B) 121 (C) 141 (D) none

13. The number of different seven digit numbers that can be written using only three digits 1, 2 & 3 under the
condition that the digit 2 occurs exactly twice in each number is -
(A) 672 (B) 640 (C) 512 (D) none

14. Number of different words that can be formed using all the letters of the word "DEEPMALA", if two vowels
are together and the other two are also together but separated from the first two is
(A) 960 (B) 1200 (C) 2160 (D) 1440

15. The number of ways of arranging the letters AAAAA, BBB, CCC, D, EE & F in a row if the letters C are
separated from one another is:
12 ! 13 !
(A) 13C3. (B)
5! 3! 2! 5! 3! 3! 2!
14 ! 15! 13! 12! 13
(C) (D) 2
– – C2
3! 3! 2! 5!(3!) 2! 5!3! 2! 5!3!

SECTION - III : ASSERTION AND REASON TYPE

1 1  1 2 
16. Statement - I : Let E =         .... upto 50 terms, then E is divisible by exactly two primes.
 3 50   3 50 
Statement - II : [x + n] = [x] + n, n  I and [x + y] = [x] + [y] if x, y  I
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

39
MATHS FOR JEE MAINS & ADVANCED

17. Statement I : The sum of the digits in the tens place of all numbers with the help of 2, 3, 4, 5 taken all at a time is 84.
Statement II: The sum of the digits in the units place of all numbers formed with the help of
(a1, a2,...., an) taken all at a time is (n –1) ! (a1 + a2 + ........+ an) (repetition of digits not allowed)
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
18. Statement - I : Let A = {x | x is a prime number and x < 30}. Then the number of different rational nubmers,
whose numerator and denominator belong to A is 93.
p
Statement - II : is a rational number  q  0, and p, q I
q
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

19. Statement -I : A five digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 with
repetition. The total number of numbers formed is 216.
Statement -II : If sum of digits of any number is divisible by 3 then the number must be divisible by 3.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

20. Statement - I : The number of ordered pairs (m, n) ; m,n  {1, 2, 3, .....20} such that 3m + 7n is a multiple of 10, is equal to 100.
Statement - II : 3m + 7n has last digit zero, when m is of 4k + 2 type and n is of 4 type where k,   W.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True

SECTION - IV : MATRIX - MATCH TYPE

21. Match the following :

Column -  Column - 
(A) The number of five - digit numbers having the (p) 77
product of digits 20 is
(B) A man took 5 space plays out of an engine to (q) 30
clean them. The number of ways in which he can
place atleast two plays in the engine from
where they came out is
(C) The number of integers between 1 & 1000 inclusive (r) 50
in which atleast two consecutive digits are equal is
1
(D) The value of   i. j
15 1  i  j  9
(s) 181
(t) 31

40
 PERMUTATION AND COMBINATION

22. Column I Column II

(A) Number of increasing permutations of m numbers (p) nm
from the n set number {a1 , a2 ,......, an}
where the order among the numbers is given by
a1 < a2 < a3 < .........an–1 < an is
(B) There are m men and n monkeys. Number of ways (q) mC
n
in which every monkey has a master, if a man can
have any number of monkeys
(C) Number of ways in which n red balls and (m –1) green (r) nC
m
balls can be arranged in a line, so that no two red
balls are together, is (balls of the same colour are alike)
(D) Number of ways in which ‘m’ different toys can be distributed (s) mn
in ‘n’ children if every child may receive any number of toys is

SECTION - V : COMPREHENSION TYPE

23. Read the following comprehensions carefully and answer the questions.
Consider the letters of the word MATHEMATICS. There are eleven letters some of them are identical. Letters
are classified as repeating and non-repeating letters. Set of repeating letters = {M, A, T}. Set of non-repeat-
ing letters = {H, E, I, C, S}
1. Possible number of words taking all letters at a time such that atleast one repeating letter is at odd position
in each word, is
9! 11! 11! 9! 9!
(A) (B) (C) – (D)
2!2!2! 2!2!2! 2!2!2! 2!2! 2!2!

2. Possible number of words taking all letters at a time such that in each word both M's are together and both T's
are together but both A's are not together, is
11! 10 ! 6!4! 9!
(A) 7 ! . 8C2 (B) 2 ! 2 ! 2 ! – 2 ! 2 ! (C) 2 ! 2 ! (D) 2 ! 2 ! 2 !

3. Possible number of words in which no two vowels are together, is

7! 8 4! 7! 8 4! 4! 7! 4!
(A) . C4 . (B) . C4 . (C) 7 ! . 8C4 . (D) . 8C4 .
2!2! 2! 2! 2! 2! 2!2!2! 2!
24. Read the following comprehensions carefully and answer the questions.
5 ball are to be placed in 3 boxes. Each box can hold all the 5 balls. Number of ways in which the balls can be
placed so that no box remains empty, if :

1. balls are identical but boxes are different

(A) 2 (B) 25 (C) 50 (D) 6

2. balls are different but boxes are identical

(A) 2 (B) 25 (C) 50 (D) 6

3. balls as well as boxes are identical

(A) 2 (B) 25 (C) 50 (D) 6

41
MATHS FOR JEE MAINS & ADVANCED

25. Read the following comprehensions carefully and answer the questions.
Counting by critical paranthesis method
Suppose we have to arrange n-pairs of paranthesis in such a way that every arrangement is matched i.e.
number of left paranthesis are always greater than or equal to number of right paranthesis in any length of the
chain from start.
2n!
S is the number of ways of arranging n - right and n - left paranthesis in a row = .
n! n!
(2n) !
Let T be the arrangement of (n + 1) right and (n – 1) left paranthesis = .
(n  1)!(n  1)!
It can be shown that set of mismatched arrangements of paranthesis in ‘S’ has bijective relation with the set
of arrangements of T.
Since the set of mismatched arrangements in S has bijective relation with the set of arrangements in T.
2n !
 number of the mismatched arrangements in S =
(n  1)!(n  1)!
2n ! 2n ! 2n !
 Number of matched arrangements in S = – =
n ! n ! (n  1)!(n  1)! n ! (n  1)!

1. The number of ways in which ‘4’ pairs of paranthesis be arranged so that every arrangement is matched is:
8
C4
(A) 3 (B) (C) 8C4 (D) 8C5
5

2. If a stamp vendor sells tickets of 1 rupee each and there are 3 persons having 1 rupee coin and 3 having 2
rupee coin standing in a row. Then the probability that stamp vendor do not run out of change if he does not
have any money to start with is:
1 1 3
(A) (B) (C) (D) None of these
4 2 4
3. Number of ways of arranging the 5 pairs of paranthesis, if first pair is matched but the next four pairs are not
matched is:
(A) 10C6 (B) 8C5 (C) 4 C2 × 4 C2 (D) None of these

SECTION - VI : INTEGER TYPE

26. In a row, there are n rooms, whose door no. are 1,2,.......,n, initially all the door are closed. A person takes n round of
the row, numbers as 1st round, 2nd round ........ nth round. In each round, he interchange the position of those door no.,
whose no is multiple of the round no. Find out after nth round,How many doors will be open.

27. 10 IIT & 2 PET students sit in a row. The total number of ways in which exactly 3 IIT students sit between 2
PET students is 10! , then find .

28. The integers from 1 to 1000 are written in order around a circle. Starting at 1, every fifteenth number is marked
(that is 1, 16, 31, .... etc.). This process in continued untill a number is reached which has already been
marked, then find number of unmarked numbers.

29. 17 persons can depart from railway station in 2 cars and 3 autos, given that 2 particular person depart by same car
15!
are . (4 persons can sit in a car and 3 persons can sit in an auto), then find the value of .
 !(3!)3

30. Find the number of positive unequal integral solution of the equation x + y + z = 20.

42
 PERMUTATION AND COMBINATION

ANSWER KEY

EXERCISE - 1
1. D 2. A 3. C 4. D 5. A 6. C 7. D 8. C 9. C 10. A 11. B 12. D 13. B
14. D 15. D 16. A 17. A 18. A 19. C 20. D 21. B 22. D 23. B 24. B 25. C 26. D
27. C 28. D 29. C 30. C

EXERCISE - 2 : PART # I

1. AD 2. AC 3. ACD 4. BC 5. BC 6. BCD 7. AB 8. ABD 9. BD

10. ABD 11. AD 12. CD 13. BD 14. AB 15. BD 16. ABC 17. CD 18. ABC
19. AD 20. BCD

PART - II
1. C 2. A 3. B 4. D 5. B 6. D 7. A

EXERCISE - 3 : PART # I
1. Aq Bs CP Dr 2. A  p B  p,q,r C  p,q,r,s D  p 3. As Bq Cp Ds
4. Aq Br Cp Ds 5. As Bp Cq Dr

PART - II
Comprehension # 1 : 1. A 2. C 3. B Comprehension # 2 : 1. B 2. A 3. D
Comprehension # 3 : 1. B 2. A 3. B Comprehension # 4 : 1. D 2. B 3. C
Comprehension # 5 : 1. A 2. A 3. B

EXERCISE - 5 : PART # I
1. B 2. A 3. A 4. A 5. B 6. B 7. C 8. B 9. D 10. A 11. A 12. B 13. A
14. B 15. A 16. C 17. B 18. A 19. C 20. B 21. B 22. D 23. C 24. C

PART - II
1. C 2. A  p B  q 3. A 4. B 6. C 7. C 8. C 9. A  p B  s C  q D  q
10. C 11. D 12. B 13. B 14. A 15. C 16. 5 17. 7 18. A

MOCK TEST

1. C 2. A 3. C 4. D 5. C 6. C 7. A 8. A 9. C 10. B 11. B, C 12. C

13. A 14. D 15. A, D 16. D 17. A 18. D 19. A 20. B
21. A  r B  t C  s D  p 22. A  r B  s C  p D  q
23. 1. B 2. A 3. A 24. 1. D 2. B 3. A 25. 1. B 2. A 3. B

26.  n  27. 16 28. 800 29. 4 30. 144

 

43
MATHS FOR JEE MAINS & ADVANCED

HINTS & SOLUTIONS

EXERCISE - 1 17. Here 21600 = 25. 33. 52
Single Choice  (2 × 5) × 2 4 × 33 × 51
Now numbers which are divisible by 10
2. Total possible words – words do not begin or terminate = (4 + 1)(3 + 1)(1 + 1) = 40
with vowel (2 × 3 × 5) × (2 4 × 32 × 5 1) now numbers which are
Total words = 5! = 120 divisible by both 10 and 15
Words which do not begin and terminate with vowel = (4 +1)(2 +1)(1+1) = 30
So the numbers which are divisible by only
= 3 × 3 × 2 × 1 × 2 = 36
40 – 30 = 10
Desired words : 180 – 36 = 84
II-Method  words which begin with vowel 18. Using multinomial theorem
(A/I) = 4! × 2 = 48 ways  say = n(A) Total no. of ways of choosing 6 chocolates out of
Similarly words terminating with vowel 8 different brand is = 8 + 6–1C6 = 13C6
= 4! × 2 = 48 ways  say = n(B)
24. For a number to be divisible by 5,5 or 0 should be at
Now exclude words which begin as well as terminates units place.
with vowel  Unit place can be filled by 2 ways
2 × 3 × 2 × 1 × 1 = 12 ways  n(A  B)
6!
Desired number of words :- Remaining digits can be filled in ways.
3 ! 2 !
48 + 48 – 12 = 84 ways
2  6!
( n(A B) = n(A) + n(B) – n(A  B)) Total ways =
3 !2 !
6. Make cases when all 5 boxes are filled by: But these arrangements also include cases where 0 is
Case I : identical 5 red balls at millions place and 5 at units place, which are
5
C5 1 way undesirable cases
Case II : 4 identical red balls and 1 blue ball 5!
5  ways (undesirable)
C1 = 5 ways 3 ! 2 !
Case III : 3 blue and 2 red balls i.e. xRxRx subtract it from total ways.
 4 gaps, for 2 blue balls 6! 5!
 4C2 = 6 ways  Desired ways = 2 × – = 110
3 ! 2 ! 3 ! 2 !
Case IV : 2 red and 3 blue balls i.e. xRxRx  3
gaps, 3 blue balls 27. Total number of positive integral solution of
 3C3 = 1 way x1 . x2 . x3 = 80 = 2 × 3 × 5 is 3 × 3 × 3 = 27
 Total number of ways are 1+5+6+1 = 13 ways
30. Number of selections of 4 consonants out of 7 is 7C4
7. First we select 3 speaker out of 10 speaker and put in Number of selections of 2 vowels from 4 is 4C2
any way and rest are no restriction i.e. total number Arrangement of words in 6! ways
10! Desired words : 7C4 × 4C2 × 6! = 151200
of ways = 10C3 . 7.2 ! =
3

8. EEQUU
4!
Words starting with E 
2!
3!
Words starting with QE 
2!
next word will be QUEEU  1
and finally QUEUE  1
Rank is 12 + 3 + 1 + 1 = 17th

44
 PERMUTATION AND COMBINATION

EXERCISE - 2 2
 n 1 
Part # I : Multiple Choice =  
 2 
3. Total number of required possibilities n2
Case - II when n is even = rmax =
5 C . 8C + 5 C . 8 C + 5 C . 8 C . 5 C 2
3 7 4 6 5 5 5
so total no. selection is
= 5C3 . 8C7 + 5C4 . 8C6 + 8C6 = 13C10 – 5C3 = 276
n2n
(n  2)/2
n(n  2) 2  2  2
6.
200
C 2 . 198C 2 . 196 C 2 ......2 C 2 200!
= 100
= 
r 1
(n  2r) =
2
–
2
100! 2 .100!
n2  n n(n  2)
101.102.103.....200 =   n   =
=  2   2 4
2100
14. Total no. of visits that a teacher goes is = 25C5
 100   102   103   100 
=   .  .  .....   (selection of 5 different kids each time & teacher goes
 2  2  2   2 
every time)
1.2.3.4.5.6.7.8.....200 Number of visits of a boy = select one particular boy &
and
2100 .100! 4 from rest 24 = 24C4
So extra visits of a teacher from a boy is
(1.3.5.7.....199)(2.4.6.8......200)
= = 25C5 – 24C4 = 24
C5
2100 .100!

(1.3.5......199).2100 .100! 17. Number of ways he can fail is either one or two, three
= = 1.3.5 .199 or four subject then total of ways.
2100 .100!
4 C + 4 C + 4 C + 4C = 2 4 – 1
1 2 3 4
9. x1 + x2 + x3 + x4  n
 x1 + x2 + x3 + x4 + y = n 19. Total number of required quadrilateral
(where y is known as pseudo variable) 7C + 7C × 5C + 7C × 5C
4 3 1 2 2
Total no. of required solution is = n + 5 –1Cn
= n+4Cn or n+4
C4 7  6  5 4 7  6  5 7 6 5 4
=  .5  
12! 1 2  3  4 1 2  3 1 2 1 2
11. We have arrange all the letter except ‘ccc’ is = 35 + 175 + 210 = 420 = 2 .7p3
5!.3!.2!
new there all 13 place where ‘i’ can be placed 13C3
Hence required number of ways is Part # II : Assertion & Reason
12! 13!
= 13 C = 11 .
5!3! 2! 3 3. Statement -1: Two circles intersect in 2 points.
6!
 Maximum number of points of intersection
12. Here given no. be 1,2,3,.........n
= 2 × number of selections of two circles from 8 circles.
Let common difference = r
Total way of selection = (1, 1 + r, 1+2r), = 2 × 8C2 = 2 × 28 = 56
(2, 2 + r, 2 + 2r), ..(n – 2r, n – r, n) Statement -2: 4 lines intersect each other in 4C2 = 6 points.
Total numbers are = (n – 2r) 4 circles intersect each other in 2 × 4C2 = 12 points.
Here rmin. = 1 and rmax. = (n – 1)/2
Further, one lines and one circle intersect in two points.
Case- I When n is odd
So, 4 lines will intersect four circles in 32 points.
(n  1)
 r max = & to tal no. of selection is  Maximum number of points = 6 + 12 + 32 = 50.
2
 n  1  n  1  5. x1 x2 x3 x4 = 2 × 5 × 7 × 11  N = 4 4
(n 1)/2 2  
n(n  1)  2  2 
=  (n  2r) = –
r 1 2 2

45
MATHS FOR JEE MAINS & ADVANCED

EXERCISE - 3 Comprehension # 04
Part # II : Comprehension 1. Since there are 5 even places and 3 pairs of repeated
Comprehension # 01 letters, therefore at least one of these must be at an odd

1. Exponent of 7 in 100! – place. Therefore, the number of ways is 11!/(2!2!2!).

2. Make a group of both M’s and another group of T’s.

100  14 
 7    7  = 14 + 2 = 16 Then except A’s we have 5 letters remaining. So M’s,
T’s, and the letters except A’s can be arranged in 7! ways.
exponent of 7 in 50! Therefore, total number of arrangements is 7! × 8C2.

 50   7  3. Consonants can be placed in 7!/(2!2!) ways. Then there

 7    7  = 8
are 8 places and 4 vowels. Therefore, number of ways is

7! 8 4!
100! 716 C .
Exponent of 7 in 100C50 = = 8 8 = 70 2!2! 4 2!
50!50! 7 7

 exponent of 7 will be 0.

2. Product of 5's & 2's constitute 0's at the end of a number

 No. of 0's in 108!
= exponent of 5 in 108!
(Note that exponent of 2 will be more than exponent of 5
in 108 !)

108   21
   = 21 + 4 = 25
 5   5 

3. As 12 = 22.3, here we have to calculate exponent of 2 and

exponent of 3 in 100!
exponent of 2

100   50   25  12   6   3 
=      =97
 2   2   2   2   2   2 

100   33  11   3 
exponent of 3 =     = 48
 3   3   3   3 

Now, 12 = 2 × 2 × 3
we require two 2's & one 3
 exponent of 3 will give us the exponent of 12 in 100!
i.e. 48

46
 PERMUTATION AND COMBINATION

EXERCISE - 4 4. First we select n grand children from 2n grand children

Subjective Type is 2nCn
Now arrangement of both group is n! × n!
1. Selecting 3 horses out of ABC A'B'C' is 6C3 ways
Now Rest all (m + 1) place where we occupy the
When AA' is always selected among (ABC A'B'C')
grandfather and m sons but grandfather refuse the
Remaining (BB'CC') can be selected in 4C1 ways similarly, sit to either side of grand children so the out of m – 1
when BB' and CC' is selected seat one seat can be selected
 Undesirable ways will be (4C1) × 3 Now required number of sitting in
using, total ways–undesirable ways 2nC × n ! × n ! × (m –1)C . m !
n 1
= desired ways we get
(6C3 – (4C1)3)  This is selection of 3 horses among 12n
= × n ! × n ! × (m –1)C1 . m ! = 2n ! . m ! . (m –1)
(ABC A'B'C') under given condition. n! n!
Remaining 3 can be selected in 10C3 ways.
6. Number of ways of distributing 200 objects into 20
Hence, desired ways will be [6C3 – 4C1 × 3]10C3 = 792
groups each containing 10 objects
Method II : Select one horse each from AA', BB' and CC'
hence 2C1 × 2C1 × 2C1 ways. Now select 3 horses from 200! 200!
= 20 × 20 = which must be an integer..
remaining 10 horses in 10C3 ways.
(10!) .20! (10!) 20 .19!
Total ways = 10C3 × 2C1× 2C1 × 2C1
7. (i) Total ways = 10!
2. Total no. of M are = 1 Total no. of  are = 4
Total no. of P are = 2 Total no. of S are = 4 undesirable cases : when 2 Americans are together (A1A2)
First we arrange all the words other than 's are or two British are together (B1B2) or two Chinese are
together (C1C2)
7! 765
  105 we plot them on Venn diagram :
2! 4! 1 2

Now, there are 8 places which can be fulfilled by I's i.e. A 1A 2 B1B2
the number of ways is 8C4
105  8  7  6  5 C1C2
Total required no. = 105 × 8C4 =
1 2  3  4
= 105 × 70 = 7350 we use,
n(A1A2  B1B2  C1C2)
st n
3. Step 1 : Select 2 lines out of n lines in C2 ways to get a = n (A1A2) + n(B1B2) + n(C1C2) – n[(A1A2)  (B1B2)]
point (say p).
– n [(B1B2)  (C1C2)] – n [(C1C2)  (A1A2)]
nd n–2
Step-2 : Now select another 2 lines in C2 ways, to get
+ n [(A1A2)  (B1B2)  (C1C2)]
another point (say Q)
where n(A 1 A 2 ) denotes  when 2 Americans
Step-3rd : When P and Q are joined we get a fresh line. are together = 9! 2! correspondingly for B1B2&C1C2
n[(A1A2)  (B1B2)] denotes when 2 Americans and 2
P Q Britishmen are together
Fresh line
= 8! × 2! × 2!
But when we select P first then Q and Q first then P we
correspondingly same for others.
get same line.
n[(A1A2)  (B1B2)  (C1C2)] denotes when 2 Americans,
n
C  n 2C 2 2 Britishmen and 2 Chinese are together
 2 Fresh lines
2
= 7! × 2! × 2! × 2! = 86

47
MATHS FOR JEE MAINS & ADVANCED

Put values we get

13  4 1 16
n(A1A2  B1B2  C1C2) R, S, G, B in = C3 ways
13 . 3
= 9! × 2! × 3 – 8! × 2 × 2 × 3 + 8!
In 16C3 ways, we have to remove undesirable ways,
= 8!(43)
when R > 5
These are undesired ways
Undesirable ways : R > 5  R  6
Desired ways = 10! – 8! (43) = 8!(47) give at least 6 to R and 2 to S and distribute remaining
between R, S, G, B
(ii) Now they are on a round table
15 – (2 + 6) = 7 remaining can be distributed between R,
Total ways = (n – 1) ! = (10 – 1) ! = 9 !
7  4 1 10
Undesired ways : S, G, B in = = C3 ways
7. 4 1
n(A1A2  B1B2  C1C2)
10
= 8! × 2! × 3 – 7! × 2! × 2!× 3 + 6! × 2! × 2! × 2! C3 are the undesirable cases
= 6! × 4 [7 × 2 × 2 × 3 – 7 × 3 + 2] Desired ways = 16C3 – 10C3 = 440

= 6! × 260 10. (i)Without changing the order of the vowels of MULTIPLE

Desired ways = 9! – 6! × 260 So we choose the first three place in 8C3 ways
= (244) 6! ways 8! 5! 8!
and the rest are arranged is  = = 3360
3!5! 2! 3!2!
8. (i) Vowels Consenents
  Hence required no. is 3360 – 1 = 3359
3 5 (ii) Keeping the position of each vowel fixed M_LT_PL_
3
C2 . 2! 6! = 6 × 6! = 4320
5!
Number of ways = = 60
2
(ii) G H other ways = 60–1 = 59
Start with G end with H
6! = 720 (iii) without changing the relative order/position of
vowels & consonants
(iii) GHT 5!
so number of ways is =  3! = 60 × 6 = 360
3! 6! = 6 × 6! = 4320 2!
Hence required number is = 360–1 = 359
(iv) – – – – –  5! . 6C3 . 3!
6 gap 13. Husband – H, Wife – W
(v) Same as above Given :
Relatives of husband (H) (a) Ladies (LH) = 4
(vi) – – – – – – – – 4C3 . 3! . 5! (b) Gentlemen (GH) = 3
(vii) – – – – – – – – 8C3 . 1 . 5! = 8C3 . 5! = 6720 Relatives of Wife (W) (a) Ladies (LW) = 3
(b) Gentlemen (GW) = 4
(viii) 3! . 5! = 720
Case 1 : Selecting (3LH) and 3(GW)
3
(ix) C2 . 5C3 . 5! = 3.10 . 5! = 120 × 30 = 3600 ways : 4
C3 × 4C3 = 16
9. Distribute 15 candies among. Case 2 : Selecting (3GH) and 3(LW)
3
ways : C3 × 3C3 = 1
Ram (R) + Shyam(S) + Ghanshyam(G) + Balram(B)
Case 3 : Selecting (2LH & 1GH) & (1Lw & 2Gw)
with condition given : R+S+G+B=15 & R 5 & S  2 ways : 4
C2 × 3C1 × 3C1 × 4C2 = 324
After giving 2 to Shyam, remaining candies 15–2=13 Case 4 : Selecting (1LH & 2GH) & (2LW & 1Gw)
4
ways : C1 × 3C2 × 3C2 × 4C1 = 144
Now distribute 13 candies in
Add all cases we get : 485 ways

48
 PERMUTATION AND COMBINATION

15. 2 clerks who prefer Bombay are to be sent to 22. Ordered pair = total – (A  B = X) = 4 n – 3n
2 different companies in Bombay, and Out of remaining 5 Subsets of X = 2 n will not repeat in both but here the
clerks (excluding 3 clerks who prefer for outside) 2 clerks whole set X has not been taken
are chosen in 5C2 ways. So subsets of x which are not repeated (2 n – 1)
Now these 4 can be sent to 2 different companies into 2 (4n  3n )  (2n  1)
groups of 2 each in 4C2 ways Hence unordered pair = + (2n – 1)
2
5
 C2 × 4C2 23. Total No. of bowlers = 6
Now for outside companies we have 6 clerks remaining Now,
we select them as (2 for each company) (i) If 4 bowlers are including the no. of ways
6 4
C2 × C2 × C2 2 selecting 11 players out of 15 players
= 6C4 × 9C7 = 15 × 36 = 540
Desired ways = (5C2 × 4C2) (6C2 × 4C2 × 2C2) = 5400 ways.
(ii) If 5 bowlers are selected = 6C5 × 9C6 = 6 × 84 = 504
16. Total students n(A  B  C) = 100 (iii) If all 6 bowlers are selected = 6C6 × 9C5 = 1 × 126 = 126
Students reading Business India = n(A) = 80 Hence total no. of ways = 540 + 504 + 126 = 1170
Students reading Business World = n(B) = 50
25. 1980 = 22 . 32 . 5 . 11,
Students reading Business Today = n(C) = 30
number of divisiors of 1980 = 36
Students reading all the three magazines = n(ABC) = 5 (i) 3.3.2 = 18
sum = 11.(1 + 2 + 22)
A B
. (1 + 3 + 32)
. (1 + 5)
(ii) 3.2 + 1.1.2 = 8
C

Hence students reading exactly two magazines

= n(A) + n(B) + n(C) – n(A  B  C) – 2 × n(A  B  C)
= 80 + 50 + 30 – 100 – 2 × 5 = 50

19. (a) Selection of r things out of n +1 different things

= Selection of r things out of n +1 different things, when
a particular thing is excluded + a particular thing is
included.
(b) Selection of r things out of not m + n different Things
can be made by selecting x thing from m and y thing
from such that x +y = r
& (x, y) = (0, r), (1, r – 1), (2, r – 2), .......(r, 0)

21. Step 1st : Arrange 5 boys in 5! ways

Step 2 nd : Select 2 gaps from 6 gaps for 4 girls
(2 girls for each gap) in 6C2 ways.
Step 3rd : Select 2 girls to sit in one of the gaps and other
2 in remaining selected gaps = 4C2 ways
Step 4th : Arrange 1st, 2 girls in 2! and other 2 in 2! ways
Hence, total ways 5! × 6C2 × 4C2 × 2 × 2 = 43200

49
MATHS FOR JEE MAINS & ADVANCED

EXERCISE - 5 14. The no. of ways to select 4 novels & 1 dictionary from
Part # I : AIEEE/JEE-MAIN 6 different novels & 3 different dictionary are 6C4 × 3C1
and to arrange these things in shelf so that dictionary is
1. Numbers greater than 1000 and less than or equal to 4000 always in middle _ _ D _ _ are 4!
Required No. of ways 6C4 × 3C1 × 4! = 1080
will be of 4 digits and will have either 1 (except 1000) or
2 or 3 in the first place with 0 in each of the remaining 15. Urn A  3 Red balls
places. Urn B  9 Blue balls
So the number of ways = selection of 2 balls from
After fixing 1st place, the second place can be filled by urn A & B each.
any of the 5 numbers. Similarly third place can be filled = 3C2 . 9C2 = 108
up in 5 ways and 4th place can be filled up in 5 ways.
Thus there will be 5 × 5 × 5 = 125 ways in which 1 will be 16. B1 + B2 + B3 + B4 = 10
in first place but this include 1000 also hence there will Statement - I
be 124 numbers having 1 in the first place. Similarly 125 B1  1, B2  1, B3  1, B4  1
for each 2 or 3. One number will be in which 4 in the first so no. of negative integers solution of equation
place and i.e. 4000. Hence the required numbers x1 + x2 + x3 + x4 = 10 – 4 = 6
6 + 4 – 1C = 9C
are 124 + 125 + 125 + 1 = 375 ways. 4–1 3
Statement - II
2. We know that a five digit number is divisible by 3, if and
Selction of 3 places from out of
only if sum of its digits ( = 15) is divisible by 3. Therefore 9 places = 9C3
we should not use 0 or 3 while forming the five digit Both statements are true and correct explaination
numbers. Now,
10  9  8 6  5  4
(i) In case we do not use 0 the five digit number can be 17. N = 10C3 – 6C3 = 
3  2 1 3  2 1
formed (from the digit 1, 2, 3, 4, 5) in 5P5 ways. = 120 – 20 = 100
(ii) In case we do not use 3, the five digit number can be N  100
formed (from the digit 0, 1, 2, 4, 5) in 18. W10, G9, B7
5
P5  4P4 = 5 !  4! = 120  24 = 96 ways. selection of one or more balls
 The total number of such 5 digit number = (10 + 1) (9 + 1) (7 + 1) – 1
= 5P5 + (5P5  4P4) = 120 + 96 = 216 = 11 × 10 × 8 – 1 = 879
4. No. of ways in which 6 men can be arranged at a round 19. (A, B)
table = (6  1)! 
Now women can be arranged in 6! ways. 2×4=8
8C + 8C + ..... + 8C = 28 – 8C – 8C – 8C
Total Number of ways = 6! × 5! 3 4 8 0 1 2
= 256 – 37 = 219
5. As for given question two cases are possible
(i) Selecting 4 out of first 5 question and 6 out of remaining 20. Tn = nC3
 n+1C3 – nC3 = 10
8 questions = 5C4 × 8C6 = 140 choices
(n + 1) n (n – 1) – n(n – 1) (n – 2) = 60
(ii) Selecting 5 out of first 5 questions and 5 out of remaining n(n – 1) = 20
8 questions = 5C5 × 8C5 = 56 choices. n=5
 Total no. of choices = 140 + 56 = 196 1 1
23. P(E1) = P(E2) =
7. Number of ways to arrange in which vowels are in 6 6
6! P(E1  E2) = P(A shows 4 and B shows 2)
alphabetical order = = 360 1
2!
= = P(E1) . P(E2)
8. Number of ways = n–1Cr–1 = 8–1C3–1 = 7C2 = 21 36
So E1, E2 are independent
10
11. C1+ 10C2 + 10C3 + 10C4 = 385 Also as E1  E2  E3 = 
12! 12! So P(E1  E2  E3)  P(E1. P(E2). P(E3)
12. Number of ways = 3 × 3! = So E1, E2, E3 are not independent.
(4!) .3! (4!)3

50
 PERMUTATION AND COMBINATION

Part # II : IIT-JEE ADVANCED 12 Balls can be distributed as 1, 1, 3 or 1, 2, 2 to each person.

When 1, 1, 3 balls are distributed to each person, then
4. x + y < 21 total number of ways :
x + y  20
x + y  18 ( x > 0 & y > 0) 5! 1
 . .3!  60
Introducing new variable t 1! 1! 3! 2 !
When 1, 2, 2 balls are distributed to each person, then
total number of ways :
x + y + t = 18 x>0 5! 1
 . .3 !  90
y>0 1! 2! 2! 2 !
Now dividing 18 identical things among 3 persons.  total = 60 + 90 = 150
18  3  1 Paragraph for Question 13 and 14 : For an
= 18 + 3 – 1 C3 – 1 = = 190
18 3  1 The first digit should be 1

5. Total number of ways of distributing n2 objects into n For bn

groups, each containing n objects 1
_ _ _ _
_.... _ _1

 n 2 Places 
(n 2 )! (n 2 )!
= .n ! = = integer Last digit is 1. so bn is equal to number of ways of
(n !) n n ! (n !) n
an–1 (i.e. remaining (n – 1) places)
(Since number of ways are always integer) bn = an–1

7. Since, r, s, t are prime numbers. For cn

 Selection of p and q are as under Last digit is 0 so second last digit must be 1
p q number of ways So cn = an–2
r0 r2 1 way bn + c n = a n
1 2
r r 1 way So an = an–1 + an–2
2 0 1 2
r r ,r ,r 3 ways Similarly bn = bn–1 + bn–2
 Total number of ways to select r = 5 13. (B)
s0 s4 1 way a1 = 1, a2 = 2
s1 s4 1 way
So a3 = 3, a 4 = 5 a5 = 8
2 4
s s 1 way
 b6 = a 5 = 8
s3 s4 1 way
4 0 1 2 3 4 14. (A)
s s,s,s,s,s 5 ways
 Total number of ways to select s = 9. an = an–1 + an–2
Similarly total number of ways to select t put n = 17
= 5 number of ways = 5 × 9 × 5 = 225. a17 = a16 + a15 (A) is correct
cn = cn–1 + cn–2
11. (D)
So put n = 17
Case- I : The number of elements in the pairs can be
1,1; 1,2; 1,3,; 2,2 c17 = c16 + c15 (B) is incorrect
C2 . 2C2 4 bn = bn–1 + bn–2
4 4 3 4 3
= C2 + C1 × C2 + C1 × C3 + = 25 put n = 17
2
Case- II : Number of pairs with  as one of subsets b17 = b16 + b15 (C) is incorrect
4
= 2 = 16 a17 = a16 + a15
 Total pairs = 25 + 16 = 41 while (D) says a17 = a15 + a15 (D) is incorrect

51
MATHS FOR JEE MAINS & ADVANCED

MOCK TEST
2 ( m + nCm) . (m  1)! (n  1)!
2 2
1. (C)
(m  n)!
B1

B2

B3
 .........
B9
  2. (m  n)! (m  1)! (n  1)! = 2.
10 10 10 10 m! n! 2 2 4mn

Selectio n of atleast one bo ok 7. (A)

(10  1)(10  1)......(10  1) 6!

 – 1 = 11 9 – 1 Number of ways = × 2 = 30
9 times 2!  4!

2. maximum size of square can be 8 x 8 8. For P  If same species are different

8 Total number of arrangements is
n
so required no. of square be  (16  r) (9  r) P 2 . (m + n – 2) !
For Q  If same species are alike then number
r 1
8 (m  n  2)!
of arrangement is
=  r 2  25 r  144  456 m!.(n  2)!
r 1 P
3. (C) Hence = nP2 . m! . (n – 2)! = nP2 . mPm . (n – 2)!
Q
arrangem 9. (C)
Dashes dots
ents S1 : 1! = 1
7
5 2 C2 1! + 2! = 3
7 1! + 2! + 3! = 9
4 3 C3 1! + 2! + 3! + 4! = 33
7
3 4 C4 1! + 2! + 3! + 4! + 5! = 153
7
further in every number unit digit is 3 so it can’t
2 5 C5 be a perfect square.
1 6 7
C6 Hence only possible values are x = 1 or x = 3
7
0 7 C7 S2 : Total – both together
11 C – 9C = 378
5 3
= 7C2 + 7C3 + 7C4 + 7C5 + 7C6 + 7C7 = 27 – 8 = 120
S3 : 1 2 3 3 4
or 5 x 9 – x
4. Team X will win if it wins (m + r)th match and  
4! 2 ways 10 ways
wins m – 1 match from the first m + r – 1 matches , 2!
ways

m 2m
Cm Hence 12 × 2 × 10 = 240
so total no. of ways =  m  r 1 C m 1
=
2 S4 : (i) One digit number
r 0
number of 3’s = 1.
5. (C) (ii) Two digit number
Maximum no. of triangle one 3’s _ 3 + 3 _ = 8 + 9 = 17
two 3’s 3 3 = 2
3p(3p  1)(3p  2)  3p(p  1)(p  2)
=  number o f 3 ’s in two digit
6 numbers is 17 + 2 = 19
p (iii) Three digit numbers one 3’s
= [9p 2 – 9p + 2 – p 2 + 3p – 2] = p [4p2 – 3p] 3__+_3_+__3
2
= 9.9 + 8.9+8.9 = 225
= p 2 [4p – 3] two 3’s 33_ + 3 _ 3 + _3 3
= 2(9+9+8) = 26.2 = 52
6. First we select m friends for one table is three 3’s 333 = 3.1 = 3
m+n
Cm and select a table by 2C1 ways.  number of 3’s in three digit numbers is
Now total number of arrangements is 225 + 52 + 3 = 280
Hence answer is 1 + 19 + 280 = 300

52
 PERMUTATION AND COMBINATION

CaseIII : If selected one card each of number –1

10. and 1 and 4 cards of no. 0.
6!
There are 2M, 2T, 2A and 1 H, E, , C, S so no. of arrangement is = 30
1! 1! 4!
First find the number of ways if odd’s no. position
place be filled is 5p 3 = 60 Case IV : If all cards selected fram the no. 0
Now Case  If even place words is same i.e no. 6!
So no. of arrangement is =1
of ways = 3 6!
Case  If even place words is different i.e no. of Hence total no. of arrangement is
ways = 3c2 × 2! = 6 20 + 90 + 30 + 1 = 141
Hence total no. of arragment is 60 × (3 + 6) = 540
13. (A)
11. First find no. of ‘2’ at the end of (125)! is
two 2’s five 1’s
125  125  125  125  125  125  125  two 2’s four 1’s one 3
(A)  2    22    23    24    25    26    27 
              two 2’s three 1’s two 3’s
= 62 + 31 + 15 + 7 + 3 + 1 + 0 = 119 two 2’s two 1’s three 3’s
Find the number of ‘5’ at the end of (125)! two 2’s one1 four 3’s
two 2’s five 3’s
125  125  125   7!
is     + ...... 7! 7! 
 5   52   53  = 2    = 672
 2! 5! 2! 4! 2! 3! 2! 
= 25 + 5 + 1 = 31
Hence no. of zero is 31 14. (D)
(B) Total no. of signals can made by each arm D P M L can be arranged in 4 ! ways & the two
= 10 so total number of different signals can be gaps out of 5 gaps can be selected in 5C2 ways.
formed = 10 10 – 1 {A A and E E} or {A E and A E} can be placed in 6
(here – 1 is because if all arms are at the ways.
Total = 4 !. 5C2. 6 = 1440
position of rest, then no signal will pass away)
15. (A,D)
(C) 4
All AAAAA BBB D EEF can be arranged in
5 12!
ways
5! 3! 2!
5! 5!
= 60 Between the gaps C can be arranged in 13C3 ways
2! 2!.2! = 30
12!
Total number of arrangement = 90 Total ways = 13C3 ×
5!  3!  2!
(D) Let number of player is n
then total number of games is nc2 = 5050 Number of ways
= without considering separation of C – in which
 n = 101
all C's are together – in which exactly two C's are
12. Here the sum of the numbers are vanishes of six 15! 13! 12! 13
together = 2 – – C2
cards i.e 5!(3!) 2! 5! 3! 2! 5! 3!
Case I If selected 3 cards each of number –1 or 1 16. (D)
6!
i.e The number of arrangement = = 20 1 
3! 3! If    = 0
 3 50 
Case II : If selected 2 cards each of no. –1, 0 or 1 i.e
1 
6!  0   <1
number of arrangement = = 90 3 50
2! 2! 2!

53
MATHS FOR JEE MAINS & ADVANCED

1  2 50 2
 number of ways = 25
 –   or –   3m gives 3 at unit place if m = 4k + 1
3 50 3 3 3
 1 33  these are 5 in number
n
7 gives 7 at unit place if n = 4 + 1
1   1   these are 5 in number
If  3  50  = 1, 1  3 + 50 < 2
   number of ways = 25
2  5 3m gives 7 at unit place if m = 4k + 3
  
3 50 3  these are 5 in number
n
7 gives 3 at unit place if n = 4 + 3
100 250
   these are 5 in number
3 3
 number of ways = 25
 34  83 m
3 gives 9 at unit place if m = 4k + 2
 E = (0 + 0 + 0 .... + 33 times ) +
 these are 5 in number
(1 + 1 + 1 + ...... + 17 times )
= 0 + 17 = 0 + 17 7n gives 1 at unit place if n = 4
= 17 (which a prime number)  these are 5 in number
17. (A)  number of ways = 25
Sum of the digits in the tens places  total number of ways = 100
= sum of the digits in the unit’s place  statement-1 is true
= (4 – 1)! (2 + 3 + 4 + 5) statement-2
= 6.14 = 84 if m = 4k + 2, then 3m gives 9 at units place
18. (D) if n = 4, then 7n gives 1 at units place
A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}  statement-2 is true but does not explain
Two different numbers for numerator and denomi- statement-1
nator from these can be obtained in 10P 2
p q 21. (A)  (r), (B)  (t), (C)  (s), (D)  (p)
= 10.9 = 90 ways and if p or q = 1
(a) Two cases
(If numerator and denominator same) 5!
 Number of ways = 90 + 1 = 91 (i) 5, 4, 1, 1, 1 = 20
3!
19. (A) 5!
Number of numbers formed by using 1, 2, 3, 4, 5 is (ii) 5, 2, 2, 1, 1 = 30
2! 2!
= 5! = 120
Total 20 + 30 = 50
Number of numbers formed by using 0, 1, 2, 4, 5 is
= 4.4.3.2.1 = 96 (b) 5 ! – D 5 – 5 . D 4.
Total number of numbers formed, which are (D5 stands for dearrangements of 5 things)
divisible by 3 (taking numbers without repitition) = 120 – 44 – 5 × 9 = 31
is = 216 (c) 1000 – 93 – 92 – 9 = 181
STATEMENT 1  Incorrect
1
STATEMENT 2  Correct (d)
15
 i . j
1 i  j  9
20. (B)
3m gives 1 at unit place if m = 4k 1  
2 2 2
 these are 5 in number =    i . j  (1  2  ......  9 
15 1 i  j 9 
7n gives 9 at unit place if n = 4 + 2
 these are 5 in number

54
 PERMUTATION AND COMBINATION

24.
1  (1  2  ........  9)2  (12  22  ...  92 ) 2 2 
=   (1  2  ...  92 ) 1. (D)
15  2  3 1 1  3C
1
2 2 1  3C
1
1   9  10  2 9  10  19   3C + 3C = 6
=     = 77 1 1
30   2  6 
2. (B)

22. (A)  (r) ; (B)  (s) ; (C)  (q) ; (D)  (p)

(A) Select m out of n = nCm 5!
number of ways of arranging in 3 1 1  = 10
3!1!1! 2!
increasing order = 1
Hence nCm 5!
2 2 1  = 15
1! 2! 2! 2!
(B) A monkey has m choice  10 + 15 = 25
m  m  ..m
  = mn 3 (A)

n times 3 1 1  1
(C) Arrange (m –1) green balls then out of 2 2 1  1
 1+1=2
m gaps select n positions for red balls
and arrange red balls = 1. mCn . 1 25.
= mCn 1. (B)
n
 n 
..n Since n = 4
(D) = nm  Number of matched arrangements
m times
8
8! C4
23. = =
4! 5! 5
1 (B)
Since there are 5 even places and 3 pairs of re- 2. (A)
peated letters therefore at least one of these must n=3
be at an odd place.  Number of matched arrangements

11 ! 6!
 the number of ways = =
2! 2 ! 2 ! 3!4!

6!
2 (A) Total number of arrangements =
3!3!
Make a bundle of both M's and another bundle
of T's. Then except A's we have 5 letters remain- 1
ing so M's, T's and the letters except A's can be  probability =
4
arranged in 7 ! ways
3. (B)
 total number of arrangements = 7 ! × 8C2
Since first pair is matched and it can be done
in 1 way
3 (A)
 for mismatched pairs n = 4
7!  Number of mismatched pairs = 8C5
Consonants can be placed in ways
2!2! 26. Here, we note the following.
1. A door will open if it face odd no of changes.
Then there are 8 places and 4 vowels
2. No. of changes faced by any door will be equal
7! 8 4! to no. of factors of the door no.
 Number of ways = . C4
2!2! 2! 3. So only those door will open, whose number is
perfect square so ans is  n ,
[where [ ] denotes the G.I.F.]

