MATHEMATICS-1A WWW.AIMSTUTORIAL.

IN

1. FUNCTIONS
DEFINITIONS, CONCEPTS AND FORMULAE: 9) Domain calculations:
Function Method for finding domain of f
1) Function: Let A and B be non empty sets and
f be a relation from A to B. If for each element f(x)
1. Delete the values of g(x) = 0 from
a A, there exists a unique b B such that g(x)
(a, b) f, then f is called a function from A to R.
B. It is denoted by f : A B.
The set A is called ‘domain of f’ and B is called 2. f(x) Solve f(x) > 0
‘codomain of f’and the set of all f images of
1
the elements of A is called the range of f which 3. Solve f(x) > 0
is denoted by f(A). f(x)
4. log f(x) Solve f(x) > 0
2) O n e - on e f un ct ion or I nject ion :-
If f : A B is such that distinct elements of A
have distinct f - images in B, then f is said to 1
5. log f(x) Solve f(x) > 0 and f(x)  1
be a one - one function.
f : A B is one- one if a 1 , a 2 A and
f(a 1 ) = f(a 2 ), then a 1 = a 2 .
LEVEL - I (VSAQ)
3) Onto function or Surjection :- Let f : A B.
If every element of B occurs as the image of 1. Define one - one function. Give an example.
atleast one element of A, then we say that f is A: If f: A  B is such that distinct elements of A have
an onto function. distinct f - images in B, then f is said to be a one -
one function.
f : A B is onto given b B, there exists Eg: f : R R defined by f(x) = 3x + 2 is one -
a A such that f(a) = b. one.

4) Bijection :- If f : A B is both one - one and 2. Define onto function. Give an example.
onto, then f is said to be a bijection from A to A: Let f : A  B. If every element of B occurs as the
B. f(a) = b  a = f -1 (b). image of atleast one element of A, then f is said to
be an onto function.
5) Constant function :- A function f : A B is
said to be a constant function, if the range of f Eg: f : R  R defined by f(x) = 3x + 2 is onto.
contains only one element. f(x) = c (a constant)
for all x domain. 3. f : N  N is defined as f(x) = 2x + 3. Is f onto ?
Explain with reason.
6) Identity function :- If A is a non - empty set, A: (Let x1, x2 domain N such that f(x) = f(x2).
f : A A defined by f(x) = x for all x A is  2x1 + 3 = 2x2 + 3
called the identity function on A and is denoted  2x1 = 2x2
by IA.  x1 = x2
 f : N  N is an injection)
7) Composite function :- If f : A B, g : B C Here codomain of f = N.
are two f un c tions, then gof : A  C is Range of f = { f(1), f(2), f(3),..........}
defined by (gof) (x) = g[f(x)]  x A. = { 5, 7, 9,..........}
N
8) Equality of two functions:- Two functions f Hence f : N  N is not a surjection (onto)
and g are said to be equal if
i) they are defined on the same domain A and
codomain B
ii) f(x) = g(x) for every x A.

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x 1
2x  1 8. If f(x) = , then find (fofof) (x).
4. f : R  R is defined as f(x) = , then x 1
3
 x  1
this function is injection or not? Justify. A: (fof) (x) = f  
A: Let x1, x2  domain R such that f(x1) = f(x2)  x  1
x 1
2x1  1 2x 2  1 1
   x 1
3 3 x 1
 2x1 + 1 = 2x2 + 1 1
x 1
 2x1 = 2x2
 x1 = x2 x  1  x 1
Hence f : R  R is an injection. 
x  1 x  1
1- x2
5. If f : R R is defined by f(x) = , then 2x
1+ x 2 
find f (tan ). 2
1- x 2 (f o f o f) (x) = f [ f o f (x)] = f(x).
A: Given that f : R R, f (x) =
1+ x 2
1 - tan2  9. If f : R  R, g : R  R are defined by
 f (tan ) = 2 = cos 2.
f(x) = 3x - 2, g(x) = x2 + 1, then find (g o f-1) (2).
1 + tan 
A: Let f(x) = y
6. If A = { -2, -1, 0, 1, 2} and f : A B is a  3x - 2 = y
surjection defined by f (x) = x2 + x + 1, then
y2
find B. x=  f 1(y)
A: f (-2) = (-2)2 + (-2) + 1 = 3 3
f (-1) = (-1)2 + (-1) + 1 = 1
x2
f (0) = 02 + 0 + 1 = 1  f 1(x) 
f (1) = 12 + 1 + 1 = 3 3
f (2) = 22 + 2 + 1 = 7
2  2
Since f : A B is a surjection,
B = f (A)
 