55
MATHS FOR JEE MAINS & ADVANCED

27. (16)
Now No. of solution of equation (2) is
10 IIT students T 1, T 2,..... T 10 can be arranged in
= Co-efficient of x20 in (x3 + x6 + x9 + ....) ×
10 ! ways. Now the number of ways in which two
(x2 + x4 + x6 + ....) × (x + x2 + x3 + ...)
PET student can be placed will be equal to the
number of ways in which 3 consecutive IIT = Co-efficient of x in (1 + x3 + x6 + x9 + ....)
14

students can be taken i.e. in 8 ways and can be (1 + x2 + x4 + -....) × (1 + x + x2 +....)

arranged in two ways  (10!) (8) (2!). = (1 + x + x + x + x5 + 2x6 + x7 + 2x8 + 2x9
2 3 4

Alternatively 3 IIT students can be selected in +2x10 + 2x11 + 3x12 + 2x13 + 3x14 + 3x15 + ....)
10
C3 ways. Now each selection of 3 IIT and 2 PET (1 + x + x2 + x3 + .......)
students in P 1 T 1 T 2 T 3 P 2 can be arranged in (2 !) = Co-efficient of x14 is 1 + 1 + 1 + 1 +1 + 2 +1
(3!) ways. Call this box X. Now this X and the + 2+2+2+2+3+2+3
remaining IIT students can be arranged in 8! ways = 24
 Total ways 10C3 (2 !) (3 !) (8 !) But x , y and z are arranged in 3! ways
So Required no of solution = 24 × 6 = 144
28. In one round, marked numbers are 1, 16, 31, ..., 991
 67 numbers
In second round marked numbers are 6, 21, 36, ..., 996
 67 numbers
In third round marked numbers are 11, 26, 41, ..., 986
 66 numbers
the next number will be 1 which has already been
marked
 total marked numbers = 67 + 67 + 66 = 200
 unmarked numbers = 1000 – 200 = 800
29. (4)
Make 1 group of 2 persons, 1 group of 4
persons and 3 group of 3 persons among 15
persons (except 2 particular persons)
15!
hence by grouping method =
2! 4! (3!)3 3!
Now we add that 2 person in the group of 2
persons and thus number of arrangement of these
groups into cars and autos is
15! 15!
× 2! × 3! =
3
2! 4! (3!) 3! 4! (3!)3
30. The given equation is x + y + z = 20 .........(1)
We have to find the number of different values
of x, y, z
Such that x  y  z and x, y, z  1
Let us assume that x < y < z
and x = x1, y – x = x2 and z – y = x3
then x = x1 ; y = x1 + x2 and z = x1 + x2 + x3
also x1 , x2 , x3  1
Substitution these values in (1) we get
3x1 + 2x2 + x3 = 20 ..........(2)
Where x1 , x2 , x3  1

56
 Dpp-1 to 7
MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) Special DPP on Permutation and Combination DPP. NO.-1
ELEMENTARY PROBLEMS ON PERMUTATION & COMBINATION
NOTE: USE FUNDAMENTAL PRINCIPLE OF COUNTING & ENJOY DOING THE FOLLOWING.
Q.11/1 In how many ways can clean & clouded (overcast) days occur in a week assuming that an entire day is
either clean or clouded. [ Ans. 27 = 128 ]
Q.22/1 Four visitors A, B, C & D arrive at a town which has 5 hotels. In how many ways can they disperse
themselves among 5 hotels, if 4 hotels are used to accommodate them. [ Ans. 5 . 4 . 3 . 2 = 120]
Q.33/1 If the letters of the word “VARUN” are written in all possible ways and then are arranged as in a
dictionary, then the rank of the word VARUN is :
(A) 98 (B) 99 (C*) 100 (D) 101
Q.44/1 How many natural numbers are their from 1 to 1000 which have none of their digits repeated.
[Hint: S D = 9 ; DD = 9 · 9 = 81 ; TD = 9 · 9 · 8 = 648] [ Ans. 738]
Q.5 3 different railway passes are allotted to 5 students. The number of ways this can be done is :
(A*) 60 (B) 20 (C) 15 (D) 10
Q.66/1 There are 6 roads between A & B and 4 roads between B & C.
(i) In how many ways can one drive from A to C by way of B ?
(ii) In how many ways can one drive from A to C and back to A, passing through B on both trips ?
(iii) In how many ways can one drive the circular trip described in (ii) without using the same road more
than once. [ Ans. (i) 24 ; (ii) 576 ; (iii) 360 ]
Q.77/1 (i) How many car number plates can be made if each plate contains 2 different letters of English
alphabet, followed by 3 different digits.
(ii) Solve the problem, if the first digit cannot be 0. (Do not simplify)
[ Ans. (i) 26 . 25 . 10 . 9 . 8 = 468000 ; (ii) 26 . 25 . 9 . 9 . 8 = 421200 ]
Q.88/1 (i) Find the number of four letter word that can be formed from the letters of the word HISTORY.
(each letter to be used at most once) [ Ans. 7 . 6 . 5 . 4 = 42 x 20 = 840 ]
(ii) How many of them contain only consonants? [ Ans. 5 . 4 . 3 . 2 = 120 ]
(iii) How many of them begin & end in a consonant? [ Ans. 5 . 4 . 4 . 4 = 400 ]
(iv) How many of them begin with a vowel? [ Ans. 240 ]
(v) How many contain the letters Y? [ Ans. 480 ]
(vi) How many begin with T & end in a vowel? [ Ans. 40 ]
(vii) How many begin with T & also contain S? [ Ans. 60 ]
(viii) How many contain both vowels? [ Ans. 240 ]
Q.99/1 If repetitions are not permitted
(i) How many 3 digit numbers can be formed from the six digits 2, 3, 5, 6, 7 & 9 ? [ Ans. 120 ]
(ii) How many of these are less than 400 ? [ Ans. 40 ]
(iii) How many are even ? [ Ans. 40 ]
(iv) How many are odd ? [ Ans. 80 ]
(v) How many are multiples of 5 ? [ Ans. 20 ]
Q.10 How many two digit numbers are there in which the tens digit and the units digit are different and odd?
[ Ans. 5 . 4 = 20 ]
Q.1111/1 Every telephone number consists of 7 digits. How many telephone numbers are there which do not
include any other digits but 2 , 3 , 5 & 7 ? [ Ans. 47 ]
[1]
Q.1212/1 (a) In how many ways can four passengers be accommodate in three railway carriages, if each
carriage can accommodate any number of passengers.
(b) In how many ways four persons can be accommodated in 3 different chairs if each person can
occupy only one chair. [ Ans. (a) 34 ; (b) 4.3.2 = 24 ]
Q.13 How many odd numbers of five distinct digits can be formed with the digits 0,1,2,3,4 ?
[ Hint:  3 × 3 × 2 × 1 × 2 = 36 Ans ]

Q.1414/1 Number of natural numbers between 100 and 1000 such that at least one of their digits is 7, is
(A) 225 (B) 243 (C*) 252 (D) none
[Hint: 9 · 10 · 10 = 900 [ 13th Test (5-12-2004)]

Total no. of numbers without 7 = 81 × 8 = 648

 required number = 900 – 648 = 252 ]

Q.1515/1 How many four digit numbers are there which are divisible by 2. [Ans. = 4500]
Q.16 The 120 permutations of MAHES are arranged in dictionary order, as if each were an ordinary five-letter
word. The last letter of the 86th word in the list is
(A) A (B) H (C) S (D*) E [12th (14-5-2006)]
[Sol. The first 24 = 4! words begins with A, the next 24 begin with E and the next 24 begin with H. So the 86th
begins with M and it is the 86 – 72 = 14th such word. The first 6 words that begin with M begin with MA
and the next 6 begin with ME. So the desired word begins with MH and it is the second such word. The
first word that begins with MH is MHAES, the second is MHASE. Thus E is the letter we seek. ]
Q.1717/1 Find the number of 7 lettered palindromes which can be formed using the letters from the English
alphabets. [Ans. 264]
[Hint: A palindrome is a word or a phase that is the same whether you read is forward or backword.
e.g. REFER]
Q.1818/1 Number of ways in which 7 different colours in a rainbow can be arranged if green is always in the
middle. [ Ans 720 ]
Q.1919/1 Two cards are drawn one at a time & without replacement from a pack of 52 cards. Determine the
number of ways in which the two cards can be drawn in a definite order. [Ans. 52 × 51= 2652]
Q.20 Find the number of ways in which the letters of the word "MIRACLE" can be arranged if vowels always
occupy the odd places.
Hint: [ Ans: 4·3·2·4·3·2·1=24·24=576 ]
Q.2121/1 Numbers of words which can be formed using all the letters of the word "AKSHI", if each word begins
with vowel or terminates in vowel. [ Ans: 2·24+2·24-2·6 = 84 ]
Q.2222/1 A letter lock consists of three rings each marked with 10 different letters. Find the number of ways in
which it is possible to make an unsuccessful attempts to open the lock. [Ans: 103 - 1 = 999]
Q.2323/1 How many 10 digit numbers can be made with odd digits so that no two consecutive digits are same.
[Ans: 5·49]
Q.24 In how many ways can the letters of the word "CINEMA" be arranged so that the order of vowels do
6!
not change. [Ans: = 120]
3!
Q.2525/1 How many natural numbers are there with the property that they can be expressed as the sum of the
cubes of two natural numbers in two different ways . [Ans. Infinitely many]
[2]
 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) Special DPP on Permutation and Combination DPP. NO.-2
Q.11/2 How many of the 900 three digit numbers have at least one even digit?
(A*) 775 (B) 875 (C) 450 (D) 750
[Sol. There are 900 three digit numbers and there are five odd digits. Thus, there are 53 = 125 three digit
numbers comprised of only odd digits. The other 900 – 125 = 775 three digit numbers must contain at
least one even digit. ] [12 & 13 05-3-2006]

Q.22/2 The number of natual numbers from 1000 to 9999 (both inclusive) that do not have all 4 different digits
is
(A) 4048 (B*) 4464 (C) 4518 (D) 4536
OR
What can you say about the number of even numbers under the same constraints?
[Hint: Total – All four different = 9 103  9 ·9 ·8 ·7
OR
Ans. 2204 ; all 4 digit even number – number of 4 digit even numbers with different digit ]

Q.33/2 The number of different seven digit numbers that can be written using only three digits
1, 2 & 3 under the condition that the digit 2 occurs exactly twice in each number is
(A*) 672 (B) 640 (C) 512 (D) none 7
[Hint: Two blocks for filling 2 can be selected in 7C2 ways and the digit 2 C 2 ·2 5
can be filled only in one way other 5 blocks can be filled in 25 ways.]
Q.44/2 Out of seven consonants and four vowels, the number of words of six letters, formed by taking four
consonants and two vowels is (Assume that each ordered group of letter is a word):
(A) 210 (B) 462 (C*) 151200 (D) 332640
[Hint : 7C4 4C2 . 6 ! = 151200 ]
Q.55/2 All possible three digits even numbers which can be formed with the condition that if 5 is one of the digit,
then 7 is the next digit is :
(A) 5 (B) 325 (C) 345 (D*) 365
5
[Hint : 5 + 8.9.5 = 365; 1st place is five + where 1st place is not five 5 7 | +| | | ]
8 9 5
Q.66/2 For some natural N , the number of positive integral ' x ' satisfying the equation ,
1 ! + 2 ! + 3 ! + ...... + (x !) = (N)2 is :
(A) none (B) one (C*) two (D) infinite
[Hint: x = 1 & x = 3 ]
Q.77/2 The number of six digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 & 7 so that digits do
not repeat and the terminal digits are even is :
(A) 144 (B) 72 (C) 288 (D*) 720
[Hint: 1 . .3. .5. .7
3C · 2! · 5C · 4! = 6 × 120 = 720 ]
2 4

[3]
Q.8 In a certain strange language, words are written with letters from the following six-letter alphabet :
A, G, K, N, R, U. Each word consists of six letters and none of the letters repeat. Each combination of
these six letters is a word in this language. The word "KANGUR" remains in the dictionary at,
(A*) 248th (B) 247th (C) 246th (D) 253rd
[Sol. beginning with A or G = 240 [18-12-2005, 13th]
beginning with 6
th
247 K A N G R U
248th K A N G U R ]
Q.9 Consider the five points comprising of the vertices of a square and the intersection point of its diagonals.
How many triangles can be formed using these points?
(A) 4 (B) 6 (C*) 8 (D) 10
5
[Hint: To form a triangle, 3 points out of 5 can be chosen in C3 = 10 ways. But of these, the three points lying
on the 2 diagonals will be collinear. So 10 – 2 = 8 triangles can be formed]

Q.1010/2 A 5 digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 & 5 without repetition.
The total number of ways this can be done is :
(A) 3125 (B) 600 (C) 240 (D*) 216
[Hint: reject 0 + reject 3  5! + 4 · 4! = 120 + 96 = 216 ]

Q.1111/2 Number of 9 digits numbers divisible by nine using the digits from 0 to 9 if each digit is used atmost
once is K · 8 ! , then K has the value equal to ______. [Ans. K = 17]
[Hint: Case-I : reject '0' ; Case-II : reject 9
Case-I: 9!; Case-II: 0, 1, 2, ......, 8; 9! – 8! = 8 · 8!; I + II = (17)8!]

Q.1212/2 Number of natural numbers less than 1000 and divisible by 5 can be formed with the ten digits, each
digit not occuring more than once in each number is ______ .
[Hint : single digit = 1 ;
two digit: 5 =8+ 0 =9  Total = 17

three digit: 5 = 8 · 8 = 64; 0 = 9 · 8 = 72  Total = 136

hence Single digit + two digit + three digit = 154 Ans.]

Q.1311/3 Number of 3 digit numbers in which the digit at hundreath's place is greater than the other two digit is
(A*) 285 (B) 281 (C) 240 (D) 204
[Sol. When all 3 are distinct 1st 2 nd 3rd
10C · 2 (largest being at the 1st place and the x y z
3
= 240 remaining two can be arranged in two ways) [11th, 16-11-2008]
when y = z
10C · 1 (largest on the 1st place and remaining two being
2
= 45 equal on the 2nd and 3rd place) e.g. (211, 100)]
Total = 285 Ans. ]
Q.1412/3 Number of permutations of 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken all at a time, such that the digit
1 appearing somewhere to the left of 2
3 appearing to the left of 4 and
5 somewhere to the left of 6, is
(e.g. 815723946 would be one such permutation)
(A*) 9 · 7! (B) 8! (C) 5! · 4! (D) 8! · 4!

[4]
Sol. Number of digits are 9 [12th, 28-09-2008] [Dpp-3]
9
select 2 places for the digit 1 and 2 in C2 ways
from the remaining 7 places select any two places for 3 and 4 in 7C2 ways
and from the remaining 5 places select any two for 5 and 6 in 5C2 ways
now, the remaining 3 digits can be filled in 3! ways
 Total ways = 9C2 · 7C2 · 5C2 · 3!
9! 7! 5! 9! 9 ·8 ·7!
= 2!·7! · 2!·5! · 2!·3! · 3! = = = 9 · 7! Ans. ]
8 8

Q.15 Number of odd integers between 1000 and 8000 which have none of their digits repeated, is
(A) 1014 (B) 810 (C) 690 (D*) 1736
[Sol. Let the last place is 9
then we have 7 · 8 · 7 = 392 (1st place only 1 to 7 can come)
If the last place has 1 | 3 | 5 | 7 then
we have (6) (8) (7) (4) = 1344 [0, 9 or 8 can not come at units place] 1/3/5/7/9
Total = 392 + 1344 = 1736] [11th, 25-01-2009, P-1]

Q.164/3 Find the number of ways in which letters of the word VALEDICTORY be arranged so that the vowels
may never be separated.
[ Hint: VLDCTRY Y or 8! × 4! = 40320 × 24 = 967680 Ans ]
[‘Valediction means farewell after graduation from a college. Valedictory : to take farewell]

Q.177/3 The number of ways in which 5 different books can be distributed among 10 people if each person can
get at most one book is :
(A) 252 (B) 105 (C) 510 (D*) 10C5.5!
10 10
[Hint: Select 5 boys in C5 and distribute 5 books in 5! ways hence C5. 5!]
Q.188/2 A new flag is to be designed with six vertical strips using some or all of the colours yellow, green, blue and
red. Then, the number of ways this can be done such that no two adjacent strips have the same colour is
(A*) 12 × 81 (B) 16 × 192 (C) 20 × 125 (D) 24 × 216
st
[Hint: 1 place can be filled in 4 ways
2nd place can be filled in 3 ways
3rd place can be filled in 3 ways and |||ly 4th, 5th and 6th each can be filled in 3 ways.
hence total ways = 4 × 35 = 12 × 81 ]

Q.194/4 5 Indian & 5 American couples meet at a party & shake hands . If no wife shakes hands with her own
husband & no Indian wife shakes hands with a male, then the number of hand shakes that takes place in
the party is :
(A) 95 (B) 110 (C*) 135 (D) 150
[Hint: 20C2  (50 + 5) = 135 ]

Q.20192/1(10/3) There are 720 permutations of the digits 1, 2, 3, 4, 5, 6. Suppose these permutations are arranged
from smallest to largest numerical values, beginning from 1 2 3 4 5 6 and ending with 6 5 4 3 2 1.
(a) What number falls on the 124th position?
(b) What is the position of the number 321546? [Ans. (a) 213564, (b) 267th]
[Sol. (a) digits 1, 2, 3, 4, 5, 6
starting with 1, number of numbers = 120

[5]
 starting with 2, 1, 3, 4 number of numbers = 2
123rd
finally 124th is = 213564
(b) N = 321546
number of numbers beginning with 1 = 120
number of numbers beginning with 2 = 120
starting with 31 .............................. = 24
starting with 3214 .......................... = 2
finally =1
th
hence N has 267 position ]
 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) Special DPP on Permutation and Combination DPP. NO.-3
Q.15/3 How many numbers between 400 and 1000 (both exclusive) can be made with the digits 2,3,4,5,6,0 if
(a) repetition of digits not allowed.
(b) repetition of digits is allowed.
[ Hint: (a) 3×5×4 = 60 ; (b) 3×6×6 = 108 – 1 = 107 ]

Q.29/3 The 9 horizontal and 9 vertical lines on an 8 × 8 chessboard form 'r' rectangles and 's' squares. The ratio
s
in its lowest terms is
r
1 17 4
(A) (B*) (C) (D) none
6 108 27
[Sol. no. of squares are [12 & 13th test (29-10-2005)]
8(9)(17)
S = 12 + 22 + 32 + ........ + 82 = = 204
6
no. of rectangles r = 9C2 · 9C2 = 1296
s 204 51 17
hence = = = ]
r 1296 324 108
Q.311/3 A student has to answer 10 out of 13 questions in an examination . The number of ways in which he can
answer if he must answer atleast 3 of the first five questions is :
(A*) 276 (B) 267 (C) 80 (D) 1200
13
[Hint : C10  number of ways in which he can reject 3 questions from the first five
or 13C10  5C3 = 286 – 10 = 276 or 5C3 . 8C7 + 5C4 . 8C6 + 5C5 . 8C5 = 276 ]
[ Note that 5 10
C3 . C7 is wrong ( cases repeat]

Q.412/3 The number of three digit numbers having only two consecutive digits identical is :
(A) 153 (B*) 162 (C) 180 (D) 161
[Hint : when two consecutive digits are 11, 22, etc = 9 . 9 = 81
when two consecutive digits are 0 0 = 9
when two consecutive digits are 11, 22, 33, ... = 9 . 8 = 72  Total ]

Q.51/4 A telegraph has x arms & each arm is capable of (x  1) distinct positions, including the position of rest.
The total number of signals that can be made is ______ . [ Ans. (x  1)x  1 ]

Q.62/4 The interior angles of a regular polygon measure 150º each . The number of diagonals of the polygon is
(A) 35 (B) 44 (C*) 54 (D) 78
[Hint : exterior angle = 30°
360
Hence number of sides = = 12
30
12(12  3)
 number of diagonals = = 54 ]
2 
[Note that sum of all exterior angles of a polygon = 2 and sum of all the interior angles of a polygon =(2n-4) ]
2

[7]
Q.73/4 Number of different natural numbers which are smaller than two hundred million & using only the digits
1 or 2 is :
(A * ) (3) . 28  2 (B) (3) . 28  1 (C) 2 (29  1) (D) none
[Hint: Two hundred million = 2 x 10 ; (2 + 2 +2 +24+25+26+27+28) + 28 = 766 ]
8 1 2 3

Q.85/4 The number of n digit numbers which consists of the digits 1 & 2 only if each digit is to be used atleast
once, is equal to 510 then n is equal to:
(A) 7 (B) 8 (C*) 9 (D) 10
[Hint: (2 x 2 x .............2) n times-(when 1 or 2 is there at all the n places] [ Ans. 2n  2 ]

Q.96/4 Number of six digit numbers which have 3 digits even & 3 digits odd, if each digit is to be used atmost
once is ______ . [ Ans. 64800 ]
[ when 0 is included ( C2 . C3 . 5 . 5 !) and when 0 is excluded (4C3 . 5C3 . 6 !) ]
4 5

[Hint: alternatively, 5C3 · 5C3 · 6! since all digits 0, 1, 2, .........8, 9 are equally likely at all places
5
C 3 5 C 3  6 !
 required number = ·9 digits
10
or required number of ways = 5C3 . 5C3 . 6 ! – 4C2 . 5C3 . (6! – 5!) ]

Q.1016/9 Find the number of 10 digit numbers using the digits 0, 1, 2, ....... 9 without repetition. How many of
these are divisible by 4.
[Sol. Digit 0, 1, 2,........8, 9
For a number to be divisible by 4 the number formed by last two digits must be divisible by 4 and can be
04, 08, 12, ........, 96 ; Total of such numbers = 24
Out of these 44 and 88 are to be rejected. (as repetition is not allowed)
Hence accepted number of cases = 22
Out of these number of cases with '0' always included are 04, 08, 20, 40, 60, 80 (six)
no. of such numbers with these as last two digits = 6 · 8! ...(1)
e.g. [ × × × × × × × × 04 ]
no. of other numbers = 16 · 7 · 7! = 14 · 8! ....(2)
e.g. [ × × × × × × × × 16 ]
 Total number = 6 · 8! + 14 · 8! = (20) · 8! Ans. ]

Q.1114/8 There are counters available in x different colours. The counters are all alike except for the colour. The
total number of arrangements consisting of y counters, assuming sufficient number of counters of each
colour, if no arrangement consists of all counters of the same colour is :
(A*) xy  x (B) xy  y (C) yx  x (D) yx  y

Q.128/4 18 points are indicated on the perimeter of a triangle ABC (see figure).
How many triangles are there with vertices at these points?
(A) 331 (B) 408
(C) 710 (D*) 711
[Hint: 18C3 – 3· 7C3 = 816 – 105 = 711 ] [08-01-2005, 12th]
Q.139/4 An English school and a Vernacular school are both under one superintendent. Suppose that the
superintendentship, the four teachership of English and Vernacular school each, are vacant, if there be
altogether 11 candidates for the appointments, 3 of whom apply exclusively for the superintendentship
and 2 exclusively for the appointment in the English school, the number of ways in which the different
appointments can be disposed of is :
(A) 4320 (B) 268 (C) 1080 (D*) 25920

[8]
[Hint :  3C1 . 4C2 . 2 ! . 6 ! ]

Q.1410/4 A committee of 5 is to be chosen from a group of 9 people. Number of ways in which it can be formed
if two particular persons either serve together or not at all and two other particular persons refuse to
serve with each other, is
(A*) 41 (B) 36 (C) 47 (D) 76
AB
[Sol. 9 C/ D separated
5 other
AB included 7C – 5C1 = 30 (7C3 denotes any 3 from (CD and 5 others – no. of ways when CD)
3
AB excluded 7C – 5C3 = 11 is taken and one 3 from remaining five)
5
——
41 ] [18-12-2005, 12th + 13th]
Q.1511/4 A question paper on mathematics consists of twelve questions divided into three parts A, B and C,
each containing four questions. In how many ways can an examinee answer five questions, selecting
atleast one from each part.
(A*) 624 (B) 208 (C) 1248 (D) 2304
4 4 4 4 4 4
[Hint: 3 ( C2 . C2 . C1) + 3 ( C1 . C1 . C3) = 432 + 194 = 624 ]
Alternative : Total – [no of ways in which he does not select any question from any one section]
12C – 3 · 8C ; Note that 4C . 4C . 4C . 9C is wrong think !]
5 5 1 1 1 2
Q.1612/4 If m denotes the number of 5 digit numbers if each successive digits are in their descending order of
magnitude and n is the corresponding figure, when the digits are in their ascending order of magnitude
then (m – n) has the value
(A) 10C4 (B*) 9C5 (C) 10C3 (D) 9C3
10 9 9
[Hint: m – n = C5 – C5 = 252 – 125 = 126 = C5 or C4 9

Alternatively: 9C5 + 9C4 – 9C5 = 9C4 = 9C5 ]

Q.171/5 There are m points on a straight line AB & n points on the line AC none of them being the point A.
Triangles are formed with these points as vertices, when
(i) A is excluded (ii) A is included. The ratio of number of triangles in the two cases is:
m n2 m n2 m  n 2 m ( n  1)
(A*) (B) (C) (D)
m n m  n 1 mn 2 ( m  1) ( n  1)
n m
m . C2  n . C2
[Hint: ]
m . n C 2  n . m C 2  mn

Q.183/5 In a certain algebraical exercise book there are 4 examples on arithmetical progressions, 5 examples on
permutation-combination and 6 examples on binomial theorem . Number of ways a teacher can select
for his pupils atleast one but not more than 2 examples from each of these sets, is ______.
[ Hint: ( 4C1 + 4C2) ( 5C1 + 5C2) ( 6C1 + 6C2) ] [ Ans. 3150 ]
5 6
[Alternatively : add one dummy exercies in each and compute C2 · C2 · C2 ] 7

[9]
 n 1 n
Cr
Q.191/9  n is equal to :
r 0 C r  nCr 1
n ( n  1) n 1 n ( n 1) n
(A) 2 (n  1) (B) (C) (D*)
2 2 2

Q.208/5 The number of 5 digit numbers such that the sum of their digits is even is :
(A) 50000 (B*) 45000 (C) 60000 (D) none
9  10 4
[Hint: ]
2
 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) Special DPP on Permutation and Combination DPP. NO.-4
Q.1 Number of ways in which 8 people can be arranged in a line if A and B must be next each other and C
must be somewhere behind D, is equal to
(A) 10080 (B*) 5040 (C) 5050 (D) 10100
[Sol.   AB     [12th 11-05-2008]
C D
AB and 6 other is 7! but A and B can be arranged in 2! ways
 Total ways = 7! · 2!
when C is behind D 7!·2!
 required number of ways = = 5040 ways Ans.]
2!
Q.22/5 Number of ways in which 7 green bottles and 8 blue bottles can be arranged in a row if exactly 1 pair of
green bottles is side by side, is (Assume all bottles to be alike except for the colour).
(A) 84 (B) 360 (C*) 504 (D) none
[Sol. GGGGGGG/BBBBBBBB [12th & 13th 03-12-2006]
one gap out of nine can be taken in 9C1 ways
now green remaining 5
gaps remaining 8
5 gaps for remaining 5 green can be selected in 8C5
8 ·7 ·6
Hence Total ways 9C1 · 8C5 = 9 · 1·2 ·3 = 72 · 7 = 504 Ans.

Alternatively: 9C6 × 6 = 504 Ans. (think ! how)

(6 bottle in 6 gaps and the remaining in any one of these 6 gaps)]
Q.34/5 The kindergarten teacher has 25 kids in her class . She takes 5 of them at a time, to zoological garden as
often as she can, without taking the same 5 kids more than once. Then the number of visits, the teacher
makes to the garden exceeds that of a kid by :
(A) 25C5  24C5 (B*) 24C5 (C) 24C4 (D) none
[Hint: Number of trips which exceeds  when one kid is never included  25C5  24C4 = 24C5 ]
Q.46/5 Seven different coins are to be divided amongst three persons . If no two of the persons receive the
same number of coins but each receives atleast one coin & none is left over, then the number of ways in
which the division may be made is
(A) 420 (B*) 630 (C) 710 (D) none
7!
[Hint: 1, 2, 4 groups] [Ans. ×3!]
1! 2! 4!
Q.57/5 Let there be 9 fixed points on the circumference of a circle. Each of these points is joined to every one
of the remaining 8 points by a straight line and the points are so positioned on the circumference that
atmost 2 straight lines meet in any interior point of the circle. The number of such interior intersection
points is :
(A*) 126 (B) 351 (C) 756 (D) none of these
[Hint : Any interior intersection point corresponds to 4 of the fixed points , namely the 4 end points of the
intersecting segments. Conversely, any 4 labled points determine a quadrilateral, the diagonals of which
intersect once within the circle . Thus the answer is A ]

[11]
Q.610/5 The number of ways in which 8 distinguishable apples can be distributed among 3 boys such that every
boy should get atleast 1 apple & atmost 4 apples is K · 7P3 where K has the value equal to
(A) 14 (B) 66 (C) 44 (D*) 22
Q.711/5 A women has 11 close friends. Find the number of ways in which she can invite 5 of them to dinner, if two
particular of them are not on speaking terms & will not attend together. [Ans. 378]
[Hint: 11 9 9 9 9
C5  C3 = C5 + 2 C4 = 3 C4 = 378 ]

5to be
  9C + 9C + 9C or 11C – 9C3 ]
selected 3 4 4 5

Q.812/5 A rack has 5 different pairs of shoes. The number of ways in which 4 shoes can be chosen from it , so
that there will be no complete pair is :
(A) 1920 (B) 200 (C) 110 (D*) 80
5C 10 . 8 . 6 . 4
[Hint: 4 . 24 or = 80 ]
4!
Q.93/6 There are 10 seats in a double decker bus, 6 in the lower deck and 4 on the upper deck. Ten passengers
board the bus, of them 3 refuse to go to the upper deck and 2 insist on going up. The number of ways
in which the passengers can be accommodated is _____. (Assume all seats to be duly numbered)
[ Ans. 4C2. 2! 6C3. 3! 5! or 172800]
Q.104/6 Find the number of permutations of the word "AUROBIND" in which vowels appear in an alphabetical
order. [Ans. 8C4 · 4 !]
8!
[Hint : A, I, O, U  treat them alike. Now find the arrangement of 8 letters in which 4 alike and 4 different = ]
4!
Q.115/6 The greatest possible number of points of intersection of 9 different straight lines & 9 different circles in
a plane is
(A) 117 (B) 153 (C*) 270 (D) none
9 9 9 9
[Hint: C2 . 1 + C1 . C1 . 2 + C2 . 2 = 270 ]
Q.126/6 An old man while dialing a 7 digit telephone number remembers that the first four digits consists of one
1's, one 2's and two 3's. He also remembers that the fifth digit is either a 4 or 5 while has no memorising
of the sixth digit, he remembers that the seventh digit is 9 minus the sixth digit. Maximum number of
distinct trials he has to try to make sure that he dials the correct telephone number, is
(A) 360 (B*) 240 (C) 216 (D) none
 4!   2 ways   10 ways  1 way        
[Hint:    for fifth place   6 th place   7 th place  
 2!      123 3 7 th
x7 = 9 – x6
x6 can take 0 to 9
= 240 ] th
[12 & 13 test (09-10-2005)]
Q.137/6 If as many more words as possible be formed out of the letters of the word "DOGMATIC" then the
number of words in which the relative order of vowels and consonants remain unchanged is ______.
[Hint: 0 A I 3! × 5! – 1 = 719 ]

Q.148/6 Number of ways in which 7 people can occupy six seats, 3 seats on each side in a first class railway
compartment if two specified persons are to be always included and occupy adjacent seats on the same
side, is 5 ! · (k) then k has the value equal to :
(A) 2 (B) 4 (C*) 8 (D) none
[Hint: including the two specified people, 4 others can be selected in 5C4 ways . The two adjacent seats can be
taken in 4 ways and the two specified people can be arranged in 2 ! ways, remaining 4 people can be
arranged in 4 ! ways .
 C4 . 4 . 2 ! 4 ! = 5 ! 8 = 8 . 5 !  C ]

[12]
Q.159/6 Number of ways in which 9 different toys be distributed among 4 children belonging to different age
groups in such a way that distribution among the 3 elder children is even and the youngest one is to
receive one toy more, is :
(A)
5 !2 (B)
9!
(C*)
9!
(D) none
3
8 2 3 ! 2 !
[Hint : distribution 2, 2, 2 and 3 to the youngest . Now 3 toys for the youngest can be selected in 9C3 ways,
remaining 6 toys can be divided into three equal groups in
6! 9C
6! 9!
way and can be distributed in 3 ! ways  3 . 3 = 3 ]
2 !3 . 3 ! 2 ! 3 ! 2 !

Q.1610/6 In an election three districts are to be canvassed by 2, 3 & 5 men respectively . If 10 men volunteer, the
number of ways they can be alloted to the different districts is :
10 ! 10 ! 10 ! 10 !
(A*) (B) (C) (D)
2 ! 3! 5 ! 2! 5! (2 !)2 5 ! (2 !) 2 3 ! 5 !
10 !
[Hint : number of groups of 2, 3, 5 = & can be deputed only in one way ]
2 ! 3! 5 !
Q.1711/6 Let Pn denotes the number of ways in which three people can be selected out of ' n ' people sitting in
a row , if no two of them are consecutive.
If , Pn + 1  Pn = 15 then the value of 'n' is :
(A) 7 (B*) 8 (C) 9 (D) 10
[Hint : Pn = n  2 C3 ; Pn + 1 = n  1 C3 ; Hence n  1 C3 - n  2 C3 = 15
n  2C + n  2C  n  2C = 15 or n  2 C2 = 15  n=8  C]
3 2 3
Q.18 The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is
(A*) 26 (B) 18 (C) 31 (D) none
[Hint: Draw a Venn diagram with circles named
A : the number of integers from 1 to 100 divisible by 2 = 50
B : the number of integers from 0 to 100 divisible by 3 = 33
C : the number of integers from 0 to 100 divisible by 5 = 20
where the overlap portions are the numbers divisible by both
A & B , B & C, etc. (i.e. by 6, 10, 15 & 30).
The total number of integers not divisible is then 26. ]
Use : n (A  B  C) = n (A) + n (B) + n(C) – n(A  B) – n (B  C) – n (C  A) + n (A  B  C) ]

Q.19203/1(2/6) In how many different ways a grandfather along with two of his grandsons and four grand daughters
can be seated in a line for a photograph so that he is always in the middle and the two grandsons are
never adjacent to each other. [Ans. 528]
[Sol. Total number of ways they can sit = 6! × × × ×××
no. of ways when the two grandsons are always adjacent = 4 · 2! · 4! = 192
where 4 denotes the no. of adjacent positions
(2!) no. of ways in which two sons can be seated
and 4! no. of ways in which the daughter can be seated in the remaining places.
 required no. of ways = 720 – 192 = 528 Ans ]
Q.209/5 A forecast is to be made of the results of five cricket matches, each of which can be win, a draw or a loss
for Indian team. Find
(i) the number of different possible forecasts
(ii) the number of forecasts containing 0, 1, 2, 3, 4 and 5 errors respectively
[ Ans: 35 = 243 ; 1, 10, 40, 80, 80, 32 ]
[Hint: N(0) = 1 ; N(1) = 2.5C4 ; N(2) 22. 5C3 ; N(3) 23. 5C2 ; N(4) = 24. 5C1 ; N(5) = 25 ]

[13]
 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) Special DPP on Permutation and Combination DPP. NO.-5
Q.14/9 There are six periods in each working day of a school. Number of ways in which 5 subjects can be
arranged if each subject is allotted at least one period and no period remains vacant is
(A) 210 (B*) 1800 (C) 360 (D) 3600
[Hint: 6C2 · 5C1 · 4! = 1800 S1 S2 S 3 S 4 S5 × × × × ×
note that at least one of the subject has to be repeated [13th Test (5-12-2004)]
5
two periods in which one subject is to be repeated C1 · 4! one subject ]
Q.21/7 There are 10 red balls of different shades & 9 green balls of identical shades. Then the number of
arranging them in a row so that no two green balls are together is
(A) (10 !) . 11P9 (B*) (10 !) . 11C9 (C) 10 ! (D) 10 ! 9 !
Q.32/7 Number of ways in which n distinct objects can be kept into two identical boxes so that no box remains
empty, is
[Hint: Consider the boxes to be different for a moment. T1 can be kept in either of the boxes in 2 ways,
silmilarly for all other things  Total ways = 2n
but this includes when all the things are in B1 or B2  number of ways = 2n  2
2n  2
Since the boxes are identical  = 2n  1  1 ]
2
Q.43/7 A shelf contains 20 different books of which 4 are in single volume and the others form sets of 8, 5 and
3 volumes respectively. Number of ways in which the books may be arranged on the shelf, if the
volumes of each set are together and in their due order is
20!
(A) (B) 7! (C*) 8! (D) 7 . 8!
8! 5! 3!
[Hint : Volume of each set may be in due order in two ways, either from left to right or from right to left. Now
we have
D1 D2 D3 D4 , = 7! × 2! . 2! . 2! = 8! ]

Q.59/7 In a certain college at the B.Sc. examination, 3 candidates obtained first class honours in each of the
following subjects: Physics, Chemistry and Maths, no candidates obtaining honours in more than one subject;
Number of ways in which 9 scholarships of different value be awarded to the 9 candidates if due regard is
to be paid only to the places obtained by candidates in any one subject is _____. [ Ans: 1680 ]
[ Hint : 3 candidates in P; 3 in C and 3 in M. Now 9 scholarships can be divided into three groups in
9! 9!
ways and distributed to P, C, M in ways ]
3
(3!) .3! (3!)3
Q.65/7 Number of rectangles in the grid shown which are not squares is

(A*) 160 (B) 162 (C) 170 (D) 185

[Sol. 7 5
Total = C2 · C2 = 210 – number of squares [13th, 19-10-2008]
24  
number of squares =  15  
8 
3 = 50
1units 2 units 3 units 4 units
 required number = 210 – 50 = 160 Ans. ]

[15]
Q.76/7 All the five digits number in which each successive digit exceeds its predecessor are arranged in the
increasing order of their magnitude. The 97th number in the list does not contain the digit
(A) 4 (B*) 5 (C) 7 (D) 8
[Sol. All the possible number are 9C5 (none containing the digit 0) = 126 [12th test (29-10-2005)]
Total numbers starting with 1 = 8C4 = 70
(using 2, 3, 4, 5, 6, 7, 8, 9)
Total starting with 23 = 6C3 = 20
(4, 5, 6, 7, 8, 9)
Total starting with 245 = 4C2 = 6
(6, 7, 8, 9)
97th number = ]
Q.87/7 The number of combination of 16 things, 8 of which are alike and the rest different, taken 8 at a time is
______. [Ans. 256]
[Hint : A A A A A A A A D1 D2 ......... D8  8C0 + 8C1 + 8C2 + ...... + 8C8 = 256 ]
Q.98/7 The number of different ways in which five 'dashes' and eight 'dots' can be arranged, using only seven of
these 13 'dashes' & 'dots' is :
(A) 1287 (B) 119 (C*) 120 (D) 1235520
7 7 7 7 7
[Hint: C0 + C1 + C2 + C3 + C4 + C5 ] 7

Q.1010/7 There are n identical red balls & m identical green balls . The number of different linear arrangements
consisting of "n red balls but not necessarily all the green balls" is xCy then
(A) x = m + n , y = m (B*) x = m + n + 1 , y = m
(C) x = m + n + 1 , y = m + 1 (D) x = m + n , y = n
[Hint: Put one more red ball & find the arrangement of n + 1 red and m green balls = m + n + 1Cm ]
Q.116/8 A gentleman invites a party of m + n (m  n) friends to a dinner & places m at one table T1 and n at
another table T2 , the table being round . If not all people shall have the same neighbour in any two
arrangement, then the number of ways in which he can arrange the guests, is
(m  n) ! 1 (m  n) ! (m  n) !
(A*) (B) (C) 2 (D) none
4 mn 2 mn mn
(m  n )! (m  1)! (n  1)!
[Hint: .  (A) ]
m! n! 2 2

Q.1211/7 Consider a determinant of order 3 all whose entries are either 0 or 1. Five of these entries are 1 and
four of them are '0'. Also aij = aji  1  i, j  3. Find the number of such determinants.
[Ans. 12]
[Sol. We have aij = aji  a12 = a21 = a (say)  a b [11th, 28-11-2009, P-2] [Dpp, p&c]
a31 = a13 = b and a23 = a32 a  c
1, 1, 1, 1, 1 and 0, 0, 0, 0 b c 
Case-I :
If diagonal elements are (1, 1, 1 )
1 0 0
Conjugate element are, (0, 0), (0, 0), (1, 1) as shown. 0 1 1
3! 0 1 1
This can be done in = 3 ways
2!

[16]
 Case-II :
If diagonal has (1, 0, 0) then the conjugate elements are
(1, 1), (1, 1), (0, 0).
1  
3!  0 
Now conjugate elements can be arranged in = 3 ways
2   0
3!
and diagonal elements can be filled in = 3 ways
2
Hence 3 × 3 = 9 ways
 Total determinant = 3 + 9 = 12 Ans.]

Q.131/8 Number of different words that can be formed using all the letters of the word "DEEPMALA" if two
vowels are together and the other two are also together but separated from the first two is
(A) 960 (B) 1200 (C) 2160 (D*) 1440
[Hint : | D | P | M | L | can be arranged in 4 ! ways & the two gaps out of 5 gaps can be selected in 5C2 ways.
4!
two A's and two E's can be arranged in 2!2! = 6 ways.

4!
 4 ! · 5C2 · 2!2! = 1440 ]

Q.14 A four digit number is called a doublet if any of its digit is the same as only one neighbour. For example,
1221 is a doublet but 1222 is not. Number of such doublets are
(A) 2259 (B*) 2268 (C) 2277 (D) 2349
[Sol. Case-1: abbc [13th, 14-02-2009] [Dpp, P&C] done
We can choose a in 9 ways, b in 9 ways, and c in 9 ways. Hence, there are 93 possibilities for this case
Case-2: bbac
There are 9 possibilities for b, and 9 possibilities for each a and c. Once again, we have that there are 93
possibilities for this case
Case-3: acbb
Once again, there are 9 possibilities for a and c, and 9 possibilities for b. Hence , there are 93 possibilities
for this case
Case-4: aabb
We have 9 possibilities for a and 9 possibilities for b. We thus have 81 possibilities for this case.
We can now sum up our answers, to get a total of 3 · 93 + 81 = 2268 total ways Ans.
Alternatively: Total four digit numbers = 9000
let us compute the number which are not doublet
(a) when no two adjacent digits are same
= 9 · 9 · 9 · 9 = 6561
(b) 3 digits in a row are same at the end of the number
if they are non zero then we have 
= 9 · 8 = 72 
if they are same then = 9 
total for (b) = 81 
(c) 3 digits at the beginning of the numbers are same, then we have
= 9 · 9 = 81
(d) All four digits are same = 9
 Number of doublets = 9000 – [6561 + 81 + 81 + 9] = 2268 Ans. ]

[17]
Q.153/8 In a unique hockey series between India & Pakistan, they decide to play on till a team wins 5 matches.
The number of ways in which the series can be won by India, if no match ends in a draw is :
(A*) 126 (B) 252 (C) 225 (D) none
[Hint : India wins exactly in 5 matches  looses in none  5C0 ways
India wins exactly in 6 matches  wins the 6th and looses anyone in the 1st five
 5C1 ways and so on . 5C0 + 5C1 + 6C2 + 7C3 + 8C4 = 126
Alternatively : Total number of ways in which Series can be won by India or Pakistan = 10C5
10
C5
 required number of ways = = 126 ]
2
Q.1612/9 Sameer has to make a telephone call to his friend Harish, Unfortunately he does not remember the 7 digit
phone number. But he remembers that the first three digits are 635 or 674, the number is odd and there is
exactly one 9 in the number. The maximum number of trials that Sameer has to make to be successful is
(A) 10,000 (B*) 3402 (C) 3200 (D) 5000
[Sol. There are 2 ways of filling the first 3 digits, either 635 or 674. Of the remaining 4 digits, one has to be 9
and the last has to be odd. If the last digit is 9 then there are 9 ways of filling each of the remaining 3
digits. thus the total number of phone numbers that can be formed are 2 × 93 = 1458.
If the last digit is not 9, then there are only 4 ways of filling the last digit. (one of 1, 3, 5 and 7). The 9
could occur in any of the 3 remaining places and the remaining 2 places can be filled in 92 ways. Thus the
total number of such numbers is : 2 × 4 × 3 × 92 = 1944  1944 + 1458 = 3402 Ans ]
Q.176/9 A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. If
internal arrangement inside the car does not matter then the number of ways in which they can travel, is
(A) 91 (B) 182 (C*) 126 (D) 3920
[Sol. They can sit in groups of either 5 and 3 or 4 and 4
8! 8!  2!
required number = 1 + = 126 ] [12th test (05-06-2005)]
5 !  3! 4!  4 !  2!
Q.183/10 One hundred management students who read at least one of the three business magazines are surveyed
to study the readership pattern. It is found that 80 read Business India, 50 read Business world, and 30
read Business Today. Five students read all the three magazines. How many read exactly two magazines?
(A*) 50 (B) 10 (C) 95 (D) 65
[Hint: n (A  B  C) = 100
n (A) = 80 ; n (B) = 50 ; n (C) = 30
now n (A  B  C) =  n (A) –  n (A  B) + n (A  B  C)
100 = 80 + 50 + 30 –  n (A  B) + 5

  n (A  B) = 65

now n (E2) =  n (A  B) – 3n(A  B  C) = 65 – 15 = 50]

Q.1913/9 Six people are going to sit in a row on a bench. A and B are adjacent, C does not want to sit adjacent
to D. E and F can sit anywhere. Number of ways in which these six people can be seated, is
(A) 200 (B*) 144 (C) 120 (D) 56
[Hint: ; C and D separated ; E and F any where [11-12-2005, 11th (PQRS)]
and E, F can be seated in 3! 2!
no. of gaps are 4 | |E |F|
4
C D can be seated in C2 · 2!
Total ways 3! · 2! · 4C2 · 2! = 144 Ans. ]

[18]
MATCH THE COLUMN:
Q.2026/10 Column-I Column-II
(A) Number of increasing permutations of m symbols are there from the n set (P) nm
numbers {a1, a2, , an} where the order among the numbers is given by
a1 < a2 < a3 <  an–1 < an is
(B) There are m men and n monkeys. Number of ways in which every monkey (Q) mC
n
has a master, if a man can have any number of monkeys
(C) Number of ways in which n red balls and (m – 1) green balls can be arranged (R) nC
m
in a line, so that no two red balls are together, is
(balls of the same colour are alike)
(D) Number of ways in which 'm' different toys can be distributed in 'n' children (S) mn
if every child may receive any number of toys, is
[Ans. (A) R; (B) S; (C) Q; (D) P] [11th pqrs & J (24-11-2006)]
[Sol. (A) n
From n elements select m in Cm ways and can be arranged only in one way.
(B) 1st working can be given to M1 M2 M3........ Mm ® 'm' men
any one of m must k1 k2 k3.......kn  'n' monkey
|||ly 2nd, 3rd , ....., nth monkey in m way
total ways = mn
(C) Number of gaps = m
select n gaps in mCn ways for n red balls  total ways = mCn
(D) 1st toy in n ways
2nd toy in n ways
and so on. Total ways = nm ]

[19]
 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) Special DPP on Permutation and Combination DPP. NO.-6
Q.14/8 Number of cyphers at the end of 2002C1001 is
(A) 0 (B*) 1 (C) 2 (D) 200
(2002)!
[Hint: 2002C = (1001)!(1001)!
1001

no. of zeros in (2002)! are

400 + 80 + 16 + 3 = 499
no. of zeroes in (1001 !)2 = 2(200 + 40 + 8 + 1) = 498
(2002)!
Hence no. of zeroes is =1 ] [12th (25-9-2005)]
(1001!) 2
Q.25/8 Three vertices of a convex n sided polygon are selected. If the number of triangles that can be constructed
such that none of the sides of the triangle is also the side of the polygon is 30, then the polygon is a
(A) Heptagon (B) Octagon (C*) Nonagon (D) Decagon
[Sol. nC – [n + n(n – 4)] = 30
3
n = 9 satisfies it  (C) ] [11th, 13-01-2008]
Q.31/10 Given 11 points, of which 5 lie on one circle, other than these 5, no 4 lie on one circle. Then the maximum
number of circles that can be drawn so that each contains atleast three of the given points is :
(A) 216 (B*) 156 (C) 172 (D) none

[Hint:  5C2 . 6C1 + 6C2 5C1 + 6C3 + 1 = 156

alternatively 11C  5C3 + 1]

3
Q.4 Number of 5 digit numbers divisible by 25 that can be formed using only the digits
1, 2, 3, 4, 5 & 0 taken five at a time is
(A) 2 (B) 32 (C*) 42 (D) 52
[ Hint : 1, 2, 3, 4, 5, 0
A number divisible by 25 if the last two digits are 25 or 50
(i) Now, if 5 is not taken then number of numbers = 0
(ii) If 0 is not taken = 6
(iii) if 2 is not taken then = 6 and
(iv) If 1 | 3 | 4 is rejected
then in each case we have, = 6
= 4
2.2 1
if 3, 4, or 1 is not taken then 10 number for each case.
Total = 30
Hence total = 30 + 6 + 6 = 42 ]

[21]
Q.58/8 There are 12 guests at a dinner party. Supposing that the master and mistress of the house have fixed
seats opposite one another, and that there are two specified guests who must always, be placed next to
one another ; the number of ways in which the company can be placed, is:
(A*) 20 . 10 ! (B) 22 . 10 ! (C) 44 . 10 ! (D) none
[Hint: 6 places on either sides  G1G2 will have 5 places each on either side and can be seated in 2 ways 
10 × 2! × 10! Ans ]
Q.69/8 Let Pn denotes the number of ways of selecting 3 people out of 'n' sitting in a row, if no two of them
are consecutive and Qn is the corresponding figure when they are in a circle. If Pn  Qn = 6, then 'n' is
equal to :
(A) 8 (B) 9 (C*) 10 (D) 12
n n4
n  2C C1 . C2
[Hint : Pn = 3 ; Qn = nC3  [ n + n (n  4) ] or Qn =
3
Pn  Qn = 6  n = 10 ]
Q.7207/1(10/8) Define a 'good word' as a sequence of letters that consists only of the letters A, B and C and in
which A never immidiately followed by B, B is never immediately followed by C, and C is never immediately
followed by A. If the number of n-letter good words are 384, find the value of n. [Ans. n = 8 ]
[Sol. There are 3 choices for the first letter and two choices for each subsequent letters.
Hence using fundamental principle
number of good words = 3 · 2n–1 = 384
2n–1 = 128 = 27
n = 8 Ans. ]
Q.811/8 Six married couple are sitting in a room. Find the number of ways in which 4 people can be selected so that
(a) they do not form a couple (b) they form exactly one couple
(c) they form at least one couple (d) they form atmost one couple
[Hint : 12C4 = 6C2 + 6C1 · 5C2 · 24 + 6C4 · 24 ] [Ans. 240, 240, 255, 480]
Q.9 In a conference 10 speakers are present. If S1 wants to speak before S2 & S2 wants to speak after S3,
then the number of ways all the 10 speakers can give their speeches with the above restriction if the
remaining seven speakers have no objection to speak at any number is
(A) 10C3 (B) 10P8 (C) 10P3 (D*) 10 !
[Hint: S1 S2 S3 or S3 S1 S2  10C3 . 2 . 7 ! ] 3

Q.102/9 Let m denote the number of ways in which 4 different books are distributed among 10 persons, each
receiving none or one only and let n denote the number of ways of distribution if the books are all alike.
Then :
(A) m = 4n (B) n = 4m (C*) m = 24n (D) none
10
[Hint: m = C4 . 4 ! and n = C4 ] 10

Q.115/9 The number of all possible selections of one or more questions from 10 given questions, each equestion
having an alternative is :
(A) 310 (B) 210  1 (C*) 310  1 (D) 210
[Hint : 1st question can be selected in three ways and so on ]
Q.12 Number of 7 digit numbers the sum of whose digits is 61 is :
(A) 12 (B) 24 (C*) 28 (D) none
[Hint: only 7, 8 and 9 can be used]

[22]
Q.139/9 The number of ways of choosing a committee of 2 women & 3 men from 5 women & 6 men, if Mr. A
refuses to serve on the committee if Mr. B is a member & Mr. B can only serve, if Miss C is the member
of the committee, is
(A) 60 (B) 84 (C*) 124 (D) none

[Hint : ; ;

(i) Miss C is taken

(a) B included  A excluded  4C . 4C2 = 24
1
(b) 4 5
B excluded  C1 . C3 = 40
(ii) Miss C is not taken
 B does not comes ; 4C . 5C = 60  Total = 124
2 3
Alt. Total  [A, B, C present + A,B present & C absent + B present & A, C absent]
Alternatively : Case 1 : Mr. ' B ' is present  ' A ' is excluded & ' C ' included
Hence number of ways = 4C2 . 4C1 = 24
Case 2 : Mr. ' B ' is absent  no constraint
Hence number of ways = 5C3 . 5C2 = 100 Total = 124 ]
Q.1410/9 Six persons A, B, C, D, E and F are to be seated at a circular table. The number of ways this can be
done if A must have either B or C on his right and B must have either C or D on his right is :
(A) 36 (B) 12 (C) 24 (D*) 18
[Hint : when A has B or C to his right we have AB or AC
when B has C or D to his right we have BC or BD
Thus  we must have ABC or ABD or AC and BD
for D, E, F on a circle number of ways = 3 ! = 6
for C, E, F on a circle number of ways = 3 ! = 6
for , E, F the number of ways = 3 ! = 6  Total = 18 ]

Q.1511/9 There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The number
of ways in which they can be arranged in a row so that atleast one ball is separated from the balls of the
same colour, is :
(A*) 6 (7 !  4 !) (B) 7 (6 !  4 !) (C) 8 !  5 ! (D) none

[Hint : 9 ! – number of ways when balls of the same colour are together = 9 !  3 ! 4 ! = 6(7!  4!) ]
2! 3! 2! 3!