 gof 1  2   g  f 1  2    g 
 3 

4
 g 
3
= {3, 1, 7}
2
4 16 25
7. If f : R  R, g : R R are defined by f (x) = 4x -1     1  1 .
3 9 9
and g (x) = x2 + 2, then find
 a + 1 10.Find the inverse of the following functions
(i) (gof)   (ii) go[fof (0)] (i) If a, b  R, f : R R defined by f (x) = ax + b (a
 4 
A: f : R R , g : R R are given by f (x) = 4x - 1,  0)
g(x) = x2 + 2 (ii) f : R ( 0 ,  ) defined by f (x) = 5x
 a +1    a + 1 (iii) f : (0,  ) R defined by f (x ) = log2x
(i) (gof)   = g f   (iv) f : Q  Q defined by f(x) = 5x + 4
 4    4 
A: (i) If a, b  R, f : R R defined by f(x) = ax + b
 4(a  1) 
= g  1 ( a  0)
 4  Let x  domain R and Y  codomain R such that f
= g [ a + 1 - 1] (x) = y
= g (a)  ax + b = y
= a2 + 2  ax = y - b
(ii) g [(fof) (0)] = g [f{f(0)}] y -b
 x = a = f -1 (y)  f is bijection
= g [f(-1)]
x - b
= g( -4 - 1)  f -1 (x) = a
= g(-5) (ii) f :R  (o,  ) defined by f(x) = 5x
= (-5)2 + 2 Let x  R and y  (0,  ) such that f(x) = y
= 27.  5x = y
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 x = log5 y = f -1 (y)
 f -1 (x) = log5x
 f is bijection

(iii) f (x) = log x + x + 1
2

(iii) f : (0,  ) R defined by f (x) = log2x
let x  (0,  ) and y  R such that f (x) = y f (-x) = log  -x + (-x) 2 + 1 
 
 log2 x = y
 x = 2y = f -1 (y) = log  x + 1 - x 
2
 f is bijection
 
 f (x) = 2 .
-1 x

(iv) f : Q  Q is defined by f(x) = 5x + 4

 ( x 2  1  x)( x 2  1  x) 
let x  domain Q and y  codomain Q such that log  
f(x) = y =
 x2  1  x 
 5x + 4 = y
 5x = y - 4  x2  1  x2 
y-4 = log  

  x = = f -1(y) 2
 x  x  1
5
x-4 = log ( x + x +1)
2 -1
 f -1 (x) = .
5
= - log (x + x +1)
2

11. Determine whether the following functions are = - f (x)

even or odd. So f (x) is an odd function.
(i) f (x) = ax - a-x + sinx
12. Find the domain of the real valued function
 ex - 1 
(ii) f (x) = x  x  1
 e + 1 f(x)  a  0 .
x  a2
2
(iii) f (x) = log (x + x 2 + 1 ).
A: To get the domain of f, x2 - a2 > 0.
A: (i) f(x) = ax - a -x + sinx  (x + a) (x - a) > 0.
Now f (-x) = a-x - a-(-x) + sin(-x)  x < - a or x > a.
= a-x - ax - sinx  x (- , - a)  (a, )
= -{ax - a-x + sinx]  Domain of f = (- , - a)  (a, ).
= - f (x)
So f (x) is an odd function. 13. Find the domain of the real valued function
 ex - 1  f(x)   x  α  x  β   0  α  β  .
(ii) f (x) = x  x 
 e +1  A: To get the domain (x - ) (x - )  0.
x   or x  .
 e-x - 1  x (-  , ]  [ )
f (-x) = (-x)  -x   Domain of f = (-  , ]  [ )
 e +1 