Q.161/9 Product of all the even divisors of N = 1000, is

(A) 32 · 102 (B) 64 · 214 (C*) 64 · 1018 (D) 128 · 106
[Sol. 1000 = 23 · 53 [11th, 28-11-2009, P-2] [Dpp, p&c]
Number of divisors = 16 which are (2 + 2 + 2 + 2 )(5 + 5 + 52 + 53)
0 1 2 3 0 1

Product of all the 16 divisors is

= (56)(56)(56)(56)(26)(26)(26)(26)
= 224 · 524
Now odd divisors are 1, 5, 25, 125
and product of odd deivisors = 56
 Product of all the even divisors = 224 · 524 = 218 · 518 · 26 = (1018)64 Ans.
Alternatively: Prodcut of all the even divisors = (24 · 56)(28 · 56) (212 · 56) (think!)
= 224 · 518 = 64 · 1018 Ans. ]

[23]
Q.17209/1(14/7) Find the number of 4 digit numbers starting with 1 and having exactly two identical digits.
[Ans. 432]
[Sol. Case-I : When the two identical digits are both unity as shown.
any one place out of 3 block for unity can be taken in 3 ways and the remaining two
blocks can be filled in 9 · 8 ways.
Total ways in this case = 3 · 9 · 8 = 216
Case-II : When the two identical digit are other than unity.
; ;
two x's can be taken in 9 ways and filled in three ways and y can be taken in 8 ways.
Total ways in this case = 9 · 3 · 8 = 216
Total of both case = 432 Ans. ]
Q.18 Consider the word W = MISSISSIPPI
(a) If N denotes the number of different selections of 5 letters from the word W = MISSISSIPPI then N
belongs to the set
(A) {15, 16, 17, 18, 19} (B) {20, 21, 22, 23, 24}
(C*) {25, 26, 27, 28, 29} (D) {30, 31, 32, 33, 34}
(b) Number of ways in which the letters of the word W can be arranged if atleast one vowel is separated
from rest of the vowels
8!·161 8!·161 8!·161 8! 165
(A) 4!·4!·2! (B*) 4 ·4!·2! (C) 4!·2! (D) 4!·2! · 4!
 10! 
(c) If the number of arrangements of the letters of the word W if all the S's and P's are separated is (K)  4!·4! 
 
then K equals
6 4 3
(A) (B*) 1 (C) (D)
5 3 2
[Sol. [13th, 28-12-2008, P-1]
(a) SSSS; IIII; PP; M
4 alike + 1 different =2·3= 6
3 alike + 2 other alike =2·2= 4
3 alike + 2 different =2·3= 6
2 alike + 2 other alike + 1 different =3·2= 6
2 alike + 3 different = 3 = 3
25
(b) Atleast one vowel is separated from the rest
= total – when all vowels together
11! 8! 11·10 ·9 ·8! 8!
= 4!·4!·2! – 4!·2! = 4!·4!·2! – 4!·2!

8! 11·10 ·9  8! 11·15  8! 165  4  8!·161

= 4!·2!  24 1 = 4!·2!  4 1 = 4!·2!  4  = 4 ·4!·2! Ans.
     

[24]
(c) Separate S first
| I | I | I | I | P | P | M | (8 gaps)
| I | I | I | I | PP | M |
7! 8 6!
· C 4  · 7 C 4 ; S's separated but P's may or may not be separated ;
4!·2! 4!
6!·7! 10!
simplifies to 4!·4! =  K = 1 Ans. ]
4!·4!
Q.19 Let A = {a, b, c, d, e, f} and B = {1, 2, 3} are two sets.
Let m denotes the number of mappings which are into from A to B
n denotes the number of mappings which are injective from B to A.
Find (m + n). [Ans. 309]
[Sol. m = into mappings from A to B
Case-I: If exactly one element of set B is not the image
of any of the elements of set A then total number of into
functions are
3C × (26 – 2) = 3 × 62 = 186
1
Case-II: If exactly two elements of set B is not the image
of any of the elements of set A then total number of into
functions are
3C × 1 = 3
2
and hence m = 186 + 3 = 189 [12th, 19-07-2009, P-1]
Alternatively:
Total mappings = 36 = 729
Surjective mappings
6!3! 720
2 2 2  3
  90
(2!) 3! 8
6! 3! 720
2 3 1    360
2!3! 2
6! 3! 720
1 1 4    90
2!4! 2  4
Total surjective mapping = 540
 not surjective mappings = 729 – 540 = 189 = m
n = number of injective mapping from B to A.
Injective mapping = 6C3 × 3! = 20 × 6 = 120 = n
 (m + n) = 189 + 120 = 309 Ans. ]

[25]
Q.20 Let a function f is defined as f : {1, 2, 3, 4}  {1, 2, 3, 4}. If f satisfy f f ( x )  = f (x),  x  {1, 2, 3, 4}
then find the number of such function. [Ans. 41]
[Sol. f f ( x )  = f (x) ; f (x) = y  f (y) = y [12th, 23-08-2009, P-2]
Case-1: range contains exactly one element
it can be done in 4C1 ways say 1
remaining 3 elements i.e. 2, 3, 4 can be mapped
only in one ways  total = 4C1 = 4
Case-2: range contains two elements this can
this can be done in 4C2 ways say 1, 2
 f (1) = 1; f (2) = 2
remaining 2 elements i.e. 3 and 4 each can be
mapped in 2 ways
Total = 4C2 · 22 = 24
Case-3: range contains 3 elements which
can be done in 4C3 ways say 1, 2, 3
 f (1) = 1; f (2) = 2 and f (3) = 3
now remaining 4 can be mapped only in
3 ways
Total = 4C3 · 3 = 12 ways
Case-4: range contains all 4 elements
f (1) = 4; f (2) = 2 ; f (3) = 3; f (4) = 4
only 1 way
Total = 4 + 24 + 12 + 1 = 41 Ans. ]

[26]
 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) Special DPP on Permutation and Combination DPP. NO.-7
Choose the correct alternative (only one is correct):
Q.1 Number of ways in which four different toys and five indistinguishable marbles can be distributed between
Amar, Akbar and Anthony , if each child receives atleast one toy and one marble, is
(A) 42 (B) 100 (C) 150 (D*) 216
4!
[Sol. Toys in group 1 1 2   3! = 36
1! 1! 2! 2!
Marbles OOO  O = 4C2 = 6
 Total ways = 36 × 6 = 216 ]
Q.2 A 3 digit palindrome is a 3 digit number (not starting with zero) which reads the same backwards as
forwards. For example 171. The sum of all even 3 digit palindromes, is
(A) 22380 (B) 25700 (C*) 22000 (D) 22400
H T U
[Sol. Number of even 3 digit palindromes = 4 · 10 = 40 [12th & 13th 03-03-2007]
4 ways to fill (hundredth and unit's place) and 10 ways to fill tenth's place
100 10(2  4  6  8) 10( 2  4  6  8) 4 ·10(1  2  3  .....  9)
sum = th +  +  
th

(100 place) ( sum of units place ) (10 place )
= 20,000 + 200 + 1800 = 22000]
Q.33/10 There are 100 different books in a shelf. Number of ways in which 3 books can be selected so that no
two of which are neighbours is
(A) 100C3 – 98 (B) 97C3 (C) 96C3 (D*) 98C3
[11th & 13th, 16-12-2007]
Q.4 A lift with 7 people stops at 10 floors. People varying from zero to seven go out at each floor. The
number of ways in which the lift can get emptied, assuming each way only differs by the number
of people leaving at each floor, is :
(A) 16C6 (B) 17C7 (C*) 16C7 (D) none
[Hint : consider each floor to be a beggar. Now distribute 7 identical coins ( 7 people) in 10 beggars each
receiving none, one or more  16C7 ]
Q.5 You are given an unlimited supply of each of the digits 1, 2, 3 or 4. Using only these four digits, you
construct n digit numbers. Such n digit numbers will be called L E G I T I M A T E if it contains the digit
1 either an even number times or not at all. Number of n digit legitimate numbers are
(A) 2n + 1 (B) 2n + 1 + 2 (C) 2n + 2 + 4 (D*) 2n – 1(2n + 1)
[Sol. × × ×............× (n places) [12th test (09-10-2005)]
Total numbers = 3n + nC2 · 3n – 2 + nC4 · 3n – 4 + .......... (nC2 indicates selection of places)
(3  1) n  (3  1) n 1
= = [4n + 2n] = 22n – 1 + 2n – 1 = 2n – 1(2n + 1) ]
2 2
Q.65/10 Two classrooms A and B having capacity of 25 and (n–25) seats respectively.An denotes the number of
possible seating arrangements of room 'A', when 'n' students are to be seated in these rooms, starting
from room 'A' which is to be filled up full to its capacity. If An – An–1 = 25! (49C25) then 'n' equals
(A*) 50 (B) 48 (C) 49 (D) 51
n n–1
[Hint: Given An = C25 · 25! ; An – 1 = C25 · 25!
hence nC25. 25! – n–1C25 . 25! = 25! 49C25

[27]
 or n–1C + n–1C24 – n–1C = 49C n – 1 = 49  n = 50
25 25 24 
 n–1C = 49C24  n = 50]
24
Q.79/10 Number of positive integral solutions satisfying the equation (x1 + x2 + x3) (y1 + y2) = 77, is
(A) 150 (B) 270 (C*) 420 (D) 1024
[Sol. (x1 + x2 + x3) (y1 + y2) = 11 · 7 or 7 · 11 [29-01-2006, 12 & 13]
In the first case (x1 + x2 + x3) = 11 and (y1 + y2) = 7, which have C2 · 6C1 solutions (using beggar)
10

In the second case (x1 + x2 + x3) = 7 and (y1 + y2) = 11, which have 6C2 · 10C1 solutions (using beggar)
 total number of solutions = 10C2 · 6C1 + 6C2 · 10C1 = 270 + 150 = 420 Ans. ]
Q.810/10 Distinct 3 digit numbers are formed using only the digits 1, 2, 3 and 4 with each digit used at most once
in each number thus formed. The sum of all possible numbers so formed is
(A*) 6660 (B) 3330 (C) 2220 (D) none
[Hint: all possible = 24
6(1 + 2 + 3 + 4)(1 + 10 + 102) = 6 · 10 · 111 = 6660
reject 1 or 2 or 3 or 4 ] [12 & 13th test (09-10-2005)]
Q.9 There are counters available in 3 different colours (atleast four of each colour). Counters are all alike
except for the colour. If 'm' denotes the number of arrangements of four counters if no arrangement
consists of counters of same colour and ' n' denotes the corresponding figure when every arrangement
consists of counters of each colour, then :
(A) m = 2 n (B*) 6 m = 13 n (C) 3 m = 5 n (D) 5 m = 3 n
[Hint : m = 34  3 = 78;  
n = 34  3 2 4  2  3   = 81  45 = 36
m 78 13
Hence = =  6 m = 13 n  B]
n 36 6
Q.1012/10 An ice cream parlour has ice creams in eight different varieties. Number of ways of choosing 3 ice
creams taking atleast two ice creams of the same variety, is
(Assume that ice creams of the same variety to be identical & available in unlimited supply)
(A) 56 (B*) 64 (C) 100 (D) none
[Hint: 10 8 10
C3  C3 = 120  56 = 64 ] [Note: C3 = all diff. + 2 alike and 1 diff. + all alike]
Q.1113/10 There are 12 books on Algebra and Calculus in our library , the books of the same subject being
different. If the number of selections each of which consists of 3 books on each topic is greatest then the
number of books of Algebra and Calculus in the library are respectively:
(A) 3 and 9 (B) 4 and 8 (C) 5 and 7 (D*) 6 and 6

[Hint: = xC3 · 12 – xC3 ]

Q.12 Three digit numbers in which the middle one is a perfect square are formed using the digits 1 to 9 . Their
sum is :
(A*) 134055 (B) 270540 (C) 170055 (D) none of these
[ Hint: Middle place 1, 4 & 9
Two terminal positions 1, 2, ...... , 9
Hence total numbers = 9 . 9 . 3 = 243 (Terminal digits in 9 ways and middle one in 3 ways)
For the middle place 1, 4 & 9 will come 81 times
 sum = 81  10 (1 + 4 + 9)  A
For units place each digit from 1 to 9 will appear 27 times
 sum = 27 (1 + 2 + ...... + 9)  B
For hundreath's place, similarly sum = 27  100 (1 + 2 + ...... + 9)  C
A + B + C gives the required sum ]
[28]
Q.13 A guardian with 6 wards wishes everyone of them to study either Law or Medicine or Engineering.
Number of ways in which he can make up his mind with regard to the education of his wards if every one
of them be fit for any of the branches to study, and atleast one child is to be sent in each discipline is :
(A) 120 (B) 216 (C) 729 (D*) 540
[Hint: Divide 6 children into groups as 123, 411 ot 222
6! 6! 6!. 3!
Now total = 1! 2! 3! 3! + 4! 1! 1! 2! 3! + 2! 2! 2! 3! = 360 + 90 + 90 = 540 ]

Q.1415/8 There are (p + q) different books on different topics in Mathematics. (p  q)

If L = The number of ways in which these books are distributed between two students X and Y such
that X get p books and Y gets q books.
M = The number of ways in which these books are distributed between two students X and Y such that
one of them gets p books and another gets q books.
N = The number of ways in which these books are divided into two groups of p books and q books
then,
(A) L = M = N (B) L = 2M = 2N (C*) 2L = M = 2N (D) L = M = 2N
Q.15 Number of ways in which 5 A's and 6 B's can be arranged in a row which reads the same backwards
and forwards, is
(A) 6 (B) 8 (C*) 10 (D) 12
[Sol. A A A A A | B B B B B B M
Middle digit must be A (think !)
         
so that even number of A's and B's are available
Take AABBB on one side of M (6th place) and then their image about M in a unique way
5!
 Number of ways = 2!·3! = 10 Ans.] [12th, 14-06-2009, P-2]

Q.16 A person writes letters to his 5 friends and addresses the corresponding envelopes. Number of ways in
which the letters can be placed in the envelope, so that atleast two of them are in the wrong envelopes,is,
(A) 1 (B) 2 (C) 118 (D*) 119
1 R and 2 R and 3 R and 4 5 R
[Hint: 5! = 4W + 3W + 2W + + all 5 wrong
none used

one way

hence at least 2 in wrong = 120 – 1 = 119 ]

Q.17 For a game in which two partners oppose two other partners, 8 men are available. If every possible pair
must play with every other pair, the number of games played is
(A) 8C2 . 6C2 (B) 8C2 . 6C2 . 2 (C*) 8C4 . 3 (D) none
[Hint: For each game 4 persons are needed . Hence select 4 from 8 in 8C4 way . Now from each selection
3 games can be had  8C4 . 3 = 210 ]
Q.18 The number 916238457 is an example of nine digit number which contains each of the digit 1 to 9
exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits
1 to 6 do not. Number of such numbers are
(A) 2268 (B*) 2520 (C) 2975 (D) 1560
[Sol. 9, 8 & 7 can be placed in 9 · 8 · 7 ways.
There are only five ways to place the 6 (any where except the right most remaining slot, think !) and the
order of 1 - 5 is fixed.

[29]
  Number of such numbers are 9 · 8 · 7 · 5 = 2520 Ans.
Alternatively-1: 6 places can be selected in 9C6 ways and 6 can be placed only at 5 places except the right
most of the 6 selected. Remaining numbers i.e. 7, 8, 9 in 3! ways.
Hence number of numbers 9C6 · 5 · 3! = 84 · 30 = 2520 Ans.
Alternatively-2: 9 C ·4!  9 C ·3!
 5   6
a b
a = selection of 5 places for 1, 2, 3, 4, 5 & 4! for 6, 7, 8, 9 = total ways
b = selection 6 places for 1, 2, 3, 4, 5, 6 & 3! for 7, 8, 9 ] [12th & 13th 07-01-2007]
Q.19 Number of functions defined from f : {1, 2, 3, 4, 5, 6}  {7, 8, 9, 10} such that the sum
f (1) + f (2) + f (3) + f (4) + f (5) + f (6) is odd, is
(A) 210 (B*) 211 (C) 212 (D) 212 – 1
[Sol. Giving any random configuration for f (1) to f (5), number of functions = 45 ways (each element from 1
to 5 can be mapped in 4 ways)
5
Now the sum  f (i) is either odd or even
i 1
if it is odd then we have only two choices for f (6) i.e. the element 8 or 10.
(as odd + even = odd).
If the sum is even then also we have only choices for f (6) i.e. 7 or 9.
hence the total functions = 45 · 2 = 211 Ans.] [12th, 24-08-2008] [TN for 11th pqrs]

Paragraph for Question Nos. 20 to 22

16 players P1, P2, P3,.......P16 take part in a tennis tournament. Lower suffix player is better than any
higher suffix player. These players are to be divided into 4 groups each comprising of 4 players and the
best from each group is selected for semifinals.
Q.20 Number of ways in which 16 players can be divided into four equal groups, is
35 8 35 8 35 8 35 8
(A*) 27  (2r  1) (B) 24  (2r  1) (C) 52  (2r  1) (D) 6  (2r  1)
r 1 r 1 r 1 r 1

Q.21 Number of ways in which they can be divided into 4 equal groups if the players P1, P2, P3 and P4 are in
different groups, is :
(11)! (11)! (11)! (11)!
(A) (B) (C*) (D)
36 72 108 216
Q.22 Number of ways in which these 16 players can be divided into four equal groups, such that when the
12!
best player is selected from each group, P6 is one among them, is (k) . The value of k is :
(4!)3
(A) 36 (B) 24 (C) 18 (D*) 20
8
28 ·8! (2r  1)
(16)! r 1 28 ·8 ·7 ·6 ·5 · 29 ·35 ·
[Sol.(i) Number of ways = (4!) 4 4! = = =
4!·4!·4!·4!·4! (24) ·(24) ·(24) ·(24) (24) ·(24) ·(24)

[30]
 8
35
= 27  (2r  1) Ans. [11th, 25-01-2009, P-1]
r 1

P1,P2,P3,P4
(ii) 16
12 others

(12)! (12) ·(11)! (11)!

12 others can be divided into 4 equal groups in each of 3 person = 6 ·6 ·6 ·6 =
4 Ans.
(3!) 108
P1,P2,P3,P4,P5 3 10!
(iii) 16 P6 ; 10 = 3! 7!
P7 – P16(10) 7

000 xxxxxxx
+P6 P1 to P5
4 players 12 players
4 12!
Now, 12 4=
4 (4!)3 ·3!

(10)! (12)!  12! 

Total ways = 3! 7! · (4!)3 ·3! =  
3  · 20  k = 20 Ans.]
 (4!) 

Choose the correct alternatives (More than one are correct):

Q.2315/10 The combinatorial coefficient C(n, r) is equal to
(A*) number of possible subsets of r members from a set of n distinct members.
(B*) number of possible binary messages of length n with exactly r 1's.
(C) number of non decreasing 2-D paths from the lattice point (0, 0) to (r, n).
(D*) number of ways of selecting r things out of n different things when a particular thing is always
included plus the number of ways of selecting 'r' things out of n, when a particular thing is always
excluded.
[Hint: In (C) number = n+rCr ; (D) nCr = n – 1Cr – 1 + n – 1Cr ][18-12-2005, 12th & 13th]

Q.2417/10 There are 10 questions, each question is either True or False. Number of different sequences of
incorrect answers is also equal to
(A) Number of ways in which a normal coin tossed 10 times would fall in a definite order if both Heads
and Tails are present.
(B*) Number of ways in which a multiple choice question containing 10 alternatives with one or more
than one correct alternatives, can be answered.
(C*) Number of ways in which it is possible to draw a sum of money with 10 coins of different
denominations taken some or all at a time.
(D) Number of different selections of 10 indistinguishable things taken some or all at a time.
[Sol. 210 – 1; (A) 210 – 2; (B) 210 – 1; (C) 210 – 1; (D) 10 ] [11th & 13th, 16-12-2007]

Q.2519/10 The continued product, 2 . 6 . 10 . 14 ...... to n factors is equal to :

(A) 2nCn (B*) 2nPn
(C*) (n + 1) (n + 2) (n + 3) ...... (n + n) (D) none

[31]
 ( 2n )! 2n
[Hint: E = 2 · 6 · 10 · 14 ........ 2n = 2n(1 · 3 · 5 ....... n) = = Pn
n!
( n  1)(n  2).....(n  n ) n!
also E= ]
n!
Q.2620/10 The Number of ways in which five different books to be distributed among 3 persons so that each
person gets at least one book, is equal to the number of ways in which
(A) 5 persons are allotted 3 different residential flats so that and each person is alloted at most one flat
and no two persons are alloted the same flat.
(B*) number of parallelograms (some of which may be overlapping) formed by one set of 6 parallel lines
and other set of 5 parallel lines that goes in other direction.
(C*) 5 different toys are to be distributed among 3 children, so that each child gets at least one toy.
(D*) 3 mathematics professors are assigned five different lecturers to be delivered , so that each professor
gets at least one lecturer.
[Hint: Given answer is 150 which comes in BCD in A, it is 5C3 · 3! = 60]
Q.2721/10 The maximum number of permutations of 2n letters in which there are only a's & b's, taken all at a time
is given by :
2nC 2 6 10 4n  6 4n  2
(A*) n (B*) . . ...... .
1 2 3 n1 n

n 1 n  2 n  3 n  4 2n  1 2n 2 n . 1 . 3 . 5 ...... (2 n  3) (2 n  1)
(C*) . . . ...... . (D*)
1 2 3 4 n 1 n n!

Q.2822/10 Number of ways in which 3 numbers in A.P. can be selected from 1, 2, 3, ...... n is :
2
 n  1 n  n  2
(A)   if n is even (B) if n is odd
 2  4

(C*)
n 12 if n is odd (D*)
n  n  2
if n is even
4 4
[Hint : n = 2m, arrange the numbers into disjoint sets
1, 3, 5, .................... (2m – 1) m number
2, 4, 6, .....................2m m numbers
m
no. of AP's = C2 + C2 m ]
Q.2923/10 The combinatorial coefficient n – 1Cp denotes
(A) the number of ways in which n things of which p are alike and rest different can be arranged in a circle.
(B*) the number of ways in which p different things can be selected out of n different thing if a particular
thing is always excluded.
(C) number of ways in which n alike balls can be distributed in p different boxes so that no box remains
empty and each box can hold any number of balls.
(D*) the number of ways in which (n – 2) white balls and p black balls can be arranged in a line if black
balls are separated, balls are all alike except for the colour. [11th & 13th, 16-12-2007]
Q.3025/10 Which of the following statements are correct?
(A*) Number of words that can be formed with 6 only of the letters of the word "CENTRIFUGAL" if
each word must contain all the vowels is 3 · 7!
(B*) There are 15 balls of which some are white and the rest black. If the number of ways in which the
balls can be arranged in a row, is maximum then the number of white balls must be equal to 7 or 8.
Assume balls of the same colour to be alike.

[32]
 (C) There are 12 things, 4 alike of one kind, 5 alike and of another kind and the rest are all different. The
total number of combinations is 240.
(D*) Number of selections that can be made of 6 letters from the word "COMMITTEE" is 35.
/ 4v 7C
[Sol. (A) 11 \  2 · 6! = 3 · 7 · 6! = 3 · 7!
7c
15! W W.....W B B B......B
(B) No. of ways = = 15cv  
r!(15  r )! r 15 r
This is maximum if r = 7 or 8

(C)  Total no. of combinations = 5 · 6 · 23 – 1 = 240 –1 = 239

(D) 2 alike + 2 other alike + 2 other different = 1

2 alike + 2 other alike + 2 different = 3C2 · 4C2 = 18
2 alike + 4 different = 3C1 · 5C1 = 15
All 6 different = 1
——
= 35 Ans. ]
Q.31 Coefficient of x2 y3 z4 in the expansion of (x + y + z)9 is equal to
(A*) the number of ways in which 9 things of which 2 alike of one kind, 3 alike of
2nd kind, and 4 alike of 3rd kind can be arranged.
(B) the number of ways in which 9 identical things can be distributed in 3 persons each receiving atleast
two things.
(C) the number of ways in which 9 identical things can be distributed in 3 persons each receiving none
one or more.
(D*) the number of ways in which 9 different books can be tied up in to three bundles one containing 2,
other 3 and third containing 4 books.
[Hint: Consider (x + y + z)9 = 9Cr x9 – r (y + z)r = 9Cr x9 – r · rCp yr – p zp [x + (y + z)]9
put r=7;p=4
9!
or 9C · 7C · 4C4 = 2! 3! 4! ]
2 3

Q.32 Number of ways in which the letters of the word 'B U L B U L' can be arranged in a line in a definite
order is also equal to the
(A*) number of ways in which 2 alike Apples and 4 alike Mangoes can be distributed in 3 children so
that each child receives any number of fruits.
(B) Number of ways in which 6 different books can be tied up into 3 bundles, if each bundle is to have
equal number of books.
(C*) coefficient of x2y2z2 in the expansion of (x + y + z)6.
(D*) number of ways in which 6 different prizes can be distributed equally in three children.
6!
[Explanation: B U L B U L; number of ways =
2!·2!·2!
(A) 2 Apples can be distributed in 3 people in 4C2 way O O Ø Ø
and 6
4 Mangoes in C2 ways O O O O Ø Ø
6! 4! 6!
 Total ways = 6C2 · 4C2 = 2!·4! · 2!·2! = 2!·2!·2!  (A) is correct

[33]
 6!
(B) 6 books in 3 bundles, two in each bundle = 2!·2!·2!·3!  (B) is incorrect
(C) Tr + 1 in (x + y + z)6 is 6Cr(x + y)6 – r · zr
put r=2
T3 = 6C2(x + y)4 · z2
= 6C2 · z2 (4Cp · x4 – p · yp) [11th pqrs & J, 21-1-2007]
put p=2
= 6C2 · z2 · 4C2·x2y2)
= 6C2· 4C2 x2y2z2
6!
=  (C) is correct
2!·2!·2!
6 !·3! 6!
(D) 6 prizes in 3 childrents, two to each = 2!·2!·2!·3! = 2!·2!·2!  (D) is correct]

MATCH THE COLUMN:

Q.3334/10 Column-I Column-II
(A) Four different movies are running in a town. Ten students go to watch (P) 11
these four movies. The number of ways in which every movie is watched
by atleast one student, is
(Assume each way differs only by number of students watching a movie) (Q) 36
(B) Consider 8 vertices of a regular octagon and its centre. If T denotes the
number of triangles and S denotes the number of straight lines that can
be formed with these 9 points then the value of (T – S) equals
(C) In an examination, 5 children were found to have their mobiles in their (R) 52
pocket. The Invigilator fired them and took their mobiles in his possession.
Towards the end of the test, Invigilator randomly returned their mobiles. The
number of ways in which at most two children did not get their own mobiles is (S) 60
(D) The product of the digits of 3214 is 24. The number of 4 digit natural
numbers such that the product of their digits is 12, is
(E) The number of ways in which a mixed double tennis game can be
arranged from amongst 5 married couple if no husband & wife plays (T) 84
in the same game, is
[Ans. (A) T; (B) R; (C) P; (D) Q; (E) S]
[Sol.(A) x + y + z + t = 10 [11th, 13-01-2008]
4 movies 10 people (people as alike objects and movies as 4 different beggars)
give one to each x, y, z and t and
hence the number of ways = the number of non-negative integral solution of the equation
x+ y+ z+ t = 6
using Beggar method 000000ØØØ
 
9
number of ways is C3 = 84 Ans.
(B) ( 9C3 – 4 ) – 8C2 = 52 Ans.
(C) at most two children got mobile of the other children
 exactly 3R and 2W + all 5 got their own mobile
5
= C3 · 1 + 1 = 11 Ans.
(D) 12 = 22 · 3
hence groups of 4 that work out
4!
1, 1, 6, 2 = = 12
2!
[34]
 4!
1, 1, 3, 4 = = 12
2!
4!
2, 2, 3, 1 = = 12
2!
36 Ans.
(E) 5C . 3C . 2 ! = 60 Ans. ]
2 2

Subjective:
Q.3428/10 A commitee of 10 members is to be formed with members chosen from the faculties of Arts, Economics,
Education, Engineering, Medicine and Science. Number of possible ways in which the faculties
representation be distributed on this committee, is ________.
(Assume every department contains more than 10 members). A Arts [4]
[Ans. 3003] B Economics
[Sol. (a) 10 members to be chosen from 6 different faculties C Education
D Engineerin g
0 0 0..........0 Ø Ø Ø Ø Ø E Medicine
      (Use Beggar)
10 5 F Science
number of ways = 15C5 = 3003 Ans. ] [11th & 13th, 16-12-2007]

Q.3532/10 On the normal chess board as shown, I1 & I2 are two insects
which starts moving towards each other. Each insect moving with
the same constant speed . Insect I1 can move only to the right or
upward along the lines while the insect I2 can move only to the left or
downward along the lines of the chess board. Find the total number
of ways the
two insects can meet at same point during their trip.
[ Hint: (8C0 . 8C0) + (8C1 . 8C1) + .... + (8C8 . 8C8) = 16C8) = 12870 ]

Q.36186/1(30/10) 10 identical ball are distributed in 5 different boxes kept in a row and labled A, B, C, D and E.
Find the number of ways in which the ball can be distributed in the boxes if no two adjacent boxes
remain empty. [Ans. 771 ways]
[Sol. Case-1 : When no box remains empty
it is equivalently distributing
10 coins in 5 beggar
00000 ØØØØ 9

  = C = 126
5 4 4
Case-2: Exactly one is empty
5C 0 0 0 0 0 0 Ø Ø Ø = 5 · 9C3 = 420
1 
6
Case-3: Exactly two remains empty
( 5C2 – two adjacent) 9C2 0000000 ØØ


7
(10 – 4) × 9C
2
6 × 36 = 216
Case-4 : Exactly three empty. There is only 1 way to select 3 if no two adjacent

[35]
 Hence 1 · 9C1 = 9 00000000Ø
Total = 771 ways ]

Q.3738/10 Find the number of ways in which 12 identical coins can be distributed in 6 different purses, if not more
than 3 & not less than 1 coin goes in each purse. [Ans. 141]
[Hint: Put one coin in each purse. Now 6 coins remaining.
case (i) 1 coin in each of six purses =1
O O O O O O
case (ii) 2 coin in each of three purses = 6C3 · 1 = 20
O
 O O  O O  O
case (iii) 2 coin each in two purses and 1 coin each in 2 purses = 6C2 . 4C2 = 90
O
 O O O O O × ×

case (iv) 2 coins in one purse and 1 coin each in 4 purses = 6C1 . 5C4 = 30
O
 O O O O O ——————
Total = 140
Alternatively : coefficient of x12 in (x + x2 + x3)6 = 141 ]

[36]
 BANSAL CLASSES MATHEMATICS
Target IIT JEE 2007 Daily Practice Problems
CLASS : XII (ABCD) DPP ON PROBABILITY DPP. NO.- 1
After 1 st Lecture
Q.146 100 cards are numbered from 1 to 100. The probability that the randomly chosen card has a digit 5 is:
(A) 0.01 (B) 0.09 (C*) 0.19 (D) 0.18
[Hint: from 50 to 60, 10 fives and 9 other fives  total 19  (C) ]

Q.24 A quadratic equation is chosen from the set of all the quadratic equations which are unchanged by
squaring their roots. The chance that the chosen equation has equal roots is :
(A*) 1/2 (B) 1/3 (C) 1/4 (D) 2/3

Q.310 If the letters of the word "MISSISSIPPI" are written down at random in a row, the probability that no
two S's occur together is :
(A) 1/3 (B*) 7/33 (C) 6/13 (D) 5/7

Q.429 A sample space consists of 3 sample points with associated probabilities given as 2p, p2, 4p – 1 then
1 1
(A*) p = 11  3 (B) 10  3 (C) <p< (D) none
4 2
[Hint: p2 + 2p + 4p – 1 = 1 (exhaustive)
p2 + 6p – 2 = 0  (A) ]

Q.539 A committee of 5 is to be chosen from a group of 9 people. The probability that a certain married couple
will either serve together or not at all is :
(A) 1/2 (B) 5/9 (C*) 4/9 (D) 2/3
7
C3  7 C 5
[Hint : 9 ]
C5

Q.653 There are only two women among 20 persons taking part in a pleasure trip. The 20 persons are divided
into two groups, each group consisting of 10 persons. Then the probability that the two women will be
in the same group is :
(A*) 9/19 (B) 9/38 (C) 9/35 (D) none
[Hint: n(S) = number of ways in which 20 people can be divided into two equal groups
20!
= 10! 10! 2!

n(A) = 18 people can be divided into groups of 10 and 8

18!
= 10! 8!

18! 10! 10! 2 10 . 9 . 2 9

P(E) = 10! 8! . 20! = 20 .19  19 Ans ]

Q.7 A bag contain 5 white, 7 black, and 4 red balls, find the chance that three balls drawn at random are all
white. [Ans. 1/56]

Bansal Classes [1]

Q.8 If four coins are tossed, find the chance that there should be two heads and two tails.
[Ans. 3/8]

Q.9 Thirteen persons take their places at a round table, show that it is five to one against two particular
persons sitting together.

Q.10 In shuffling a pack of cards, four are accidentally dropped, find the chance that the missing cards should
(13) 2 2197
be one from each suit. [Ans. 52 = ]
C4 20825
Q.11 A has 3 shares in a lottery containing 3 prizes and 9 blanks, B has 2 shares in a lottery containing 2 prizes
and 6 blanks. Compare their chances of success. [Ans. 952 to 715]

Q.12 There are three works, one consisting of 3 volumes, one of 4 and the other of one volume. They are placed
3
on a shelf at random, prove that the chance that volumes of the same works are all together is .
140

Q.13 The letter forming the word Clifton are placed at random in a row. What is the chance that the two
vowels come together? [Ans. 2/7]

Q.14 Three bolts and three nuts are put in a box. If two parts are chosen at random, find the probability that
one is a bolt and one is a nut. [Ans. 3/5]

Q.15 There are 'm' rupees and 'n' ten nP's, placed at random in a line. Find the chance of the extreme coins
n (n  1)
being both ten nP's. [Ans. ]
(m  n )(m  n  1)

Q.16 A fair die is tossed. If the number is odd, find the probability that it is prime. [Ans. 2/3]

Q.17 Three fair coins are tossed. If both heads and tails appear, determine the probability that exactly one
head appears. [Ans. 1/2]

Q.18 3 boys and 3 girls sit in a row. Find the probability that (i) the 3 girls sit together. (ii) the boys are girls sit
in alternative seats. [Ans. 1/5, 1/10]

Q.19 A coin is biased so that heads is three times as likely to appear as tails. Find P (H) and P (T).
[Ans. 3/4, 1/4]

Q.20 In a hand at "whist" what is the chance that the 4 kings are held by a specified player?
4
C 4 · 48C 9
[Ans. 52 ]
C13

Bansal Classes [2]

 BANSAL CLASSES
Target IIT JEE 2007 Daily Practice Problems
CLASS : XII (ABCD) DPP ON PROBABILITY DPP. NO.- 2
After 2 nd Lecture
Q.1 Given two independent events A, B such that P (A) = 0.3, P (B) = 0.6. Determine
(i) P (A and B) (ii) P (A and not B) (iii) P (not A and B)
(iv) P (neither A nor B) (v) P (A or B)
[Ans. (i) 0.18, (ii) 0.12, (iii) 0.42, (iv) 0.28, (v) 0.72]

Q.2 A card is drawn at random from a well shuffled deck of cards. Find the probability that the card is a
(i) king or a red card (ii) club or a diamond (iii) king or a queen
(iv) king or an ace (v) spade or a club (vi) neither a heart nor a king.
7 1 2 2 1 9
[Ans. (i) , (ii) , (iii) , (iv) , (v) , (vi) ]
13 2 13 13 2 13

Q.3 A coin is tossed and a die is thrown. Find the probability that the outcome will be a head or a number
2
greater than 4. [Ans. ]
3
~
Q.4 Let A and B be events such that P ( A ) = 4/5, P(B) = 1/3, P(A/B) = 1/6, then
(a) P(A  B) ; (b) P(A  B) ; (c) P(B/A) ; (d) Are A and B independent?
[Ans. (a) 1/18, (b) 43/90, (c) 5/18, (d) NO]
P(A  B) 1 P ( B) 1
[Sol. (a) P(A/B) =   P(A  B) = = Ans.]
P(B) 6 6 18
1 1 1 18  30  5 43
(b) P(A  B) = + – = = Ans.
5 3 18 90 90
P(B  A) 1 5 5
(c) P(B/A) = = · = Ans.
P(A) 18 1 18
1 1
(d) P(A) · P(B) = · = 15  P(A  B). A & B are not independent ]
5 3
1 1 1
Q.5 If A and B are two events such that P (A) = , P (B) = and P (A and B) = , find
4 2 8
5 3
(i) P (A or B), (ii) P (not A and not B) [Ans. (i) , (ii) ]
8 8

Q.6168 A 5 digit number is formed by using the digits 0, 1, 2, 3, 4 & 5 without repetition. The probability that the
number is divisible by 6 is :
(A) 8 % (B) 17 % (C*) 18 % (D) 36 %
[Hint: Number should be divisible by 2 and 3.
n(S) = 5 · 5! ; n (A) : reject '0' = 2 · 4!
reject 3, 4! + 2 · 3 · 3!
Total n(A) = 3 · 4! + 6 · 3! = 18 · 3!
18 · 3!
 p = 5 · 5! = 18% ]

Bansal Classes [3]

Q.754 An experiment results in four possible out comes S1, S2, S3 & S4 with probabilities p1, p2, p3 & p4
respectively. Which one of the following probability assignment is possbile.
[Assume S1 S2 S3 S4 are mutually exclusive]
(A) p1 = 0.25 , p2 = 0.35 , p3 = 0.10 , p4 = 0.05
(B) p1 = 0.40 , p2 =  0.20 , p3 = 0.60 , p4 = 0.20
(C) p1 = 0.30 , p2 = 0.60 , p3 = 0.10 , p4 = 0.10
(D*) p1 = 0.20 , p2 = 0.30 , p3 = 0.40 , p4 = 0.10

Q.888 In throwing 3 dice, the probability that atleast 2 of the three numbers obtained are same is
(A) 1/2 (B) 1/3 (C*) 4/9 (D) none
[Hint : P(E) = 1 – P(all different) = 1 – (6/6) · (5/6) · (4/6) = 1 – (120/216) = 4/9 ]

Q.9173 There are 4 defective items in a lot consisting of 10 items. From this lot we select 5 items at random. The
probability that there will be 2 defective items among them is
1 2 5 10
(A) (B) (C) (D*)
2 5 21 21

[Hint: [12th (26-12-2004)]

4
C 2 · 6 C3 10
p= 10 =  (D) ]
C5 21
Q.10134 From a pack of 52 playing cards, face cards and tens are removed and kept aside then a card is drawn
at random from the ramaining cards. If
A : The event that the card drawn is an ace
H : The event that the card drawn is a heart
S : The event that the card drawn is a spade
then which of the following holds ?
(A*) 9 P(A) = 4 P(H) (B) P(S) = 4P (A  H)
(C) 3 P(H) = 4 P(A S) (D) P(H) = 12 P(A S)

face cards
[Hint: 52   36 ;
10's

1 1 1 1 1 1
P(A) = ; P(H) = ; P(S) = ; P(A  H) = ; P(A  S) = ; P (A  S) = ]
9 4 4 36 36 3

Q.11220 6 married couples are standing in a room. If 4 people are chosen at random, then the chance that exactly
one married couple is among the 4 is :
16 8 17 24
(A*) (B) (C) (D)
33 33 33 33
[Hint: n(S) = 12C4 = 55 × 9
n(A) = 6C1. 5C2. 22 = 6 × 10 × 4
6  10  4 2.2.4 16
P(E) = 55  9  11. 3  33 Ans]

Bansal Classes [4]

Q.1245 The chance that a 13 card combination from a pack of 52 playing cards is dealt to a player in a game of
bridge, in which 9 cards are of the same suit, is
4 · 13C9 · 39 C 4 4!· 13C9 · 39C 4 13
C9 · 39 C 4
(A*) 52 (B) 52 (C) 52 (D) none
C13 C13 C13

Q.1364 If two of the 64 squares are chosen at random on a chess board, the probability that they have a side in
common is :
(A) 1/9 (B*) 1/18 (C) 2/7 (D) none
4 · 2  6 · 4 · 3  36 · 4
[Hint: n (S) = 64C2 · 2 ; n (A) = .
64 · 63
Alternatively : n (A) = 7 · 8 + 7 · 8 = 112
Ask: Prob that they have a corner in common ]

Q.14182 Two red counters, three green counters and 4 blue counters are placed in a row in random order. The
probability that no two blue counters are adjacent is
7 7 5
(A) (B) (C*) (D) none
99 198 42
[Sol. R R G G G B B B B when counters are alike [14-8-2005, 13th]
9!
n (S) =
2!3!4!
5! 6
n(A) = · C4 |R|R|G|G|G|
3! 2!
5!·15 2!3!4! 6!· 60 60 15 5
 P(A) = · = 9 · 8 · 7 · 6! = 7 · 8 · 9 = =
3!2! 9! 7 · 2 · 9 42
Alternatively : n (S) = 9! R1R1G1G2G3B1B2B3B4
n(A) = 5! · 6C4 · 4! when counters are different
5!· 6 · 5 · 4 · 3 5 · 4 · 3 5
p= = = ]
9! 9 · 8 · 7 42

Q.15 The probabilities that a student will receive A, B, C or D grade are 0.40, 0.35, 0.15 and 0.10 respectively.
Find the probability that a student will receive
(i) not an A grade (ii) B or C grade (iii) at most C grade
[Ans. (i) 0.6, (ii) 0.5, (iii) 0.25]
Q.16 In a single throw of three dice, determine the probability of getting
(i) a total of 5 (ii) a total of at most 5 (iii) a total of at least 5.
1 5 53
[Ans. (i) , (ii) , (iii) ]
36 108 54
Q.17 A die is thrown once. If E is the event "the number appearing is a multiple of 3" and F is the event "the
number appearing is even", find the probability of the event "E and F". Are the events E and F independent?
1 1 1
[Ans. P(E) = , P(F)= , P(E and F) = ; Yes] es]
3 2 6
Q.18 In the two dice experiment, if E is the event of getting the sum of number on dice as 11 and F is the event
of getting a number other than 5 on the first die, find P (E and F). Are E and F independent events?
2 30 1
[Ans. P(E)= , P(F)= , P(E  F) = ; Not independent]
36 36 36
Bansal Classes [5]
Q.19 A natural number x is randomly selected from the set of first 100 natural numbers. Find the probability
100 55 11
that it satisfies the inequality. x + > 50 [Ans: = ]
x 100 20
 1 27 53 
Note: {1, 2, 48, 49, 50, ........ ,100 }  wrong Ans given by students 50 , 50 , 100 
 

Q.20 3 students A and B and C are in a swimming race. A and B have the same probability of winning and each
is twice as likely to win as C. Find the probability that B or C wins. Assume no two reach the winning
point simultaneously.
[Sol. P(C) = p ; P(A) = 2p ; P(B) = 2p
 5p = 1  p = 1/5
2 1 3
P(B or C) = P(B) + P(C) =   ]
5 5 5
Q.21 A box contains 7 tickets, numbered from 1 to 7 inclusive. If 3 tickets are drawn from the box, one at a
time, determine the probability that they are alternatively either odd-even-odd or even-odd-even.
4·3·3  3·4·2 6 2
[Ans. p = = = ]
7·6·5 210 7
Q.22 5 different marbles are placed in 5 different boxes randomly. Find the probability that exactly two boxes
remain empty. Given each box can hold any number of marbles.
[Sol. n(S) = 55 ; For computing favourable outcomes.
2 boxes which are to remain empty, can be selected in 5C2 ways and 5 marbles can be placed in the
remaining 3 boxes in groups of 221 or 311 in
 5! 5!  5
3!    = 150 ways  n (A) = C2· 150
 2!2!2! 3!2!
150 60 12
Hence P(E) = 5C2 · 5
= = Ans.]
5 125 25
Q.23132 South African cricket captain lost the toss of a coin 13 times out of 14. The chance of this happening was
7 1 13 13
(A*) 13 (B) 13 (C) 14 (D)
2 2 2 213
14
[Hint: L and W can be filled at 14 places in 2 ways.
 n(S) = 214.
Now 13 L's and 1W can be arranged at 14 places in 14 ways.
Hence n(A) = 14
14 7
 p= = ] [14-8-2005, 13th]
214 213
Q.24 There are ten prizes, five A's, three B's and two C's, placed in identical sealed envelopes for the top ten
contestants in a mathematics contest. The prizes are awarded by allowing winners to select an envelope
at random from those remaining. When the 8th contestant goes to select the prize, the probability that the
remaining three prizes are one A, one B and one C, is
(A*) 1/4 (B) 1/3 (C) 1/12 (D) 1/10
10
[Hint: n(S) = C7 = 120 [08-01-2006, 12 & 13]
n(A) = 5C4 · 3C2 · 2C1
5 ·3· 2 1
P(E) = = Ans. ]
120 4

Bansal Classes [6]

 BANSAL CLASSES MATHEMATICS
Target IIT JEE 2007 Daily Practice Problems
CLASS : XII (ABCD) DPP ON PROBABILITY DPP. NO.- 3
After 3rd
Lecture
Q.133 Whenever horses a, b, c race together, their respective probabilities of winning the race are 0.3, 0.5 and
0.2 respectively. If they race three times the probability that “the same horse wins all the three races” and
the probablity that a, b, c each wins one race, are respectively
8 9 16 3 12 15 10 8
(A*) ; (B) , (C) ; (D) ;
50 50 100 100 50 50 50 50
[Sol. P(a) = 0.3 ; P(b) = 0.5 ; P(c) = 0.2  a, b, c are exhaustive
P(same horse wins all the three races) = P(aaa or bbb or ccc)
27  125  8 160 4
= (0.3)3 + (0.5)3 + (0.2)3 =  =
1000 1000 25
P(each horse wins exactly one race)
9
= P(abc or acb or bca or bac or cab or cba) = 0.3× 0.5 × 0.2 × 6 = 0.18 = ]
50

Q.263 Let A & B be two events. Suppose P(A) = 0.4 , P(B) = p & P(A  B) = 0.7. The value of p for which
A & B are independent is :
(A) 1/3 (B) 1/4 (C*) 1/2 (D) 1/5
[Sol. P(A  B) = P(A) + P(B) – P(A) · P(B)
0.7 = 0.4 + p – 0.4p
1
 0.6p = 0.3  p = ]
2

Q.378 A & B are two independent events such that P ( A ) = 0.7, P ( B ) = a & P(A  B) = 0.8, then, a =
(A) 5/7 (B*) 2/7 (C) 1 (D) none

Q.472 A pair of numbers is picked up randomly (without replacement) from the set
{1, 2, 3, 5, 7, 11, 12, 13, 17, 19}. The probability that the number 11 was picked given that the sum of
the numbers was even, is nearly :
(A) 0.1 (B) 0.125 (C*) 0.24 (D) 0.18
7
P (A  B) C1 7
[Hint : P A B = P (B) = 8 = ; A : 11 is picked , B : sum is even ]
C 2 1 29
Q.542 For a biased die the probabilities for the diffferent faces to turn up are given below :
Faces : 1 2 3 4 5 6
Probabilities : 0.10 0.32 0.21 0.15 0.05 0.17
The die is tossed & you are told that either face one or face two has turned up. Then the probability that
it is face one is :
(A) 1/6 (B) 1/10 (C) 5/49 (D*) 5/21
P A  ( AB)  P (A ) 0.10
[Hint : P(A/A  B) = PAB = = ]
P (A )  P (B) 0.10  0.32

Bansal Classes [7]

Q.6170 A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only.
The probability that the determinant chosen has the value non negative is :
(A) 3/16 (B) 6/16 (C) 10/16 (D*) 13/16
[Hint : 1  P (Determinant has negative value)
3 13  1 1 0 1 0 1 
1 =  1 0 ; 1 1 ; 1 0  ]
16 16  

Q.799 15 coupons are numbered 1, 2, 3,..... , 15 respectively. 7 coupons are selected at random one at a time
with replacement. The probability that the largest number appearing on a selected coupon is 9 is :
6 7 7
9 8 3 9 7  87
(A)   (B)   (C)   (D*)
 16   15  5 157
7
[Hint: n(S) = × × × × × × × = 15 ; came of the number must be 9 ]

Q.8229 A card is drawn & replaced in an ordinary pack of 52 playing cards. Minimum number of times must a
card be drawn so that there is atleast an even chance of drawing a heart, is
(A) 2 (B*) 3 (C) 4 (D) more than four
1 3 1 3 3 1
[Hint:  .  . . ]
4 4 4 4 4 4

Q.9 An electrical system has open-closed switches

S1, S2 and S3 as shown.