 1  14. Find the domain of the real valued function

 x
- 1 
e
= (-x)

 1 + 1 
 2x 2  5x  7
f(x) 

 e x

  x  1 x  2 x  3  .
 1- e x  A: To get the domain of f, (x - 1)(x - 2)(x - 3)  0.
= (-x)  x   x  1, 2, 3
 1+ e   Domain of f = R - {1, 2, 3}

 ex - 1 
= x  x 
 e  1
So f (x) is an even function.

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15.Find the domain of the function 17. Find the domain of the function
(i) f (x) = log (x2 - 4x + 3)
1 1
(i) f (x) = 2 (ii) f (x) = log(2 - x ) (ii) f (x) = lo g 0 .3 (x - x 2 )
(x - 1)(x + 3)

1 A: (i) f (x) = log (x2 - 4x + 3)  R

A: f (x) = R  x2 - 4x + 3 > 0
(x 2 - 1)(x + 3)
 (x - 1) (x - 3) > 0
 (x2 - 1) ( x + 3)  0
 x  (-  , 1) U (3,  )
 (x + 1) ( x - 1) ( x + 3)  0 Domain of f = (-  , 1) U (3,  ).
 x -3, -1, 1
 Domain of f = R - {-3, -1, 1} (ii) f (x) = log0.3 (x - x 2 )  R
1  x - x2 > 0
(ii) f (x) = log(2 - x)  x (1 - x) > 0
 x (x - 1) < 0
 2 - x > 0 and 2 - x  1
 x  (0, 1)
 x - 2 < 0 and x  1.
Domain of f = (0, 1).
 x < 2 and x  1.
 x  (-  , 2) and x  1 18. Find the range of the function
Domain of f = (-  , 1)  (1, 2). (i) f (x) = log |4 - x2|
(ii) f (x) = [x] - x
16. Find the domain of the function
A: (i) f (x) = log |4 -x2|  R
1 Let f (x) = y
(i) f (x) = x 2 - 25 (ii) f (x) =
1- x 2  log |4 - x2| = y
(iii) f (x) =  |4 - x2| = ey > 0  y  R
4x - x 2
 Range of f is R.
A: f (x) = x 2  25  R (ii) f (x) = [x]  x  R
 x2 - 25  0  [x] - x  0
 (x + 5) (x - 5)  0  x  [x]
 x (-  , -5] U [5,  ) x Z
 Domain of f = (-  , -5] U [5,  )  Domain of f = Z
1  Range of f = {0}.
(ii) f (x) = R
19. Find the range of
1  x2
x2 - 4
 1 - x2 > 0 (i) f(x) = (ii) f(x) = 9 + x 2
x-2
 x2 - 1 < 0
 (x + 1) ( x - 1) < 0 x2 - 4
A: (i) f(x) = R
 x  (-1, 1) x-2
 Domain of f = (-1, 1) x-2 0
(iii) f (x) = 4x  x 2  R x 2
 4x - x2  0
 Domain of f = R - {2}
 x(4 - x)  0
Then y = x + 2 x  2 y 4
Range of f = R - {4}.
 x(x - 4)  0
 x  [0, 4] (ii) f(x) = 9 + x 2
 Domain of f = [0 4] Let y = f(x) =9 + x2  R
 Domain of f = R
When x = 0, f (0) = 9 = 3
When x  R - {0}, f(x) > 3
 Range of f = [3,  ).
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20. Find the domain and range of
(iii) f x = 9- x2 ∈ R

(
)
2+x x
(i) f(x) = (ii) f(x) =  9 - x2  0
2- x 1+ x 2
x  x2 - 9  0
(iii) f(x) = 9- x2 (iv) f(x) =  (x + 3) (x - 3)  0
2-3x
 Domain of f = [-3, 3]
2+x Let f (x) = y
A: (i) f x = ∈R
(
)

2-x
2-x  0 9 - x2 = y
 x 2 9 - x2 = y2
Domain of f = R - {2}.
Let f (x) = y
x= 9 - y2

2 x  9 - y2  0
 2x  y  y2 - 9  0
y  [-3, 3]
 2 + x = 2y - xy
 x(1+y) = 2(y-1) Since y takes only non negative values
2(y  1)  Range of f = [0, 3].
x = y 1
x
Clearly x is not defined for y + 1 = 0 (iv) f(x) = ∈R
2- 3x
 Range of f = R - {-1}  2 - 3x  0
.
 x  2/3
x Domain of f is R - {2/3}
(ii) f(x) = Let f (x) = y
1+ x 2
x x
f(x) = ∈R  2 - 3x = y
1+ x 2
 1 + x2  0  x = 2y - 3xy
 Domain of f = R
 x( 1+3y) = 2y
Let f(x) = y
2y

x
=y  x = 1  3y
1+ x 2
 x = y + yx2  1 + 3y  0
 yx2 - x + y = 0  y  -1/3
1± 1  4y2  Range of f = R - {1/3}
x = 2y
∈R
21. If a function is defined as
 1 - 4y2  0 and y  0
 ( 1 + 2y) (1-2y)  0 and y  0  x + 2, x > -1
 ( y + ½) (y - ½)  0 and y  0 
f x = 2,-1 ≤ x ≤ 1
(
)

 y  [-½, ½] and y  0  x - 1,-3 < x < -1


Also x = 0  y = 0 Find the values of (i) f(0) (ii) f(2) + f(-2)
 Range of f = [-½, ½] A: (i) f (0) = 2

(ii) f(2) + f(-2) = {2 + 2} + {-2 -1]

=4-3

= 1.

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Part 2:- To prove that gof: A C is onto.
22.If f : R R and g: R R are defined by
Now f : A B, g : B  C are onto functions.
f (x) = 3x - 1 and g (x) = x 2 + 1, then find
(i) (fog)(x) (ii) (gof)(x)  gof : A C is a function.
A: Given that f : R R, g: R R are defined by
Let c  C.
f(x) = 3x - 1, g(x) = x2 + 1
Since g : B C is onto, there exists atleast one
(i) (fog)(x) = f [ g(x)] element b  B such that g(b) = c.