The switches operate independently of one another and the current will flow from A to B either if S1 is
1
closed or if both S2 and S3 are closed. If P(S1) = P(S2) = P(S3) = , find the probability that the circuit
2
will work.
1 5
[Sol. P(S1) = P(S2) = P(S3) = [Ans. ]
2 8
E : event that the current will flow.
1 1 1 5
P(E)= P (S2  S3 ) or S1  = P (S2  S3) + P (S3) – P (S1  S2  S3 ) = + – = ]
4 2 8 8
Q.10 A certain team wins with probability 0.7, loses with probability 0.2 and ties with probability 0.1. The
team plays three games. Find the probability
(i) that the team wins at least two of the games, but lose none.
(ii) that the team wins at least one game. [Ans. (i) 0.49 ; (ii) 0.973 ]
[Sol. P (W) = 0.7 ; P (L) = 0.2 ; P (T) = 0.1
E : winning at least 2 games but lose none
P (E) = P (W W T or W T W or T W W or W W W)
= 3 × 0.7 × 0.7 × 0.1 + (0.7)3 = 0.7 × 0.7 [0.3 + 0.7] = 0.49
F : wining at least 1 game
A = L or T  P (A) = 0.3 ; P (F) = 1 – P (A A A) = 1 – (0.3)3 = 1 – 0.027 = 0.973 ]

Bansal Classes [8]

Q.11 An integer is chosen at random from the first 200 positive integers. Find the probability that the integer is
divisible by 6 or 8. 1
[Sol. n (S) = 200 [Ans. ]
4
n (A) = divisible by 6 or 8
n (6) = 6 + 12 + ..... + 198 = 33
n (8) = 8 + 16 + ......+ 200 = 25
n (6  8) = 24 + 48 + ..... + 192 = 8
n (A) = 33 + 25 – 8 = 50
1
P (A) = ]
4

Q.12 A clerk was asked to mail four report cards to four students. He addresses four envelops that unfortunately
paid no attention to which report card be put in which envelope. What is the probability that exactly one
of the students received his (or her) own card?
8 1
[Ans. = ]
24 3
Q.13 Find the probability of at most two tails or at least two heads in a toss of three coins.
7
[Sol. A = at most two tails n (s) – {T T T} [Ans. ]
8
B = at least two heads H H H, T H H, H T H, H H T
7 4 4
P (A) = ; P (B) = ; P (A  B) =
8 8 8
7 4 4 7
P (A  B) = + – = Ans. ]
8 8 8 8

Q.14 What is the probability that in a group of

(i) 2 people, both will have the same date of birth.
(ii) 3 people, at least 2 will have the same date of birth.
Assume the year to be ordinarry consisting of 365 days.
[Sol.(i) A : both have the same date of birth
1 1 1 1 1
P (A) = ·  ·  .......365 times = Ans.
365 365 365 365 365
(ii) B : at least 2 in a group of 3 will have the same birthday
P (B) = 1 – P (all 3 have different birthday )
364 363 364  363
=1–1× · =1– Ans. ]
365 365 (365) 2
2
Q.15 The probability that a person will get an electric contract is and the probability that he will not get
5
4 2
plumbing contract is . If the probability of getting at least one contract is , what is the probability that
7 3
he will get both ?
2 4 3
[Sol. P (E) = ; P (F) = P (plumbing) = 1 – =
5 7 7
P (E  F) = P (E) + P (F) – P (E  F)
2 2 3 17
= + –x  x= Ans. ]
3 5 7 105

Bansal Classes [9]

Q.16 Five horses compete in a race. John picks two horses at random and bets on them. Find the probability
that John picked the winner. Assume dead heal. 4
[Sol. n (S) = 5C2 = 10   p 2 [Ans. ]
10
n (A) = 1 · 4C1 = 4  5 ]

Q.17 Two cubes have their faces painted either red or blue. The first cube has five red faces and one blue face.
When the two cubes are rolled simultaneously, the probability that the two top faces show the same
colour is 1/2. Number of red faces on the second cube, is
(A) 1 (B) 2 (C*) 3 (D) 4
[Sol. nd
Let the number of red faces on the 2 cube = x [08-01-2006, 12 & 13]
number of blue faces = (6 – x)
P (R R or B B) = 1/2
5 x 1 6x 1
· + · =
6 6 6 6 2
5x + 6 – x = 18
4x = 12  x=3 Ans. ]

Q.18 A H and W appear for an interview for two vaccancies for the same post.
P(H) = 1/7 ; P(W) = 1/5. Find the probability of the events
(a) Both are selected (b) only one of them is selected (c) none is selected.
1 2 24
[Ans. , , ]
35 7 35
Q.19 A bag contains 6R, 4W and 8B balls. If 3 balls are drawn at random determine the probability of the
event
(a) all 3 are red ; (b) all 3 are black ; (c) 2 are white and 1 is red ;
(d) at least 1 is red ; (e) 1 of each colour are drawn
(f) the balls are drawn in the order of red, white, blue.
5 7 3 149 4 2
[Ans. (a) , (b) , (c) , (d) , (e) , ]
204 102 68 204 17 51
Q.20 The odds that a book will be favourably reviewed by three independent critics are 5 to 2, 4 to 3, and 3
to 4 respectively. What is the probability that of the three reviews a majority will be favourable?
209
[Ans. ]
343
Q.21 In a purse are 10 coins, all five nP's except one which is a rupee, in another are ten coins all five nP's.
Nine coins are taken from the former purse and put into the latter, and then nine coins are taken from the
latter and put into the former. Find the chance that the rupee is still in the first purse.
10
[Ans. ]
19
Q.22 A, B, C in order cut a pack of cards, replacing them after each cut, on condition that the first who cuts
16 12 9
a spade shall win a prize. Find their respective chances. [Ans. , , ]
37 37 37
Q.23 A and B in order draw from a purse containing 3 rupees and 4 nP's, find their respective chances of first
22 13
drawing a rupee, the coins once drawn not being replaced. [Ans. , ]
35 35

Bansal Classes [10]

 BANSAL CLASSES MATHEMATICS
Target IIT JEE 2007 Daily Practice Problems
CLASS : XII (ABCD) DPP ON PROBABILITY DPP. NO.- 4
After 4 th Lecture
Q.137 There are n different gift coupons, each of which can occupy N(N > n) different envelopes, with the
same probability 1/N
P1: The probability that there will be one gift coupon in each of n definite envelopes out of N given envelopes
P2: The probability that there will be one gift coupon in each of n arbitrary envelopes out of N given envelopes
Consider the following statements n! N!
(i) P1 = P2 (ii) P1 = n (iii) P2 = n
N N ( N  n )!
n! N!
(iv) P2 = n (v) P1 = n
N ( N  n )! N
Now, which of the following is true
(A) Only (i) (B*) (ii) and (iii) (C) (ii) and (iv) (D) (iii) and (v)
n !
[Sol. From the given data n (S) = Nn ; n (A) = n !  P1 = n
N
n!
P1 = n Since the n different gift coupans can be placed in the n definite (Out of N)
N
envelope in nPn = n ! ways
N! N
P2 = n As n arbitrary envelopes out of N given envelopes can be chosen in Cn
(N  n)! N
ways and the n gift coupans can occupy these envelopes in n ! ways. ]

N!
= n ]
N ( N  n )!
Q.261 The probability that an automobile will be stolen and found withing one week is 0.0006. The probability that
an automobile will be stolen is 0.0015. The probability that a stolen automobile will be found in one week is
(A) 0.3 (B*) 0.4 (C) 0.5 (D) 0.6
[Hint: P (S  F) = 0.0006, where S : moter cycle is stolen ; F : moter cycle found
P (S) = 0.0015
P(F  S) 6 104 2
P (F/S) = = =  (B) ]
P(S) 15 104 5
Q.325 One bag contains 3 white & 2 black balls, and another contains 2 white & 3 black balls. A ball is drawn
from the second bag & placed in the first, then a ball is drawn from the first bag & placed in the second.
When the pair of the operations is repeated, the probability that the first bag will contain 5 white balls is:
(A) 1/25 (B) 1/125 (C*) 1/225 (D) 2/15

[Hint:

2 2 1 1 1
P(E) = P [W B W B] = · · · = ]
5 6 5 6 225
Q.475 A child throws 2 fair dice. If the numbers showing are unequal, he adds them together to get his final
score. On the other hand, if the numbers showing are equal, he throws 2 more dice & adds all 4 numbers
showing to get his final score. The probability that his final score is 6 is:
145 146 147 148
(A) (B) (C) (D*)
1296 1296 1296 1296
[Hint: P (6) = (51, 15, 24, 42) or 11 & (22 or 13 or 31) or (22 & 11) ]
Bansal Classes [11]
Q.5118 A person draws a card from a pack of 52 cards, replaces it & shuffles the pack. He continues doing this
till he draws a spade. The probability that he will fail exactly the first two times is :
(A) 1/64 (B*) 9/64 (C) 36/64 (D) 60/64
[Hint: P(E) = P(FFS) = 3/4.3/4.1/4]

Q.6 Indicate the correct order sequence in respect of the following :

I. If the probability that a computer will fail during the first hour of operation is 0.01, then if we turn
on 100 computers, exactly one will fail in the first hour of operation.
II. A man has ten keys only one of which fits the lock. He tries them in a door one by one discarding
the one he has tried. The probability that fifth key fits the lock is 1/10.
III. Given the events A and B in a sample space. If P(A) = 1, then A and B are independent.
IV. When a fair six sided die is tossed on a table top, the bottom face can not be seen. The probability
that the product of the numbers on the five faces that can be seen is divisible by 6 is one.
(A) FTFT (B*) FTTT (C) TFTF (D) TFFF
100
100C
 1   99 
[Hint: I. P(X = 1) =  
1  100  100
 [18-12-2005, 12 & 13]
 
II. Every key that fits have the same probability = 1/10
III. Consider P(A  B) = P(A) + P(B) – P(A  B)
but P(A  B) = P(A) = 1
1 = 1 + P(B) – P(A  B)
P(A  B) = P(B) = P(B) · P(A) ( P(A) = 1 )
IV. Each product 1 2 3 4 5 ; 1 2 3 4 6 ; 1 2 3 5 6 ; 1 2 4 5 6 ; 1 3 4 5 6 ; 2 3 4 5 6 is divisible by six.]

Q.7156 An unbaised cubic die marked with 1, 2, 2, 3, 3, 3 is rolled 3 times. The probability of getting a total
score of 4 or 6 is
16 50 60
(A) (B*) (C) (D) none
216 216 216
[ Hint: 1 , 2 , 2 , 3 , 3 , 3 (thrown 3 times)
1 2 3
P(1) = ; P(2) = ; P(3) =
6 6 6
P(S) = P(4 or 6) = P( 112 (3 cases) or 123 (6 cases) or 222 )
1 1 2 1 2 3 2 2 2 6  36  8 50 25
= 3. . .  6 . .  . . =   ]
6 6 1 6 6 1 6 6 6 216 216 108

Q.8200 A bag contains 3 R & 3 G balls and a person draws out 3 at random. He then drops 3 blue balls into the
bag & again draws out 3 at random. The chance that the 3 later balls being all of different colours is
(A) 15% (B) 20% (C*) 27% (D) 40%

3
C1 · 3C 2 2
C1 · 1C1 · 3C1 3
C 2 · 3C1 1
C1 · 2 C1 · 3C1
[Sol. ; 6
· 6 + 6
· 6 ]
C2 C3 C3 C3

Q.9169 A biased coin with probability P, 0 < P < 1, of heads is tossed until a head appears for the first time. If the
probability that the number of tosses required is even is 2/5 then the value of P is
(A) 1/4 (B) 1/6 (C*) 1/3 (D) 1/2

Bansal Classes [12]

Q.10202 If a, b  N then the probability that a2 + b2 is divisible by 5 is
9 7 11 17
(A*) (B) (C) (D)
25 18 36 81
[Hint : Square of a number ends in 0, 1, 4, 5, 6 and 9 favourable ordered pairs of
(a2, b2) can be (0, 0) ; (5, 5) ; (1, 4) , (4, 1) ; (1, 9) , (9, 1) ; (4, 6) , (6, 4) ;
(6, 9) , (9, 6) and P(0) = 1/10 = P(5) ; P(1) = P(4) = P(6) = P(9) = 2/10 ]
Q.1152 In an examination, one hundred candidates took paper in Physics and Chemistry. Twenty five candidates
failed in Physics only. Twenty candidates failed in chemistry only. Fifteen failed in both Physics and
Chemistry. A candidate is selected at random. The probability that he failed either in Physics or in
Chemistry but not in both is
9 3 2 11
(A*) (B) (C) (D)
20 5 5 20
th
[ 13 Test (5-12-2004)]
Q.12 In a certain game A's skill is to be B's as 3 to 2, find the chance of A winning 3 games at least out of 5.
3 2 2133
[Hint: odd in favour of A 3 : 2  P(A) = =p ;q= [Ans. ]
5 5 3125
3 2 4 5
5C 3  2 5 3  2 3
P(E) = 3     + C4
·   ·  +   ]
5 5
    5  5 5
Q.13 In each of a set of games it is 2 to 1 in favour of the winner of the previous game. What is the chance that
the player who wins the first game shall wins three at least of the next four? [Ans. 4/9]
2 1 1 2
[Hint: P(W/W) = ; P(L/W) = ; P(W/L) = ; P(L/L) = ]
3 3 3 3
Q.14 A coin is tossed n times, what is the chance that the head will present itself an odd number of times?
n
C1  n C 3  n C5  ....... 2 n 1 1 1
[Hint: P(E) = n = n = 2 ] [Ans. ]
2 2 2
Q.15 A fair die is tossed repeatidly. A wins if it is 1 or 2 on two consecutive tosses and B wins if it is 3, 4, 5 or 6
on two consecutive tosses. The probability that A wins if the die is tossed indefinitely, is
1 5 1 2
(A) (B*) (C) (D)
3 21 4 5
[Sol. Let P(S) = P(1 or 2) = 1/3 [29-10-2005, 12th Jaipur]
P(F) = P(3 or 4 or 5 or 6) = 2/3
P(A wins) = P[( S S or S F S S or S F S F S S or .......) or (F S S or F S F S S or ........)]
12
27
9 1 9 2 9 1 2 3 2 5
= + 2
=  +  = + = =
2 9 7 27 7 7 21 21 21
1 1
9 9
5 16
P (A winning) = ; P (B winning) = Ans. ]
21 21
Q.16 Counters marked 1, 2, 3 are placed in a bag, and one is withdrawn and replaced. The operation being
repeated three times, what is the chance of obtaining a total of 6? [Ans. 7 / 27]
( 6 cases) 6 1 7
[Hint: P(1 2 3 or 2 2 2) = 3 = ]
3 27
Bansal Classes [13]
Q.17 A normal coin is continued tossing unless a head is obtained for the first time. Find the probability that
(a) number of tosses needed are at most 3. 7 1
(b) number of tosses are even. [Ans. (a) , (b) ]
8 3
1 1 1 7
[Sol. (a) P (H or T H or T T H);   
2 4 8 8
14 1 4 1
(b) T H or T T T H or ....... ; P (E) = 1  1 4 =   ]
4 3 3
Q.18 A purse contains 2 six sided dice. One is a normal fair die, while the other has 2 ones, 2 threes, and 2
fives. A die is picked up and rolled. Because of some secret magnetic attraction of the unfair die, there is
75% chance of picking the unfair die and a 25% chance of picking a fair die. The die is rolled and shows
up the face 3. The probability that a fair die was picked up, is
1 1 1 1
(A*) (B) (C) (D)
7 4 6 24
[Sol. N = Normal die ; P(N) = 1/4 [08-01-2006, 12 & 13]
M = magnetic die ; P(M) = 3/4
A = die shows up 3
P(A) = P(A  N) + P(A  M)
= P(N) P(A/N) + P(M) · P(A/M)
1 1 3 2 7
= · + · =
4 6 4 6 24
P( N  A) 1 4 · 1 6  1
P(N/A) = = 7 24 = Ans. ]
P (A ) 7
Q.19 Before a race the chance of three runners, A, B, C were estimated to be proportional to 5, 3, 2, but
during the race A meets with an accident which reduces his chance to 1/3. What are the respective
chance of B and C now? [Ans. B = 2/5 ; C = 4/15]
Q.20 A fair coin is tossed a large number of times. Assuming the tosses are independent which one of the
following statement, is True?
(A) Once the number of flips is large enough, the number of heads will always be exactly half of the total
number of tosses. For example, after 10,000 tosses one should have exactly 5,000 heads.
(B*) The proportion of heads will be about 1/2 and this proportion will tend to get closer to 1/2 as the
number of tosses inreases
(C) As the number of tosses increases, any long run of heads will be balanced by a corresponding run of
tails so that the overall proportion of heads is exactly 1/2
(D) All of the above [29-10-2005, 12th]
Q.21 A and B each throw simultaneously a pair of dice. Find the probability that they obtain the same score.
73
Hint: [ P [ (2&2) or (3&3) or (4&4) ...] [ Ans: ]
648
Q.22 A is one of the 6 horses entered for a race, and is to be ridden by one of two jockeys B or C. It is 2 to
1 that B rides A, in which case all the horses are equally likely to win; if C rides A, his chance is trebled,
what are the odds against his winning? [Ans. 13 to 5]

Bansal Classes [14]

 Direction for Q.23 to Q.25
Let S and T are two events defined on a sample space with probabilities
P(S) = 0.5, P(T) = 0.69, P(S/T) = 0.5
Q.23 Events S and T are:
(A) mutually exclusive (B*) independent
(C) mutually exclusive and independent (D) neither mutually exclusive nor independent
Q.24 The value of P(S and T)
(A*) 0.3450 (B) 0.2500 (C) 0.6900 (D) 0.350
Q.25 The value of P(S or T)
(A) 0.6900 (B) 1.19 (C*) 0.8450 (D) 0
P(S  T) P(S  T )
[Sol. P(S/T) =  0.5 =  P(S  T) = 0.5 × 0.69 = P(S)P(T)
P (T ) 0.69
 S and T are independent Ans.
 P (S and T) = P (S) · P(T) = 0.69 × 0.5 = 0.345 Ans.
P (S or T) = P (S) + P (T) – P (S  T) = 0.5 + 0.69 – 0.345 = 0.8450 Ans. ]
[29-10-2005, 12th]

Bansal Classes [15]

 BANSAL CLASSES MATHEMATICS
Target IIT JEE 2007 Daily Practice Problems
CLASS : XII (ABCD) DPP ON PROBABILITY DPP. NO.- 5
After 5 th Lecture
Q.1113 The first 12 letters of the english alphabets are written down at random. The probability that there are
4 letters between A & B is :
(A) 7/33 (B) 12/33 (C) 14/33 (D*) 7/66
10
[Hint: n(S) = 12 ! ; n(A) = C4 · 4! 2! 7!
 p = 7 / 66 ]
Q.2223 Events A and C are independent. If the probabilities relating A, B and C are P (A) = 1/5;
P (B) = 1/6 ; P (A  C) = 1/20 ; P (B  C) = 3/8 then
(A*) events B and C are independent
(B) events B and C are mutually exclusive
(C) events B and C are neither independent nor mutually exclusive
(D) events B and C are equiprobable
[Hint: P (A  C) = P (A) · P (C)
1 1 1
= · P (C)  P(C) =
20 5 4
1 1 3 1 1
now P (B  C) = + – P (B  C), hence P(BC) = – = = P(B) · P(C)  (A) ]
6 4 8 3 24
Q.3 Assume that the birth of a boy or girl to a couple to be equally likely, mutually exclusive, exhaustive and
independent of the other children in the family. For a couple having 6 children, the probability that their
"three oldest are boys" is
20 1 2 8
(A) (B) (C) (D*)
64 64 64 64
[Hint: E : B1 B2 B3 × × × where × means B or G [27-11-2005, 12th]
1 1 1 1 8
P(E) = · · = = Ans. ]
2 2 2 8 64
Q.4 A and B play a game. A is to throw a die first, and is to win if he throws 6, If he fails B is to throw, and
to win if he throws 6 or 5. If he fails, A is to throw again and to win with 6 or 5 or 4, and so on, find the
169 155
chance of each player. [Ans. A : ;B: ]
324 324
Q.569 Box A contains 3 red and 2 blue marbles while box B contains 2 red and 8 blue marbles. A fair coin is
tossed. If the coin turns up heads, a marble is drawn from A, if it turns up tails, a marble is drawn from
bag B. The probability that a red marble is chosen, is
1 2 3 1
(A) (B*) (C) (D)
5 5 5 2

[Hint: ;
R = event that a red marble is drawn
P (R) = P (R  H) + P (R  T)
= P (H) P (R / H) + P (T) · P (R / T)
1 3 2  8 1 2
=    = · = ]
2  5 10  10 2 5
Bansal Classes [17]
Q.674 A examination consists of 8 questions in each of which one of the 5 alternatives is the correct one. On the
assumption that a candidate who has done no preparatory work chooses for each question any one of
the five alternatives with equal probability, the probability that he gets more than one correct answer is
equal to :
(A) (0.8)8 (B) 3 (0.8)8 (C) 1  (0.8)8 (D*) 1  3 (0.8)8
1
[Hint: p = = 0.2 ; q = 0.8 ; P (E) = 1 – P (0 or 1) ]
5
Q.7181 The germination of seeds is estimated by a probability of 0.6. The probability that out of 11 sown seeds
exactly 5 or 6 will spring is :
11 11
11
C5 . 65 C 6 35 25
   5
(A*) (B) (C) 11C   (D) none of these
510 511 5  6

Q.8152 The probability of obtaining more tails than heads in 6 tosses of a fair coins is :
(A) 2/64 (B*) 22/64 (C) 21/64 (D) none
6
C 4  6 C 5  6 C 6 22
[Hint : P (4 or 5 or 6) =  ]
64 64
Q.9192 An instrument consists of two units. Each unit must function for the instrument to operate. The reliability
of the first unit is 0.9 & that of the second unit is 0.8. The instrument is tested & fails. The probability that
"only the first unit failed & the second unit is sound" is :
(A) 1/7 (B*) 2/7 (C) 3/7 (D) 4/7
[ Hint: A : the instrument has failed
B1 : first unit fails and second is healthy
B2 : first unit healthy and second unit fails
B3 : both fails
B4 : both healthy
P(B1) = 0.1 × 0.8 = 0.08
P(B2) = 0.2 × 0.9 = 0.18
P(B3) = 0.1 × 0.2 = 0.02
P(B4) = 0.9 × 0.8 = 0.72

Now compute P(B1/A) ]

Q.10196 Lot A consists of 3G and 2D articles. Lot B consists of 4G and 1D article. A new lot C is formed by
taking 3 articles from A and 2 from B. The probability that an article chosen at random from C is
defective, is
1 2 8
(A) (B) (C*) (D) none
3 5 25
3 3
[Hint : A = event that the item came from lot A ; P(A) = =
3 2 5
B = item came from B ; P (B) = 2/5
D = item from mixed lot ' C ' is defective
P(D) = P (D  A) + P (D  B)
= P(A). P(D/A) + P(B). P(D/A)
3 2 2 1 8
=     Ans. ]
5 5 5 5 25

Bansal Classes [18]

Q.1128 A die is weighted so that the probability of different faces to turn up is as given :
Number 1 2 3 4 5 6
Probability 0.2 0.1 0.1 0.3 0.1 0.2
If P (A / B) = p1 and P (B / C) = p2 and P (C / A) = p3 then the values of p1, p2, p3 respectively are
Take the events A, B & C as A = {1, 2, 3}, B = {2, 3, 5} and C = {2, 4, 6}
2 1 1 1 1 1 1 1 1 2 1 1
(A) , , (B) , , (C) , , (D*) , ,
3 3 4 3 3 6 4 3 6 3 6 4

Q.12 If mn coins have been distributed into m purses, n into each find
(1) the chance that two specified coins will be found in the same purse, and
(2) what the chance becomes when r purses have been examined and found not to contain either of
the specified coins.
n 1 n 1
[Ans. (1) , (2) ]
mn  1 mn  rn  1
Q.13 A box has four dice in it. Three of them are fair dice but the fourth one has the number five on all of its
faces. A die is chosen at random from the box and is rolled three times and shows up the face five on all
the three occassions. The chance that the die chosen was a rigged die, is
216 215 216
(A) (B) (C*) (D) none
217 219 219
3 normal die
[Sol. 4 [27-11-2005, 12th]
1 rigged die
A : die shows up the face 5
B1 : it is a rigged die ; P(B1) = 1/4
B2 : it is a normal die ; P(B2) = 3/4
1
P(A/B1) = 1 ; P(A/B2) =
216
1
·1 216
P(B1/A) = 4 = Ans. ]
1 3 1 219
·1  ·
4 4 216
Q.14 On a Saturday night 20% of all drivers in U.S.A. are under the influence of alcohol. The probability that
a driver under the influence of alcohol will have an accident is 0.001. The probability that a sober driver
will have an accident is 0.0001. If a car on a saturday night smashed into a tree, the probability that the
driver was under the influence of alcohol, is
(A) 3/7 (B) 4/7 (C*) 5/7 (D) 6/7
[Hint: A : car met with an accident [29-10-2005, 12th Jaipur]
B1: driver was alcoholic, P(B1) = 1/5
B2: driver was sober, P(B2) = 4/5
P(A/B1) = 0.001; P(A/B2) = 0.0001
(.2)(.001)
P(B1/A) = = 5/7 Ans.]
(.2)(.001)  (.8)(.0001)

Bansal Classes [19]

 Direction for Q.15 to Q.17 (3 Questions)
A JEE aspirant estimates that she will be successful with an 80 percent chance if she studies 10 hours per
day, with a 60 percent chance if she studies 7 hours per day and with a 40 percent chance if she studies
4 hours per day. She further believes that she will study 10 hours, 7 hours and 4 hours per day with
probabilities 0.1, 0.2 and 0.7, respectively
Q.15 The chance she will be successful, is
(A) 0.28 (B) 0.38 (C*) 0.48 (D) 0.58
Q.16 Given that she is successful, the chance she studied for 4 hours, is
6 7 8 9
(A) (B*) (C) (D)
12 12 12 12
Q.17 Given that she does not achieve success, the chance she studied for 4 hour, is
18 19 20 21
(A) (B) (C) (D*)
26 26 26 26
[Sol. A : She get a success [18-12-2005, 12 & 13]
T : She studies 10 hrs : P(T) = 0.1
S : She studies 7 hrs : P(S) = 0.2
F : She studies 4 hrs : P(F) = 0.7
P(A/T) = 0.8 ; P(A/S) = 0.6 ; P(A/F) = 0.4
P(A) = P(A  T) + P (A  S) + P (A  F)
= P(T) · P(A/T) + P(S) · P(A/S) + P(F) · P(A/F)
= (0.1)(0.8) + (0.2)(0.6) + (0.7)(0.4)
= 0.08 + 0.12 + 0.28 = 0.48 Ans.(15)
P(F  A) (0.7)(0.4) 0.28 7
P(F/A) = = = = Ans.(16)
P ( A) 0.48 0.48 12

P(F  A ) P( F)  P( F  A) (0.7)  0.28 0.42 21

P(F/ A ) = = = = = Ans.(17) ]
P( A ) 0.52 0.52 0.52 26

Q.18 There are four balls in a bag, but it is not known of what colour they are ; one ball is drawn at random and
found to be white. Find the chance that all the balls are white. Assume all number of white ball in the bag
to be equally likely. [Ans. 2/5]
Q.19 A letter is known to have come either from London or Clifton. On the postmark only the two consecutive
letters ON are legible. What is the chance that it came from London? [Ans. 12/17]
Q.20 A purse contains n coins of unknown value, a coin drawn at random is found to be a rupee, what is the
chance that is it the only rupee in the purse? Assume all numbers of rupee coins in the purse is equally likely.
2
[Ans. ]
n (n  1)
Q.21 One of a pack of 52 cards has been lost, from the remainder of the pack two cards are drawn and are
found to be spades, find the chance that the missing card is a spade. [Ans. 11/50]
Q.22 A, B are two inaccurate arithmeticians whose chance of solving a given question correctly are (1/8) and
(1/12) respectively. They solve a problem and obtained the same result. If it is 1000 to 1 against their
making the same mistake, find the chance that the result is correct. [Ans. 13/14]
[Hint: A : they obtained the same result

Bansal Classes [20]

 1 11
B1 : A  B ; P(B1) = · Now P(A/B1) = 0
18 12 

7 1 
B2 : A  B ; P(B1) = ·  P(A/B2) = 0
8 12

1 1 
B3 : A  B ; P(B3) = · P(A/B3) = 1
8 12 

7 11  7 11 1
B4 : A  B ; P(B4) = · P(A/B4) = · ·
8 12 8 12 1001
1 1
· 13
Now P(B3/A) = 8 12 = ]
1 1 7 11 1 14
·  · ·
8 12 8 12 1001
Q.23 We conduct an experiment where we roll a die 5 times. The order in which the number read out is
important.
(a) What is the total number of possible outcomes of this experiment?
(b) What is the probability that exactly 3 times a "2" appears in the sequence (say event E)?
(c) What is the probability that the face 2 appears at least twice (say event F)?
(d) Which of the following are true : E  F, F  E?
(e) Compute the probabilities : P(E  F), P(E/F), P(F/E)
(f) Are the events E and F independent?
5
C3 · 52 10 · 52
[Hint: (a) 65 ; (b) P(E) = =
65 65
5 5
C1 · 54  55 5
(c) P(F) = 1– P(face two appears exact once or no two) = 1 – ; P(F) = 1 – 2 ·  
65 6
P(E  F) P(E ) P(E  F)
(d) E  F; (e) P(E  F) =P(E) ; P(E/F) = = ; P(F/E) = =1
P(F) P (F) P(E)
(f) No]

Bansal Classes [21]

 BANSAL CLASSES MATHEMATICS
Target IIT JEE 2007 Daily Practice Problems
CLASS : XII (ABCD) DPP ON PROBABILITY DPP. NO.- 6
After 6th Lecture
Q.1 A bowl has 6 red marbles and 3 green marbles. The probability that a blind folded person will draw a red
marble on the second draw from the bowl without replacing the marble from the first draw, is
2 1 1 5
(A*) (B) (C) (D)
3 4 2 8
6R
[Hint: 9 [27-11-2005, 12th]
3G
E : Event that the 2nd drawn marble is red; R : 1st drawn is red; G = 1st drawn is green
P(E) = P(E  R) + P(E  G)
= P(R) · P(E/R) + P(E) · P(E/G)
6 5 3 6 48 2
= · + · = = ]
9 8 9 8 72 3
Q.2114 5 out of 6 persons who usually work in an office prefer coffee in the mid morning, the other always drink
tea. This morning of the usual 6, only 3 are present. The probability that one of them drinks tea is :
(A*) 1/2 (B) 1/12 (C) 25/72 (D) 5/72
[ Hint: 6 persons
Total number of ways in which 3 persons one of which drinks tea and 2 others can be selected
= 1C1 · 5C2 ways
number of ways any 3 can be selected 6C3
5
C2 10 1
 P(E) = 6
  ]
C3 20 2

Q.329/5 Pal’s gardner is not dependable , the probability that he will forget to water the rose bush is 2/3. The
rose bush is in questionable condition . Any how if watered, the probability of its withering is 1/2 & if
not watered then the probability of its withering is 3/4. Pal went out of station & after returning he finds
that rose bush has withered. What is the probability that the gardner did not water the rose bush.
[Ans: 3/4 ]
[Sol. A = Rose bush has withered
B1 = Gardener did not water the rose bush P(B1) = 2/3
B2 = Gardener watered the rose bush P(B2) = 1/3
3 1
P(A/B1) = ; P(A/B2) =
4 2
2 3
P( B1 ).P( A / B1 ) . 6 3
P(B1/A) = P( B ).P( A / B )  P( B ).P( A / B ) = 3 4 =  Ans ]
1 1 2 2 2 3 1 1 62 4
.  .
3 4 3 2
Q.466 The probability that a radar will detect an object in one cycle is p. The probability that the object will be
detected in n cycles is :
(A) 1  pn (B*) 1  (1  p)n (C) pn (D) p(1 – p)n–1

Bansal Classes [23]

[Hint: P (A) = p
p (object is not detected in one cycle) = 1 – p
p (object is not detected in n cycle) = (1 – p)n
p (object will be detected) = 1 – (1 – p)n ]
Q.5139 Nine cards are labelled 0, 1, 2, 3, 4, 5, 6, 7, 8. Two cards are drawn at random and put on a table in a
successive order, and then the resulting number is read, say, 07 (seven), 14 (fourteen) and so on. The
probability that the number is even, is
5 4 1 2
(A*) (B) (C) (D)
9 9 2 3
[Sol. n (S) = number of ways in which two numbers are drawn in a definite order
= 9 × 8 = 72
n (A) = any one number from 0, 2, 4, 6, 8 can be taken in 5C1 ways and any one can be taken from the
remaining 8 in 8C1 ways.
40 5
Hence total ways = 8 × 5 = 40 ; p = = ]
72 9

Q.6140 Two cards are drawn from a well shuffled pack of 52 playing cards one by one. If
A : the event that the second card drawn is an ace and
B : the event that the first card drawn is an ace card.
then which of the following is true?
4 1 1 1
(A) P (A) = ; P (B) = (B*) P (A) = ; P (B) =
17 13 13 13
1 1 16 4
(C) P (A) = ; P (B) = (D) P (A) = ; P (B) =
13 17 221 51
[Sol. P (A) = P { (B  A)  ( B  A) } = P (B  A) + P ( B  A)
4 3 48 4 1
= P (B) P (A / B) + P ( B ) P ( A / B ) = · + · =
52 51 52 51 13
1
P (B) = ]
13

(1  3p) (1  p) (1  2p)
Q.7213 If , & are the probabilities of three mutually exclusive events defined on a
3 4 2
sample space S, then the true set of all values of p is
1 1  1  1 1 1 1
(A*)  ,  (B)  ,1 (C)  ,  (D)  , 
3 2  3   4 3 4 2
[Hint: P(A)  0 ; P(B)  0 ; P(C)  0 ; P(A) + P(B) + P(C)  1 ]

Q.848/5 A lot contains 50 defective & 50 non defective bulbs . Two bulbs are drawn at random, one at a time,
with replacement . The events A, B, C are defined as :
A = { the first bulb is defective}; B = { the second bulb is non defective}
C = { the two bulbs are both defective or both non defective}
Determine whether (i) A,B,C are pair wise independent (ii) A,B,C are independent
[Ans: (i) A,B,C are pairwise independent (ii) A,B,C are not independent. ]

Bansal Classes [24]

[Sol.
A : first bulb is defective
B : second bulb is good
C : two bulbs are either both good or both defective
1 1 1
P (A) = P (B) = P (C) =
2 2 4
1 1 1
P (A  B) = · = 
2 2 4 
1 1 1 
P (B  C) = · = 
2 2 4 
1 1 1 
P (C  A) = · = 
2 2 4
Note :
P (A  B) = P(A)·P(B)
P (B  C) = P(B)·P(C)
P (C  A) = P(C)·P(A)
Hence the events are pairwise independent.
P (A  B  C) = 0  Hence, A, B, C are non independent]

Q.9150 An Urn contains 'm' white and 'n' black balls. All the balls except for one ball, are drawn from it. The
probability that the last ball remaining in the Urn is white, is
m n 1 mn
(A*) (B) (C) ( m  n ) ! (D) ( m  n ) !
mn mn
[Hint: n(S) = mCm – 1 · nCn
n(A) = m + nCm +n – 1
m
 P= ]
mn

Q.10151 A Urn contains 'm' white and 'n' black balls. Balls are drawn one by one till all the balls are drawn.
Probability that the second drawn ball is white, is
m n (m  n  1)
(A*) (B)
mn (m  n )(m  n  1)
m(m  1) mn
(C) (D)
(m  n )(m  n  1) (m  n )(m  n  1)
[Hint: E : event that 2nd drawn is white
n m m m 1
P (E) = P (B W or W W) = · + ·
m  n m  n 1 m  n m  n 1
m(m  n  1) m
= =  (A) ]
(m  n )(m  n  1) mn

Bansal Classes [25]

Q.11 Mr. Dupont is a professional wine taster. When given a French wine, he will identify it with probability
0.9 correctly as French, and will mistake it for a Californian wine with probability 0.1. When given a
Californian wine, he will identify it with probability 0.8 correctly as Californian, and will mistake it for a
French wine with probability 0.2. Suppose that Mr. Dupont is given ten unlabelled glasses of wine, three
with French and seven with Californian wines. He randomly picks a glass, tries the wine, and solemnly
says : "French". The probability that the wine he tasted was Californian, is nearly equal to
(A) 0.14 (B) 0.24 (C*) 0.34 (D) 0.44
[Sol. P(F/F) = 0.9 ; P(C/F) = 0.1 ; P(C/C) = 0.8 ; P(F/C) = 0.2
3 7
P(F) = ; P(C) =
10 10
A : Wine tasted was French
7
B1 : It is a Californian wine ; P(B1) =
10
3
B2 : It is a French wine ; P(B2) =
10
P(A/B1) = 0.2 ; P(A/B2) = 0.9
0.7  0.2 0.14 14
P(B1/A) = = = Ans. ]
0.7  0.2  0.3  0.9 0.14  0.27 41

Q.12120 Let A, B & C be 3 arbitrary events defined on a sample space 'S' and if,
P(A) + P(B) + P(C) = p1 , P(A  B) + P(B  C) + P(C  A) = p2 & P(A  B  C) = p3, then the
probability that exactly one of the three events occurs is given by :
(A) p1  p2 + p3 (B) p1  p2 + 2p3 (C) p1  2p2 + p3 (D*) p1  2p2 + 3p3
Q.13226 Three numbers are chosen at random without replacement from {1, 2, 3,...... , 10}. The probability that
the minimum of the chosen numbers is 3 or their maximum is 7 is
1 1 1 11
(A) (B) (C) (D*)
2 3 4 40
[Hint: N = {1, 2,.......10}  3 are drawn
A = minimum of the chosen number is 3
B = maximum number of the chosen number is 7.
7
C 2  6 C 2  3C1
P(A or B) = P(A) + P(B) – P(A  B) = 10 ]
C3

Q.14 A biased coin which comes up heads three times as often as tails is tossed. If it shows heads, a chip is
drawn from urn–I which contains 2 white chips and 5 red chips. If the coin comes up tails, a chip is
drawn from urn–II which contains 7 white and 4 red chips. Given that a red chip was drawn, what is the
probability that the coin came up heads? [Ans. 165/193]
[Sol. A = red chip was drawn
H = coin shows up head ; P(H) = 3/4
T = coin shows up tail ; P(T) = 1/4
2W 7W
Urn-I ; Urn-II
5R 4R
now A = (A  H) + (A  T)
3 5 1 4 15 1 193
P(A) = P(H) P(A/T) + P(T) P(T/T) = · + · = + = 11· 28
4 7 4 11 28 11
P(H  A) 15 28 ·11 165
now P(H/A) = = × = Ans. ]
P ( A) 28 193 193
Bansal Classes [26]
Q.1568/5 In a college, four percent of the men and one percent of the women are taller than 6 feet. Further
60 percent of the students are women. If a randomly selected person is taller than 6 feet, find the
probability that the student is a women. [Ans. 3/11]
[Sol. Let A = person is taller than 6 feet
P (A / M) = 0.04 ; P (A / W) = 0.01
P (M) = 0.4 ; P (W) = 0.6
P ( W ) · P (A / W ) (0.60) · (0.01)
P (W / A) = P( W ) · P( A / W )  P( M ) · P( A / M ) = (0.60) · (0.01)  (0.40) · (0.04)

60 6 3
= = = Ans.]
60  160 22 11

Q.1681 If at least one child in a family with 3 children is a boy then the probability that 2 of the children are boys,
is
3 1 1 3
(A*) (B) (C) (D)
7 4 3 8
[Hint: n (S) = B G G (3) ; B B G (3) ; B B B (1) ; hence n(S) = 7
3
n (A) = B B G (3)  p= ]
7
Q.17116 The probabilities of events, A  B, A, B & A  B are respectively in A.P. with probability of second
term equal to the common difference. Therefore the events A and B are
(A) compatible (B) independent
(C) such that one of them must occur (D*) such that one is twice as likely as the other
[Hint: P(A  B), P(A), P(B), P(A  B) are in A.P. with d = P(A)
 P(A) – P(A  B) = P(A)  P(A  B) = 0  A & B are ME or incompatible
also P(B) – P(A) = P (A)  2P(A) = P(B)
 if P(A) = p ; P(B) = 2p  (D) compatible means whcih can happen simultaneously ]
Q.18211 From an urn containing six balls, 3 white and 3 black ones, a person selects at random an even number
of balls (all the different ways of drawing an even number of balls are considered equally probable,
irrespective of their number). Then the probability that there will be the same number of black and white
balls among them
4 11 11 2
(A) (B*) (C) (D)
5 15 30 5
[Sol. Total number of possible cases = 3 (either 2 or 4 or 6 are drawn)
3 3
1  C1  C1
3
C 2  3C 2 3 C3  3C3  11
Hence required probability = 3  6  6
 6 =  (B)]
 C 2 C 4 C6  15

Q.19241 One purse contains 6 copper coins and 1 silver coin ; a second purse contains 4 copper coins. Five
coins are drawn from the first purse and put into the second, and then 2 coins are drawn from the second
and put into the first. The probability that the silver coin is in the second purse is
1 4 5 2
(A) (B) (C*) (D)
2 9 9 3
[Sol. (4C + 1S) goes from A to B and 2C return from B to A.
6
C 4 · 1C1 8
C2 15 28 5
 P(E) = 7 · 7 = · = ]
C5 C2 21 36 9
Bansal Classes [27]
Q.2031 7 persons are stopped on the road at random and asked about their birthdays. If the probability that 3 of
K
them are born on Wednesday, 2 on Thursday and the remaining 2 on Sunday is 6 , then K is equal to
7
(A) 15 (B*) 30 (C) 105 (D) 210
3 2 2
K 7C
1 4 1 1
[Hint : = .   . C2   .    K = 30 ]
76 3 7
  7 7

Q.2195 Two buses A and B are scheduled to arrive at a town central bus station at noon. The probability that bus
A will be late is 1/5. The probability that bus B will be late is 7/25. The probability that the bus B is late
given that bus A is late is 9/10. Then the probabilities
(i) neither bus will be late on a particular day and
(ii) bus A is late given that bus B is late, are respectively
(A) 2/25 and 12/28 (B) 18/25 and 22/28 (C*) 7/10 and 18/28 (D) 12/25 and 2/28
1 7 9
[Hint : (i) P(A) = ; P(B) = ; P(B/A) =
5 25 10
P (A  B) = 1 – P(AUB)
= 1 – [P(A) + P(B) – P(A  B) ]

1 7 
= 1 –   P(A)  P(B / A)
 5 25 

1 7 1 7  7
=1–      = Ans.]
 5 25 5 25  10
P(A  B) P(A)  P(B / A)
(ii) P(A/B) = =
P(B) P(B)

1 9

= 5 10 = 9 x 25 = 9 = 18 Ans. ]
7 50 7 14 28
25

Q.2248 A box contains a normal coin and a doubly headed coin. A coin selected at random and tossed twice, fell
headwise on both the occasions. The probability that the drawn coin is a doubly headed coin is
2 5 3 4
(A) (B) (C) (D*)
3 8 4 5
[Sol. A  Normal coin
B  DH coin
P(B  HH)
P(B/HH) =
P( HH)

1 1
2 2 1 8 4
= = = x =  D]
1 1 1 1 4 2 5 5
  1 
2 4 2 8 8

Bansal Classes [28]

Q.2356 A box contains 5 red and 4 white marbles. Two marbles are drawn successively from the box without
replacement and the second drawn marble drawn is found to be white. Probability that the first marble is
also while is
3 1 1 1
(A*) (B) (C) (D)
8 2 3 4
[Sol. Method-1 : since the 2 is known to be W, there are only 3 ways of the remaining 8 in which the 1st
nd

can be white, so that probability = 3/8

Method-2 : A – 2nd drawen found to be white
B1 - 1st drawn is W ; B2 – 1st drawn is R
4 5
P (B1) = ; P (B2) =
9 9
3 4
P (A / B1) = ; P (A / B2) =
8 8
4 3
· 12 12 3
P (B1 / A) = 9 8 = = =
4 3 5 4 12  20 32 8
·  ·
9 8 9 8

12 3
Alternatively : P [ (W W) / (R W or W W) ] = = Ans. ]
32 8

Q.24 A and B in order draw a marble from bag containing 5 white and 1 red marbles with the condition that
whosoever draws the red marble first, wins the game. Marble once drawn by them are not replaced into
the bag. Then their respective chances of winning are
2 1 3 2 2 3 1 1
(A) & (B) & (C) & (D*) &
3 3 5 5 5 5 2 2

[Sol. [29-10-2005, 12th Jaipur]

P(A) = P(R or W W R or W W W W R)
1 5 4 1 5 4 3 2 1 1 1 1 1 1
= + · · + · · · · = + + = ;  P(B) = ]
6 6 5 4 6 5 4 3 2 6 6 6 2 2
Q.25112 In a maths paper there are 3 sections A, B & C. Section A is compulsory. Out of sections B & C a
student has to attempt any one. Passing in the paper means passing in A & passing in B or C. The
probability of the student passing in A, B & C are p, q & 1/2 respectively. If the probability that the
student is successful is 1/2 then :
1 1
(A) p = q = 1 (B) p = q = (C) p = 1, q = 0 (D*) p = 1, q =
2 2
1  1
[Hint: p (S) = P A and (B or C)  = p · q  
2 2

1 p 1  1
= q   ; 1 = p  q    (D) ]
2 2 2  2

Bansal Classes [29]

Q.26210 A box contains 100 tickets numbered 1, 2, 3,.... ,100. Two tickets are chosen at random. It is given that
the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5,
with probability
1 2 3
(A*) (B) (C) (D) none
9 11 19
[Hint: N = {1, 2, ....5, .......10, ..........100}
two tickets are drawn
A : maximum number on the two chosen ticket is  10  n(S) = 10
B : minimum number on the two chosen ticket is 5
P(A  B) 5
C1 5 1
P(B / A) = = = = [one of the ticket is 5 and one is frm 6, 7, 8, 9, 10] ]
P(A) 10
C2 45 9

Q.27240 Sixteen players s1 , s2 ,..... , s16 play in a tournament. They are divided into eight pairs at random. From
each pair a winner is decided on the basis of a game played between the two players of the pair. Assume
that all the players are of equal strength. The probability that "exactly one of the two players s1 & s2 is
among the eight winners" is
4 7 8 9
(A) (B) (C*) (D)
15 15 15 15
14 th
[Hint: 7 players (leaving s1 & s2 ) out of 14 can be selected in C7 and the 8 player can be chosen
in two ways i.e. either s1 or s2. Hence the total ways = 14C7. 2
2 . 14 C 7 8
Therefore p = 16
= ]
C8 15
[ Alternatively : Let
E1 : S1 and S2 are in the same group
E2 : S1 and S2 are in the different group
E : exactly one of the two players S1 & S2 is among the eight winners.
E = (EE1) + (E E2)
P(E) = P(EE1) + p (E E2)
P(E) = P(E1).P(E/E1) + p(E2).P(E/E2) ....(1)
(14)!
(2) 7 . 7! 1
Now P(E1) = 
16! 15
8
2 . 8!