= f[x2 + 1] Since f : A  B is also onto, there exists atleast

one element a  A such that f(a) = b
= 3(x2 + 1) - 1
Now (gof) (a) = g[f(a)]
= 3x + 2.
2
= g(b)
(ii) (gof(x) = g[ f(x)] = c
For c  C, there is an element a  A such that
= g[3x - 1]
= (3x - 1)2 + 1 (gof) (a) = c.
so gof : A C is onto.
= 9x2 - 6x + 2
23. If f and g are real valued functions defined by since gof : A  C is both one-one and onto, hence
f (x) = 2x - 1 and g (x) = x2, then find gof : A  C is a bijection.
(i) (fg) (x) (ii) (f + g + 2) (x)
A: f(x) = 2x - 1, g(x) = x2 2. If f: A  B, g : B C are bijections, then
prove that (gof) -1 = f -1og -1.
(i) (fg)(x) = f(x) g(x) A: Given that f : A  B, g : B  C are bijections.
= (2x -1) (x2)  f -1 : B  A, g -1 : C B
Now gof : A  C is also a bijection.
= 2x3 - x2
 (gof) -1 : C  A
(ii) (f + g + 2) (x) = f(x) + g(x) + 2 Also g -1 : C  B, f -1 B  A  f -1og -1 : C  A.
= 2x - 1 + x2 + 2 Thus (gof) -1 and f- 1og -1 both the functions exist
and have the same domain C and the same
= x2 + 2x + 1 codomain A.
= (x + 1)2 Let c be any element in C.
Since g : B  C is onto, there exists atleast one
LEVEL - I (LAQ) element b  B such that g(b) = c
1. If f : A  B, g : B  C are two bijections, then  b = g -1 (c)  g is a bijection
prove that gof: A  C is also a bijection.
A: Given : f : A  B, g : B C are bijections. Since f : A  B is onto, there exists atleast one
element a  A such that f(a) = b.
Part 1 :- To prove that gof : A  C is one-one.
 a = f -1(b)  f is a bijection
Now f : A  B, g : B C are one-one functions.
 gof: AC is a function. Consider (gof) (a) = g[f(a)]
Let a1, a2  A  f(a1), f(a2)  B and (gof) (a1), = g(b)
(gof)(a2) C. (gof) (a) = c.
 a = (gof) -1 (c)  gof is a bijection
Suppose that (gof)(a1) = (gof)(a2)
Also (f og ) (c) = f- 1 [g -1(c)]
-1 -1
 g[f(a1)] = g[f(a2)] = f- 1 (b)
 f(a1) = f(a2)  g is one-one =a
 a1 = a2  f is one-one  (gof) (c) = (f- 1og -1) (c)  c C.
-1

 gof : A  C is one-one. Hence (gof) -1 = f -1og -1

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3. If f : A  B is a bijection, then show that Thus foIA and f both the functions exist and have
fof-1 = IB and f-1of = IA. the same domain A and the same codomain B.
A: Given that f : A  B is a bijection
 f -1 : B  A. Let a  A
Since f : A  B, there exists a unique element
Part 1:- To show that fof-1 = IB b  B such that f(a) = b
Now f -1 : B  A, f : A  B  fof -1 : B  B.
Also IB : B  B Consider (foIA) (a) = f[IA(a)]
= f(a)
Thus fof-1 and IB have the same domain B and  (foIA) (a) = f(a) for all a  A
the same codomain B. Hence foIA = f .............(1)
Let a be any element in A. Part 2:- To show that IB of = f
Since f : A  B, there is a unique element b B. Now f : A  B, IB : B B  IBof : A  B
such that f(a) = b Also f : A  B
 a = f- 1 (b)  f is a bijection Thus IBof and f both the functions exist and have
the same domain A and codomain B.
Consider (fof -1) (b) = f[f -1(b)] Consider (IBof)(a) = IB[f(a)]
= f (a) = IB(b)
=b =b
= IB (b)  IB : B B  IB(b) = b = f(a)
(fof- 1) (b) = IB(b)  b  B
Thus fof -1 = IB  (IB of) (a) = f(a) for all a  A
 IB of = f ...............(2)
Part 2:- To prove that f -1of = IA From (1) & (2) foIA = f = IB of.
Now f : A  B, f -1 : B  A  f -1of : A  A
Also IA : A  A 5. If f : A  B, g : B  A are two functions such
Thus f-1of and IA have the same domain A and the that gof = IA and fog = IB then prove that g = f-1.
same codomain A. A: Given that f : A  B, g : B  A are two functions
such that gof = IA and fog = IB.
Now (f -1of)(a) = f -1 [f(a)] Part 1:- To prove that f is one-one.
= f -1 (b) Let a1, a2 A  f(a1), f(a2)  B
=a
Consider f(a1) = f(a2)
= IA(a)  IA : A  A  IA(a) = a
 g[f(a1)] = g[f(a2)]
(f- 1of) (a) = IA(a)  aA
 f -1of = IA  (gof) (a1) = (gof) (a2)
 IA (a1) = IA(a2)  gof = IA
Hence fof -1 = IB and f -1of = IA.  a1 = a 2
Thus f : A  B is one-one.
4. If f : A  B, IA and IB are identity functions on A Part 2:- To prove that f is onto.
and B respectively, then prove that foIA = IB of = f. Let b  B.
A: Given that f : A  B  g : B  A, there exists a unique element a  A
IA : A  A is defined by IA (a) = a  a  A. such that g(b) = a.
IB : B  B is defined by IB (b) = b  b  B. Now f(a) = f[g(b)]
Part 1:- To prove that foIA = f = (fog) (b)
Now IA : A  A, f : A  B  foIA : A  B = IB (b)  fog = IB
Also f : A  B =b

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So f : A  B is onto. Now, suppose that f -1(b1) = f - 1(b2)
Since f is both one-one and onto, so f is a bijection.  a1 = a 2
 f -1 : B  A f(a1) = f(a2)  f : A  B is a funciton
Also g : B  A  b1 = b 2
Thus both the functions f -1 and g have the same So f : B  A is a one-one function.
-1

domain B and same codomain A. Part 2: To prove that f -1 : B  A is onto.

Part 3:- To show that g = f -1 Let a  A.
From previous part, f(a) = b Since f : A  B, there exists a unique element b
 a = f -1 (b) B such that f(a) = b
Also g(b) = a  f -1(b) = a  f is a bijection
 g(b) = f -1(b)  b  B.
So, for every a  A, there is an element bB
Hence g = f -1. such that f -1 (b) = a
So f -1 : B A is onto
6. If f: AB, g: B  C, h: C D are functions, Since f -1: B  A is both one-one and onto,
then prove that ho(gof) = (hog)of. hence f -1 : B  A is a bijection.
A: Given that f : A  B, g : B C, h : C D
8. Let A = {1, 2, 3}, B = {a, b, c}, C = {p, q, r}.
Now f : A  B, g : B  C  gof : A  C If f : A  B, g : B C are defined by
Also gof : A  C, h: C  D  ho(gof) : A  D f = {(1, a), (2, c), (3, b)}, g = {(a, q), (b, r),
Now g : B  C, h: C  D hog : B  D (c, p)}, then show that f -1og -1 = (gof) -1.
Also f : A  B, hog : B  D  (hog)of : A D A: Given that A = {1, 2, 3}, B = {a, b, c}, C = {p, q, r}
Thus ho(gof) and (hog)of both the functions exist f : A  B, g : B  C are given by
and have the same domain and the same f = {(1, a), (2, c), (3, b)} f -1 = {(a, 1), (b, 3), (c, 2)} and
codomain. g = {(a, q), (b, r), (c, p)}g -1 = {(q, a), (r, b), (p, c)}
Let a be any element in A.
[ho(gof)] (a) = h[(gof) (a)] Now (f -1og -1)(p)= f -1[g -1(p)]
= h[g{f(a)}] = f -1 (c)
Also [(hog)of] (a) = (hog)[f(a)] =2
= h[g{f(a)] Similarly (f -1og -1) (q) =1, (f -1og -1) (r) = 3
Thus [ho(gof)](a) = [hog)of] (a) for all a  A  f -1og -1 = {(p, 2), (q, 1), (r, 3)} ............(1)
Hence ho(gof) = (hog)of.
Also (gof)(1) = g[f(1)]
7. If f : A  B is a bijection, then prove that = g(a)
f -1 : B A is a bijection =q
A: Given that f : A  B is a bijection
 f -1 : B  A is a function Similarly (gof) (2) = p, (gof) (3) = r
Part 1: To prove that f -1 : B  A is one-one. gof = {(1, q), (2, p), (3, r)}
Let b1, b2  B. (gof) -1 = {(q, 1), (p, 2), (r, 3)} ....... (2)
 f : A  B is onto, there exist a1, a2 A such
From (1) and (2) f -1og -1 = (gof) -1.
that
f(a1) = b1, f(a2) = b2
 a1 = f -1 (b1), a2 = f -1 (b2)  f: A  B is a
bijection
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9. If f : Q  Q defined by f(x) = 5x+4 for all x  Q,
 3x - 2, x 1
show that f is a bijection and find f- 1. 
2
Given : f : Q  Q is defined by f(x) = 5x + 4  x -2, - 2  x  2
2. If f(x) =  then find f(4), f(2.5),
Part 1:- To prove that f is one-one 2x + 1, x<-3