1 14
P(E2) = 1  
15 15
1 14
P(E) = .1  · P ( exactly one of either S1 & S2 wins)
15 15
1 14  1 1 1 1  1 1 1 1 7 8
= + . .  .  =  .    Ans ]
15 15  2 2 2 2  15 14 2 15 15 15

Bansal Classes [30]

Q.28 The number 'a' is randomly selected from the set {0, 1, 2, 3, ...... 98, 99}. The number 'b' is selected
from the same set. Probability that the number 3a + 7b has a digit equal to 8 at the units place, is
1 2 4 3
(A) (B) (C) (D*)
16 16 16 16

3a ends in 
1 3 7 9
7b ends in 
1 8
[Hint: 3 [27-11-2005, 12th]
7 8
9 8
3
Out of 16 case 3 are favorable  p= ]
16
Q.29 We are given two urns as follows :
Urn A contains 5 red marbles, 3 white marbles and 8 blue marbles.
Urn B contains 3 red marbles and 5 white marbles
A fair dice is tossed if 3 or 6 appears, a marble is chosen from B, otherwise a marble is chosen from A. Find
the probability that (i) a red marble is chosen, (ii) a white marble is chosen, (iii) a blue marble is chosen.
(Use Tree Diagram)
1 1 1
[Ans. (i) ; (ii) ; (iii) ]
3 3 3
Q.30 We are given two Urns as follows :
Urn A contains 5 red marbles, 3 white marbles.
Urn B contains 1 red marbles and 2 white marbles.
A fair die is tossed, if a 3 or 6 appears, a marble is drawn from B and put into A and then a marble is
drawn from A ; otherwise, a marble is drawn from A and put into B and then a marble is drawn from B.
(Use Tree Diagram)
(i) What is the probability that both marbles are red?
(ii) What is the probability that both marbles are white?
61 371
[Ans. (i) ; (ii) ]
216 1296
Q.31 Two boys A and B find the jumble of n ropes lying on the floor. Each takes hold of one loose end. If the
1
probability that they are both holding the same rope is then the number of ropes is equal to
101
(A) 101 (B) 100 (C*) 51 (D) 50
[Sol. The n strings have a total of 2n ends. One boy picks up one end, this leaves (2n – 1) ends for the second
boy to choose, of which only one is correct.
1 1 1
 p=  =  2n – 1 = 101  n = 51 ] [08-01-2006, 12 & 13]
2n  1 2n  1 101

Bansal Classes [31]

 Direction for Q.32 to Q.35 (4 Questions)
Read the passage given below carefully before attempting these questions.
A standard deck of playing cards has 52 cards. There are four suit (clubs, diamonds, hearts and spades),
each of which has thirteen numbered cards (2, ....., 9, 10, Jack, Queen, King, Ace)
In a game of card, each card is worth an amount of points. Each numbered card is worth its number (e.g.
a 5 is worth 5 points) ; the Jack, Queen and King are each worth 10 points ; and the Ace is either worth
your choice of either 1 point or 11 points. The object of the game is to have more points in your set of cards
than your opponent without going over 21. Any set of cards with sum greater than 21 automatically loses.
Here's how the game played. You and your opponent are each dealt two cards. Usually the first card for
each player is dealt face down, and the second card for each player is dealt face up. After the initial cards
are dealt, the first player has the option of asking for another card or not taking any cards. The first
player can keep asking for more cards until either he or she goes over 21, in which case the player loses,
or stops at some number less than or equal to 21. When the first player stops at some number less than
or equal to 21, the second player then can take more cards until matching or exceeding the first player's
number without going over 21, in which case the second player wins, or until going over 21, in which
case the first player wins.
We are going to simplify the game a little and assume that all cards are dealt face up, so that all cards are
visible. Assume your opponent is dealt cards and plays first.
Q.32 The chance that the second card will be a heart and a Jack, is
4 13 17 1
(A) (B) (C) (D*)
52 52 52 52
Q.33 The chance that the first card will be a heart or a Jack, is
13 16 17
(A) (B*) (C) (D) none
52 52 52
Q.34 Given that the first card is a Jack, the chance that it will be the heart, is
1 4 1 1
(A) (B) (C*) (D)
13 13 4 3
Q.35 Your opponent is dealt a King and a 10, and you are dealt a Queen and a 9. Being smart, your opponent
does not take any more cards and stays at 20. The chance that you will win if you are allowed to take as
many cards as you need, is
97 25 15 1
(A) (B) (C) (D*)
564 282 188 6
51 1 1
[Sol. P (2nd card is J of H) = · = Ans.(32) [29-01-2006, 12th & 13th]
52 51 52
There are 16 ways to get a Jack or a hearts : get one of the thirteen hearts (Ace through King of hearts),
or get one of the Jack of clubs, Jack of spades, or Jack of diamonds.
Hence, probability (Jack or hearts) = 16/52. Ans.(33)

P(H  J ) 1
P (H / J) = = Ans.(34)
P(J) 4
4 4 1
P(win) = P(any 2's or Ace or Ace and Ace) = + = Ans.(35)]
48 48 6

Bansal Classes [32]

More than one alternative are correct:
Q.36230 If A & B are two events such that P(B)  1, BC denotes the event complementry to B, then
P (A )  P (A  B)
(A*) P A BC =
  1  P (B)
(B*) P (A  B)  P(A) + P(B)  1
(C*) P(A) > < P A B according as P A BC > < P(A)
 
(D*) P A BC + P A C BC = 1
   
[Sol. (B) 1 > P(A) + P(B) – P(AB) or P(A B)  P(A) + P(B) – 1 Þ (B)
(C) Let P(A) > P(A/B)
P(A  B)
or P(A) >
P(B)
P(A). P(B) > P(AB) ....(1)
TPT P(A/BC) > P(A)
P ( A  Bc )
> P(A)
P ( Bc )
P(A) – P(AB) > P(A) [ 1–P(B)]
– P(AB) > – P(A). P(B)
or P(A). P(B) > P(AB) ....(2)
from (1) and (2) P(A) > P(A/B)  P(A/Bc) > P(A) ]

Q.37225 A bag initially contains one red & two blue balls. An experiment consisting of selecting a ball at random,
noting its colour & replacing it together with an additional ball of the same colour. If three such trials are
made, then :
(A*) probability that atleast one blue ball is drawn is 0.9
(B*) probability that exactly one blue ball is drawn is 0.2
(C*) probability that all the drawn balls are red given that all the drawn balls are of same colour is 0.2
(D*) probability that atleast one red ball is drawn is 0.6.
[Hint : (i) P(E1) = 1  P(R R R)
1 2 3
= 1   . .  = 0.9
3 4 5
2 1 2
(ii) P(E2) = 3 P(B R R) = 3. . . = 0.2
3 4 5
P ( R R R)
(iii) P(E3) = P(R R R/R R R  B B B) =
P ( R R R)  P (BBB)
2 3 4 8 0.1
but P(B B B) = . . =  P(E3) = = 0.2
3 4 5 20 0.1  0.4
2
(iv) P(E4) = 1  P(B B B) = 1  = 0.6 ]
5

Bansal Classes [33]

Q.38227 Two real numbers, x & y are selected at random. Given that 0  x  1 ; 0  y  1. Let A be the event
that y2  x ; B be the event that x2  y, then :
1
(A*) P (A  B) = (B*) A & B are exhaustive events
3
(C) A & B are mutually exclusive (D) A & B are independent events.
Q.39221 For any two events A & B defined on a sample space ,
P (A)  P (B)  1
(A*) P A B  , P (B)  0 is always true
P (B)
(B*) P AB  = P (A) - P (A  B)
(C*) P (A  B) = 1 - P (Ac). P (Bc) , if A & B are independent
(D) P (A  B) = 1 - P (Ac). P (Bc) , if A & B are disjoint
P( A  B)
[ Hint: For A P(A/B) =
P( B)
 T.P.T. P(AB) > P(A) + P(B) – 1
or T.P.T. , 1 > P(A) + P(B) – P(AB)
or T.P.T. , 1 > P(AB)
which is true  (A) is correct
(B) and (C) are obvious ]

Q.40205 If E1 and E2 are two events such that P(E1) = 1/4, P(E2/E1) =1/2 and P(E1/ E2) = 1/4
(A*) then E1 and E2 are independent
(B) E1 and E2 are exhaustive
(C*) E2 is twice as likely to occur as E1
(D*) Probabilities of the events E1  E2 , E1 and E2 are in G.P.
P E1  E 2 
[Hint : P  E 2 E1  =
P (E 1 )

1 P  E1  E 2  1
=  P (E1  E2) = = P (E2). P  E1 E 2 
2 14 8

1 1
= P (E2).  P (E2) =
4 2
1
Since P (E1  E2) = = P (E1). P (E2)  events are independent
8
1 1 1 5
Also P (E1  E2) = +  =  E1 & E2 are non exhaustive ]
2 4 8 8

Q.41179 Let 0 < P(A) < 1 , 0 < P(B) < 1 & P(A  B) = P(A) + P(B)  P(A). P(B), then :
(A) P(B/A) = P(B)  P(A) (B) P(AC  BC) = P(AC) + P(BC)
C C
(C*) P((A  B) ) = P(A ). P(B )C (D*) P(A/B) = P(A)

Q.4227 If M & N are independent events such that 0 < P(M) < 1 & 0 < P(N) < 1, then :
(A) M & N are mutually exclusive (B*) M & N are independent
(C*) M & N are independent (D*) P M N + P  M N = 1
 ALLEN® Matrices 1
é0 1 0 ù
MATRICES
6. Let A = êê1 0 0 úú . Then the number of 3 × 3
1.
é2
Let A = ê
3ù
, a Î R be written as P + Q where êë0 0 1 úû
ëa 0 úû
matrices B with entries from the set {1, 2, 3, 4, 5}
P is a symmetric matrix and Q is skew symmetric
and satisfying AB = BA is _______.
matrix. If det(Q) = 9, then the modulus of the sum of
ì æa bö ïü
all possible values of determinant of P is equal to : 7. Let M = ïí A = ç ÷ : a, b,c, d Î {±3, ±2, ±1, 0}ý .
îï è c d ø þï
(1) 36 (2) 24 (3) 45 (4) 18
Define f : M ® Z, as f(A) = det(A), for all A Î M,
æ 1 -1 0 ö
ç ÷ where Z is set of all integers. Then the number
2. Let A = ç 0 1 -1 ÷ and B = 7A20 – 20A7 + 2I,
ç0 0 1 ÷ø
è of A Î M such that f(A) = 15 is equal to ________.
where I is an identity matrix of order 3 × 3.
é1 0ù
If B = [bij], then b13 is equal to _________. 8. If P = ê 1 ú , then P50 is:
1
ëê 2 úû
dy
3. Let y = y(x) satisfies the equation - A =0,
dx é1 0 ù é1 50 ù
(1) ê ú (2) ê ú
ë25 1 û ë0 1 û
é y sin x 1ù
for all x > 0 , where A = ê 0 -1 1 úú .
ê é1 25ù é1 0 ù
ê 1ú (3) ê ú (4) ê ú
ê2 0 ú ë0 1 û ë50 1 û
ë xû
é 1 2ù
p
If y(p) = p +2, then the value of y æç ö÷ is :
–1
9. Let A = ê
-1 4 ú . If A = aI + bA, a, b Î R, I is
è2ø ë û

p 4 p 1 a 2 × 2 identity matrix, then 4(a – b) is equal to :

(1) + (2) -
2 p 2 p
3p 1 p 4 8
(1) 5 (2) (3) 2 (4) 4
(3) - (4) - 3
2 p 2 p
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices

10. Let A and B be two 3 × 3 real matrices such that

4. Let A = {a ij} be a 3 × 3 matrix, where
(A2 –B2) is invertible matrix. If A5 = B5 and
ì( -1) if i < j ,
j- i
ï A3B2 = A2B3, then the value of the determinant
a ij = í 2 if i = j ,
ï of the matrix A3+B3 is equal to :
î(-1) if i > j ,
i+ j

then det (3Adj(2A–1)) is equal to ______. (1) 2 (2) 4 (3) 1 (4) 0

5. Let A = [aij] be a real matrix of order 3 × 3, é1 1 1ù

ê ú
11. If A = ê0 1 1ú and M = A + A2 + A3 + ... + A20,
such that ai1 + ai2 + ai3 = 1, for i = 1, 2, 3. Then,
êë0 0 1úû
3
the sum of all the entries of the matrix A is
then the sum of all the elements of the matrix M
equal to :
is equal to______.
(1) 2 (2) 1 (3) 3 (4) 9

E
2 Matrices ALLEN®
æ 1 2 ö 18. Let A and B be 3 × 3 real matrices such that A
ç 5 5÷ æ 1 0ö is symmetric matrix and B is skew-symmetric
12. If A = ç ÷ , B=ç ,i= -1 , and
ç -2 1 ÷ è i 1 ÷ø
çè ÷ matrix. Then the system of linear equations
5 5ø
(A2B2 – B2A2)X = O, where X is a 3 × 1 column
T
Q = A BA, then the inverse of the matrix A
matrix of unknown variables and O is a
Q2021 AT is equal to :
3 × l null matrix, has :
æ 1 ö
- 2021 (1) no solution
ç 5 ÷ æ 1 0ö
(1) ç ÷ (2) ç (2) exactly two solutions
ç 2021 1 ÷ è -2021 i 1 ÷ø
çè ÷
5 ø (3) infinitely many solutions
(4) a unique solution
æ 1 0ö æ 1 -2021 i ö
(3) ç (4) ç
è 2021 i 1 ÷ø è0 1 ÷ø 19. Let M be any 3 × 3 matrix with entries from the

æ1 0 0ö set {0, 1, 2}. The maximum number of such

13. Let A = ç 0 1 1 ÷ . Then A2025 – A2020 is equal to : matrices, for which the sum of diagonal
ç1 0 0÷
è ø elements of MTM is seven, is _______.
6 5
(1) A – A (2) A
5
é 3 - 1 -2 ù
(4) A6
P = êê 2 0 a úú ,
(3) A – A
20. Let where aÎR.
14. Let A be a 3 × 3 real matrix. êë 3 -5 0 úû
If det(2Adj(2 Adj(Adj(2A)))) = 241, then the
Suppose Q = [qij] is a matrix satisfying PQ = kI3
value of det(A2) equal _____.
k
15. If the matrix A = æç 0 2ö 3
÷ satisfies A(A + 3I) = 2I, for some non-zero k Î R. If q 23 = - and
èK -1 ø 8
then the value of K is : k2
Q = , then a2 + k2 is equal to _______.
1 1 2
(1) (2) - (3) –1 (4) 1
2 2 21. Let A be a 3 × 3 matrix with det(A) = 4. Let Ri
16. The number of elements in the set
denote the ith row of A. If a matrix B is obtained

node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices

ìï æa bö üï by performing the operation R2 ® 2R2 + 5R3 on
÷ : a,b,d Î {-1,0,1} and ( I - A ) = I - A
3
íA = ç ý,
îï è 0 d ø þï
2A, then det(B) is equal to :
where I is 2 × 2 identity matrix, is : (1) 16 (2) 80 (3) 128 (4) 64
1
2
xn é 1 -a ù
17. Let J n,m = ò m dx, " n > m and n, m Î N . 22. If for the matrix, A = ê ú , AAT = I2, then
x -1 ëa b û
0
the value of a4 + b4 is :
Consider a matrix A = [aij]3 × 3

{
(1) 4 (2) 2 (3) 3 (4) 1
J -J , i £ j
where a ij = 6+ i,3 i +3,3 . Then
éx y z ù
0 , i>j
23. Let A = êê y z x úú , where x, y and z are real
adjA –1 is : êë z x y úû
(1) (15)2 × 242 (2) (15)2 × 234
numbers such that x + y + z > 0 and xyz = 2.
(3) (105)2 × 238 (4) (105)2 × 236
If A2 = I3, then the value of x3 + y3 + z3 is_____.

E
 ALLEN® Matrices 3

é 0 æ q öù 30. If x, y, z are in arithmetic progression with

ê - tan ç ÷ ú
24. If A =
ê tan æ q ö è 2 ø ú and common difference d, x ¹ 3d, and the
ç
êë è 2 ø ÷ 0 úû
é3 4 2 xù
éa -b ù ê ú
(I2 + A) (I2 – A)–1 =ê ú , then 13 (a2 + b2) determinant of the matrix ê 4 5 2 y ú is
ëb aû ê ú
ë5 k zû
is equal to ________ .
zero, then the value of k2 is
é1 0 0 ù
25. If the matrix A = êê0 2 0 úú satisfies the (1) 72 (2) 12 (3) 36 (4) 6
êë3 0 -1úû
éa b ù éa ù é0 ù
31. Let A = ê ú and B = ê ú ¹ ê ú such that
é1 0 0 ù ëc d û ë b û ë0 û
equation A20 + aA19 + bA = êê0 4 0 úú for some
AB = B and a + d = 2021, then the value of
êë0 0 1 úû
ad – bc is equal to ________.
real numbers a and b, then b – a is equal to ______.
æ 0 sin a ö æ 2 1 ö
26. Let A be a symmetric matrix of order 2 with integer 32. If A = ç ÷ and det ç A - I ÷ = 0, then
è sin a 0 ø è 2 ø
entries. If the sum of the diagonal elements of A2 is
a possible value of a is
1, then the possible number of such matrices is
(1) 4 (2) 1 (3) 6 (4) 12 p p p p
(1) (2) (3) (4)
2 3 4 6
éa ù éb ù
27. Let A = ê 1 ú and B = ê 1 ú be two 2 × 1
ëa 2 û ëb2 û é2 3 ù
33. If A = ê ú , then the value of
matrices with real entries such that A = XB, ë0 -1û

where X=
1 é1 -1ù
ê ú, and k Î R. If ( )
det(A4) + det A10 - (Adj(2A))10 is equal to
3 ë1 k û
_________ .
a12 + a 22 = (
2 2
3
b1 + b22 ) 2 2
and (k + 1)b2 ¹ -2b1b2 , é1 2 0ù
34. Let A + 2B = ê 6 -3 3 úú
ê
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices

then the value of k is _______.

êë -5 3 1 úû
é i -i ù
28. Let A = ê ú , i = -1 .Then, the system of
ë-i i û é 2 -1 5 ù
and 2A - B = êê2 -1 6 úú . If Tr(A) denotes the
éx ù é 8 ù
linear equations A8 ê ú = ê ú has : êë 0 1 2 úû
ë y û ë64 û
(1) A unique solution sum of all diagonal elements of the matrix A,

(2) Infinitely many solutions then Tr(A) – Tr(B) has value equal to
(3) No solution (1) 1 (2) 2 (3) 0 (4) 3
(4) Exactly two solutions
35. Let I be an identity matrix of order 2 × 2 and
29. The total number of 3 × 3 matrices A having
é2 -1ù
enteries from the set (0, 1, 2, 3) such that the P= ê ú . Then the value of n Î N for
ë 5 -3û
sum of all the diagonal entries of AAT is 9, is
which Pn = 5I – 8P is equal to _______.
equal to _____.

E
4 Matrices ALLEN®
SOLUTION æ0 0 1ö
2 ç ÷
C = ç 0 0 0 ÷,
1. Official Ans. by NTA (1)
ç0 0 0÷
è ø
é2 3 ù
Sol. A=ê ú, a ÎR æ0 0 0ö
ëa 0 û ç ÷
C = ç 0 0 0 ÷ = C 4 = C5 = ......
3

é 3+aù ç0 0 0÷
è ø
A+A ê 2T
2 ú
and P = =ê ú B = 7 A20 – 20 A7 + 2I
2 êa + 3 0 ú
êë 2 úû = 7 (I + C)20 – 20 (I + C)7 + 2I
= 7(I + 20C + 20C2 C2) – 20 (I + 7C + 7C2 C2) + 2I
é 3-aù
0
A - AT ê 2 ú So b13 = 7 × 20C2 – 20 × 7C2 = 910
and Q = =ê ú
2 êa - 3 0 ú
êë 2 úû 3. Official Ans. by NTA (1)
y
As, det (Q) = 9 Sol. | A |= - + 2 sin x + 2
x
Þ (a – 3)2 = 36 dy
=| A |
dx
Þ a=3±6
dy y
\ a = 9, - 3 = - + 2 sin x + 2
dx x
dy y
+ = 2 sin x + 2
\ dx x
1
ò x dx
I.F. = e =x

Þ yx = ò x(2 sin x + 2)dx

2
(a - 3)
=0- = 0, for a = -3 xy = x2 – 2x cosx + 2sinx + c .....(i)
4
Now x = p, y = p + 2
(a - 3)2 1
=0- = - (12)(12), for a = 9 Use in (i)
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices
4 4
c=0
\ Modulus of the sum of all possible values of
Now (i) be comes
det. (P) = -36 + 0 = 36 Ans. xy = x2 – 2x cosx + 2 sinx
put x = p/2
Þ Option (1) is correct
2
2. Official Ans. by NTA (910) p æ pö p p p
y = ç ÷ - 2. cos + 2sin
2 è 2ø 2 2 2
æ 1 -1 0 ö
ç ÷ p p2
Sol. Let A = ç 0 1 -1÷ = I + C y= +2
ç0 0 1 ÷ 2 4
è ø

æ1 0 0ö æ 0 -1 0 ö
ç ÷ ç ÷
where I = ç 0 1 0 ÷ ,C = ç 0 0 -1 ÷
ç0 0 1÷ ç0 0 0 ÷
è ø è ø

E
 ALLEN® Matrices 5
4. Official Ans. by NTA (108) 6. Official Ans. by NTA (3125)
é 2 -1 1ù éa b c ù
Sol. A = êê-1 2 -1úú ê
Sol. Let matrix B = êd e f ú
ú
êë 1 -1 2 úû êëg n i úû

|A| = 4 Q AB = BA
3adj ( 2A -1 ) = 3.2 2 adj ( A -1 ) é0 1 0 ù éa b c ù é a b c ù é0 1 0 ù
ê 1 0 0 ú ê d e f ú = êd e f ú ê1 0 0 ú
123 123 ê úê ú ê úê ú
= 123 adj ( A-1 ) = 123 A-1 =
2
2
= = 108 êë0 0 1 úû êëg h i úû êëg h i úû êë0 0 1 úû
A 16
éd e f ù é b a c ù
5. Official Ans. by NTA (3) êa b cú = ê e d f ú
ê ú ê ú
é a11 a12 a13 ù êëg h i úû êë h g i úû
Sol. A = êêa 21 a 22 a 23 úú
Þ d = b, e = a, f = c, g=h
êëa 31 a 32 a 33 úû
éa b c ù
é1ù ê ú
\ Matrix B = ê b a c ú
ê ú
Let x = ê1ú êë g g i úû
êë1úû
No. of ways of selecting a, b, c, g, i
éa11 + a12 + a13 ù é1ù =5×5×5×5×5
AX = êêa 21 + a 22 + a 23 úú = êê1úú = 55 = 3125
êëa 31 + a 32 + a 33 úû êë1úû
\ No. of Matrices B = 3125
Þ AX = X 7. Official Ans. by NTA (16)
Replace X by AX Sol. |A| = ad – bc = 15

A2X = AX = X where a,b,c,d Î {± 3, ±2, ±1, 0}

Replace X by AX
Case I ad = 9 & bc = – 6
A3X = AX = X
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices

For ad possible pairs are (3,3), (–3,–3)

é x1 x2 x3 ù
Let A = êê y1
3
y2 y3 úú For bc possible pairs are (3,–2), (–3,2), (–2,3),(2,–3)
êë z1 z2 z3 úû So total matrix = 2 × 4 = 8

é1ù é x1 + x 2 + x 3 ù é1ù Case II ad = 6 & bc = –9

A êê1úú = êê y1 + y 2 + y3 úú = êê1úú
3

êë1úû êëz1 + z 2 + z 3 úû êë1úû Similarly total matrix = 2 × 4 = 8

Sum of all the element = 3 Þ Total such matrices are = 16

E
6 Matrices ALLEN®
8. Official Ans. by NTA (1) 11. Official Ans. by NTA (2020)
é1 0ù é n2 + n ù
Sol. P = ê1 ú ê1 n ú
ê 2 ú
1ú ê
êë 2 úû Sol. A n = ê0 1 n ú
ê0 0 1 úû
é 1 0ù é 1 0ù ë
P = ê1
2 ú ê1 ú = é1 0 ù So, required sum
ê ê ú
1ú ê 1 ú ë1 1 û
ëê 2 ûú ëê 2 úû
æ 20 ´ 21 ö 20
æ r2 + r ö
é 1 0ù é 1 0ù
= 20 × 3 + 2 × ç
è 2 ø
÷ + å çè
r =1 2 ø
÷
é1 0 ù ê ú = ê3 ú
P =ê3
ú 1
ë1 1 û êê 1ú ê 1ú = 60 + 420 + 105 + 35 × 41 = 2020
ë2 ûú ëê 2 ûú

é1 0 ù é1 0 ù é1 0 ù 12. Official Ans. by NTA (2)

P4 = ê úê ú=ê ú
ë1 1 û ë1 1 û ë2 1 û æ 1 2 ö æ 1 -2 ö
ç ÷ ç ÷
M Sol. AA T = ç
5 5÷ ç 5 5÷
ç -2 1 ÷ ç 2 1 ÷
é 1 0ù ç ÷ ç ÷
\ P 50 = ê ú è 5 5ø è 5 5ø
ë25 1 û
9. Official Ans. by NTA (4) æ1 0ö
AAT = ç ÷=I
è0 1ø
é 1 2ù
Sol. A=ê ú , |A| = 6
ë -1 4 û Q2 = ATBA ATBA = ATBIBA
Þ Q2 = AT B2A
é2 1ù
- ú
adjA 1 é 4 -2 ù ê 3 3 Q3 = ATB2AATBA Þ Q3 = ATB3A
A -1 = = ê =ê ú
A 6 ë 1 1 úû ê 1 1 ú Similarly : Q2021 = AT B2021 A ...... (1)
êë 6 6 úû
æ1 0 ö æ1 0 ö æ 1 0ö
é2 1ù Now B2 = ç ÷ç ÷=ç ÷
ê3 - ú è i 1 ø è i 1 ø è 2i 1ø
3 é a 0 ù é b 2b ù
ê ú=ê ú+ê ú
ê1 1 ú ë 0 a û ë -b 4b û æ 1 0 öæ 1 0 ö 3 æ 1 0ö
ëê 6 6 ûú B3 = ç ÷ç ÷ Þ B = ç 3i 1 ÷
è 2i 1 øè i 1 ø è ø
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices
2ü
a+b= ï
3ï 2 1 5 æ 1 0ö
ýÞ a = + = Similarly B2021 = ç ÷
1 ï 3 6 6 è 2021i 1 ø
b=-
6 þ ï
\ AQ2021 AT = AAT B2021 AAT = IB2021I
4(a - b) = 4(1) = 4
10. Official Ans. by NTA (4) æ 1 0ö
Þ AQ2021 AT = B2021 = ç ÷
Sol. C = A - B ;| C| ¹ 0
2 2
è 2021i 1 ø

A5 = B5 and A3B2 = A2B2 æ 1æ 0ö ö

-1
1 0
\ (AQ2021 AT)–1 = ç ÷ = ç ÷
5
Now, A – A B = B – A B 3 2 5 2 3
è 2021i 1 ø è -2021i 1 ø

Þ A3 ( A2 - B2 ) + B3 ( A2 - B2 ) = 0

Þ ( A3 + B3 )( A2 - B2 ) = 0
Post multiplying inverse of A2 – B2 :
A3 + B3 = 0

E
 ALLEN® Matrices 7
13. Official Ans. by NTA (1) 15. Official Ans. by NTA (1)
é1 0 0 ù é1 0 0 ù é0 2ù
Sol. Given matrix A = ê
Sol. A = ê0 1 1 ú Þ A = êê1 1 1 úú
ê ú 2
ëk -1úû
êë1 0 0 úû êë1 0 0 úû
A4 + 3 IA = 2I

é1 0 0 ù é1 0 0 ù Þ A4 = 2I – 3A
A = ê2 1 1 ú Þ A = êê3 1 1 úú
3 ê ú 4
Also characteristic equation of A is
êë1 0 0 úû êë1 0 0 úû |A – lI| = 0
0 -l 2
é 1 0 0ù Þ =0
k -1 - l
A = ê n - 1 1 1 úú
n ê
êë 1 0 0 úû Þ l + l2 – 2k = 0
Þ A + A2 = 2K.I
é0 0 0 ù
Þ A2 = 2KI – A
A 2025
-A 2020
= êê 5 0 0 úú
êë0 0 0 úû Þ A4 = 4K2I + A2 – 4AK
Put A2 = 2KI – A
é0 0 0 ù and A4 = 2I – 3A
A - A = êê 5 0 0 úú
6
2I – 3A = 4K2I + 2KI – A – 4AK
êë0 0 0 úû
Þ I(2 – 2K – 4K2) = A(2– 4K)
14. Official Ans. by NTA (4)
Þ -2I ( 2K2 + K -1) = 2A (1 - 2K)
Sol. adj (2A) = 22 adjA Þ -2I ( 2K - 1 )( K + 1 ) = 2A (1 - 2K )

Þ adj(adj (2A)) = adj(4 adjA) = 16 adj (adj A) Þ ( 2K - 1 )( 2A ) - 2I ( 2K - 1 )( K + 1 ) = 0

= 16 |A| A Þ ( 2K - 1) [ 2A - 2I ( K + 1)] = 0

Þ adj (32 |A| A) = (32 |A|)2 adj A 1

Þ K=
2
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices

12(32|A|)2 |adj A| = 23 (32|A|)6 |adj A| 16. Official Ans. by NTA (8)

23.230 |A|6 . |A|2 = 241 Sol. (I – A)3 = I3 – A3 – 3A(I – A) = I – A3

8 8 Þ 3A(I – A) = 0 or A2 = A
|A| = 2 Þ |A| = ±2
éa 2 ab + bd ù éa b ù
2
|A| = |A| = 4 2 Þ ê ú=ê ú
ë0 d 2 û ë0 d û

Þ a2 = a, b(a + d – 1) = 0, d2 = d

If b ¹ 0, a + d = 1 Þ 4 ways

If b = 0, a = 0, 1 & d = 0, 1 Þ 4 ways

Þ Total 8 matrices

E
8 Matrices ALLEN®
17. Official Ans. by NTA (3) 1
| A |=
Sol. 210.218

n -1 2 1
adjA-1 = A -1 = A -1 =
(A)
2

J6+i, 3–Ji+3,3 ; i £ j
Þ (210.218)2
x6+ i xi+3
1 1

Þ ò
0
2
x 3 - 1 ò0 x 3 - 1
- 2
(105)2 × 238

1/2 (
x i+3 x3 - 1 ) 18. Official Ans. by NTA (3)
Þ ò 0 x3 - 1 Sol. Let AT = A and BT = –B
1/ 2
x 3 + i +1 æ x 4+i ö C = A2B2 – B2A2
Þ =
3 + i + 1 çè 4 + i ÷ø 0
CT = (A2B2)T – (B2A2)T
4 +i
æ1ö = (B2)T(A2)T – (A2)T(B2)T
ç2÷
aij = j6 + i, 3 – ji + 3, 3 = è ø
4+i = B2A2 – A2B2
5
æ 1ö CT = –C
çè 2 ÷ø 1
a11 = = C is skew symmetric.
5 5.2 5
1 So det(C) = 0
a12 =
5.25 so system have infinite solutions.
1
a13 = 19. Official Ans. by NTA (540)
5.25
1 é a b cù éa d g ù
a 22 = Sol. êd e f ú ê b e h ú
6.26 ê úê ú
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices
êë g h i úû êë c f i úû
1
a 23 =
6.26 a2 + b2 + c2 + d2 + e2 + f2 + g2 + h2 + i2 = 7
1
a 33 = Case-I : Seven (1's) and two (0's)
7.2 7
é 1 1 1 ù 9C
2 = 36
ê 5.25 5.25 5.25 ú
ê ú Case-II : One (2) and three (1's) and five (0's)
1 1 ú
A=ê 0
ê 6.26 6.26 ú 9!
ê ú = 504
ê 0 1 ú 5!3!
0
êë 7.2 7 úû
\ Total = 540
1 é 1 1 ù
| A |= 5 ê 6
´ 7ú
5.2 ë 6.2 7.2 û

E
 ALLEN® Matrices 9
20. Official Ans. by NTA (17) 23. Official Ans. by NTA (7)
Sol. PQ = kI Sol. A2 = I
|P|.|Q| = k3 Þ AA' = 1 (as A' = A)
Þ |P| =2k ¹ 0 Þ P is an invertible matrix Þ A is orthogonal

Q PQ = kI So, x2 + y2 + z2 = 1 and xy + yz + zx = 0
Þ (x + y + z)2 = 1 + 2 × 0
\ Q = kP–1I Þx+y+z=1
adj.P Thus,
\ Q= x3 + y3 + z3 = 3 × 2 + 1 × (1 – 0)
2 =7
24. Official Ans. by NTA (13)
k
Q q23 = - Sol. a 2 + b2 = I2 + A I2 - A
-1

8
q q
= sec2 ´ cos2 = 1
- ( 3a + 4 ) k 2 2
\ =- Þk=4
2 8 25. Official Ans. by NTA (4)
é1 0 0ù
\ |P| = 2k Þ k = 10 + 6a ...(i)
Sol. A = êê0 2 0 úú
Put value of k in (i).. we get a =–1
êë3 0 -1úû
21. Official Ans. by NTA (4)
é1 0 0ù é1 0 0 ù
A = êê0 4 0 ú , A = êê0 8 0 úú
ú
Sol. |A| = 4 2 3

Þ |2A| = 23 × 4 = 32 êë00 1 úû êë3 0 -1úû

Q B is obtained by R2 ® 2R2 + 5R3 é1 0 0 ù
A = êê0 16 0 úú
4

Þ |B| = 2 × 32 = 64 êë0 0 1 úû
option (4) Hence
22. Official Ans. by NTA (4) é1 0 0ù é1 0 0ù
A = êê0 2 20
20
0 ú , A = êê0 219
ú 19
0 úú
é 1 -a ù
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices

Sol. A=ê ú AAT = I2 êë0 0 1 úû êë3 0 -1úû

ëa b û é1 + a + b 0 0 ù
So
A 20 + aA19 + b A = êê 0 2 20 + a.219 + 2b 0 úú
é 1 -a ù é 1 a ù é1 0 ù êë 3a + 3b 0 1 - a - b úû
Þ ê úê ú=ê ú
ëa b û ë-a b û ë 0 1 û é1 0 0 ù
= êê0 4 0 úú
é 1 + a 2 a - ab ù é 1 0 ù
Þ ê 2ú
=ê ú êë0 0 1 úû
ë a - ab a + b û ë 0 1 û
2

Therefore a + b = 0 and 220 + 219a – 2a = 4

Þ a2 = 0 & b2 = 1
Þa=
(
4 1 - 218 ) = -2
\ a4 + b4 = 1 2 (2 18
- 1)

hence b = 2
so (b – a) = 4

E
10 Matrices ALLEN®
26. Official Ans. by NTA (1) On comparing we get
æa bö
Sol. A=ç ÷ , a, b, c Î I k2 + 1 2 2
èb cø = Þ k +1 = 2
3 3

æ a b öæ a b ö æ a + b
2 2
b(a + c) ö Þ k = ±1 ...(3)
A2 = ç ÷ç =
÷ çç 2 ÷÷
è b c øè b c ø è b ( a + c ) b + c ø
2

2
& (k - 1) = 0 Þ k = 1 ...(4)
Sum of the diagonal entries of 3

A2 = a2 + 2b2 + c2 From both we get k = 1

Given a2 + 2b2 + c2 = 1, a, b, c Î I 28. Official Ans. by NTA (3)

b = 0 & a2 + c2 = 1 é i -i ù
Sol. A=ê ú
ë -i i û
Case–1 : a = 0 Þ c = ±1 (2-matrices)
é -2 2 ù é-1 1 ù
Case-2 : c = 0 Þ a = ±1 (2-matrices) A2 = ê ú = 2ê ú
ë 2 -2 û ë 1 -1û
Total = 4 matrices
é 2 -2 ù é 1 -1ù
27. Official Ans by NTA (1) A4 = 22 ê ú = 8ê ú
ë -2 2 û ë -1 1 û
Sol. A = XB
é 2 -2 ù é 1 -1ù
A8 = 64 ê ú = 128 ê ú
é a1 ù 1 é1 -1ù é b1 ù ë -2 2 û ë -1 1 û
ê ú= ê úê ú
ëa2 û 3 ë1 k û ë b 2 û
éxù é 8 ù
A8 ê ú = ê ú
é 3 a1 ù é b1 - b2 ù ë y û ë64 û
ê ú=ê ú
êë 3 a 2 ûú ë b1 + kb2 û
é 1 -1ù éx ù é 8 ù
Þ 128 ê úê ú = ê ú
ë-1 1 û ëy û ë64 û
b1 – b2 = 3 a1 ...(1)
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices
é x-y ù é 8 ù
b1 + kb2 = 3a2 ...(2) Þ 128 ê ú=ê ú
ë - x + y û ë64 û

Given, a12 + a22 =

2 2
3
(
b1 + b22 ) Þ x-y=
1
.....(1)
16
(1)2 + (2)2
1
& -x + y = .....(2)
(b1 + b2)2 + (b1 + kb2)2 = 3(a12 + a22) 2

Þ From (1) & (2) : No solution.

2 2 (1 + k 2 ) 2 2
a12 + a22 = b + b 2 + b1b 2 (k - 1)
3 1 3 3

2 2
Given, a12 + a22 = b12 + b22
3 3

E
 ALLEN® Matrices 11
29. Official Ans. by NTA (766) 31. Official Ans. by NTA (2020)

éa b cù éa b ù éa ù
Sol. Let A = ê d e f ú Sol. A=ê ú , B=ê ú
êë g h i úû ëc d û ëb û

diagonal elements of AB = B

AAT, a2 + b2 + c2, d2 + e2 + f2, g2 + b2 + c2 Þ (A – I) B = O

Sum = a2 + b2 + c2 + d2 + e2 + f2 + g2 + h2 + i2 = 9 Þ |A – I | = O, since B ¹ O
a, b, c, d, e, f, g, h, i Î {0, 1, 2, 3}
(a - 1) b
Case No. of Matrices =0
c (d - 1)
(1) All – 1s 9!
= 1
9! ad – bc = 2020
(2) One ® 3 9! 32. Official Ans. by NTA (3)
=9
remaining-0 1!´ 8!
Sol. A2 = sin2a I
(3) One-2 9!
= 8 ´ 63
five-1s 1!´ 5!´ 3! I æ 2 1ö
2

So, A2 - = ç sin a - ÷ = 0
three-0s 2 è 2ø
(4) two – 2's 9!
= 63 ´ 4 1
one-1 2!´ 6! Þ sin a = ±
2
six-0's
33. Official Ans. by NTA (16)
Total no. of ways = 1 + 9 + 8 × 63 + 63 × 4

= 766 Sol. 2A adj (2A) = |2A|I

30. Official Ans. by NTA (1) Þ A adj (2A) = –4I ....(i)

3 4 2 x Now, E = |A4| + |A10 – (adj(2A))10|

node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices

Sol. 4 5 2 y =0
| A 20 - A 10 (adj 2A)10 |
5 k z = (–2)4 +
| A |10
R2 ® R1 + R3 – 2R2
| A 20 - (A adj(2A))10 |
= 16 +
3 4 2 x | A |10
Þ 0 k -6 2 0 = 0
5 k z | A 20 - 210 I |
= 16 + (from (1))
210
Þ ( k - 6 2 ) ( 3z - 5x ) = 0
Now, characteristic roots of A are 2 and –1.
if 3z – 5x = 0 Þ 3(x + 2d) – 5x = 0 So, characteristic roots of A20 are 210 and 1.
Þ x = 3d (Not possible) Hence, (A20 – 210 I) (A20 – I) = 0
Þ k=6 2 Þ k2 = 72 Option (1) Þ |A20 – 210I| = 0 (as A20 ¹ I)
Þ E = 16 Ans.

E
12 Matrices ALLEN®
34. Official Ans. by NTA (2) 35. Official Ans. by NTA (6)
æ 1 2 0ö é2 -1ù
ç ÷ Sol. P=ê ú
Sol. A + 2B = ç 6 -3 3 ÷ ...(1) ë 5 -3û
ç -5 3 1 ÷
è ø
é5 0 ù é16 -8 ù é -11 8 ù
æ 2 -1 5 ö
5I - 8P = ê ú-ê ú =ê ú
ç ÷ ë0 5 û ë 40 -24 û ë -40 29 û
2A - B = ç 2 -1 6 ÷
ç0 1 2÷ é -1 1 ù
è ø P2 = ê ú
ë-5 4 û
æ 4 -2 10 ö
ç ÷ é 3 -2ù é -11 8 ù
Þ 4A – 2B = ç 4 -2 12 ÷ ...(2)
P3 = ê ú Þ P6 = ê ú = Pn
ç0 2 4 ÷ ë10 -7û ë -40 29û
è ø
æ 5 0 10 ö Þn=6
ç ÷
(1) + (2) Þ 5A = ç 10 -5 15 ÷
ç -5 5 5 ÷
è ø
æ 1 0 2ö æ 2 0 4ö
ç ÷ ç ÷
A = ç 2 -1 3 ÷ and 2A = ç 4 -2 6 ÷
ç -1 1 1 ÷ ç -2 2 2 ÷
è ø è ø
æ 2 0 4 ö æ 2 -1 5 ö
ç ÷ ç ÷
\ B = ç 4 -2 6 ÷ - ç 2 -1 6 ÷
ç -2 2 2 ÷ ç 0 1 2 ÷
è ø è ø
æ 0 1 -1ö
ç ÷
B = ç 2 -1 0 ÷
ç -2 1 0 ÷
è ø
tr ( A ) = 1 - 1 + 1 = 1
tr ( B ) = -1

tr(A) = 1 and tr(B) = –1

node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Matrices

\ tr(A) – tr(B) = 2

E
 Question bank on Parabola, Ellipse & Hyperbola
Select the correct alternative : (Only one is correct)

Q.1 Two mutually perpendicular tangents of the parabola y2 = 4ax meet the axis in P1 and P2. If S is the focus
1 1
of the parabola then  is equal to
l (SP1 ) l (SP2 )

4 2 1 1
(A) (B) (C) (D)
a a a 4a

Q.2 Which one of the following equations represented parametrically, represents equation to a parabolic
profile ?
t
(A) x = 3 cos t ; y = 4 sin t (B) x2  2 =  2 cos t ; y = 4 cos2
2
t t
(C) x = tan t ; y = sec t (D) x = 1  sin t ; y = sin + cos
2 2

x 2 y2
Q.3 The magnitude of the gradient of the tangent at an extremity of latera recta of the hyperbola 2  2  1
a b
is equal to (where e is the eccentricity of the hyperbola)
(A) be (B) e (C) ab (D) ae

x2 y2
Q.4 Let 'E' be the ellipse + = 1 & 'C' be the circle x2 + y2 = 9. Let P & Q be the points (1 , 2) and
9 4
(2, 1) respectively. Then :
(A) Q lies inside C but outside E (B) Q lies outside both C & E
(C) P lies inside both C & E (D) P lies inside C but outside E.

Q.5 Let S be the focus of y2 = 4x and a point P is moving on the curve such that it's abscissa is increasing at
the rate of 4 units/sec, then the rate of increase of projection of SP on x + y = 1 when P is at (4, 4) is
3
(A) 2 (B) – 1 (C) – 2 (D) –
2

x 2 y2
Q.6 Eccentricity of the hyperbola conjugate to the hyperbola   1 is
4 12
2 4
(A) (B) 2 (C) 3 (D)
3 3

Q.7 The points of contact Q and R of tangent from the point P (2, 3) on the parabola y2 = 4x are
1
(A) (9, 6) and (1, 2) (B) (1, 2) and (4, 4) (C) (4, 4) and (9, 6) (D) (9, 6) and ( , 1)
4

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [2]

 y2
Q.8 The eccentricity of the ellipse (x – 3)2 + (y – 4)2 = is
9
3 1 1 1
(A) (B) (C) (D)
2 3 3 2 3
x2 y2
Q.9 The asymptote of the hyperbola  = 1 form with any tangent to the hyperbola a triangle whose
a 2 b2
area is a2tan  in magnitude then its eccentricity is :
(A) sec (B) cosec (C) sec2 (D) cosec2

Q.10 A tangent is drawn to the parabola y2 = 4x at the point 'P' whose abscissa lies in the interval [1,4]. The
maximum possible area of the triangle formed by the tangent at 'P' , ordinate of the point 'P' and the
x-axis is equal to
(A) 8 (B) 16 (C) 24 (D) 32

Q.11 From an external point P, pair of tangent lines are drawn to the parabola, y2 = 4x. If 1 & 2 are the

inclinations of these tangents with the axis of x such that, 1 + 2 = , then the locus of P is :
4
(A) x  y + 1 = 0 (B) x + y  1 = 0 (C) x  y  1 = 0 (D) x + y + 1 = 0
x2 y2
Q.12 The equation + = 1 (p  4, 29) represents
29  p 4  p
(A) an ellipse if p is any constant greater than 4.
(B) a hyperbola if p is any constant between 4 and 29.
(C) a rectangular hyperbola if p is any constant greater than 29.
(D) no real curve if p is less than 29.
x 2 y2
Q.13 For an ellipse   1 with vertices A and A', tangent drawn at the point P in the first quadrant meets
9 4
the y-axis in Q and the chord A'P meets the y-axis in M. If 'O' is the origin then OQ2 – MQ2 equals to
(A) 9 (B) 13 (C) 4 (D) 5

Q.14 Length of the normal chord of the parabola, y2 = 4x, which makes an angle of with the axis of x is:
4
(A) 8 (B) 8 2 (C) 4 (D) 4 2

Q.15 An ellipse and a hyperbola have the same centre origin, the same foci and the minor-axis of the one is the
same as the conjugate axis of the other. If e1, e2 be their eccentricities respectively, then e12  e 22
equals
(A) 1 (B) 2 (C) 3 (D) 4

Q.16 The coordiantes of the ends of a focal chord of a parabola y2 = 4ax are (x1, y1) and (x2, y2) then
x1x2 + y1y2 has the value equal to
(A) 2a2 (B) – 3a2 (C) – a2 (D) 4a2

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [3]

 x2 y 2
Q.17 The line, lx + my + n = 0 will cut the ellipse + = 1 in points whose eccentric angles differ by
a 2 b2
/2 if :
(A) a2l2 + b2n2 = 2 m2 (B) a2m2 + b2l2 = 2 n2
(C) a2l2 + b2m2 = 2 n2 (D) a2n2 + b2m2 = 2 l2

Q.18 Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola
16y2 – 9x2 = 1 is
(A) x2 + y2 = 9 (B) x2 + y2 = 1/9 (C) x2 + y2 =7/144 (D) x2 + y2 = 1/16

Q.19 If the normal to a parabola y2 = 4ax at P meets the curve again in Q and if PQ and the normal at Q makes
angles  and  respectively with the x-axis then tan (tan  + tan ) has the value equal to
1
(A) 0 (B) – 2 (C) – (D) – 1
2

Q.20 If the normal to the parabola y2 = 4ax at the point with parameter t1 , cuts the parabola again at the point
with parameter t2 , then
(A) 2 < t 22 < 8 (B) 2 < t 22 < 4 (C) t 22 > 4 (D) t 22 > 8

Q.21 The locus of the point of instruction of the lines 3 x  y  4 3 t = 0 & 3 tx + ty  4 3 = 0

(where t is a parameter) is a hyperbola whose eccentricity is
2 4
(A) 3 (B) 2 (C) (D)
3 3
x2 y2
Q.22 The equation to the locus of the middle point of the portion of the tangent to the ellipse + =1
16 9
included between the co-ordinate axes is the curve :
(A) 9x2 + 16y2 = 4 x2y2 (B) 16x2 + 9y2 = 4 x2y2
2 2
(C) 3x + 4y = 4 x y2 2 (D) 9x2 + 16y2 = x2y2

Q.23 A parabola y = ax2 + bx + c crosses the x  axis at ( , 0) ( , 0) both to the right of the origin. A circle
also passes through these two points. The length of a tangent from the origin to the circle is :
bc b c
(A) (B) ac2 (C) (D)
a a a

Q.24 Two parabolas have the same focus. If their directrices are the x  axis & the y axis respectively, then
the slope of their common chord is :
(A) ± 1 (B) 4/3 (C) 3/4 (D) none

Q.25 The locus of a point in the Argand plane that moves satisfying the equation,
z  1 + i  z  2  i = 3
(A) is a circle with radius 3 & centre at z = 3/2
(B) is an ellipse with its foci at 1  i and 2 + i and major axis = 3
(C) is a hyperbola with its foci at 1  i and 2 + i and its transverse axis = 3
(D) is none of the above .