Let x1, x2 Q (domain) and f(-2), f(-4), f(0), f(-7)

f(x1) = f(x2) A: i) f(4) = 3(4) - 2 = 10
 5x1 + 4 = 5x2 + 4 ii) f(2.5) is not defined
 5x1 = 5x2 iii) f(- 2) = (- 2)2 - 2 = 4 - 2 = 2
 x1 = x 2 iv) f(- 4) = 2 (- 4) + 1 = - 8 + 1 = - 7
 f : Q  Q is one-one. v) f(0) = 02 - 2 = - 2
vi) f(- 7) = 2 (- 7) + 1 = - 14 + 1 = - 13
Part 2:- To prove that f is onto
Let y  the codomain Q and x  domain Q such 1
3. If f(x) = x + then prove that [f(x)]2 = f(x2) +
that x
f(x) = y f(1).
 5x + 4 = y 1
A: Given f(x) = x +
y-4 x
 x=
5 2 1 1 1
 1
So for every y  codomain Q, there is a preimage [f(x)]2 =  x   = x2 + 2 + 2 x = x2 + + 2.
 x x x x2
y-4
 domain Q such that 1 1
5
 
y-4 and f(x2) = x2 + 2 , f(1) = 1 + 2 = 2.
x 1
f 5 =y
1
Thus f : Q  Q is onto. f (x2) + f(1) = x2 + 2 + 2 = [f(x)]2
x
Hence proved.
Part 3:- To find f-1(x)
Since f is both one-one, onto, so it is a bijection. 4. If f : R - {+1}  R is defined by f(x) = log
f(x) = y  x = f-1 (y)
 
y - 4 -1 1+ x 2x 
5x + 4 = y  x = =f (y) f
5 1- x , then show that  1 + x 2  = 2f(x).
 f-1(x) = x 5- 4 . 1+ x
A: Given f(x) = log 1- x
LEVEL - II (VSAQ)
2x
1+

1. If f : R - {0}  R is defined by f(x) = x3 - 1/x3, 2x  1+ x 2
Now, f = log 2x = log
then show that f(x) + f(1/x) = 0.  1+ x 2  1-
 
A: f (x) = x3 - 1/x3 1+ x 2

Now f (x) + f(1/x) = x3 - 1/x3 + 1/x3 - x3 = 0.

1+ x 2 + 2x
1+ x 2
1+ x 2 - 2x
1+ x 2
2
(1+ x)2 1+ x 1+ x
= log = log = 2log 1- x = 2f(x).
(1- x)2 1- x
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9. If f(x) = 4x - 1, g(x) = x2 + 2 then find
2 4
cos x + sin x i) gof (x) ii) fof (x)
5. If f(x) = 2 4 x  R then show that
sin x + cos x A: Given f(x) = 4x - 1, g(x) = x2 + 2
f(2012) = 1 i) gof(x) = g[f(x)] = g[4x - 1] = (4x - 1)2 + 2
2 4
cos x + sin x = 16x2 + 1 - 8x + 2 = 16x2 - 8x + 3
A: Given that f(x) = 2 4
sin x + cos x ii) (fof (x) = f[f(x)] = f(4x - 1)
= 4(4x - 1) - 1 = 16x - 5
2 4 2 2
1- sin x + sin x 1- sin x (1  sin x)
f(x) = 2 4 = 2 2 10.If f(x) = 2, g(x) = x2, h(x) = 2x, then find fo(goh)
1- cos x + cos x 1- cos x (1 - cos x)
(x).
2 2
1- sin x cos x A: Given f(x) = 2, g(x) = x2, h(x) = 2x
= 2 2 .
1- cos x sin x fo(goh)(x) = f[g(h(x)] = f[g(2x)]
= 1 = f[(2x)2] = f[4x2]
 f(2012) = 1 =2