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [4]

Q.26 A circle has the same centre as an ellipse & passes through the foci F1 & F2 of the ellipse, such that the
two curves intersect in 4 points. Let 'P' be any one of their point of intersection. If the major axis of the
ellipse is 17 & the area of the triangle PF1F2 is 30, then the distance between the foci is :
(A) 11 (B) 12 (C) 13 (D) none

Q.27 The straight line joining any point P on the parabola y2 = 4ax to the vertex and perpendicular from the
focus to the tangent at P, intersect at R, then the equaiton of the locus of R is
(A) x2 + 2y2 – ax = 0 (B) 2x2 + y2 – 2ax = 0
2 2
(C) 2x + 2y – ay = 0 (D) 2x2 + y2 – 2ay = 0

Q.28 A normal chord of the parabola y2 = 4x subtending a right angle at the vertex makes an acute angle  with
the x-axis, then  equals to
(A) arc tan 2 (B) arc sec 3 (C) arc cot 2 (D) none

Q.29 If the eccentricity of the hyperbola x2  y2 sec2  = 5 is 3 times the eccentricity of the ellipse
x2 sec2  + y2 = 25, then a value of  is :
(A) /6 (B) /4 (C) /3 (D) /2

Q.30 Point 'O' is the centre of the ellipse with major axis AB & minor axis CD. Point F is one focus of the
ellipse. If OF = 6 & the diameter of the inscribed circle of triangle OCF is 2, then the product
(AB) (CD) is equal to
(A) 65 (B) 52 (C) 78 (D) none

Q.31 Locus of the feet of the perpendiculars drawn from vertex of the parabola y2 = 4ax upon all such chords
of the parabola which subtend a right angle at the vertex is
(A) x2 + y2 – 4ax = 0 (B) x2 + y2 – 2ax = 0
2 2
(C) x + y + 2ax = 0 (D) x2 + y2 + 4ax = 0

Q.32 For all real values of m, the straight line y = mx + 9 m2  4 is a tangent to the curve :
(A) 9x2 + 4y2 = 36 (B) 4x2 + 9y2 = 36 (C) 9x2  4y2 = 36 (D) 4x2  9y2 = 36
Q.33 C is the centre of the circle with centre (0, 1) and radius unity. P is the parabola y = ax2. The set of values
of 'a' for which they meet at a point other than the origin, is
 1 1 1 1 
(A) a > 0 (B) a   0,  (C)  ,  (D)  ,  
 2 4 2 2 
4 x2 y2
Q.34 A tangent having slope of  to the ellipse + = 1 intersects the major & minor axes in points A
3 18 32
& B respectively. If C is the centre of the ellipse then the area of the triangle ABC is :
(A) 12 sq. units (B) 24 sq. units (C) 36 sq. units (D) 48 sq. units

x 2 y2 x 2 y2 1
Q.35 The foci of the ellipse  2  1 and the hyperbola   coincide. Then the value of b2 is
16 b 144 81 25
(A) 5 (B) 7 (C) 9 (D) 4

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [5]

Q.36 TP & TQ are tangents to the parabola, y2 = 4ax at P & Q. If the chord PQ passes through the fixed point
( a, b) then the locus of T is :
(A) ay = 2b (x  b) (B) bx = 2a (y  a)
(C) by = 2a (x  a) (D) ax = 2b (y  b)

Q.37 Through the vertex O of the parabola, y2 = 4ax two chords OP & OQ are drawn and the circles on OP
& OQ as diameters intersect in R. If 1, 2 &  are the angles made with the axis by the tangents at P &
Q on the parabola & by OR then the value of, cot 1 + cot 2 =
(A)  2 tan  (B)  2 tan () (C) 0 (D) 2 cot 

Q.38 Locus of the middle points of the parallel chords with gradient m of the rectangular hyperbola xy = c2 is
(A) y + mx = 0 (B) y  mx = 0 (C) my  x = 0 (D) my + x = 0

Q.39 If the chord through the point whose eccentric angles are  &  on the ellipse,
(x2/a2) + (y2/b2) = 1 passes through the focus, then the value of (1 + e) tan(/2) tan(/2) is
(A) e + 1 (B) e  1 (C) 1  e (D) 0

Q.40 The given circle x2 + y2 + 2px = 0, p  R touches the parabola y2 = 4x externally, then
(A) p < 0 (B) p > 0 (C) 0 < p < 1 (D) p < – 1

Q.41 The locus of the foot of the perpendicular from the centre of the hyperbola xy = c2 on a variable tangent
is :
(A) (x2  y2)2 = 4c2 xy (B) (x2 + y2)2 = 2c2 xy
2 2 2
(C) (x + y ) = 4x xy (D) (x2 + y2)2 = 4c2 xy

Q.42 The tangent at P to a parabola y2 = 4ax meets the directrix at U and the latus rectum at V then SUV
(where S is the focus) :
(A) must be a right triangle (B) must be an equilateral triangle
(C) must be an isosceles triangle (D) must be a right isosceles triangle.

Q.43 Given the base of a triangle and sum of its sides then the locus of the centre of its incircle is
(A) straight line (B) circle (C) ellipse (D) hyperbola

x2 y2
Q.44 P is a point on the hyperbola  = 1, N is the foot of the perpendicular from P on the transverse
a 2 b2
axis. The tangent to the hyperbola at P meets the transverse axis at T . If O is the centre of the hyperbola,
the OT. ON is equal to :
(A) e2 (B) a2 (C) b2 (D)b2/a2

Q.45 Two parabolas y2 = 4a(x - l1) and x2 = 4a (y – l2) always touch one another, the quantities l1 and l2 are
both variable. Locus of their point of contact has the equation
(A) xy = a2 (B) xy = 2a2 (C) xy = 4a2 (D) none

Q.46 If a normal to a parabola y2 = 4ax make an angle  with its axis, then it will cut the curve again at an angle
1  1 
(A) tan–1(2 tan) (B) tan1  tan  (C) cot–1  tan  (D) none
2  2 

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [6]

Q.47 If PN is the perpendicular from a point on a rectangular hyperbola x2  y2 = a2 on any of its asymptotes,
then the locus of the mid point of PN is :
(A) a circle (B) a parabola (C) an ellipse (D) a hyperbola
x2 y2 x2 y2
Q.48 Which one of the following is the common tangent to the ellipses,  = 1 &  =1?
a 2  b2 b2 a 2 a 2  b2
(A) ay = bx + a 4  a 2 b 2  b 4 (B) by = ax  a 4  a 2 b 2  b 4
(C) ay = bx  a 4  a 2 b 2  b 4 (D) by = ax + a 4  a 2 b 2  b 4

Q.49 The vertex of a parabola is (2,2) and the co-ordinates of its two extrimities of the latus rectum are (–2,0)
and (6,0). The equation of the parabola is
(A) y2 – 4y + 8x – 12 = 0
(B) x2 + 4x – 8y – 12 = 0
(C) x2 – 4x + 8y – 12 = 0
(D) x2 – 8y – 4x + 20 = 0

Q.50 The equation to the chord joining two points (x1, y1) and (x2, y2) on the rectangular hyperbola xy = c2
is
x y x y
(A) + =1 (B) + =1
x1  x 2 y1  y 2 x1  x 2 y1  y 2
x y x y
(C) + =1 (D) + =1
y1  y 2 x1  x 2 y1  y 2 x1  x 2

Q.51 The length of the chord of the parabola y2 = x which is bisected at the point (2, 1) is
(A) 2 3 (B) 4 3
(C) 3 2 (D) 2 5
x2 y2
Q.52 The normal at a variable point P on an ellipse  = 1 of eccentricity e meets the axes of the ellipse
a 2 b2
in Q and R then the locus of the mid-point of QR is a conic with an eccentricity e  such that :
(A) e  is independent of e (B) e  = 1
(C) e  = e (D) e  = 1/e
Q.53 If the tangents & normals at the extremities of a focal chord of a parabola intersect at (x1, y1) and
(x2, y2) respectively, then :
(A) x1 = x2 (B) x1 = y2 (C) y1 = y2 (D) x2 = y1

Q.54 If P(x1, y1), Q(x2, y2), R(x3, y3) & S(x4, y4) are 4 concyclic points on the rectangular hyperbola
x y = c2, the co-ordinates of the orthocentre of the triangle PQR are :
(A) (x4,  y4) (B) (x4, y4) (C) ( x4,  y4) (D) ( x4, y4)

Q.55 If the chord of contact of tangents from a point P to the parabola y2 = 4ax touches the parabola x2 = 4by,
the locus of P is :
(A) circle (B) parabola (C) ellipse (D) hyperbola

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [7]

Q.56 An ellipse is drawn with major and minor axes of lengths 10 and 8 respectively. Using one focus as
centre, a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse.
The radius of the circle is
(A) 3 (B) 2 (C) 2 2 (D) 5

Q.57 The latus rectum of a parabola whose focal chord PSQ is such that SP = 3 and SQ = 2 is given by
(A) 24/5 (B) 12/5 (C) 6/5 (D) none of these

Q.58 The chord PQ of the rectangular hyperbola xy = a2 meets the axis of x at A ; C is the mid point of PQ
& 'O' is the origin. Then the  ACO is :
(A) equilateral (B) isosceles
(C) right angled (D) right isosceles.

Q.59 The circle x2 + y2 = 5 meets the parabola y2 = 4x at P & Q. Then the length PQ is equal to
(A) 2 (B) 2 2 (C) 4 (D) none

Q.60 A common tangent to 9x2 + 16y2 = 144 ; y2  x + 4 = 0 & x2 + y2  12x + 32 = 0 is

(A) y = 3 (B) x =  4 (C) x = 4 (D) y =  3

Q.61 A conic passes through the point (2, 4) and is such that the segment of any of its tangents at any point
contained between the co-ordinate axes is bisected at the point of tangency. Then the foci of the conic
are

  
(A) 2 2 , 0 &  2 2 , 0    
(B) 2 2 , 2 2 &  2 2 ,  2 2 
(C) (4, 4) & ( 4,  4) (D)  4 2,4 2 &  4 2 , 4 2

Q.62 If two normals to a parabola y2 = 4ax intersect at right angles then the chord joining their feet passes
through a fixed point whose co-ordinates are
(A) ( 2a, 0) (B) (a, 0) (C) (2a, 0) (D) none

Q.63 The equation of a straight line passing through the point (3, 6) and cutting the curve y = x orthogonally
is
(A) 4x + y – 18 =0 (B) x + y – 9 = 0 (C) 4x – y – 6 = 0 (D) none

Q.64 Latus rectum of the conic satisfying the differential equation, x dy + ydx = 0 and passing through the
point (2, 8) is
(A) 4 2 (B) 8 (C) 8 2 (D) 16

Q.65 The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the
tangent and normal at its point whose eccentric angle is /4 is

(A)
a 2

 b 2 ab
(B)
a 2
 b2  (C)
a  b 
2 2

(D)
a 2  b2
a 2  b2 a 2

 b 2 ab ab  a  b 
2 2
a 2

 b 2 ab

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [8]

Q.66 PQ is a normal chord of the parabola y2 = 4ax at P, A being the vertex of the parabola. Through P a line
is drawn parallel to AQ meeting the xaxis in R. Then the length of AR is :
(A) equal to the length of the latus rectum
(B) equal to the focal distance of the point P
(C) equal to twice the focal distance of the point P
(D) equal to the distance of the point P from the directrix.

Q.67 If the normal to the rectangular hyperbola xy = c2 at the point 't' meets the curve again at 't1' then t3 t1
has the value equal to
(A) 1 (B) – 1 (C) 0 (D) none

Q.68 Locus of the point of intersection of the perpendicular tangents of the curve
y2 + 4y  6x  2 = 0 is :
(A) 2x  1 = 0 (B) 2x + 3 = 0 (C) 2y + 3 = 0 (D) 2x + 5 = 0
a2 x2 y2
Q.69 If tan 1. tan 2 =  then the chord joining two points 1 & 2 on the ellipse 2  2 = 1 will subtend
b2 a b
a right angle at :
(A) focus (B) centre
(C) end of the major axis (D) end of the minor axis

x2 y2
Q.70 With one focus of the hyperbola   1 as the centre , a circle is drawn which is tangent to the
9 16
hyperbola with no part of the circle being outside the hyperbola. The radius of the circle is
11
(A) less than 2 (B) 2 (C) (D) none
3

Q.71 Length of the focal chord of the parabola y2 = 4ax at a distance p from the vertex is :
2a2 a3 4a3 p2
(A) (B) (C) (D)
p p2 p2 a

Q.72 The locus of a point such that two tangents drawn from it to the parabola y2 = 4ax are such that the slope
of one is double the other is :
9 9
(A) y2 = ax (B) y2 = ax (C) y2 = 9 ax (D) x2 = 4 ay
2 4
x 2 y2
Q.73 AB is a double ordinate of the hyperbola 2  2  1 such that AOB (where 'O' is the origin) is an
a b
equilateral triangle, then the eccentricity e of the hyperbola satisfies
2 2 2
(A) e > 3 (B) 1 < e < (C) e = (D) e >
3 3 3

Q.74 An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability that
this point lies outside the ellipse is 2/3 then the eccentricity of the ellipse is :
2 2 5 8 2
(A) (B) (C) (D)
3 3 9 3

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [9]

Q.75 The triangle PQR of area 'A' is inscribed in the parabola y2 = 4ax such that the vertex P lies at the vertex
of the parabola and the base QR is a focal chord. The modulus of the difference of the ordinates of the
points Q and R is :
A A 2A 4A
(A) (B) (C) (D)
2a a a a

x 2 y2
Q.76 If the product of the perpendicular distances from any point on the hyperbola 2  2  1 of eccentricity
a b
e = 3 from its asymptotes is equal to 6, then the length of the transverse axis of the hyperbola is
(A) 3 (B) 6 (C) 8 (D) 12

Q.77 The point(s) on the parabola y2 = 4x which are closest to the circle,
x2 + y2  24y + 128 = 0 is/are :
(A) (0, 0) 
(B) 2 , 2 2  (C) (4, 4) (D) none

Q.78 A point P moves such that the sum of the angles which the three normals makes with the axis drawn from
P on the standard parabola, is constant. Then the locus of P is :
(A) a straight line (B) a circle (C) a parabola (D) a line pair

Q.79 If x + iy =   i where i =  1 and  and  are non zero real parameters then  = constant and
 = constant, represents two systems of rectangular hyperbola which intersect at an angle of
   
(A) (B) (C) (D)
6 3 4 2
Q.80 Three normals drawn from any point to the parabola y2 = 4ax cut the line x = 2a in points whose
ordinates are in arithmetical progression. Then the tangents of the angles which the normals make the
axis of the parabola are in :
(A) A.P. (B) G.P. (C) H.P. (D) none
Q.81 A circle is described whose centre is the vertex and whose diameter is three-quarters of the latus rectum
of the parabola y2 = 4ax. If PQ is the common chord of the circle and the parabola and L1 L2 is the latus
rectum, then the area of the trapezium PL1 L2Q is :
 2 1 2  2  2
(A) 3 2 a2 (B)  2  a (C) 4a2 (D)   a2
   2 

Q.82 The tangent to the hyperbola xy = c2 at the point P intersects the x-axis at T and the y-axis at T. The
normal to the hyperbola at P intersects the x-axis at N and the y-axis at N. The areas of the triangles
1 1
PNT and PN'T' are  and ' respectively, then  is
 '
(A) equal to 1 (B) depends on t (C) depends on c (D) equal to 2

Q.83 If y = 2 x  3 is a tangent to the parabola y2 = 4a  x  1 , then ' a ' is equal to :

 3
22 14  14
(A) (B)  1 (C) (D)
3 3 3

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [10]

Q.84 An ellipse having foci at (3, 3) and (– 4, 4) and passing through the origin has eccentricity equal to
3 2 5 3
(A) (B) (C) (D)
7 7 7 5

Q.85 The ellipse 4x2 + 9y2 = 36 and the hyperbola 4x2 – y2 = 4 have the same foci and they intersect at right
angles then the equation of the circle through the points of intersection of two conics is
(A) x2 + y2 = 5 (B) 5 (x2 + y2) – 3x – 4y = 0
(C) 5 (x2 + y2) + 3x + 4y = 0 (D) x2 + y2 = 25

Q.86 Tangents are drawn from the point ( 1, 2) on the parabola y2 = 4 x. The length , these tangents will
intercept on the line x = 2 is :
(A) 6 (B) 6 2 (C) 2 6 (D) none of these

Q.87 The curve describes parametrically by x = t2 – 2t + 2, y = t2 + 2t + 2 represents

(A) straight line (B) pair of straight lines
(C) circle (D) parabola

Q.88 At the point of intersection of the rectangular hyperbola xy = c2 and the parabola y2 = 4ax tangents to
the rectangular hyperbola and the parabola make an angle  and  respectively with the axis of X, then
(A)  = tan–1(– 2 tan) (B)  = tan–1(– 2 tan)
1 1
(C)  = tan–1(– tan) (D)  = tan–1(– tan)
2 2

Q.89 The tangent and normal at P(t), for all real positive t, to the parabola y2 = 4ax meet the axis of the
parabola in T and G respectively, then the angle at which the tangent at P to the parabola is inclined to the
tangent at P to the circle passing through the points P, T and G is
(A) cot–1t (B) cot–1t2 (C) tan–1t (D) tan–1t2
x 2 y2 x 2 y2
Q.90 Area of the quadrilateral formed with the foci of the hyperbola 2   1 and 2  2  1 is
a b2 a b
1 2
(A) 4(a2 + b2) (B) 2(a2 + b2) (C) (a2 + b2) (D) (a + b2)
2

Q.91 A bar of length 20 units moves with its ends on two fixed straight lines at right angles. A point P marked
on the bar at a distance of 8 units from one end describes a conic whose eccentricity is
5 2 4 5
(A) (B) (C) (D)
9 3 9 3

Q.92 In a square matrix A of order 3, ai i = mi + i where i = 1, 2, 3 and mi's are the slopes (in increasing order
of their absolute value) of the 3 normals concurrent at the point (9, – 6) to the parabola y2 = 4x. Rest all
other entries of the matrix are one. The value of det. (A) is equal to
(A) 37 (B) – 6 (C) – 4 (D) – 9
x2
Q.93 An equation for the line that passes through (10, –1) and is perpendicular to y =  2 is
4
(A) 4x + y = 39 (B) 2x + y = 19 (C) x + y = 9 (D) x + 2y = 8

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [11]

 Direction for Q.94 to Q.97. (4 questions together)
A quadratic polynomial y = f (x) with absolute term 3 neither touches nor intersects the abscissa axis and
is symmetric about the line x = 1. The coefficient of the leading term of the polynomial is unity. A point
A(x1, y1) with abscissa x1 = 1 and a point B(x2, y2) with ordinate y2 = 11 are given in a cartisian
rectangular system of co-ordinates OXY in the first quadrant on the curve y = f (x) where 'O' is the
origin. Now answer the following questions:
Q.94 Vertex of the quadratic polynomial is
(A) (1, 1) (B) (2, 3) (C) (1, 2) (D) none
 
Q.95 The scalar product of the vectors OA and OB is
(A) –18 (B) 26 (C) 22 (D) –22
Q.96 The area bounded by the curve y = f(x) and a line y = 3 is
(A) 4/3 (B) 5/3 (C) 7/3 (D) 28/3
Q.97 The graph of y = f(x) represents a parabola whose focus has the co-ordinates
(A) (1, 7/4) (B) (1, 5/4) (C) (1, 5/2) (D) (1, 9/4)
Direction for Q.98 to Q.66. (3 questions together)
The graph of the conic x2 – (y – 1)2 = 1 has one tangent line with positive slope that passes through the
origin. the point of tangency being (a, b). Then
a
Q.98 The value of sin–1   is
b
5   
(A) (B) (C) (D)
12 6 3 4
Q.99 Length of the latus rectum of the conic is
(A) 1 (B) 2 (C) 2 (D) none
Q.100 Eccentricity of the conic is
4
(A) (B) 3 (C) 2 (D) none
3

Select the correct alternatives : (More than one are correct)

Q.101 Consider a circle with its centre lying on the focus of the parabola, y2 = 2 px such that it touches the
directrix of the parabola. Then a point of intersection of the circle & the parabola is :
p  p   p   p 
(A)  , p (B)  ,  p (C)   , p (D)   ,  p
2  2   2   2 

Q.102 Identify the statements which are True.

(A) the equation of the director circle of the ellipse, 5x2 + 9y2 = 45 is x2 + y2 = 14.
x2 y 2
(B) the sum of the focal distances of the point (0 , 6) on the ellipse + = 1 is 10.
25 36
(C) the point of intersection of any tangent to a parabola & the perpendicular to it from the focus lies on
the tangent at the vertex.
x2 y2
(D) P & Q are the points with eccentric angles  &  +  on the ellipse  = 1, then the area of the
a 2 b2
triangle OPQ is independent of .
Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [12]
 x2 y2
Q.103 For the hyperbola  = 1 the incorrect statement is :
9 3
(A) the acute angle between its asymptotes is 60º
(B) its eccentricity is 4/3
(C) length of the latus rectum is 2
(D) product of the perpendicular distances from any point on the hyperbola on its asymptotes is less than
the length of its latus rectum .

Q.104 The locus of the mid point of the focal radii of a variable point moving on the parabola, y2 = 4ax is a
parabola whose
(A) Latus rectum is half the latus rectum of the original parabola
(B) Vertex is (a/2, 0)
(C) Directrix is y-axis
(D) Focus has the co-ordinates (a, 0)

Q.105 P is a point on the parabola y2 = 4ax (a > 0) whose vertex is A. PA is produced to meet the directrix in
D and M is the foot of the perpendicular from P on the directrix. If a circle is described on MD as a
diameter then it intersects the xaxis at a point whose coordinates are :
(A) ( 3a, 0) (B) ( a, 0) (C) ( 2a, 0) (D) (a, 0)

Q.106 If the circle x2 + y2 = a2 intersects the hyperbola xy = c2 in four points P(x1, y1), Q(x2, y2), R(x3, y3),
S(x4, y4), then
(A) x1 + x2 + x3 + x4 = 0 (B) y1 + y2 + y3 + y4 = 0
(C) x1 x2 x3 x4 = c4 (D) y1 y2 y3 y4 = c4

x2 y2
Q.107 Extremities of the latera recta of the ellipses 2  2 1 (a > b) having a given major axis 2a lies on
a b
2
(A) x = a(a – y) 2
(B) x = a (a + y) (C) y2 = a(a + x) (D) y2 = a (a – x)

Q.108 Let y2 = 4ax be a parabola and x2 + y2 + 2 bx = 0 be a circle. If parabola and circle touch each other
externally then :
(A) a > 0, b > 0 (B) a > 0, b < 0 (C) a < 0, b > 0 (D) a < 0, b < 0

Q.109 The tangent to the hyperbola, x2  3y2 = 3 at the point  

3 , 0 when associated with two asymptotes
constitutes :
(A) isosceles triangle (B) an equilateral triangle
(C) a triangles whose area is 3 sq. units (D) a right isosceles triangle .

Q.110 Let P, Q and R are three co-normal points on the parabola y2 = 4ax. Then the correct statement(s) is/are
(A) algebraic sum of the slopes of the normals at P, Q and R vanishes
(B) algebraic sum of the ordinates of the points P, Q and R vanishes
(C) centroid of the triangle PQR lies on the axis of the parabola
(D) circle circumscribing the triangle PQR passes through the vertex of the parabola

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [13]

Q.111 A variable circle is described to pass through the point (1, 0) and tangent to the curve
y = tan (tan 1 x). The locus of the centre of the circle is a parabola whose :
(A) length of the latus rectum is 2 2
(B) axis of symmetry has the equation x + y = 1
(C) vertex has the co-ordinates (3/4, 1/4)
(D) none of these

Q.112 Which of the following equations in parametric form can represent a hyperbola, where 't' is a parameter.
a  1 b  1 tx y x ty
(A) x = t   & y = t   (B)  +t=0 & + 1=0
2  t 2  t a b a b

t
(C) x = et + et & y = et  et (D) x2  6 = 2 cos t & y2 + 2 = 4 cos2
2

Q.113 The equations of the common tangents to the ellipse, x2 + 4y2 = 8 & the parabola y2 = 4x can be
(A) x + 2y + 4 = 0 (B) x – 2y + 4 = 0 (C) 2x + y – 4 = 0 (D) 2x – y + 4 = 0

Q.114 Variable chords of the parabola y2 = 4ax subtend a right angle at the vertex. Then :
(A) locus of the feet of the perpendiculars from the vertex on these chords is a circle
(B) locus of the middle points of the chords is a parabola
(C) variable chords passes through a fixed point on the axis of the parabola
(D) none of these

x2 y2 y2 x2
Q.115 Equations of a common tangent to the two hyperbolas  = 1 &  = 1 is :
a 2 b2 a 2 b2
(A) y = x + a 2  b 2 (B) y = x  a 2  b 2
(C) y =  x + a 2  b 2 (D)  x  a 2  b 2

Q.116 The equation of the tangent to the parabola y = (x  3)2 parallel to the chord joining the points (3, 0) and
(4, 1) is :
(A) 2 x  2 y + 6 = 0 (B) 2 y  2 x + 6 = 0
(C) 4 y  4 x + 13 = 0 (D) 4 x  4 y = 13

Q.117 Let A be the vertex and L the length of the latus rectum of the parabola, y2  2 y  4 x  7 = 0. The
equation of the parabola with A as vertex, 2L the length of the latus rectum and the axis at right angles to
that of the given curve is :
(A) x2 + 4 x + 8 y  4 = 0 (B) x2 + 4 x  8 y + 12 = 0
2
(C) x + 4 x + 8 y + 12 = 0 (D) x2 + 8 x  4 y + 8 = 0

dx 3y
Q.118 The differential equation dy = represents a family of hyperbolas (except when it represents a pair
2x
of lines) with eccentricity :
3 5 2 5
(A) (B) (C) (D)
5 3 5 2

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [14]

Q.119 If a number of ellipse be described having the same major axis 2a but a variable minor axis then the
tangents at the ends of their latera recta pass through fixed points which can be
(A) (0, a) (B) (0, 0) (C) (0, – a) (D) (a, a)

Q.120 The straight line y + x = 1 touches the parabola :

(A) x2 + 4 y = 0 (B) x2  x + y = 0
(C) 4 x2  3 x + y = 0 (D) x2  2 x + 2 y = 0

Q.121 Circles are drawn on chords of the rectangular hyperbola xy = c2 parallel to the line y = x as
diameters. All such circles pass through two fixed points whose co-ordinates are :
(A) (c, c) (B) (c,  c) (C) ( c, c) (D) ( c,  c)

Bansal Classes Q. B. on Parabola, Ellipse, Hyperbola [15]

[16] Q. B. on Parabola, Ellipse, Hyperbola Bansal Classes
Select the correct alternative : (Only one is correct)
Q.1 C Q.2 B Q.3 B Q.4 D Q.5 C Q.6 A
Q.7 B Q.8 B Q.9 A Q.10 B Q.11 C Q.12 B
Q.13 C Q.14 B Q.15 B Q.16 B Q.17 C Q.18 D
Q.19 B Q.20 D Q.21 B Q.22 A Q.23 D Q.24 A
Q.25 D Q.26 C Q.27 B Q.28 B Q.29 B Q.30 A
Q.31 A Q.32 D Q.33 D Q.34 B Q.35 B Q.36 C
Q.37 A Q.38 A Q.39 B Q.40 B Q.41 D Q.42 C
Q.43 C Q.44 B Q.45 C Q.46 B Q.47 D Q.48 B
Q.49 C Q.50 A Q.51 D Q.52 C Q.53 C Q.54 C
Q.55 D Q.56 B Q.57 A Q.58 B Q.59 C Q.60 C
Q.61 C Q.62 B Q.63 A Q.64 C Q.65 A Q.66 C
Q.67 B Q.68 D Q.69 B Q.70 B Q.71 C Q.72 A
Q.73 D Q.74 A Q.75 C Q.76 B Q.77 C Q.78 A
Q.79 D Q.80 B Q.81 D Q.82 C Q.83 D Q.84 C
Q.85 A Q.86 B Q.87 D Q.88 A Q.89 C Q.90 B
Q.91 D Q.92 C Q.93 D Q.94 C Q.95 B Q.96 A
Q.97 D Q.98 D Q.99 C Q.100 D
Select the correct alternatives : (More than one are correct)
Q.101 A,B Q.102 A,C,D Q.103 B,D Q.104 A,B,C,D
Q.105 A,D Q.106 A,B,C,D Q.107 A,B Q.108 A,D
Q.109 B,C Q.110 A,B,C,D Q.111 B,C Q.112 A,C,D
Q.113 A,B Q.114 A,B,C Q.115 A,B,C,D Q.116 C,D
Q.117 A,B Q.118 B,D Q.119 A,C Q.120 A,B,C
Q.121 A,D
ANSWER KEY
 Special Dpp on Conic Section (Parabola, Ellipse and Hyperbola)
MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 1
Q.156/para If on a given base, a triangle be described such that the sum of the tangents of the base angles is a
constant, then the locus of the vertex is :
(A) a circle (B*) a parabola (C) an ellipse (D a hyperbola
k k
[Sol. tan A = ; tan B =
ah ah
tan A + tan B = 'C' (a constant)
k k k (a  h )  k (a  h )
 C  C
a h a h a2  h2
2a
 a2 – h2 = ·k  locus of (h, k) will be
C
2a
x2 = a2 – ·y  A parabola Ans.]
C

Q.2 The locus of the point of trisection of all the double ordinates of the parabola y2 = lx is a parabola whose
latus rectum is
l 2l 4l l
(A*) (B) (C) (D)
9 9 9 36
[Sol. Let y2 = 4ax; 4a = l (at2,2at)
A
A(at2, 2at)
hence h = at2 P(h,k)
3k = 2at
Q
h
9k2 = 4a2 ·
a B

4a l
y2 = x  y2 = x  (A) ]
9 9

Q.3 Let a variable circle is drawn so that it always touches a fixed line and also a given circle, the line not
passing through the centre of the circle. The locus of the centre of the variable circle, is
(A*) a parabola (B) a circle
(C) an ellipse (D) a hyperbola

[Sol. Case-I: PS = R + r1 = PM
 locus of P is a parabola
Case-II: R – r1 = PM
 locus of P is a parabola. ]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [1]

Q.4 The vertex A of the parabola y2 = 4ax is joined to any point P on it and PQ is drawn at right angles to AP
to meet the axis in Q. Projection of PQ on the axis is equal to
(A) twice the latus rectum (B*) the latus rectum
(C) half the latus rectum (D) one fourth of the latus rectum
[Sol. 2
A(0, 0), P(at , 2at), Q(x, 0)
Slope of AP × slope of PQ = – 1
2at  2at
2
  1
at x1  at 2
(x1 – at2)(at2) = 4a2t2
x = 4a + at2 = AQ
 projection QM = AQ – AM = 4a = Latus rectum Ans.]

Q.5 Two unequal parabolas have the same common axis which is the x-axis and have the same vertex which
is the origin with their concavities in opposite direction. If a variable line parallel to the common axis meet
the parabolas in P and P' the locus of the middle point of PP' is
(A*) a parabola (B) a circle (C) an ellipse (D) a hyperbola
[Sol. P(at12, 2at1)
P'(– bt22, – 2bt2)
Slope of PP' = 0
at1 + bt2 = 0 ....(1)
 at12  bt 22 
Mid-point of PP' =  , at1  bt 2  = (h, k)
2 
 

 at12
put t2 =
b

 at 
2 at1  1 · b  4h
k b  4at1
    ;  t1 =
a  at
2
a ( b  a ) t 2 k (b  a )
  1
at12  b 1

 b 

4h 8a
 k = at1 – bt2 = at1 + at1 = 2a ·  k2 = ·h
k (b  a ) ba
8a
 locus of (h, k) is y2 = x , a parabola Ans.]
ba
Q.61002/hyp The straight line y = m(x – a) will meet the parabola y2 = 4ax in two distinct real points if
(A) m  R (B) m  [–1, 1]
(C) m  (– , 1]  [1, )R (D*) m  R – {0}
[Sol. y = m(x – a) passes through the focus (a, 0) of the parabola. Thus for this to be focal chord
m  R – {0} Ans.] [11th, 14-02-2009]

 x
Q.71007/hyp All points on the curve y 2  4a  x  a sin  at which the tangent is parallel to x-axis lie on
 a
(A) a circle (B*) a parabola (C) an ellipse (D) a line

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [2]

  x
[Sol. y 2  4a  x  a sin  [12th, 04-01-2009, P-1]
 a

dy  1 x
2y = 4a 1  a · cos 
dx  a a
 tangent is parallel to x-axis.
dy x
 = 0,  cos =–1
dx a
x 2 x
 sin =  1  cos =0
2 a
 x
 y2 = 4a  x  a sin  = 4ax  a parabola Ans.]
 a

Q.8 Locus of trisection point of any arbitrary double ordinate of the parabola x2 = 4by, is
(A) 9x2 = by (B) 3x2 = 2by (C*) 9x2 = 4by (D) 9x2 = 2by
[Sol. Let A  (2bt, bt2), B  (– 2bt, bt2) be the extremities on the double ordinate AB.
If C(h, k) be it's trisection point, then
3h = 4bt – 2bt, 3k = 2bt2 + bt2
3h 2 k k 9h 2
 t= , t =  
2b b b 4b 2
Thus locus of C is 9x2 = 4by Ans.]

Q.9103/para The equation of the circle drawn with the focus of the parabola (x  1)2  8 y = 0 as its centre and
touching the parabola at its vertex is :
(A) x2 + y2  4 y = 0 (B) x2 + y2  4 y + 1 = 0
2 2
(C) x + y  2 x  4 y = 0 (D*) x2 + y2  2 x  4 y + 1 = 0
[Hint: Put X2 = 8Y ; when x – 1 = X and y = Y
 (X – 0)2 + (Y – 2)2 = 4  (x – 1)2 + (y – 2)2 = 4  (D) ]

Q.10 The length of the latus rectum of the parabola, y2  6y + 5x = 0 is

(A) 1 (B) 3 (C*) 5 (D) 7
[Sol. 2
y = – 6y + 5x = 0
 9
 (y – 3)2 = 9 – 5x = – 5  x    (C) Ans.]
 5

Q.115/para Which one of the following equations represented parametrically, represents equation to a parabolic
profile?
t
(A) x = 3 cos t ; y = 4 sin t (B*) x2  2 =  2 cos t ; y = 4 cos2
2
t t
(C) x = tan t ; y = sec t (D) x = 1  sin t ; y = sin + cos
2 2

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [3]

 2 2
 x  y 2 2
[Sol. (A)       cos t  sin t  1  An ellipse
3
   4

 2 t 
(B) x2 – 2 = – 2 cos t = – 2  2 cos 1  A parabola
 2 
(C) x  tan x , y  sec t (x  0, y  0)
sec2t – tan2t = 1  y – x = 1  a line tangent
2
 t t
(D) x2 = 1 – sin t = 2 – (1 + sin t) = 2 –  sin  cos  = 2 – y2  a circle]
 2 2

Q.128/para The length of the intercept on y  axis cut off by the parabola, y2  5y = 3x  6 is
(A*) 1 (B) 2 (C) 3 (D) 5
2
[Sol. y  5y = 3x  6 [11th, 14-02-2009]
for point of intersection with y-axis, i.e. x = 0
 y2 – 5y + 6 = 0
y = 2, 3
 A(0, y1), B(0, y2)
 AB = y 2  y1 = | 3 – 2 | = 1 Ans.]

Q.13 A variable circle is described to pass through (1, 0) and touch the line y = x. The locus of the centre of
the circle is a parabola, whose length of latus rectum, is
1
(A) 2 (B*) 2 (C) (D) 1
2
[Sol. CF = CN Þ locus of C is a parbola with focus at (1, 0) and directrix y = x N
 length of latus rectum = 2(distance from focus to directrix) C(h,k)

 1  (1,0)F S(a,0)
= 2  = 2 Ans.]
 2

Q.14 Angle between the parabolas y2 = 4b (x – 2a + b) and x2 + 4a (y – 2b – a) = 0

at the common end of their latus rectum, is
1 1
(A) tan–1(1) (B*) cot–11 + cot–1 + cot–1
2 3
(C) tan–1 3
  (D) tan–1(2) + tan–1(3)
[Hint: y2 = 4b (x – (2a – b) ) or y2 = 4bX where x – (2a – b) = X [Q.37, Ex-26, Loney]
x2 + 4a (y – (a + 2b) ) or x2 = – 4aY where y – (a + 2b) = Y
for y2 = 4bX, extremities of latus rectum (b, 2b) and (b, – 2b) w.r.t. X Y axis
i.e. (2a, 2b) and (2a, – 2b) w.r.t. xy axis
for x2 = – 4aY, extremities of latus rectum (2a, – a) and (–2a, –a) w.r.t. XY axis
i.e. (2a, 2b) and (–2a, 2b)
Hence the common end of latus rectum (2a, 2b)
dy dy 2b
now for 1st parabola 2y = 4b  = y = 1 at (2a, 2b)
dx dx 1
 dy dy x
also for 2nd parabola 2x = – 4a or =– = – 1 at (2a, 2b)
dx dx 2a
Hence parabolas intersect orthogonally at (2a, 2b)  (B)]

Q.15 A point P on a parabola y2 = 4x, the foot of the perpendicular from it upon the directrix, and the focus are
the vertices of an equilateral triangle, find the area of the equilateral triangle. [Ans. 4 3 ]
[Sol. PM = 1 + t2
PS = 1 + t2
MS = 1 + t2
 22 + 4t2 = (1 + t2)2
 PM = 1 + t2 = 4
3 2
 Area of PMS = (4 ) = 4 3 Ans.]
4

Q.16 Given y = ax2 + bx + c represents a parabola. Find its vertex, focus, latus rectum and the directrix.
  b 4ac  b 2    b 1 4ac  b 2  1 2
[Ans.  2a , ;  ,   ; , y = 4ac  b  1 ]
4 a   2a 4a 4a  a
   4a 4a
[Sol. y = ax2 + bx + c
2
 b  1  4ac  b 2 
 x    y
 2a  a 4a 

  b 4ac  b 2    b 1 4ac  b 2 
 vertex :  2a , 4 a
;
 focus :  2a , 4a  4 a


   
1 4ac  b 2 1
Latus rectum : and directrix : y =  ]
a 4a 4a

Q.17 Prove that the locus of the middle points of all chords of the parabola y2 = 4ax passing through the vetex
is the parabola y2 = 2ax.
[Sol. P(at2, 2at)
Mid-point of AP
a 2 
M  t , at  = M(h, k)
2 
 k2 = a2t2 = 2ah
 y2 = 2ax Ans.]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [5]

Q.18 Prove that the equation to the parabola, whose vertex and focus are on the axis of x at distances a and
a' from the origin respectively, is y2 = 4(a' – a)(x – a) [Q.10, Ex-25, Loney]

[Sol. Case-I:

 equation of parabola : y2 = – 4(a – a')(x – a) = 4(a' – a)(x – a) Hence proved.

Case-II:

 equation of parabola : y2 = 4(a' – a)(x – a) Hence proved. ]

Q.19 Prove that the locus of the centre of a circle, which intercepts a chord of given length 2a on the axis of
x and passes through a given point on the axis of y distant b from the origin, is the curve
x2 – 2yb + b2 = a2. [Q.14, Ex-25, Loney]
[Sol. (x – x1) + (y – y1)2 = r2
2

x-axis intercept = 2a  2 x12  ( x12  y12  r 2 = 2a

r2 – y12 = a2
 y12 = r2 – a2
Passes through (0, b)
 x12 + (b – y1)2 = r2  x12 + b2 – 2by1 + y12 – r2 = 0  x12 + b2 – 2by1 – a2 = 0
 Locus of (x1, y1) will be x – 2by + b2 = a2 Hence proved. ]
2

Q.20 A variable parabola is drawn to pass through A & B, the ends of a diameter of a given circle with centre
at the origin and radius c & to have as directrix a tangent to a concentric circle of radius 'a' (a >c) ; the
axes being AB & a perpendicular diameter, prove that the locus of the focus of the parabola is the
x2 y2
standard ellipse  1 where b2 = a2 – c2.

a 2 b2
[Sol. (h – c)2 + k2 = (c cos – a)2 ....(1) [Q.22, Ex-25, Loney]
sub. (h + c)2 + k2 = (c cos + a)2 ....(2)
—————————————
4ch = 4ca cos   h = a cos .....(3)
add 2(c2 + h2 + k2) = 2 (c2cos2 + a2) .....(4)
h
put cos  = in equation (4)
a

h2
we get c2 + h2 + k2 + a2
= c2 ·
a2
c2 a2 + h2a2 + k2a2 = c2 h2 + a4
(a2 – c2)h2 + k2a2 = a2(a2 – c2)
h2 k2
+ =1
a2 a 2  c2

x2 y2
 =1 ]
a2 b2
Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [6]
 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 2
 
Q.1 If a focal chord of y2 = 4ax makes an angle ,   0,  with the positive direction of x-axis, then
 4
minimum length of this focal chord is
(A) 6a (B) 2a (C*) 8a (D) None
[Sol. 2
Length of focal chord making an angle  with x-axis is 4a cosec .
 
For    0,  , it's minimum length = (4a)(2) = 8a units. Ans.]
 4

Q.2 OA and OB are two mutually perpendicular chords of y2 = 4ax, 'O' being the origin. Line AB will always
pass through the point
(A) (2a, 0) (B) (6a, 0) (C) (8a, 0) (D*) (4a, 0)
[Sol. Let A  (at12, 2at1), B  (at22, 2at2)
Thus t1t2 = – 4
Equation of line AB is
y(t1 + t2) = 2(x + at1t2),
i.e. y(t1 + t2) = 2(x – 4a)
which clearly passes through a fixed point (4a, 0) Ans. ]

Q.34/para ABCD and EFGC are squares and the curve y = k x passes through the origin D and the points B
FG
and F. The ratio is
BC
5 1 3 1
(A*) (B)
2 2
5 1 3 1
(C) (D)
4 4
[Sol. 2 2
y =k x  y2 = 4ax where k2 = 4a [13th 15-10-2006]
B = at12 , 2at1 ; F = at 22 , 2at 2
    {t1 > 0, t2 > 0}

FG 2at 2 t2
to find = 2at = t
BC 1 1

now DC = BC  at12 = 2at1  t1 = 2

also at 22 – at12 = 2at2

t 22 – 4 = 2t2

t 22 – 2t2 – 4 = 0

2  4  16 2  20
t2 = but t2 > 0  t2 
2 2

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [7]

 t2 =  5 1
t2 5 1
 t1 = 2
Ans. ]

Q.413/para From an external point P, pair of tangent lines are drawn to the parabola, y2 = 4x. If 1 & 2 are the

inclinations of these tangents with the axis of x such that, 1 + 2 = , then the locus of P is :
4
(A) x  y + 1 = 0 (B) x + y  1 = 0 (C*) x  y  1 = 0 (D) x + y + 1 = 0
1
[Hint: y = mx +
m
or 2
m h – mk + 1 = 0
k 1
m1 + m2 = ; m1 m2 =
h h
 m1  m 2 k 1
given 1 + 2 =  1
4
 1  m1m 2 = 1  h h
 y = x – 1]

Q.514/para Maximum number of common chords of a parabola and a circle can be equal to
(A) 2 (B) 4 (C*) 6 (D) 8
[Sol. A circle and a parabola can meet at most in four points. Thus maximum number of common chords in
4C i.e. 6 Ans.] [13th, 14-02-2009]
2

Q.653/para PN is an ordinate of the parabola y2 = 4ax. A straight line is drawn parallel to the axis to bisect NP and
meets the curve in Q. NQ meets the tangent at the vertex in apoint T such that AT = kNP, then the value
of k is (where A is the vertex)
(A) 3/2 (B*) 2/3 (C) 1 (D) none
[Sol. Equation of PN : x = at 2

y = c bisects PN
 c = at
which cuts the parabola at Q
c2
 c2 = 4ax  x=
4a
 c2   at 2 
 Q  4a  = Q  4 , at 
 , c 
   
at  0
2
 Equation of NQ: y – 0 = at  at 2 (x – at2)
4
4
y= ( x  at 2 )
3t
 4at 
which cuts x = 0 at  0, 
 3 

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [8]

 4at
 T= and NP = 2at
3
AT 4at 3 2
  = Ans.]
NP 2at 3

Q.724/para Let A and B be two points on a parabola y2 = x with vertex V such that VA is perpendicular to VB and
| VA |
 is the angle between the chord VA and the axis of the parabola. The value of is
| VB |
(A) tan  (B) tan3 (C) cot2 (D*) cot3
2
[Hint: tan  = t ....(1) [12 & 13 05-3-2006]
1

2 2 y (at12 , 2at1)
Also t1 × t2 = – 1 A 2
y =x
t1t2 = – 4 
V x
| VA | = a 2 t14  4a 2 t12 = at1 t12  4
using t1t2 = – 4 B
(at22 , 2at2)
4a 16 8a
| VB | =  4 = 2
4  t12
t1 t12 t1

3 2
| VA | at1 4  t1 t13
 = =
| VB | 8a 4  t12 8

2
Also tan  = t ;  t13 = 8 cot3
1

| VA |
 = cot3 Ans. ]
| VB |
Q.825/para Minimum distance between the curves y2 = x – 1 and x2 = y – 1 is equal to
3 2 5 2 7 2 2
(A*) (B) (C) (D)
4 4 4 4
[Sol. Both curve are symmstrical about the line y = x. If line AB is the line of shortest distance then at A and B
slopes of curves should be equal to one [13th, 14-02-2009]
dy 1
for y2 = x – 1, = =1
dx 2y
1 5
 y= ,x=
2 4
1 5 5 1
 B   ,  and A   , 
2 4 4 2
hence minimum distance AB,

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [9]

 2 2
5 1 5 1 3 2
=       = Ans.]
4 2 4 2 4
Q.928/para The length of a focal chord of the parabola y2 = 4ax at a distance b from the vertex is c, then
(A) 2a2 = bc (B) a3 = b2c (C) ac = b2 (D*) b2c = 4a3
[Sol. Equation focal chord PQ : 2x – (t1 + t2)y – 2a = 0 ....(1)
l(PQ) = c
a
cosec  =
b
now c = 4a cosec2
a2
c = 4a · 2 ; 4a3 = b2c ]
b
Q.1034/para The straight line joining any point P on the parabola y2 = 4ax to the vertex and perpendicular from the
focus to the tangent at P, intersect at R, then the equaiton of the locus of R is
(A) x2 + 2y2 – ax = 0 (B*) 2x2 + y2 – 2ax = 0
2 2
(C) 2x + 2y – ay = 0 (D) 2x2 + y2 – 2ay = 0
[Sol. T : ty = x + at2 ....(1) [12th, 04-01-2009, P-1]
line perpendicular to (1) through (a,0)
tx + y = ta ....(2)
2
equation of OP : y – x = 0 ....(3)
t
from (2) & (3) eleminating t we get locus ]
Q.1141/para Locus of the feet of the perpendiculars drawn from vertex of the parabola y2 = 4ax upon all such
chords of the parabola which subtend a right angle at the vertex is
(A*) x2 + y2 – 4ax = 0 (B) x2 + y2 – 2ax = 0
(C) x2 + y2 + 2ax = 0 (D) x2 + y2 + 4ax = 0
[Hint: Chord with feet of the perpendicular as (h, k) is hx + ky = h2 + k2 .....(1)
homogenise y = 4ax with the help of (1) and use coefficient of x + coefficient of y2 = 0
2 2

2 2
Alternatively-1: tan 1 = t ; tan 2 = t
1 2

2 2
·  1 ,  t 1t 2 = – 4
t1 t 2
 equation of chord PQ 2x – (t1 + t2)y – 8a = 0
slope of AR × slope of PQ = – 1
k 2 
   =–1
h  t1  t 2 

 2k
 t1 + t2 =
h
  2k 
 equation of chord PQ will be 2x –   y – 8a = 0
 h 
hx + ky = 4ah
(h, k) lies on this line

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [10]

  h2 + k2 = 4ah
 locus of R(h, k) is x2 + y2 = 4ax Ans.
Alternatively-2:
1 h
Slope of PQ = slope of OR =
k
 equation of PQ will be
h
y–k= (x – h)  hx + ky = h2 + k2
k
This chord subtends a right angle at vertex O(0, 0)
 by homogenisation we get equation of pair of straight line OP and OQ
 hx  ky 
y2 = 4ax  2 
 h  k2 
 OP ^ OQ
 coefficient of x2 + coefficient of y2 = 0
 h2 + k2 = 4ah
 locus of (h, k) wil be
x2 + y2 = 4ax Ans.]

More than one are correct:

Q.12503/para Consider a circle with its centre lying on the focus of the parabola, y2 = 2 px such that it touches the
directrix of the parabola. Then a point of intersection of the circle & the parabola is
p  p   p   p 
(A*)  , p (B*)  ,  p (C)   , p (D)   ,  p
2  2   2   2 
2
 p 2 2
[Sol. Equation of circle will be  x    y  p
 2
which intersects y2 = 2px
2
 p 2
  x    2px  p
 2

3p 2
x2 + px – =0
4
2
 x  3p  x  p  = 0

 4  2

3p
x 0
4
p
 x= only
2
p
 y2 = 2p  y=±p
2
p  p 
Hence  ,  p  and  , p  Ans.]
2  2 
Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [11]
 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 3
Q.1 y-intercept of the common tangent to the parabola y2 = 32x and x2 = 108y is
(A) – 18 (B*) – 12 (C) – 9 (D) – 6
8
[Sol. Tangent to y2 = 32x is y = mx + and tangent to x2 = 108y is y = mx – 27m2
m
8 8
 = – 27m2,  m3 =
m 27
2
 m=
m
8  3
 y-intercept = = 8  = – 12 Ans.]
m  2 

Q.29/para The points of contact Q and R of tangent from the point P (2, 3) on the parabola y2 = 4x are
1
(A) (9, 6) and (1, 2) (B*) (1, 2) and (4, 4) (C) (4, 4) and (9, 6) (D) (9, 6) and ( , 1)
4
t 1t 2  2 
[Hint:   t1 = 1 and t2 = 2
t1  t 2  3 

Hence point t12 , 2t1 and t 22 , 2 t 2

   
i.e. (1, 2) and (4, 4) ] [13th Test, 24-03-2005]


Q.318/para Length of the normal chord of the parabola, y2 = 4x , which makes an angle of with the axis of x is:
4
(A) 8 (B*) 8 2 (C) 4 (D) 4 2
[Sol. 3
N : y + tx = 2t + t ; slope of the normal is – t
hence – t = 1  t = – 1  coordinates of P are (1, –2)
Hence parameter at Q, t2 = – t1 – 2/t1 = 1 + 2 = 3
 Coordinates at Q are (9, 6)
 l (PQ) = 64  64  8 2 ]

Q.421/para If the lines (y – b) = m1(x + a) and (y – b) = m2(x + a) are the tangents to the parabola y2 = 4ax, then
(A) m1 + m2 = 0 (B) m1m2 = 1 (C*) m1m2 = – 1 (D) m1 + m2 = 1
[Sol. Clearly, both the lines passes through (–a, b) which is a point lying on the directrix of the parabola
Thus, m1m2 = – 1 [13th, 14-02-2009]
Because tangents drawn from any point on the directrix are always mutually perpendicular]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [13]

Q.523/para If the normal to a parabola y2 = 4ax at P meets the curve again in Q and if PQ and the normal at Q
makes angles  and  respectively with the x-axis then tan (tan  + tan ) has the value equal to
1
(A) 0 (B*) – 2 (C) – (D) – 1
2
[Sol. tan  = – t1 and tan  = – t2
2
also t2 = – t1 –
t1

t1 t2 + t12 = – 2
tan  tan  + tan2 = – 2  (B) ]

Q.643/para C is the centre of the circle with centre (0, 1) and radius unity. P is the parabola y = ax2. The set of
values of 'a' for which they meet at a point other than the origin, is
 1 1 1 1 
(A) a > 0 (B) a   0,  (C)  ,  (D*)  ,  
 2 4 2 2 
y
[Hint: put x2 = in circle, x2 + (y – 1)2 =1, we get (Note that for a < 0 they cannot intersect other than origin)
a
y 1
+ y2 – 2y = 0 ; hence we get y = 0 or y = 2 –
a a
1
substituting y=2– in y = ax2, we get
a
1 2a  1 1
ax2 = 2 – a ; x2 = >0  a> ] [12th&13th (14-8-2005)]
a2 2

Q.780/para PQ is a normal chord of the parabola y2 = 4ax at P, A being the vertex of the parabola. Through P a
line is drawn parallel to AQ meeting the xaxis in R. Then the length of AR is :
(A) equal to the length of the latus rectum
(B) equal to the focal distance of the point P
(C*) equal to twice the focal distance of the point P
(D) equal to the distance of the point P from the directrix.
2
[Hint : t2 =  t1   t1t2 + t12 =  2
t1
Equation of the line through P parallel to AQ
2
y  2 at1 = (x  at12)
t2
put y = 0  x = at12  at1t2
= at 12  a ( 2  t 12 )
= 2a + 2 at 12 = 2(a + a t 12 )
= twice the focal distance of P ]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [14]

Q.884/para Length of the focal chord of the parabola y2 = 4ax at a distance p from the vertex is :
2a2 a3 4a3 p2
(A) (B) 2 (C*) 2 (D)
p p p a
2 P
2a
2
a
2 a 1  t2
 
[Hint : Length =  2 at     at 2  2  =
 t   t  t2
Now equation of focal chord, 2 tx + y (1  t2)  2 at = 0 p
2
2
2 at 4a 2 1  t  –
 p=  = .
1  t2 p2 t2

a 2 4a3
Alternatively : cosec  =  Length of focal chord = 4a cosec  = 2 ]
p p

Q.988/para The triangle PQR of area 'A' is inscribed in the parabola y2 = 4ax such that the vertex P lies at the
vertex of the parabola and the base QR is a focal chord. The modulus of the difference of the ordinates
of the points Q and R is :
A A 2A 4A
(A) (B) (C*) (D)
2a a a a

2a 1
[Hint : d = 2 at  =2at . R
t t

a t2 2a t 1
1 a 2a  1
Now A =  t 1 = a2  t  
2 t2  t
0 0 1 –

 1 2A
 2a  t   = ]
 t a

Q.10127/para The roots of the equation m2 – 4m + 5 = 0 are the slopes of the two tangents to the parabola
y2 = 4x. The tangents intersect at the point
4 1 1 4  1 4
(A)  ,  (B*)  ,  (C)   , 
 5 5 5 5  5 5
(D) point of intersection can not be found as the tangents are not real
1
[Sol. y = mx + [29-01-2006, 12th & 13th]
m
it passes through (h, k)
1
k = mh +  h m2 – k m + 1 = 0 ....(1)
m
k 1
m1 + m2 = and m1m2 =
h h
but m1 and m2 are the roots of m2 – 4m + 5 = 0
k 1
 m1 + m2 = 4 = and m1m2 = 5 =
h h
1 4 1 4
h= and  k =   ,  Ans. ]
5 5 5 5
Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [15]
Q.1117/para Through the focus of the parabola y2 = 2px (p > 0) a line is drawn which intersects the curve at
y1y 2
A(x1, y1) and B(x2, y2). The ratio equals
x1x 2
(A) 2 (B) – 1 (C*) – 4 (D) some function of p
2
[Sol. y = 4ax, 4a = 2p > 0 th
[13 (27-8-2006)]
x1 = at12 , y1 = 2at1

x2 = at 22 , y2 = 2at2
and t1t2 = – 1
4a 2 t 1 t 2
ratio = = – 4 Ans. ]
a 2 t12 t 22
Q.1215/para If the line 2x + y + K = 0 is a normal to the parabola, y2 + 8x = 0 then K =
(A)  16 (B)  8 (C)  24 (D*) 24
[Sol. m = – 2, a = – 2
 equation of normal
y = – 2x – 2(–2)(–2) – (–2)(–2)3
2x + y + 24 = 0
 k = 24 Ans.]

Q.1389/para The normal chord of a parabola y2 = 4ax at the point whose ordinate is equal to the abscissa, then
angle subtended by normal chord at the focus is :
 
(A) (B) tan 1 2 (C) tan 1 2 (D*)
4 2
[Sol. a t12 = 2at1  t1 = 2 ; P(4a , 4a)
2
t2 = – t 1 – t = – 3
1
 Q (9a ,– 6a )
4a 4
 mSP = 
4a  a 3
 6a 3
mSQ =   ]
9a  a 4

Q.1490/para The point(s) on the parabola y2 = 4x which are closest to the circle,
x2 + y2  24y + 128 = 0 is/are :
(A) (0, 0) (B) 2 , 2 2
  (C*) (4, 4) (D) none

[Hint : centre (0, 12) ; slope of tangent at (t2, 2 t) is 1/t, hence slope of
normal is  t. This must be the slope of the line joining centre
(0, 12) to the point (t2, 2 t)  t=2]

[Sol. slope at normal at P = mCP ]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [16]

More than one are correct:
Q.15508/para Let y2 = 4ax be a parabola and x2 + y2 + 2 bx = 0 be a circle. If parabola and circle touch each
other externally then :
(A*) a > 0, b > 0 (B) a > 0, b < 0 (C) a < 0, b > 0 (D*) a < 0, b < 0
[Hint : For externally touching a & b must have the same sign

Q.16516/para The straight line y + x = 1 touches the parabola :

(A*) x2 + 4y = 0 (B*) x2  x + y = 0
(C*) 4x2  3x + y = 0 (D) x2  2x + 2y = 0
[Hint: put y = 1 – x and see that the resulting exprassion is a perfect square]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [17]

 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 4
Q.145/para TP & TQ are tangents to the parabola, y2 = 4ax at P & Q. If the chord PQ passes through the fixed
point ( a, b) then the locus of T is :
(A) ay = 2b (x  b) (B) bx = 2a (y  a)
(C*) by = 2a (x  a) (D) ax = 2b (y  b)

[Hint : Chord of contact of (h, k) [12th, 04-01-2009, P-1]

ky = 2a (x + h). It passes through ( a, b)
 bk = 2a ( a + h)
 Locus is by = 2a (x  a) ]

Q.246/para Through the vertex O of the parabola, y2 = 4ax two chords OP & OQ are drawn and the circles on
OP & OQ as diameters intersect in R. If 1, 2 &  are the angles made with the axis by the tangents at
P & Q on the parabola & by OR then the value of, cot 1 + cot 2 =
(A*)  2 tan  (B)  2 tan () (C) 0 (D) 2 cot 
1
[Hint : Slope of tangant at P is
t1

1
and at Q =
t2
 cot 1 = t1 and cot 2 = t2
2
Slope of PQ =
t1  t 2
 t1  t 2 
 Slope of OR is    = tan 
 2 
(Note angle in a semicircle is 90º)
1
 tan  =  (cot 1 + cot 2)  cot 1 + cot 2 =  2 tan  ]
2

Q.351/para If a normal to a parabola y2 = 4ax makes an angle  with its axis, then it will cut the curve again at an
angle
1  1 
(A) tan–1(2 tan) (B*) tan1  tan  (C) cot–1  tan  (D) none
2  2 
[Sol. equation of normal at t : y + tx = 2at + at3
 mN at A = – t = tan 
t = – tan  = m1
1
Now tangent at B t1y = x + at12 with m2 =
tan 
2
also t1 =  t 
t

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [19]

 1
t
t1 1  tt1 1 t2 sec 2  . tan 
 tan  = t = =  1 = [As t t1 = – t2 – 2]
1 t  t1 2 t   2(sec 2 )
t1  t

tan  1  tan  
Hence tan =  = tan  
2  2 
Alterntively: Equation of normal at A
y + tx = 2at + at3
 Slope of normal at A, mA = – t  tan  = – t
T
Equation of tangent at B: t1y = xt + at12
1 2
slope, tan a = t where t1 =  t 
1 t

1 2 tan 
  tan    tan  =
tan  tan  2  tan 2 

tan 
 tan   1  2  tan 2  
tan   tan  2  tan 2  tan  
 tan  = = =  2 2 
1  tan  tan  tan   2  tan   tan  
1 2
tan 
2  tan 

tan   tan  
=   = tan–1   ]
2  2 
Q.452/para Tangents are drawn from the points on the line x  y + 3 = 0 to parabola y2 = 8x. Then the variable
chords of contact pass through a fixed point whose coordinates are :
(A) (3, 2) (B) (2, 4) (C*) (3, 4) (D) (4, 1)
[Hint: Let P (a, (a + 3)) be a point on the line and chord of contact is
(a + 3)y = 4 (x + a)  4x  3y + a (4  y) = 0  line passes through a fixed point (3,4) ]

Q.561/para If the tangents & normals at the extremities of a focal chord of a parabola intersect at (x1, y1) and
(x2, y2) respectively, then :
(A) x1 = x2 (B) x1 = y2 (C*) y1 = y2 (D) x2 = y1
[Hint: x1 = at1t2, y1 = a(t1 + t2) ; x2 = a( t12 + t 22 + t1t2 + 2), y2 = – at1t2(t1 + t2) with t1t2 = – 1

x1 = – a, y1 = a(t1 + t2) ; x2 = a( t12  t 22  1 ) ; y2 = a(t1 + t2) ]

Q.674/para If two normals to a parabola y2 = 4ax intersect at right angles then the chord joining their feet passes
through a fixed point whose co-ordinates are :
(A) ( 2a, 0) (B*) (a, 0) (C) (2a, 0) (D) none
[Hint : t1 t2 =  1 ]
[Sol. N : y + tx = 2at + at3 ; passes through (h, k)
k
Hence at3 + (2a – h)t + k = 0 ; t1 t2 t3 = – ; t1 t2 = – 1
a
chord joining t1 and t2 is 2x – (t1 + t2)y + 2at1 t2 = 0
(2x – 2a) – (t1 + t2)y = 0 x = a & y = 0
Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [20]
Alternatively: If the normal intersect at right angles then their corresponding tangents will also intersect at right
angles hence the chord joining their feet must be a focal chord
 it will always pass through (a, 0)]
Q.775/para The equation of a straight line passing through the point (3, 6) and cutting the curve y = x orthogonally is
(A*) 4x + y – 18 =0 (B) x + y – 9 = 0 (C) 4x – y – 6 = 0 (D) none
m m3
[Hint: Normal to the parabola y2 = x is y = mx   ; pass through the point
2 4
(3, 6)  m3  10m + 24 = 0 ; m =  4 is a root  required equation 4x + y  18 = 0
dy 1 1 t 6
alt. (t2, t) be a point on y = x  dx = 2 x = 2t  t 2  3 =  2t (slope of normal)
 2 t3  5t  6 = 0
= (t – 2) (2t2 + 4t + 3)  t = 2  slope of normal is  4]
Q.8121/para The tangent and normal at P(t), for all real positive t, to the parabola y2 = 4ax meet the axis of the
parabola in T and G respectively, then the angle at which the tangent at P to the parabola is inclined to the
tangent at P to the circle passing through the points P, T and G is
(A) cot–1t (B) cot–1t2 (C*) tan–1t (D) tan–1t2
1
[Sol. slope of tangent = (m1) at P on parabola
t

2at 2t
slope of PS = 2
 2 (normal to the circle)
a ( t  1) t  1

1 t2
 slope of tangent at P on circle = (m 2 )
2t

1 1 t2
 2 2
 tan= t

2t = 2  1  t 2 t = t 
1 t2 2 t (1  t 2 )
1 2
2t
  = tan–1t  (C) ]
Q.9131/para A circle with radius unity has its centre on the positive y-axis. If this circle touches the parabola
y = 2x2 tangentially at the points P and Q then the sum of the ordinates of P and Q, is
15 15
(A*) (B) (C) 2 15 (D) 5
4 8
dy
[Sol. = 4t [18-12-2005, 12th]
dx P
(0,a)
 2t 2  a  P(t, 2t2)
 (4t)   =–1 (–t, 2t2)Q

 t 
1
2t2 – a = – ....(1)
4
Also t2 + (2t2 – a)2 = 1 [(t, 2t2) satisfies the circle x2 + (y – a)2 = 1 ]
15 15
t2 =  4t2 = Ans. ]
16 4
Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [21]
Q.1029/para Normal to the parabola y2 = 8x at the point P (2, 4) meets the parabola again at the point Q. If C is
the centre of the circle described on PQ as diameter then the coordinates of the image of the point C in
the line y = x are
(A*) (– 4, 10) (B) (– 3, 8) (C) (4, – 10) (D) (– 3, 10)
[Sol. at12 = 2; y2 = 8x  a=2
t1 = 1  t1 = 1 or – 1
2
t2 = – t 1 – =–3
t1

Q is 2(3) 2 , 2( 2)(3) i.e. (18, – 12)

 
 C is (10, – 4)
The image is (– 4, 10) ] [12th, 06-01-2008]

Q.1149/para Two parabolas y2 = 4a(x - l1) and x2 = 4a (y – l2) always touch one another, the quantities l1 and l2
are both variable. Locus of their point of contact has the equation
(A) xy = a2 (B) xy = 2a2 (C*) xy = 4a2 (D) none
[Sol. 2 2
y = 4a (x – l1) ; x = 4a(y – l2)
dy dy
2y = 4a and 2x = 4a
dx dx
dy 2a dy x1
= y and =
dx ( x1 , y1 ) 1 dx ( x1 , y1 ) 2a

2a x
  1  x y = 4a2  R.H. ]
y1 2a 1 1

Q.1298/para A pair of tangents to a parabola is are equally inclined to a straight line whose inclination to the axis
is . The locus of their point of intersection is :
(A) a circle (B) a parabola (C*) a straight line (D) a line pair
[Sol. Let P at1t 2 , a ( t1  t 2 )  = P(h, k)
1 1
slope of tangents at A, m1 = t and at B, m2 = t
1 2
Let m = tan 
m1  m m  m2
then 1  mm  1  mm
1 2
tan(1 – ) = tan( – 2)
 (1 – ) = n + 2a
1 1 k

tan 1  tan  2 t1 t 2 2( t1  t 2 ) k
 tan 2 = tan (1 + 2) = 1  tan  tan  =  ; tan 2 = a 
1 1 t1t 2  1 h
1 2
1 1 h  a
t1 t 2 a
 locus of P(h, k) will be y = (x – a)tan 2 Ans.]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [22]

Q.13110/para In a parabola y2 = 4ax the angle  that the latus rectum subtends at the vertex of the parabola is
(A) dependent on the length of the latus rectum
5
(B) independent of the latus rectum and lies between &
6
3 5
(C) independent of the latus rectum and lies between &
4 6
2 3
(D*) independent of the latus rectum and lies between &
3 4
x
[Hint: Equation of latus rectum is x = a  =1 [11th, 14-02-2009]
a
x
 angle subtended at the vertex of y2 = 4ax will be y2 = 4ax  y2 = 4x2
a
 slopes of OA and OB will be 2 and – 2 respectively
2  (2) 4 4
 tan  = =   =  – tan–1  (D) ]
1  2(2)  3 3

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [23]

 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 5
Q.1para Normals are drawn at points A, B, and C on the parabola y2 = 4x which intersect at P(h, k).
The locus of the point P if the slope of the line joining the feet of two of them is 2 , is
 1
(A) x + y = 1 (B*) x – y = 3 (C) y2 = 2(x – 1) (D) y2 = 2 x  
 2
[Sol. The equation of normal at (at2, 2at) is [12th, 20-12-2009, complex]
y + tx = 2at + at3 ....(1)
As (1) passes through P(h, k), so
t1 y t1
at3 + t(2a – h) – k = 0 t2 ....(2)
t3
Here a=1 P(h, k)
t1 + t2 + t3 = 0 ....(3)
2 x
Also = 2  t1 + t2 = 1 ....(4)
t1  t 2
t2
From (3) and (4)  t3 = – 1
Put t3 = – 1 in (2), we get
– 1 – 1 (2 – h) – k = 0 t3
 –1–2+h–k=0
 Locus of P(h, k), is x – y = 3 Ans.]

Q.2115/para Tangents are drawn from the point ( 1, 2) on the parabola y2 = 4 x. The length , these tangents will
intercept on the line x = 2 is :
(A) 6 (B*) 6 2 (C) 2 6 (D) none of these
[Sol. SS1 = T 2

(y2  4 x) (y12  4 x1) = (y y1  2 (x + x1))2

(y2  4 x) (4 + 4) = [ 2 y  2 (x  1) ]2 = 4 (y  x + 1)2
2 (y2  4 x) = (y  x + 1)2 ;
solving with the line x = 2 we get ,
2 (y2  8) = (y  1)2 or 2 (y2  8) = y2  2 y + 1
or y2 + 2 y  17 = 0
where y1 + y2 =  2 and y1 y2 =  17
Now y1  y22 = (y1 + y2)2  4 y1 y2
or y1  y22 = 4  4 ( 17) = 72
 (y1  y2) = 72 = 6 2 ]

Q.3123/para Which one of the following lines cannot be the normals to x2 = 4y ?

(A) x – y + 3 = 0 (B) x + y – 3 = 0 (C) x – 2y + 12 = 0 (D*) x + 2y + 12 = 0
1
[Hint: equation of the normal to x2 = 4y in terms of slope y = mx+ +2]
m2

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [25]

 x2
Q.4129/para An equation of the line that passes through (10, –1) and is perpendicular to y =  2 is
4
(A) 4x + y = 39 (B) 2x + y = 19 (C) x + y = 9 (D*) x + 2y = 8
2
[Sol. 4y = x – 8 [29-01-2006, 12&13]
dy
4 = 2x
dx
dy x1
dx =
x1 , y1 2

2 y1  1
 slope of normal = – ; but slope of normal = x  10
x1 1

y1  1 2
 x1  10 = – x  x1y1 + x1 = – 2x1 + 20  x1y1 + 3x1 = 20
1

x12  8
substituting y1 = (from the given equation)
4
 x2  8 
 1
x1  4  3  = 20

 x1( x12 – 8 + 12) = 80  x1( x12 + 4) = 80
 
x13 + 4x1 – 80 = 0

x12 (x1 – 4) + 4x(x1 – 4) + 20(x1 – 4) = 0

(x1 – 4)( x12 + 4x1 + 20) = 0

Hence x1 = 4 ; y1 = 2
 P = (4, 2)
equation of PA is
1
y+1=– (x – 10)  2y + 2 = – x + 10  x + 2y – 8 = 0 Ans. ]
2

Paragraph for question nos. 5 to 6

Consider the parabola y2 = 8x
Q.5408/para Area of the figure formed by the tangents and normals drawn at the extremities of its latus rectum is
(A) 8 (B) 16 (C*) 32 (D) 64

Q.6409/para Distance between the tangent to the parabola and a parallel normal inclined at 30° with the x-axis, is
16 16 3 2 16
(A*) (B) (C) (D)
3 9 3 3
[Sol.
(i) For y2 = 4ax
(4a )(4a )
A= = 8a2
2
Here a = 2
 A = 32 sq. units [08-01-2006, 12th & 13th]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [26]

 a
(ii) y = mx +
m
y = mx – 2am – am3

(a m )  2am  am 3 ( m 2  1) 2 (1  m 2 ) 1  m 2
d= =a =a m ; put m = tan
1  m2 m 1 m2

sec 2  · sec 
=a tan  = a (sec2 · cosec )

put a = 2 and  = 30°

4 16
d=2· ·2= Ans. ]
3 3

Paragraph for question nos. 7 to 9

Tangents are drawn to the parabola y2 = 4x from the point P(6, 5) to touch the parabola at Q and R.
C1 is a circle which touches the parabola at Q and C2 is a circle which touches the parabola at R.
Both the circles C1 and C2 pass through the focus of the parabola.
Q.7para Area of the PQR equals
1 1
(A*) (B) 1 (C) 2 (D)
2 4
Q.8para Radius of the circle C2 is
(A) 5 5 (B*) 5 10 (C) 10 2 (D) 210
Q.9para The common chord of the circles C1 and C2 passes through the
(A) incentre (B) circumcentre
(C*) centroid (D) orthocentre of the PQR
[Sol. Equation of tangent of slope m to y2 = 4x is
1 y (6,5)P
y = mx + ....(1) [12th, 03-01-2010, P-1] (4,4) R(9,6)
m Q
(i) As (1) passes through P(6, 5), so C1
S(1,0) x
1 O
5 = 6m +
m
1 1
 6m2 – 5m + 1 = 0  m= or m =
2 3
 1 2   1 2 
Points of contact are  2 m  and  2 , m 
 , 
 m1 1  m2 2

Hence P (4, 4) and Q (9, 6)

6 5 1
1 1
Area of PQR = 2 4 4 1 =  (A)
9 6 1 2

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [27]

 1
(ii) y= x + 2  x – 2y + 4 = 0 ....(2)
2
1
and y= x + 3  x – 3y + 9 = 0
3
Now equation of circle C2 touching x – 3y + 9 = 0 at (9, 6), is
(x – 9)2 + (y – 6)2 + ( x – 3y + 9) = 0
As above circle passes through (1, 0), so
64 + 36 + 10 = 0   = – 10
Circle C2 is x2 + y2 – 28x + 18y + 27 = 0 .....(3)
Radius of C2 is
r22 = 196 + 81 – 27 = 277 – 27 = 250  r2 = 5 10  (B)
(iii) Equation of C1
(x – 4)2 + (y – 4)2 + (x – 2y + 4) = 0
As above circle passes through (1, 0)
9 + 16 + (5) = 0  =–5
2 2
Now C1 is x + y – 13x + 2y + 12 = 0 ....(4)
 Common chord of (3) and (4) is P(6,5)
15x – 16y – 15 = 0 ....(5)
 19 
Also centroid (G) of PQR is  , 5  G  19 ,5 
 3  3 
Q(4,4) R(9,6)
 19 
Clearly  , 5  satisfies equation (5)
 3 
Hence (C) ]

Paragraph for question nos. 10 to 12

Tangents are drawn to the parabola y2 = 4x at the point P which is the upper end of latus rectum.
Q.10410/para Image of the parabola y2 = 4x in the tangent line at the point P is
(A) (x + 4)2 = 16y (B) (x + 2)2 = 8(y – 2)
(C*) (x + 1)2 = 4(y – 1) (D) (x – 2)2 = 2(y – 2)
Q.11 Radius of the circle touching the parabola y2 = 4x at the point P and passing through its focus is
(A) 1 (B*) 2 (C) 3 (D) 2
Q.12 Area enclosed by the tangent line at P, x-axis and the parabola is
2 4 14
(A*) (B) (C) (D) none
3 3 3
[Sol. Point P is (1, 2) [13th, 17-02-2008] [Illustration]
Tangent is 2y = 2(x + 1)
i.e. y=x+1 ....(1)
2
hence image of y = 4x in (2) can be written as
(x + 1)2 = 4(y – 1)  (C)
note find the image of (t2, 2t) in the tangent line and then eliminate t to get the image
now family of circle touching the parabola y2 = 4x at (1, 2)

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [28]

 (x – 1)2 + (y – 2)2 + (x – y + 1) = 0 ....(2)
it passes through (1, 0)
4 + 2 = 0  =–2
hence circle is
x2 + y2 – 4x – 2y + 3 = 0
r= 4 1 3 = 2  (B)
2 2
 y2  y3 y 2 
Area =    ( y  1)  dy =   y
0
4  12 2 0
8 2
= –2+2=  (A) ]
12 3

More than one are correct:

Q.13509/para Let P, Q and R are three co-normal points on the parabola y2 = 4ax. Then the correct statement(s)
is/are
(A*) algebraic sum of the slopes of the normals at P, Q and R vanishes
(B*) algebraic sum of the ordinates of the points P, Q and R vanishes
(C*) centroid of the triangle PQR lies on the axis of the parabola
(D*) circle circumscribing the triangle PQR passes through the vertex of the parabola

Q.14511/para Variable chords of the parabola y2 = 4ax subtend a right angle at the vertex. Then :
(A*) locus of the feet of the perpendiculars from the vertex on these chords is a circle
(B*) locus of the middle points of the chords is a parabola
(C*) variable chords passes through a fixed point on the axis of the parabola
(D) none of these
[Hint : A = x2 + y2  4ax = 0 ; B = y2 = 2a (x  4a) ; C  (4a, 0) ]

Q.15517/para Through a point P (– 2, 0), tangents PQ and PR are drawn to the parabola y2 = 8x. Two circles
each passing through the focus of the parabola and one touching at Q and other at R are drawn. Which
of the following point(s) with respect to the triangle PQR lie(s) on the common chord of the two circles?
(A*) centroid (B*) orthocentre
(C*) incentre (D*) circumcentre
[Sol. (–2, 0) is the foot of directrix.
Hence Q and R are the extremities of the latus rectum and angle
QPR = 90° with PQR as right isosceles.
Hence by symmetric the common chord of the two circles will
be the x-axis which will be the median, altitude, angle bisector
and also the perpendicular bisector.
Hence centroid, orthocentre, incentre and circumcentre all will
lie on it. ] [13th, 09-03-2008]
Q.16para TP and TQ are tangents to parabola y2 = 4x and normals at P and Q intersect at a point R on
the curve. The locus of the centre of the circle circumscribing TPQ is a parabola whose
7 
(A*) vertex is (1, 0). (B*) foot of directrix is  , 0  .
8 
1 9 
(C) length of latus-rectum is . (D*) focus is  , 0 .
4 8 

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [29]

[Sol. We have 2h = t32 + 2 ....(1) y (2a, –at3) Q(t2 )
2k = t3 ....(2) T
Here a 1
 2h = 4k2 + 2 t1t 2  2
P(t1) (h,k)
t1  t 2  t 3  0
x
 2y2 = x – 1
1
y2 = (x – 1) (Parabola)
2
2
Now interpret. ] [12th, 20-12-2009] R(at3 , 2at3)

Match the column:

Q.1771 Consider the parabola y2 = 12x
Column-I Column-II
(A) Tangent and normal at the extremities of the latus rectum intersect (P) (0, 0)
the x axis at T and G respectively. The coordinates of the middle
point of T and G are
(B) Variable chords of the parabola passing through a fixed point K on (Q) (3, 0)
the axis, such that sum of the squares of the reciprocals of the two
parts of the chords through K, is a constant. The coordinate of the
point K are (R) (6, 0)
(C) All variable chords of the parabola subtending a right angle at the
origin are concurrent at the point
(D) AB and CD are the chords of the parabola which intersect at a point (S) (12, 0)
E on the axis. The radical axis of the two circles described on AB
and CD as diameter always passes through
[Ans. (A) Q; (B) R; (C) S; (D) P] [13th, 09-03-2008]

Subjective:
Q.18para Let L1 : x + y = 0 and L2 : x – y = 0 are tangent to a parabola whose focus is S(1, 2).
m
If the length of latus-rectum of the parabola can be expressed as (where m and n are coprime)
n
then find the value of (m + n). [Ans. 0011 ]

[Sol. [12th, 20-12-2009, complex]

Feet of the perpendicular (N1 and N2) from focus upon any tangent to parabola lies on the tangent line
at the vertex.
Now equation of SN1 is x + y =  passing through (1, 2)  =3
Equation of SN1 is x + y = 3

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [30]

 3 3
Solving x + y = 3 and y = x, we get N1   , 
2 2
|||ly equation of SN2 is x – y =  passing through (1, 2)  =–1
Equation of SN2 is y – x = 1
 1 1 
Solving y – x = 1 and y = – x, we get N2   , 
 2 2
Now equation of tangent line at vertex is, 2x – 4y + 3 = 0
Distance of S(1, 2) from tangent at vertex is
| 2 83| 3 1
= = =  latus rectum .
20 2 5 4
6 m
and hence length of latus rectum = =
5 n
Hence m + n = 6 + 5 = 11 Ans.]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [31]

 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 6

x2 y2
Q.12/elli Let 'E' be the ellipse + = 1 & 'C' be the circle x2 + y2 = 9. Let P & Q be the points (1 , 2) and
9 4
(2, 1) respectively. Then :
(A) Q lies inside C but outside E (B) Q lies outside both C & E
(C) P lies inside both C & E (D*) P lies inside C but outside E.

y2
Q.23/elli The eccentricity of the ellipse (x – 3)2 + (y – 4)2 = is
9

3 1 1 1
(A) (B*) (C) (D)
2 3 3 2 3
[Sol. 2 2
9(x – 3) + 9(y – 4) = y2  9(x – 3)2 + 8y2 – 72y + 144 = 0
2 2
9(x – 3) + 8(y – 9y) + 144 = 0
 2
9  81 9
2
9(x – 3)2 +8   y     + 144 = 0  9(x – 3) 2 + 8 y   = 162 – 144 = 18
 2 4  2


 9  9
8 y  
2 2y  2·4 1 1
9( x  3) 2 ( x  3) 2
  1    1; e2 = 1 – = ;  e=
18 18 2 94 9 9 3
Alternatively: put x–3=X and y – 4 = Y]

Q.347/ellipse An ellipse has OB as a semi minor axis where 'O' is the origin. F, F  are its foci and the angle FBF
is a right angle. Then the eccentricity of the ellipse i
1 1 3 1
(A*) (B) (C) (D)
2 2 2 4
[Sol. BF1 = OA  b 2  a

2 b2 2 b2 1 1
e  1   e 1    e  ]
a2 2b 2 2 2

x2 y2
Q.410/ellipse There are exactly two points on the ellipse   1 whose distance from the centre of the
a2 b2
a 2  2b 2
ellipse are greatest and equal to . Eccentricity of this ellipse is equal to
2
3 1 1 2
(A) (B) (C*) (D)
2 3 2 3

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [33]

[Sol. The given distance is clearly the length of semi major axis [11th, 25-02-2009, P-1]

a 2  2b 2
Thus, =a  2b2 = a2  2a2(1 – e2) = a2
2
1 1
 e2 =  e= Ans.]
2 2

Q.512/ellipse A circle has the same centre as an ellipse & passes through the foci F1 & F2 of the ellipse, such that
the two curves intersect in 4 points. Let 'P' be any one of their point of intersection. If the major axis of
the ellipse is 17 & the area of the triangle PF1F2 is 30, then the distance between the foci is :
(A) 11 (B) 12 (C*) 13 (D) none
[Hint : x + y = 17 ; xy = 60, To find x 2  y 2 ] [11th, 14-02-2009]
now, x2 + y2 = (x + y)2 – 2xy
= 289 – 120 = 169
 x 2  y 2  13 ]

Q.624/ellipse The latus rectum of a conic section is the width of the function through the focus. The positive
difference between the lengths of the latus rectum of 3y = x2 + 4x – 9 and x2 + 4y2 – 6x + 16y = 24
is
1 3 5
(A*) (B) 2 (C) (D)
2 2 2
[Hint: 3y = (x + 2)2 – 13 [12 & 13th 03-03-2007]
th

3y + 13 = (x + 2)2
 13 
 (x + 2)2 = 3  y    Latus Rectum = 3
 3
The other conic is, (x – 3)2 + 4(y2 + 4y) = 24 + 9
(x – 3)2 + 4(y + 2)2 = 49
( x  3) 2 ( y  2) 2
+ = 1 which is an ellipse
72 (7 2) 2

2b 2 2 · 49 7
Latus Rectum = = 4·7 =
a 2
7 1
 positive difference – 3 = Ans. ]
2 2

Q.731/ellipse Imagine that you have two thumbtacks placed at two points, A and B. If the ends of a fixed length of
string are fastened to the thumbtacks and the string is drawn taut with a pencil, the path traced by the
pencil will be an ellipse. The best way to maximise the area surrounded by the ellipse with a fixed length
of string occurs when
I the two points A and B have the maximum distance between them.
II two points A and B coincide.
III A and B are placed vertically.
IV The area is always same regardless of the location of A and B.
(A) I (B*) II (C) III (D) IV

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [34]

[Sol. A = ab 'a' is constant and b varies [12 & 13 05-3-2006]
A2 = 2a2(a2 – c2)
for A to be maximum c must be minimum; A & B  centre
as A  B  c0
ellipse becomes circle ]

Q.837/ellipse An ellipse having foci at (3, 3) and (– 4, 4) and passing through the origin has eccentricity equal to
3 2 5 3
(A) (B) (C*) (D)
7 7 7 5
[Hint : PS1 + PS2 = 2a [11th, 14-02-2009]
3 2  4 2  2a
2 a  7 2
Also 2ae = S1S2 = 1  49  5 2

2ae 5 2 5
   = e  (C) ]
2a 7 2 7

Q.9ellipse Let S(5, 12) and S'(– 12, 5) are the foci of an ellipse passing through the origin.
The eccentricity of ellipse equals
1 1 1 2
(A) (B) (C*) (D)
2 3 2 3
[Sol. We have 2ae = 13 2 = focal length ...(1) [12th, 20-12-2009, complex]
 2a = 26  a = 13 (By focus-directrix property)
 On putting a = 13 in equation (1), we get
1
2(13)e = 13 2  e = Ans. ]
2

More than one are correct:

x2 y2
Q.10501/ellipse Consider the ellipse  = 1 where   (0, /2).
tan 2  sec2 
Which of the following quantities would vary as  varies?
(A*) degree of flatness (B*) ordinate of the vertex
(C) coordinates of the foci (D*) length of the latus rectum
tan 2 
[Hint: e2 =1– = cos2 (as sec2 > tan2) [12 & 13 05-3-2006]
sec 2 
hence e = cos  ; vertex (0, ± sec )
2b 2 2 tan 2 
foci = (0, 1) ; l(LR) = = = 2 sin  · tan ]
a sec 

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [35]

 x2 y2
Q.11504/ellipse Extremities of the latera recta of the ellipses   1 (a > b) having a given major axis 2a
a2 b2
lies on
(A * ) x 2 = a(a – y) (B*) x2= a (a + y) (C) y2 = a(a + x) (D) y2 = a (a – x)
b2
[Sol. h = + ae ; k = +
a
 h2   h 2 
k = +a(1 – e2) = + a 1  2  = +  a  a 
  
 a   

h2 h2
+ ve sign , k = a    a  k h2 = a ( a – k) (A)
a a
h2
– ve sign , k =  a   h2 = a (a + k) (B) ]
a

Subjective:
Q.12elli Consider two concentric circles S1 : | z | = 1 and S2 : | z | = 2 on the Argand plane. A parabola is
drawn through the points where 'S1' meets the real axis and having arbitrary tangent of 'S2' as its
directrix. If the locus of the focus of drawn parabola is a conic C then find the area of the quadrilateral
formed by the tangents at the ends of the latus-rectum of conic C. [Ans. 0016]
[Sol. Clearly the parabola should pass through (1, 0) and (–1,0). Let directrix of this parabola be
x cos + y sin = 2. If M (h,k) be the focus of this parabola, then distance of (±1, 0) from 'M' and from
the directrix should be same. y
 (h – 1)2 + k2 = (cos – 2)2 ....(1)
and (h + 1)2 + k2 = (cos + 2)2 ....(2)
|z|=2
h
Now (2) – (1) cos = ....(3) |z| P(2cos, 2sin)
=1
2
Also (2) + (1)  (h2 + k2 + 1) = (cos2 + 4) ....(4) O x
(–1,0) (0,0) (1,0)
From (3) and (4), we get
M(h,k)
h2 3h 2
h2 + k2 + 1 = 4 +  + k2 = 3
4 4
x 2 y2
Hence locus of focus M(h, k) is   1 (Ellipse)
4 3
Also we know that area of the quadrilateral formed by the tangents at the ends of the latus-rectum is
2a 2
(where e is eccentricity of ellipse) [12th, 20-12-2009]
e
2( 4) 3 1 1
 Requred area = = 16 (square units) (As e2 = 1 – =  e= ) ]
1 4 4 2
2

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [36]

 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 7
Q.113/ellipse Point 'O' is the centre of the ellipse with major axis AB & minor axis CD. Point F is one focus of the
ellipse. If OF = 6 & the diameter of the inscribed circle of triangle OCF is 2, then the product
(AB) (CD) is equal to
(A*) 65 (B) 52 (C) 78 (D) none
2 2 2 2
[Hint : a e = 36  a  b = 36 ....(1); 4ab = ?
A'
Using r = (s  a) tan in  OCF
2
1 = (s  a) tan 45º where a = CF
2 = 2 (s  a)
or 2 = 2s  2a = 2s  AB
or 2 = (OF + FC + CO)  AB
AB C D
2=6+ +  AB
2 2
AB  CD
= 4  2 (a  b) = 8  a  b = 4  (2)
2
From (1) & (2) a + b = 9  2a = 13 ; 2b = 5  (AB) (CD) = 65 ]

Q.217/ellipse The y-axis is the directrix of the ellipse with eccentricity e = 1/2 and the corresponding focus is at
(3, 0), equation to its auxilary circle is
(A*) x2 + y2 – 8x + 12 = 0 (B) x2 + y2 – 8x – 12 = 0
(C) x2 + y2 – 8x + 9 = 0 (D) x2 + y2 = 4
[Sol. Directrix : x = 0 [12th & 13th 19-3-2006]
e = 1/2
Focus = (3, 0)
1
 ( x  3) 2  y 2 = 2 · | x |

1 2
 (x – 3)2 + y2 = ·x  4(x – 3)2 + 4y2 = x2  3x2 – 24x + 4y2 + 36 = 0
4
( x  4) 2 y2
 3(x – 4)2 + 4y2 = 12  + =1 ....(1)
4 3
 a=2;b= 3 ; centre (4, 0)  auxillary circle is (x – 4)2 + y2 = 4 Ans. ]

x2 y2 x2 y2
Q.320/elliWhich one of the following is the common tangent to the ellipses, 2 2  = 1 and 2  = 1?
a b b2 a a 2  b2
(A) ay = bx + a 4  a 2 b 2  b 4 (B*) by = ax  a 4  a 2 b 2  b 4
(C) ay = bx  a 4  a 2 b 2  b 4 (D) by = ax + a 4  a 2 b 2  b 4

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [37]

 x2 y2
[Sol. Equation of a tangent to  1
a 2  b2 b2

y = mx  (a 2  b 2 ) m 2  b 2 ....(1)

x2 y2
If (1) is also a tangent to the ellipse 1 then 
a2 a 2  b2
(a2 + b2)m2 + b2 = a2m2 + a2 + b2 (using c2 = a2m2 + b2)
a2 a
b2m2 = a2  m2 = 2  m=+
b b

a a2
y=+ x + (a 2  b 2 ) 2
 b2
b b

by = + ax + a 4  a 2b2  b 4
Note : Although there can be four common tangents but only one of these appears in B]

x2 y2
Q.426/ellise x  2y + 4 = 0 is a common tangent to y2 = 4x &  2 = 1. Then the value of b and the other
4 b
common tangent are given by :
(A*) b = 3 ; x + 2y + 4 = 0 (B) b = 3 ; x + 2y + 4 = 0
(C) b = 3 ; x + 2y  4 = 0 (D) b = 3 ; x  2y  4 = 0
[Sol. y = x/2 + 2 is tangent on the ellipse then 4 = 4.(1/4) + b2  b2 = 3
parabola is , y = mx + 1/m
1
using condition of tangency, = 4m2 + 3
m2
4y2 + 3y – 1 = 0 (when m2 = y)
2
4y + 4y – y – 1 = 0 4y(y + 1) – (y + 1) = 0
 y = 1/4 ; y = –1
m = + 1/2
y = x/2 + 2 or y = –x/2 – 2 2y + x + 4 = 0 (other tangent) ]

Q.533/ellipse If  &  are the eccentric angles of the extremities of a focal chord of an standard ellipse,
then the eccentricity of the ellipse is :
cos   cos  sin   sin  cos   cos  sin   sin 
(A) (B) (C) (D*)
cos(  ) sin (  ) cos(  ) sin (   )

x   y      ae    
[Hint : cos + sin = cos ; cos = cos
a 2 b 2 2 a 2 2
   
cos 2 2 sin 2 sin   sin 
 e= . = ]
 
cos 2
 
2 sin 2 sin (   )

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [38]

Q.634/ellipse An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability
that this point lies outside the ellipse is 2/3 then the eccentricity of the ellipse is :
2 2 5 8 2
(A*) (B) (C) (D)
3 3 9 3

2  a 2   ab b 8 2 2
[Hint : = = 1  = 1  1  e 2  e2 =  e = ]
3 a 2 a 9 3

x 2 y2
Q.741/elli Consider the particle travelling clockwise on the elliptical path  = 1. The particle leaves the
100 25
orbit at the point (–8, 3) and travels in a straight line tangent to the ellipse. At what point will the particle
cross the y-axis?
 25   25   7
(A*)  0,  (B)  0,   (C) (0, 9) (D)  0, 
 3   3   3
[12th & 13th 11-3-2007]
x2 y2
Q.8ellipse The Locus of the middle point of chords of an ellipse   1 passing through P(0, 5)
16 25
is another ellipse E. The coordinates of the foci of the ellipse E, is
 3  3
(A)  0,  and  0,  (B) (0, – 4) and (0, 1)
 5  5 
 11   1 
(C*) (0, 4) and (0, 1) (D)  0,  and  0, 
 2  2 
[Sol. We have 4 cos  = 2h and 5(1 + sin ) = 2k [12th, 20-12-2009, complex]
As 2 2
cos  + sin  = 1
2
h2  2k  x2 4y2 4y x2 4
 +   1 = 1  + – =0  + (y2 – 5y) = 0
4  5  4 25 5 4 25
2
x2 4  5   25  y
   y    =0 (0, 5)
4 25  2  4 

2 M
 5
2 y  (h, k)
x  2 x
 + = 1 ....(1) (–4, 0) (4, 0)
4 25 P
4 (4cos, 5sin)

5
Put X = x, y– =Y (0, –5)
2
 Equation (1) becomes
X2 Y2 Y
  1 (Ellipse )
4 25 (0, 5/2)
4 •

4·4 9 3 O (2,0) X
e2 = 1 –   e= (–2,0)
25 25 5
(0,–5/2)

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [39]

  3  3
 F1 =  0,  , F2 =  0, 
 2  2 
Hence in x–y system, foci are (0, 4), (0, 1)  (C) ]

Paragraph for question nos. 9 to 11

Consider the curve C : y2 – 8x – 4y + 28 = 0. Tangents TP and TQ are drawn on C at P(5, 6) and
Q(5, – 2). Also normals at P and Q meet at R.

Q.966MB The coordinates of circumcentre of PQR, is

(A) (5, 3) (B*) (5, 2) (C) (5, 4) (D) (5, 6)

Q.1067MB The area of quadrilateral TPRQ, is

(A) 8 (B) 16 (C*) 32 (D) 64

Q.1168MB Angle between a pair of tangents drawn at the end points of the chord y + 5t = tx + 2
of curve C  t  R , is
   
(A) (B) (C) (D*)
6 4 3 2
2
[Sol.(i) Given curve is a parabola (y – 2) = 8(x – 3) whose focus is (5, 2).
As P(5, 6) and Q (5, – 2) are the coordinates of the end points of the latus-rectum of the parabola.
 Normals at P & Q are perpendicular to each other and meeting on the axis of the parabola
 PQR is right angled at R
 Circumcentre of PQR is focus of the parabola i.e. (5, 2)
 8  8 
(ii) Area of quadrilateral TPRQ = Area of square TPRQ =     32 (square units)
 2  2 
(iii) Also y + 5t = tx + 2 is a focal chord of the given parabola

 Angle between a pair of tangents = .]
2

More than one are correct:

Q.12506/ellipse If a number of ellipse be described having the same major axis 2a but a variable minor axis then the
tangents at the ends of their latera recta pass through fixed points which can be
(A*) (0, a) (B) (0, 0) (C*) (0, – a) (D) (a, a)
[Hint: e is a variable quantity
xae yb2
  1  ex + y = a  y – a + ex = 0
a 2 ab 2
it passes through (0, a).
|||ly other point is (0, – a) ]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [40]

 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 8
x2 y2
Q.121/ellipse The normal at a variable point P on an ellipse 2  2 = 1 of eccentricity e meets the axes of the
a b
ellipse in Q and R then the locus of the mid-point of QR is a conic with an eccentricity e  such that :
(A) e  is independent of e (B) e  = 1
(C*) e  = e (D) e  = 1/e

Q.228/ellipse The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the
tangent and normal at its point whose eccentric angle is /4 is :
2
a  b 2 ab
 a 2
 b2  2
a  b  2
a 2  b2
(A*) (B) (C) (D)
a 2  b2 a 2
 b 2 ab
 2
ab  a  b  2
a 2
 b 2 ab

 a b  2 ab a 2  b2
[Hint : P  ,  p1 = 2 ; p 2 =  p1p2 = result ]
 2 2 a  b2 2 a 2  b2  
x cos  y sin 
[Sol. T:  1
a b

ab
p1 = ....(1)
b 2 cos 2   a 2 sin 2 

ax by
N1 :   a 2  b2
cos  sin 

(a 2  b 2 ) sin  cos 
p2 = ....(2)
a 2 sin 2   b 2 cos 2 

ab(a 2  b 2 ) ab(a 2  b 2 )
p1p2 = when  = /4; p1p2 = = Ans ]
 a 2 b2  a 2  b2
2  
 2 2 
1
Q.371MB If P is any point on ellipse with foci S1 & S2 and eccentricity is such that
2
  
 PS1S2 =  PS2S1 = , S1PS2 =  , then cot , cot , cot are in
2 2 2
(A*) A.P. (B) G.P. (C) H.P. (D) NOT A.P., G.P. & H.P.
2ae S1P S2 P 2a
[Sol. By sine rule in PS1S2, we get sin (  ) = sin  = sin  = sin   sin 

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [41]

        
2 sin   cos   Y
sin (  ) e  2   2 
 e = sin   sin   
1        
2 sin   cos   P
 2   2 
–(+)
1 1 S2   S1
1 e 1 X
Now
 
= tan tan = 2  2 1 2ae
1 e 2 2 1 3 3
1
2 2
  1
 tan tan = ... (1)
2 2 3
Also we know that
     
cot + cot + cot = cot cot cot
2 2 2 2 2 2
     
 2 cot = cot + cot  cot , cot , cot are in A.P.. ]
2 2 2 2 2 2

Paragraph for question nos. 4 to 6

x 2 y2
Consider the ellipse  = 1 and the parabola y2 = 2x. They intersect at P and Q in the first and
9 4
fourth quadrants respectively. Tangents to the ellipse at P and Q intersect the x-axis at R and tangents to
the parabola at P and Q intersect the x-axis at S.
Q.4ellipse The ratio of the areas of the triangles PQS and PQR, is
(A) 1 : 3 (B) 1 : 2 (C*) 2 : 3 (D) 3 : 4
Q.5 The area of quadrilateral PRQS, is
3 15 15 3 5 3 5 15
(A) (B*) (C) (D)
2 2 2 2
Q.6 The equation of circle touching the parabola at upper end of its latus rectum and passing through its
vertex, is
9
(A) 2x2 + 2y2 – x – 2y = 0 (B) 2x2 + 2y2 +4x – y=0
2
(C) 2x2 + 2y2 + x – 3y = 0 (D*) 2x2 + 2y2 – 7x + y = 0
x 2 y2
[Sol. Solving the curves y2 = 2x and  = 1 for the points of intersection, we have
9 4
y 3 , 3
3 2
4x2 + 18x – 36 = 0 x = , – 6 P
2
2
But from y = 2x we have x > 0 –3 ,0
2
R
3 x
 x= S O (0,0) M 3 ,0 (6, 0)
2 2

3 Q
at which y2 = 2 · 3 ,– 3
2
2

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [42]

  y=± 3
3  3 
 P , 3  and Q  ,  3 
2  2 

x 2 y2 x 3 y
 
Now equation of tangents at P and Q to ellipse   1 is     3  1 which intersect at
9 4 9 2 4
R(6, 0) [12th, 03-01-2010, P-2]
3  3 
Equation of tangents at P and Q to parabola y2 = 2x will be y  3 = x +
  which cut x-axis S , 0
2  2 

1 3  3
PQ · MS MS
 
2
Area PQS 2 2  2  3
 = = = 3 = = Ans.(i)
Area PQR 1 MR 9 3
PQ · MR 6
2 2 2

1 1 15 3
Area of quadrilateral PRQS = PQ(MS + MR) = · 2 3 (6 – ( 3 2) ) = Ans.(ii)
2 2 2
1 
(iii) Clearly upper end of latus rectum of parabola is  ,1 .
2 
1 
And equation of tangent at  ,1 to
2 
1
y2 = 2x is y = x +
2
 The equation of circle is
2
 1 2  1
 x    ( y  1)   y  x   = 0
 2  2 y
As above circle passes through V (0,0), so
1  5 1 ,1
1 = 0   = 2
4 2 2
x
 The equation of required circle is V(0,0)
2
 1 2 5 1
 x    ( y  1)   y  x   = 0
 2 2 2
2 2
 2x + 2y – 7x + y = 0 ]

Paragraph for question nos. 7 to 11

Let the two foci of an ellipse be (– 1, 0) and (3, 4) and the foot of perpendicular from the focus
(3, 4) upon a tangent to the ellipse be (4, 6).
Q.774MB The foot of perpendicular from the focus (– 1, 0) upon the same tangent to the ellipse is
 12 34   7 11   17 
(A*)  ,  (B)  ,  (C)  2,  (D) (– 1, 2)
5 5  3 3   4
Q.8 The equation of auxiliary circle of the ellipse is
(A) x2 + y2 – 2x – 4y – 5 = 0 (B*) x2 + y2 – 2x – 4y – 20 = 0
(C) x2 + y2 + 2x + 4y – 20 = 0 (D) x2 + y2 + 2x + 4y – 5 = 0

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [43]

Q.9 The length of semi-minor axis of the ellipse is
(A) 1 (B) 2 2 (C*) 17 (D) 19
Q.10 The equations of directrices of the ellipse are
21 17
(A) x – y + 2 = 0, x – y – 5 = 0 (B) x + y – = 0, x + y + =0
2 2
3 5 31 19
(C) x – y + = 0, x – y – = 0 (D*) x + y – = 0, x + y + =0
2 2 2 2
Q.11 The point of contact of the tangent with the ellipse is
 40 68  4 8  8 17   41 83 
(A*)  ,  (B)  ,  (C)  ,  (D)  , 
 11 11  7 7 5 5   13 13 
[Sol.
 43
(i) Equation tangent is (y – 6) = –   (x – 4) i.e., x + 2y – 16 = 0
64
  (1) 0 [1( 1)  2(0)  16]
So, = =–
1 2 (1) 2  ( 2) 2

 12 34 
 ()   , 
5 5 
 3 1 4  0 
(ii) Centre   ,   (1, 2) and radius = (4  1) 2  (6  2) 2 = 5
 2 2 
So, circle is x2 + y2 – 2x – 4y – 20 = 0
(iii) a = radius = 5. Also 2ae = (3  1) 2  (4  0) 2 = 4 2 ,
So b2 = a2 – a2e2
2
 b2 = 25 – 2 2
  = 17
 b2 = 17 gives b = 17
a 5 25
(iv) The directrices are at distances i.e. = = from centre (1, 2) and perpendicular to the
e 2 2 /5 2 2
|1 2  k | 25 19 31
line joining foci. Let its equation be x + y + k = 0, so =  k= , 
2 2 2 2 2
Ans.5 Let the point of contact of tangent be P  (16 – 2, ). Now SP = ePM, (focus-directrix property),
2
2  31 
2 2 16  2    
 2
 (16 – 2 – 3)2 + ( – 4)2 =  

 5  2
2 2
 25 (5 – 60 + 185) = 4 – 4 + 1
68 40
 (11 – 68)2 = 0   = , So 16 – 2 = .]
11 11

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [44]

Subjective:
x2 y2
Q.1270MBFind the number of integral values of parameter 'a' for which three chords of the ellipse + =1
2a 2 a2
 a2 
(other than its diameter) passing through the point P11a ,   are bisected by the parabola y2 = 4ax.
 4 
[Ans. 0002 ]
x2 y2
[Sol. Any point on the parabola y2 = 4ax is (at2, 2at). Equation of chord of the ellipse   1 , whose
2a 2 a2

x · at 2 y · 2at a 2t 4 4a 2 t 2
mid-point is (at2, 2at) is + = +
2a 2 a2 2a 2 a2
 tx + 4y = at3 + 8at ( t  0)
 a 2 
As it passes through 11a ,  4  ,

 

 a2 
 11at – 4  4  = at3 + 8at  at3 – 3at + a2 = 0
 
3
 t – 3t + a = 0 (a  0)
Now, three chords of the ellipse will be bisected by the parabola if the equation (1) has three real and
distinct roots.
Let f(t) = t3 – 3t + a
f '(t) = 3t2 – 3 = 0  t=±1
So, f (1) f(–1) < 0
 a  (– 2, 2)
But a  0, so a  (– 2, 0)  (0, 2)
 Number of integral values of 'a' = 2. ]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [45]

 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 9
Q.1 2 2
Consider the hyperbola 9x – 16y + 72x – 32y – 16 = 0. Find the following:
(a) centre (b) eccentricity (c) focii (d) equation of directrix
(e) length of the latus rectum (f) equation of auxilary circle
(g) equation of director circle
5 9
[Ans. (a) (–4, – 1); (b) ; (c) (1, – 1), (– 9, – 1); (d) 5x + 4 = 0, 5x + 36 = 0, (e) ;
4 2
2 2 2
(f) (x + 4) + (y + 1) = 16; (g) (x + 4) + (y + 1) = 7]2

[Sol.
(i) The equation of the hyperbola can be written as
9(x2 + 8x) – 16(y2 + 2y) = 16 i.e. 9{(x + 4)2 – 16} – 16{(y + 1)2 – 1} = 16
( x  4) 2 ( y  1) 2
i.e. 9(x + 4)2 – 16(y + 1)2 = 144 i.e.  1
16 9
X2 Y2
Shifting the origin to (– 4, – 1), the equation of the hyperbola becomes   1.
16 9
 The centre of the hyperbola is the point (– 4, – 1) Ans.(i)
(ii) The semi-transverse axis a = 4, the semi-conjugate axis b = 3
b2 = a2(e2 – 1)
5
 9 = 16(e2 – 1)  e= Ans.(ii)
4
(iii) The transverse axis lies along the new x-axis and the conjugate axis lies along the new y-axis.
CA = 4, CA' = 4.
45
CS = ae = =5
4
 AS = 1
 The coordinates of S are (1, – 1). Ans.(iii)
CS' = ae = 5
 The coordinates of S' are (–9, – 1) Ans.(iii)
If the directrix corresponding to S meet the transverse axis at Z,
a 16
CZ = =
e 5
16 4
 AZ = 4 – =
5 5
4
The equation of the directrix is x = i.e. 5x + 4 = 0 Ans.(iv)
5
|||ly the equation of the directrix orresponding to S' is 5x + 36 = 0 Ans.(iv)]

Q.21/hypThe area of the quadrilateral with its vertices at the foci of the conics
9x2 – 16y2 – 18x + 32y – 23 = 0 and
25x2 + 9y2 – 50x – 18y + 33 = 0, is
(A) 5/6 (B*) 8/9 (C) 5/3 (D) 16/9

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [47]

[Sol. 1st is a hyperbola [12th, 04-01-2009, P-1]
9(x – 1)2 – 16(y – 1)2 = 16 with e = 5/4
nd
and 2 is an ellipse
25(x – 1)2 + 9(y – 1)2 = 1 with e = 4/5
with x – 1 = X and y – 1 = Y
1 1 10 8 8
area = d1d2 = · · = Ans.
2 2 3 15 9
Note that eE · eH = 1 ]

x 2 y2
Q.35/hyp Eccentricity of the hyperbola conjugate to the hyperbola   1 is
4 12
2 4
(A*) (B) 2 (C) 3 (D)
3 3

b2 12 1 1
[Hint: e12  1  2 =1+ =4  e1 = 2 ; now  = 1 [12th, 04-01-2009, P-1]
a 4 e12 e 22
1 1 3 4 2
e 22 = 1 – 4 = 4  e 22 =  e2 = ]
3 3

Q.413/hyper The locus of the point of intersection of the lines 3 x  y  4 3 t = 0 & 3 tx + ty  4 3 = 0

(where t is a parameter) is a hyperbola whose eccentricity is
2 4
(A) 3 (B*) 2 (C) (D)
3 3

x 2 y2
[Hint: hyperbola   1] [11th, 14-02-2009]
16 48

Q.515/hyp If the eccentricity of the hyperbola x2  y2 sec2  = 5 is 3 times the eccentricity of the ellipse
x2 sec2  + y2 = 25, then a value of  is :
(A) /6 (B*) /4 (C) /3 (D) /2
x2 y2
[Sol.  1
5 5 cos2 

b2 5 cos 2 
e12  1  =1+ = 1 + cos2 ; |||ly eccentricity of the ellipse
a2 5

x2 y2 2 25 cos 2  2 2
  1 is e 2  1  = sin2 ; put e1 = 3 e2  e1 = 3 e 2
25 cos 2  25 25
1
 1 + cos2 = 3sin2  2 = 4 sin2  sin  = ]
2

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [48]

 x 2 y2 x 2 y2 1
Q.617/hyp The foci of the ellipse  2
 1 and the hyperbola   coincide. Then the value of b2
16 b 144 81 25
is
(A) 5 (B*) 7 (C) 9 (D) 4
5 3 9 b2
[Hint: eH = ; eE =  =1–  b2 = 7 ] [12th Test(16-1-2005)]
4 4 16 16

More than one are correct:

Q.7505hyp Which of the following equations in parametric form can represent a hyperbola, where 't' is a parameter.
a  1 b  1 tx y x ty
(A*) x = t   & y = t   (B)  +t=0 & + 1=0
2  t 2  t a b a b

t
(C*) x = et + et & y = et  et (D*) x2  6 = 2 cos t & y2 + 2 = 4 cos2
2
Q.8hyper Let p and q be non-zero real numbers. Then the equation (px2 + qy2 + r)(4x2 + 4y2 – 8x – 4) = 0
represents
(A*) two straight lines and a circle, when r = 0 and p, q are of the opposite sign.
(B*) two circles, when p = q and r is of sign opposite to that of p.
(C*) a hyperbola and a circle, when p and q are of opposite sign and r  0.
(D*) a circle and an ellipse, when p and q are unequal but of same sign and r is of sign opposite to that
of p.
[Sol. (px2 + qy2 + r) (4x2 + 4y2 – 8x – 4) = 0 [12th, 03-01-2010, P-1]
 2 2 2
4x + 4y – 8x – 4 = 0 (x – 1) + y = 1 2

or
px2 + qy2 + r = 0 will represents
(i) two straight lines if r = 0 and p, q are of opposite sign.
(ii) a circle if p = q and r is of opposite sign that of p.
(iii) a hyperbola if p and q are of opposite sign & r  0.
(iv) an ellipse if p and q are unequal but of same sign and r is of sign opposite to that of p.]
Match the column:
Q.970 Match the properties given in column-I with the corresponding curves given in the column-II.
Column-I Column-II
(A) The curve such that product of the distances of any of its tangent (P) Circle
from two given points is constant, can be
(B) A curve for which the length of the subnormal at any of its point is (Q) Parabola
equal to 2 and the curve passes through (1, 2), can be
(C) A curve passes through (1, 4) and is such that the segment joining (R) Ellipse
any point P on the curve and the point of intersection of the normal
at P with the x-axis is bisected by the y-axis. The curve can be (S) Hyperbola
(D) A curve passes through (1, 2) is such that the length of the normal
at any of its point is equal to 2. The curve can be
[Ans. (A) R, S; (B) Q; (C) R; (D) P]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [49]

[Sol. [13th, 17-02-2008]
(A) Very important property of ellipse and hyperbola (p1p2 = b2)  (R), (S)
dy y2
(B) y =2  = 2x + C
dx 2
x = 1, y = 2  C=0
 2
y = 4x  parabola  (Q)
(C) Equation of normal at P
1
Y – y = – (X  x )
m
Y = 0, X = x + my
x
X = 0, Y = y –
m dy
hence x + my + x = 0  2x + y =0
dx
2x dx + y dy = 0
y2
x2 + = C passes through (1, 4)
2
1+8=C
y2 x 2 y2
hence x2 + =9   =1  ellipse  (R)
2 9 18
(D) length of normal
(x + my – x)2 + y2 = 4
m2y2 + y2 = 4
4  y2 dy 4  y2 y dy
  dx
m2 = 2 ; = ;  2
y dx y 4y

– 4  y2 = x + C
x = 1, y = 4  C=–1
 (x – 1)2 = 4 – y2
(x – 1)2 + y2 = 4  circle  (P)]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [50]

 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 10

x2 y2
Q.12/hyp The magnitude of the gradient of the tangent at an extremity of latera recta of the hyperbola  1
a2 b2
is equal to (where e is the eccentricity of the hyperbola)
(A) be (B*) e (C) ab (D) ae
xx1 yy1 x · ae y · b 2 ex y
[Sol. T: 2 – ; – =1 or – =1 or ex – y = a  m = e Ans. ]
a b2 a 2 a · b2 a a
[12th, 04-01-2009, P-1]

Q.28/hyp The number of possible tangents which can be drawn to the curve 4x2  9y2 = 36, which are perpendicular
to the straight line 5x + 2y 10 = 0 is :
(A*) zero (B) 1 (C) 2 (D) 4
[Hint: y =  (5/2) x + 5  m = 2/5  a2m2  b2 = 9 . 4/25  4 = (36  100)/25 < 0
Note that the slope of the tangent (2/5) is less than the slope of the asymptote which is 2/3 which is not
possible ] [12th, 04-01-2009, P-1]


Q.333/hyp Locus of the point of intersection of the tangents at the points with eccentric angles  and  on
2
x 2 y2
the hyperbola 2  2 = 1 is :
a b
(A) x = a (B*) y = b (C) x = ab (D) y = ab
x sec  y tan 
[Sol. Tangent at  ,  1
a b
 x cos ec y cot 
at   1
2 a b
 (bsech – (a tan)k = ab
(bcosec)h – (a cot)k = ab

b sec  ab
b cos ec ab b(sec   cos ec) b(sec   cos ec)
K=– = cot  sec   tan  cos ec = – = b  (B) ]
b sec  a tan  cos ec  sec 
b cos ec a cot 

x2 y2
Q.49/hyp The equation + = 1 (p  4, 29) represents
29  p 4  p
(A) an ellipse if p is any constant greater than 4.
(B*) a hyperbola if p is any constant between 4 and 29.
(C) a rectangular hyperbola if p is any constant greater than 29.
(D) no real curve if p is less than 29.
[Hint: For ellipse 29 – p > 0 and 4–p>0  p<4
for hyperbola 29 – p > 0 and 4–p<0  p  (4, 29)]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [51]

 x2 y2
Q.546/hyp If – = 1 represents family of hyperbolas where ‘’ varies then
cos 2  sin 2 
(A*) distance between the foci is constant
(B) distance between the two directrices is constant
(C) distance between the vertices is constant
(D) distances between focus and the corresponding directrix is constant
[Hint: d2 = 4 a2 e2
= 4(a2 + b2) = 4  d = 2  (A) ]

Q.635/hyp Number of common tangent with finite slope to the curves xy = c2 & y2 = 4ax is :
(A) 0 (B*) 1 (C) 2 (D) 4
3 2 2
[Hint: m =  (a /4c ) ]

x2 y2 x2 y2
Q.752hyp Area of the quadrilateral formed with the foci of the hyperbola   1 and   1 is
a2 b2 a2 b2
1 2
(A) 4(a2 + b2) (B*) 2(a2 + b2) (C) (a2 + b2)
(a + b2) (D)
2
[Hint: Given hyperbolas are conjugate and the quadrilateral formed by their foci is a square
x2 y2 x2 y2
now   1 and   1
a2 b2 a2 b2

b2 a2 (a 2  b 2 ) 2 a 2  b2
e12  1  2
; e2  1 
2 2
; e1 e 2  ; e1e2 =
a2 b2 a 2b2 ab
(2ae1 )(2be 2 ) 2ab(a 2  b 2 )
A= = 2abe1e2 = ] [13th test (24-3-2005)]
2 ab

Q.855hyp For each positive integer n, consider the point P with abscissa n on the curve y2 – x2 = 1. If dn
represents the shortest distance from the point P to the line y = x then Lim(n · d n ) has the value equal to
n 

1 1 1
(A*) (B) (C) (D) 0
2 2 2 2
[Sol. Curve is rectangular hyperbola. [13th, 16-12-2007]

n  n2 1
perpendicular distance, dn = 2

n  2
Lim(n · d n ) = Lim  n  1  n 
n  n  2  

n 1 1
= Lim = Ans. ]
n  2 n2 1  n 2 2

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [52]

 Paragraph for question nos. 9 to 11
The graph of the conic x2 – (y – 1)2 = 1 has one tangent line with positive slope that passes through the
origin. the point of tangency being (a, b). Then
a
Q.9407/hyp The value of sin–1   is
b
5   
(A) (B) (C) (D*)
12 6 3 4
Q.10 Length of the latus rectum of the conic is
(A) 1 (B) 2 (C*) 2 (D) none
Q.11 Eccentricity of the conic is
4
(A) (B) 3 (C) 2 (D*) none
3
[Sol.(i) differentiate the curve [13th test (09-10-2005)]
dy
2x – 2(y – 1) =0
dx
dy  a b b

dx  a , b b  1 = a
 (mOP =
a
)

a2 = b2 – b ....(1)
Also (a, b) satisfy the curve
a2 – (b – 1)2 = 1
a2 – (b2 – 2b + 1) = 1
a2 – b2 + 2b = 2
 – b + 2b = 2  b = 2 { putting a2 – b2 = – b from (1) }
 a= 2 (a  – 2)
a 
 sin–1   = Ans.
b 4
2b 2
Sol.(ii) Length of latus rectum = = 2a = distance between the vertices = 2
a
(note that the hyperbola is rectangular)
Sol.(iii) Curve is a rectangular hyperbola  e = 2 Ans. ]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [53]

 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) DPP. NO.- 11
Q.144/hyp If x + iy=   i where i =  1 and  and  are non zero real parameters then  = constant and
 = constant, represents two systems of rectangular hyperbola which intersect at an angle of
   
(A) (B) (C) (D*)
6 3 4 2

[Hint : x2 – y2 + 2xyi =  + i  ; x2 – y2 =  and xy =  ; which intersects at  (D) ]
2
Q.212/hyp Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola
16y2 – 9x2 = 1 is
(A) x2 + y2 = 9 (B) x2 + y2 = 1/9 (C) x2 + y2 =7/144 (D*) x2 + y2 = 1/16
y2 x2
[Sol.  1 [12th, 04-01-2009, P-1]
1 / 16 1 / 9
Locus will be the auxilary circle
x2 + y2 = 1/16 ]

Q.345/hyp PQ is a double ordinate of the ellipse x2 + 9y2 = 9, the normal at P meets the diameter through Q at R,
then the locus of the mid point of PR is
(A) a circle (B) a parabola (C*) an ellipse (D) a hyperbola
x2 y2
[Sol.  1 ; a = 3 , b = 1
9 1

a 2x b2 y
Equation of PR :   a 2  b2
a cos  b sin 

3x y
 8 ....(1)
cos  sin 
b sin  sin 
Equation of CQ : y = – x  y=– x
a cos  3 cos 
y x
 ....(2) put in (1)
sin  3 cos 
3x x 10 x 12 cos   4 cos 
  8   8  x1 = ; y1 = from (2)
cos  3 cos  3 cos  5 5
12 cos  27 cos  10h
we have , 2h =  3 cos  =  cos =
5 5 27
4 sin  sin 
2k = sin – =  sin = 10 k
5 5
100 x 2
sin2+ cos2 =1  100y2 + =1  Ellipse ]
729

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [55]

 x2 y2
Q.440/hyp With one focus of the hyperbola   1 as the centre , a circle is drawn which is tangent to the
9 16
hyperbola with no part of the circle being outside the hyperbola. The radius of the circle is
11
(A) less than 2 (B*) 2 (C) (D) none
3
16 25 5
[Hint : e2 = 1 + =  e=
9 9 3
 focus = (5, 0)
Use reflection property to prove that circle cannot touch at two points. It can
only be tangent at the vertex
r=5–3=2]
x2 y2
Q.549hyp If the tangent and normal at any point of the hyperbola  1 , meets the conjugate axis at Q

a2 b2
and R, then the circle described on QR as diameter passes through the
(A) vertices (B*) focii
(C) feet of directrices (D) ends of latera recta

Q.628/hyp Let the major axis of a standard ellipse equals the transverse axis of a standard hyperbola and their
director circles have radius equal to 2R and R respectively. If e1 and e2 are the eccentricities of the
ellipse and hyperbola then the correct relation is
(A) 4e12 – e22 = 6 (B) e12 – 4e22 = 2 (C*) 4e22 – e12 = 6 (D) 2e12 – e22 = 4

x2 y2 x 2 y2
[Sol.  1 ....(1);  1 ....(2)
a2 b2 a 2 b12

R= a 2  b12

2R = a 2  b2
 2 b2 2 b12 
 2 a 2
 b12 = a 2  b2 e1  1  2 ; e 2  1  2 
 a a 

4 a 2  b12 = a2 + b2
 
 b12  2
41  2  = 1 + b
 a 
  a2
4[(1 – (e22 – 1)] = 1 + 1 – e12
8 – 4e22 = 2 – e12
4e22 – e12 = 6 Ans. ] [12th, 06-01-2008]

Q.739/hyp If the normal to the rectangular hyperbola xy = c2 at the point 't' meets the curve again at 't1'
then t3 t1 has the value equal to
(A) 1 (B*) – 1 (C) 0 (D) none

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [56]

 dx
[Sol. x = ct  =c
dt
c dy c
y=  =  2
t dt t
dy 1
 2
dx t
 mN = t2
1
 t2 = mAB = – t t
1
 t3 t1 = – 1 ]

x2 y2
Q.823/hyp P is a point on the hyperbola  = 1, N is the foot of the perpendicular from P on the transverse
a 2 b2
axis. The tangent to the hyperbola at P meets the transverse axis at T . If O is the centre of the hyperbola,
the OT. ON is equal to :
(A) e2 (B*) a2 (C) b2 (D)b2/a2
[Hint: OT = a cos  ; N = a sec   OT . ON = a2 ]

More than one are correct:

dy
Q.9513/hyp Solutions of the differential equation (1 – x2) + xy = ax where a  R, is
dx
(A*) a conic which is an ellipse or a hyperbola with principal axes parallel to coordinates axes.
(B*) centre of the conic is (0, a)
(C) length of one of the principal axes is 1.
(D*) length of one of the principal axes is equal to 2.
dy x ax
[Sol. + y = [12th, 06-01-2008]
dx 1 x 2
1 x2
x 1
 1 x 2 dx  log |1 x 2 | 1
I.F. e  e 2 =
|1 x2 |
y x
= a dx  C
|1 x |2
| 1  x 2 |3 2
let |1 – x2| = v2; – 2x dx = 2v dv; x dx = – v dv
y v dv a
hence  a  3 = + +C
v v v
y = a + Cv
y = a + C |1 x2 |
(y – a)2 = C2 |1 – x2| = C2(1 – x2) or C2(x2 – 1)
(y – a)2 + C2x2 = C2 or (y – a)2 – C2x2 = – C2
(y  a)2 x2 (y  a)2 x2
+ =1 or – =–1  centre (0, a) ]
C2 1 C2 1

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [57]

Q.10514/hyp In which of the following cases maximum number of normals can be drawn from a point P lying in the
same plane
(A*) circle (B) parabola (C) ellipse (D) hyperbola
[13th, 20-01-2008]
Q.11515/hyp If  is eliminated from the equations
a sec  – x tan  = y and b sec  + y tan  = x (a and b are constant)
then the eliminant denotes the equation of
x2 y2
(A) the director circle of the hyperbola  1
a2 b2

x2 y2
(B) auxiliary circle of the ellipse  1
a2 b2

x2 y2
(C*) Director circle of the ellipse  1
a2 b2
a 2  b2
(D*) Director circle of the circle x2 + y2 = .
2
[Sol. a sec  = y + x tan  [13th, 10-08-2008, P-2]
b sec  = x – y tan 
————————
(a2 + b2)sec2 = x2(1 + tan2) + y2(1 + tan2)
 x2 + y2 = a2 + b2  (C) and (D)]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [58]

 MATHEMATICS
Daily Practice Problems
Target IIT JEE 2010
CLASS : XIII (VXYZ) MISCELLANEOUS DPP. NO.- 12
Q.129/hyp If P(x1, y1), Q(x2, y2), R(x3, y3) & S(x4, y4) are 4 concyclic points on the rectangular hyperbola
x y = c2, the co-ordinates of the orthocentre of the triangle PQR are :
(A) (x4,  y4) (B) (x4, y4) (C*) ( x4,  y4) (D) ( x4, y4)
[Hint: A rectangular hyperbola circumscribing a  also passes through its orthocentre
 c   c 
 ct , 
if  i t  where i = 1, 2, 3 are the vertices of the  then therefore orthocentre is  t t t ,  ct t t 
1 2 3 ,
 i  12 3 

  c 

where t1 t2 t3 t4 = 1. Hence orthocentre is   ct ,

4
t 4  = (– x4 , – y4) ]

x 2 y2
Q.24/hyp Let F1, F2 are the foci of the hyperbola  = 1 and F3, F4 are the foci of its conjugate hyperbola.
16 9
If eH and eC are their eccentricities respectively then the statement which holds true is
(A) Their equations of the asymptotes are different.
(B) eH > eC
(C*) Area of the quadrilateral formed by their foci is 50 sq. units.
(D) Their auxillary circles will have the same equation.
[Hint: eH = 5/4; eC = 5/3 [12th & 13th 11-3-2007]
d1d 2 100
area = = = 50
2 2
AC: x2 + y2 = 16; AH = x2 + y2 = 9 ]

Q.331/hyp The chord PQ of the rectangular hyperbola xy = a2 meets the axis of x at A ; C is the mid point of PQ
& 'O' is the origin. Then the  ACO is :
(A) equilateral (B*) isosceles (C) right angled (D) right isosceles.
[Sol. Chord with a given middle point
x y
 2
h k
obv. OCA is isosceles with OC = CA.]

x2 y2
Q.47/hyp The asymptote of the hyperbola  = 1 form with any tangent to the hyperbola a triangle whose
a 2 b2
area is a2tan  in magnitude then its eccentricity is :
(A*) sec (B) cosec (C) sec2 (D) cosec2
[ Hint : A = ab = a tan   b/a = tan , hence e = 1 + (b /a )  e = 1 + tan2 e = sec ]
2 2 2 2 2

Q.534/hyp Latus rectum of the conic satisfying the differential equation, x dy + y dx = 0 and passing through the
point (2, 8) is :
(A) 4 2 (B) 8 (C*) 8 2 (D) 16
dy dx
[Sol.  0  ln xy = c  xy = c
y x
Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [59]
 passes through (2,8)  c = 16
xy =16 LR = 2a(e2 – 1) = 2a (e = 2)
solving with y = x
vertex is (4, 4)
distance from centre to vertex = 4 2
L.R. = length of TA = 8 2 Ans ]
x 2 y2
Q.641/hyp AB is a double ordinate of the hyperbola   1 such that AOB (where 'O' is the origin) is an
a 2 b2
equilateral triangle, then the eccentricity e of the hyperbola satisfies
2 2 2
(A) e > 3 (B) 1 < e < (C) e = (D*) e >
3 3 3
x 2 y2
[Sol.  1 where y = l
a 2 b2
x2 l2 a2
1  x2 = (b2 + l2) ....(1)
a2 b2 b2
now x2 + l2 = 4l2  x2 = 3l2 ....(2)
a 2 (b 2  l 2 )
from (1) and (2) 2
 3l 2  a2b2 + a2l2 = 3b2l2
b
l2 (3b2 – a2) = a2 b2

a 2b2 b2 1 b2 4 4 2
l2 =  0  3b2 – a2 > 0  2
 ; 1+ 2   e2 >  e >
3b 2  a 2 a 3 a 3 3 3
b 1 b2 4 2 4 2
Note:   1    e   e ]
a 3 a2 3 3 3

Q.747/hyp The tangent to the hyperbola xy = c2 at the point P intersects the x-axis at T and the y-axis at T. The
normal to the hyperbola at P intersects the x-axis at N and the y-axis at N. The areas of the triangles
1 1
PNT and PN'T' are  and ' respectively, then  is
 '
(A) equal to 1 (B) depends on t (C*) depends on c (D) equal to 2
x yt
[Sol. Tangent :  2
ct c
put y = 0; x = 2ct (T)
2c
x = 0; y = (T')
t
c
|||ly normal is y – = t2(x – ct)
t
c
put y = 0; x = ct – (N)
t3
c
x = 0; – ct3 (N')
t
Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [60]
 c  c c 2 (1  t 4 )
Area of PNT =  ct  3   =
2t  t  2t 4
c 3 c 2 (1  t 4 )
area of  PN'T' = ct   ct   ' =
t  2

1 1 2t 4 2 2 2
= 2 4
4 = 2 4 (t + 1) = 2
  4
 2
 ' c (1  t ) c (1  t ) c (1  t ) c
which is independent of t. ]

Q.850hyp At the point of intersection of the rectangular hyperbola xy = c2 and the parabola y2 = 4ax tangents
to the rectangular hyperbola and the parabola make an angle  and  respectively with the axis of X,
then
(A*)  = tan–1(– 2 tan) (B)  = tan–1(– 2 tan)
1 1
(C)  = tan–1(– tan) (D)  = tan–1(– tan)
2 2
[Sol. Let (x1, y1) be the point of intersection  y12  4ax1 and x1y1 = c2
y2 = 4ax xy = c2
dy 2a dy y
  
dx y dx x
dy 2a dy y
 tan    tan    1
dx ( x1 , y1 ) y1 dx ( x1, y1 ) x1
2
tan   y1 / x1  y1 4ax1
     2
tan  2a / y1 2ax1 2ax1
  = tan–1(– 2 tan) ]

Q.919/hyp Locus of the middle points of the parallel chords with gradient m of the rectangular hyperbola
xy = c2 is
(A*) y + mx = 0 (B) y  mx = 0 (C) my  x = 0 (D) my + x = 0
x y k
[ Hint : equation of chord with mid point (h, k) is  =2;m=–  y + mx = 0 ]
h k h

Q.1020/hyp The locus of the foot of the perpendicular from the centre of the hyperbola xy = c2 on a variable
tangent is :
(A) (x2  y2)2 = 4c2 xy (B) (x2 + y2)2 = 2c2 xy
(C) (x2 + y2) = 4x2 xy (D*) (x2 + y2)2 = 4c2 xy
[Hint: 2 2 2
hx + ky = h + k . Solve it with xy = c & D = 0
or compare these with tangent at t and eliminate ‘t’. ]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [61]

Q.1125/hyp The equation to the chord joining two points (x1, y1) and (x2, y2) on the rectangular hyperbola xy =
c2 is :
x y x y
(A*) + =1 (B) + =1
x1  x 2 y 1  y2 x1  x 2 y 1  y2

x y x y
(C) + =1 (D) + =1
y1  y 2 x 1  x2 y1  y 2 x 1  x2
x y
[Hint : note that chord of xy = c2 whose middle point is (h, k) in  2
h k
further, now 2h = x1 + x2 and 2k = y1 + y2 ]

x 2 y2
Q.12 A tangent to the ellipse   1 meets its director circle at P and Q. Then the product of the slopes
9 4
of CP and CQ where 'C' is the origin is
9 4 2 1
(A) (B*) (C) (D) –
4 9 9 4
x 2 y2
[Sol. The equation of the tangent at (3 cos , 2 sin ) on   1 is
9 4
x y
cos   sin   1 ... (i)
3 2
The equation of the director circle is
x2 + y2 = 9 + 3 = 13 ... (ii)
The combined equation of CP and CQ is obtained by homogenising equation (ii) with (i). Thus combined
equation is
2
x y 
x2 + y2 = 13  cos   sin  
 3 2 
 13 2  2 13  13 2  2
  cos   1 x  sin  cos  xy +  sin   1 y  0
9  3 4 
 Product of the slopes of CP and CQ
13
cos 2   1
coefficient of x 2 9 13 cos 2   9 4 13 cos 2   9 4 4
   
coefficient of y 2 13 = 2 = 2 ]
sin 2   1 13 sin   4 9 9  13 cos   4 9 9
4

x 2 y2
Q.13 The foci of a hyperbola coincide with the foci of the ellipse   1 . Then the equation of the
25 9
hyperbola with eccentricity 2 is
x 2 y2 x 2 y2
(A)   1 (B*)  1 (C) 3x2 – y2 + 12 = 0 (D) 9x2 – 25y2 – 225 = 0
12 4 4 12
[Sol. For the ellipse, a2 = 25, b2 = 9
16 4
 9 = 25(1 – e2)  e2 =  e=
25 5

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [62]

  One of the foci is (ae, 0) i.e. (4, 0)
 For the hyperbola
a'e' = 4  2a' = 4  a' = 2
and 2 2
b' = 4(e' – 1) = 4 × 3 = 12
x 2 y2
 equation of the hyperbola is   1 Ans.]
4 12

Paragraph for question nos. 14 to 16

From a point 'P' three normals are drawn to the parabola y2 = 4x such that two of them make angles with
the abscissa axis, the product of whose tangents is 2. Suppose the locus of the point 'P' is a part of a
conic 'C'. Now a circle S = 0 is described on the chord of the conic 'C' as diameter passing through the
point (1, 0) and with gradient unity. Suppose (a, b) are the coordinates of the centre of this circle. If L1
and L2 are the two asymptotes of the hyperbola with length of its transverse axis 2a and conjugate axis
2b (principal axes of the hyperbola along the coordinate axes) then answer the following questions.
Q.14404/hyp Locus of P is a
(A) circle (B*) parabola (C) ellipse (D) hyperbola
Q.15 Radius of the circle S = 0 is
(A*) 4 (B) 5 (C) 17 (D) 23
Q.16 The angle   (0, /2) between the two asymptotes of the hyperbola lies in the interval
(A) (0, 15°) (B) (30°, 45°) (C) (45°, 60°) (D*) (60°, 75°)
[Sol. Equation of a normal y = mx – 2m – m3 [12th & 13th 03-03-2007]
passes through (h, k)]
m3 + (2 – h)m + k = 0
m1m2m3 = – k
but m1 m2 = 2
 m3 = – k/2
this must satisfy equation (1)
k3 k
– (2 – h) + k = 0
8 2
3
k – 4k(2 – h) + 8k = 0 (k  0)
k2 – 8 – 4h + 8 = 0
locus of 'P' is y2 = 4x which is a parabola Ans.
now chord passing through (1, 0) is the focal chord.
Given that gradient of focal chord is 1
2
 t1  t 2 = 1  t1 + t2 = 2, Also t1t2 = – 1
equation of circle described on t1t2 as diameter is
(x – t12 )(x – t 22 ) + (y – 2t1)(y – 2t2) = 0
x2 + y2 – x( t12 + t 22 ) + t12 t 22 – 2y(t1 + t2) + 4t1t2 = 0
x2 + y2 – x[4 + 2] + 1 – 2y(2) – 4 = 0
x2 + y2 – 6x – 4y – 3 = 0
centre a = 3 and b = 2; r = 4 Ans.

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [63]

 x2 y2
now the hyperbola is – =1
9 4
2x 2x
asymptotes are y = and y = –
3 3
now tan  = 2/3
  = 2
2 · ( 2 3) 12  12 
tan  = 1  ( 4 9) ; tan  = ;  = tan–1  
5 5
hence   (60°, 75°) Ans. ]

Paragraph for question nos. 17 to 19

A conic C passes through the point (2, 4) and is such that the segment of any of its tangents at any point
contained between the co-ordinate axes is bisected at the point of tangency. Let S denotes circle
described on the foci F1 and F2 of the conic C as diameter.
Q.17 Vertex of the conic C is
(A) (2, 2), (–2, – 2) (B*) 2 2 , 2 2 ,  2 2 ,  2 2
   
(C) (4, 4), (–4, – 4) (D)  2, 2 ,  2,  2
 
Q.18 Director circle of the conic is
(A) x2 + y2 = 4 (B) x2 + y2 = 8 (C) x2 + y2 = 2 (D*) None
Q.19 Equation of the circle S is
(A) x2 + y2 = 16 (B) x2 + y2 = 8 (C*) x2 + y2 = 32 (D) x2 + y2 = 4
[Sol. Y  y = m (X  x) ; if Y = 0 then
(0, y–mx)
y
X=x and if X = 0 then Y = y  m x. P(x,y)
m
y dy y O y
Hence x  = 2x  = (x– ,0)
m
m dx x
dy dx
   = c  xy = c
y x (4,4)

passes through (2, 4)

 equation of conic is xy = 8
which is a rectangular hyperbola with e = 2 .
2 2, 2 2
Hence the two vertices are 2 2 , 2 2 ,  2 2 ,  2 2
    (–4, –4)

focii are (4, 4) & ( 4, 4)

 2 2,  2 2
 Equation of S is x2 + y2 = 32 Ans. ]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [64]

Assertion and Reason
Q.20 Statement-1: Diagonals of any parallelogram inscribed in an ellipse always intersect at the centre of
the ellipse.
Statement-2: Centre of the ellipse is the only point at which two chords can bisect each other and
every chord passing through the centre of the ellipse gets bisected at the centre.
(A*) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement -1 is True, Statement -2 is False
(D) Statement -1 is False, Statement -2 is True
[Sol. Statement-2 is correct as ellipse is a central conic and it also explains Statement-1.
Hence, code (A) is the correct answer.]

Q.21 Statement-1: The points of intersection of the tangents at three distinct points A, B, C on the parabola
y2 = 4x can be collinear.
Statement-2: If a line L does not intersect the parabola y2 = 4x, then from every point of the line two
tangents can be drawn to the parabola.
(A) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement -1 is True, Statement -2 is False
(D*) Statement -1 is False, Statement -2 is True
[Sol. Area of the triangle made by the intersection points of tangents at point A(t1), B(t2) and C(t3) is
1
t t t t t  t 0
2 1 2 2 3 3 1
Hence, Statement-1 is wrong. Statement-2 is correct.
Hence, code (D) is the correct answer. ]

Q.22 Statement-1: The latus rectum is the shortest focal chord in a parabola of length 4a.
because
2
1
Statement-2: As the length of a focal chord of the parabola y  4ax is a  t   , which is minimum
2
 t
when t = 1.
(A*) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement -1 is True, Statement -2 is False
(D) Statement -1 is False, Statement -2 is True
[Sol. Let P(at2, 2at) be the end of a focal chord PQ of the parabola y2 = 4ax. Thus, the coordinate of the
a 2a 
other end point Q is  2 ,  
t t 

2 2 2 2
 2 a  2a   1  1
 PQ   at  2    2at     t 2  2   4 t  
 t   t   t   t

2 2
 1  1  1 2 1  1
 at   t   4  at   t  2 24  at  
 t  t  t t  t

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [65]

 2
1  1
 Length of focal chord is, a  t   , where  t    2 for all t  0.
 t  t
2
 1
 a  t    4a  PQ  4a
 t
Thus, the length of the focal chord of the parabola is 4a which is the length of its latus rectum.
Hence, the latusrectum of a parabola is the shortest focal chord.
Thus, Statement-1 and Statement-2 is true and Statement-2 s correct explanation of Statement-1 ]
Q.23 Statement-1: If P(2a, 0) be any point on the axis of parabola, then the chord QPR, satisfy
1 1 1
2
 2
 .
(PQ) (PR ) 4a 2
Statement-2: There exists a point P on the axis of the parabola y2 = 4ax (other than vertex), such that
1 1
2
 = constant for all chord QPR of the parabola.
( PQ ) ( PR ) 2
(A*) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement -1 is True, Statement -2 is False
(D) Statement -1 is False, Statement -2 is True
[Sol. Let P(h, 0) (where h  0) be a point on the axis of parabola y2 = 4ax the straight line passing through
P cuts the parabola at a distance r.
 (r sin )2 = 4a (h + r cos )
 r2 sin2 – (4a cos )r – 4ah = 0 ... (i)
4a cos  4ah
where, r1 + r2 = 2 and r1r2 = – .
sin  sin 2 

1 1 1 1 r12  r22 cos 2  sin 2 

      
PQ 2 PR 2 r12 r22 r12 r22 h2 2ah
which is constant only, if h2 = 2ah i.e., h = 2a
1 1 cos 2  sin 2  1
     2
PQ 2 PR 2 4a 2 4a 2 4a
1 1
Thus, 2
 = constant for all chords QPR,
PQ PR 2
if h = 2a.
Hence, (2a, 0) is the required point on the axis of parabola.
 Statement-1 and Statement-2 are true and Statement-2 is correct explanation of Statement-1 ]
Q.24 Statement-1: The quadrilateral formed by the pair of tangents drawn from the point (0, 2) to the
parabola y2 – 2y + 4x + 5 = 0 and the normals at the point of contact of tangents in a
square.
Statement-2: The angle between tangents drawn from the given point to the parabola is 90°.
(A) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement -1 is True, Statement -2 is False
(D*) Statement -1 is False, Statement -2 is True
Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [66]
 x=0
[Sol. (y – 1)2 = – 4(x + 1)
Directrix x+1=1
x=0 (0,2)
If tangents are drawn from (0, 2) to the parabola (i.e. from directrix)
then length of tangent will be unequal hence the quadrilateral formed by
pair of tangents and normals at the point of contact is rectangle. ]

More than one are correct:

Q.25502hyp If the circle x2 + y2 = a2 intersects the hyperbola xy = c2 in four points P(x1, y1), Q(x2, y2),
R(x3, y3), S(x4, y4), then
(A*) x1 + x2 + x3 + x4 = 0 (B*) y1 + y2 + y3 + y4 = 0
(C*) x1 x2 x3 x4 = c 4 (D*) y1 y2 y3 y4 = c4
[Sol. solving xy = c2 and x2 + y2 = a2
c4
x2 + = a2
x2
x4– ax3– a2x2 + ax + c4 = 0
  xi  0 ;  yi  0
x1 x2 x3 x4 = c4  y1 y2 y3 y4 = c4 ]

Q.26503hyp The tangent to the hyperbola, x2  3y2 = 3 at the point  3 , 0 when associated with two asymptotes

constitutes :
(A) isosceles triangle (B*) an equilateral triangle
(C*) a triangles whose area is 3 sq. units (D) a right isosceles triangle .
2 2
[Hint: area of the  = ab sq units ; H : x /3 – y / 1 = 1 ]

Q.27 The locus of the point of intersection of those normals to the parabola x2 = 8 y which are at right
angles to each other, is a parabola. Which of the following hold(s) good in respect of the locus?
(A*) Length of the latus rectum is 2.
 11 
(B) Coordinates of focus are  0, 
 2
(C*) Equation of a directro circle is 2y – 11 = 0
(D) Equation of axis of symmetry y = 0.
[Hint: Locus is x2  2 y + 12 = 0 ] [REE '97, 6]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [67]

Match the column:
Q.28105 Column-I Column-II
(A) If the chord of contact of tangents from a point P to the (P) Straight line
2 2
parabola y = 4ax touches the parabola x = 4by, the locus of P is
(B) A variable circle C has the equation (Q) Circle
2 2 2 2
x + y – 2(t – 3t + 1)x – 2(t + 2t)y + t = 0, where t is a parameter.
The locus of the centre of the circle is
x2 y2
(C) The locus of point of intersection of tangents to an ellipse  = 1 (R) Parabola
a 2 b2
at two points the sum of whose eccentric angles is constant is
(D) An ellipse slides between two perpendicular straight lines. (S) Hyperbola
Then the locus of its centre is
[Ans. (A) S; (B) R; (C) P; (D) Q]
[Sol.
(A) yy1 = 2a (x + x1) ; x2 = 4by = 4b [(2a/y1) (x + x1)]  y1x2  8 abx  8 abx1 = 0 ;
D = 0 gives xy =  2ab  Hyperbola
(B) 2
centre is x = t – 3t + 1 ....(1) [18-12-2005, 12th]
y = t2 + 2t ....(2)
(2) – (1) gives – x + y = 5t – 1
1 x  y
or t=
5
Substituting the value of t in (2)
2
 y  x 1  y  x 1
y=   +2  
 5   5 
25y = (y – x + 1)2 + 10(y – x + 1)
25y = y2 + x2 + 1 – 2xy – 2x + 2y + 10y – 10x + 10
x2 + y2 – 2xy – 12x – 13y + 11 = 0
which is a parabola
as   0 and h2 = ab ]
 
a cos b sin
2 2
(C) h= ; k=
   
cos cos
2 2
 
given = constant = C
2
  a cos C b sin C b 
 cos =   y =  tan C  x
2 h k a 
Locus of (h, k) is a straight line
(D) y1y2 = x1x2 = b2 ....(1)
and (x2 – x1) + (y2 – y1)2 = 4(a2 – b2)
2 ....(2)
Also 2h = x1 + x2
2k = y1 + y2
from (2) (x1 + x2)2 + (y1+y2)2– 4(x1 x2 + y1y2) = 4(a2 – b2)
4 (h2 + k2) – 4 (2b2) = 4 (a2 – b2)
 x2 + y2 = a2 +b2  Circle

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [68]

Alternative: Equation of director circle with centre (h, k)
(x – h)2 + (y – k)2 = a2 + b2
(0, 0) lies on it  h2 + k2 = a2 + b2  locus is x2 + y2 = a2 + b2 ]
Q.29106 Column-I Column-II
x 2 y2
(A) For an ellipse   1 with vertices A and A', tangent drawn at the (P) 2
9 4
point P in the first quadrant meets the y-axis in Q and the chord A'P meets
the y-axis in M. If 'O' is the origin then OQ2 – MQ2 equals to
(B) If the product of the perpendicular distances from any point on the (Q) 3
x2 y2
hyperbola   1 of eccentricity e = 3 from its asymptotes
a2 b2
is equal to 6, then the length of the transverse axis of the hyperbola is
(C) The locus of the point of intersection of the lines (R) 4
3 x  y  4 3 t = 0 and 3 tx + ty  4 3 = 0
(where t is a parameter) is a hyperbola whose eccentricity is
(D) If F1 & F2 are the feet of the perpendiculars from the foci S1 & S2 (S) 6
x 2 y2
of an ellipse  = 1 on the tangent at any point P on the ellipse,
5 3
then (S1F1). (S2F2) is equal to [Ans. (A) R; (B) S; (C) P; (D) Q]
[Sol.(A) a=3;b=2
x cos  y sin 
T:  1
3 2
x = 0 ; y = 2 cosec
2 sin 
chord A'P, y= ( x  3)
3(cos   1)
2 sin 
put x = 0 y = = OM
1  cos 
Now OQ2 – MQ2 = OQ2 – (OQ – OM)2 = 2(OQ)(OM) – OM2 = OM{ 2(OQ) – (OM) }

2 sin   4 2 sin   4 sin   2(1  cos )  (1  cos 2 )  4(1  cos )(2  1  cos )
=  =  = =4
1  cos   sin  1  cos   1  cos   sin (1  cos )  (1  cos )(1  cos )

a 2b2 a 2 . a 2 (e 2  1)
(B) p1p2 = = = 6;
a 2  b2 a 2 e2

2a 2
 6  a2 = 9  a = 3
3
hence 2a = 6
x 2 y2
(C) hyperbola  1
16 48
(D) Product of the feet of the perpendiculars is equal to the square of its semi minor axes.]

Dpp's on Conic Section (Parabola, Ellipse, Hyperbola) [69]