6. If f(x + y) = f(xy)  x,y  R then prove that ‘f’

11. If f(x) = x2, g(x) = 2x then solve the equation
is a constant function. fog(x) = gof(x)
A: Let f(0) = k
 
2
A: fog(x) = f[g(x)] = f(2x) = 2x = 22x
Given that f(x + y) = f(xy)
2
Now, f(x) = f(x + 0) = f(x.0) = f(0) = k. x
and gof(x) = g[f(x)] = g[x2] = 2 2
which is a constant,  x  R Since, fog(x) = gof(x)  22x = 2
x

Hence, f(x) is a constant function.

 2x = x2  x2 - 2x = 0  x(x - 2) = 0
7. If the function f : {-1, 1}  {0, 2} is defined by  x = 0 or 2
f(x) = ax + b is a surjection, then find a, b.
0, if x  Q
A: Here f is a surjection, so two cases arise. 12.If f, g: R  R are defined by f(x) =  1, if x  Q

case i) f (-1) = 0, f(1) = 2
  1, if x  Q
- a + b = 0, a + b = 2  a = 1, b = 1 and g (x) =  0, = then find (fog) () +
 if x  Q
case ii) f(1) = 0, f(-1) = 2
(gof) (e)
 a + b = 0, - a + b = 2  a = - 1, b = 1
A: (fog) () = f[g()] = f(0) = 0
Hence, a   1, b  1 (gof) (e) = g[f(e)] = g(1) = - 1 [  Q]
 (fog) () + (gof) (e) = 0 - 1 = - 1
8. If f: R  R is defined by f(x) = 2x2 + 3 and
g(x) = 3x - 2 then find i) fog(x) ii) gof (x), 13.If f(x) = ex and g(x) = logex then show that
iii) fof (0), iv) [go(fof)](3) fog = gof and find f-1, g-1
A: Given that f(x) = 2x2 + 3 and g(x) = 3x - 2 A: Given that f(x) = ex and g(x) = logex
i) fog(x) = f[g(x)] = f[3x - 2] = 2(3x - 2)2 + 3
= 2(9x2 + 4 -12x) + 3 = 18x2 - 24x + 11 take fog(x) = f[g(x)] = f[logex] = eloge x = x
ii) gof(x) = g[f(x)] = g[2x2 + 3] = 3(2x2 + 3) - 2 gof(x) = g[f(x)] = g(ex) = logeex = x logex = x
= 6x2 + 9 - 2 = 6x2 + 7
iii) fof(0) = f[f(0)] = f[3] = 2(3)2 + 3 = 21 Clearly, fog(x) = gof(x)
iv) [go(fof)](3) = g[fof(3)] = g[f{f(3)}] Hence, f-1(x) = g(x) = logex and g-1 (x) = f(x) = ex.
= g[f(21)] = g[2(21)2 + 3]
= g[885] = 3(885) - 2 = 2653.

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14.If f(x) = 1 + x + x + ........ for |x| < 1 then show
2
(f + 4) (6) = f(6) + 4 = - 4 + 4 = 10
x-1
that f-1 (x) = .  f + 4 = {(4, 9), (5, 10), (6, 0)}
x
1 ii) take (fg) (4) = [f(4)] [g(4)] = (5) (-4) = - 20
A: Given that f(x) = 1 + x + x2 + ....... =
1- x (fg) (6) = [f(6)] [g(6)] = (- 4) (5) = - 20
2 a  fg = {(4, - 20), (6, - 20)}
 a + ar + ar + ................. = ,r < 1
1- r
Let f(x) = y  x = f-1 (y) f f(4) 5  5
iii) take   (4) = = = and
1 g g(4)  4 4
1 1
  y  1- x =  x = 1- .
1- x y y
f f(6)  4 f   5    4  
y -1 y -1 x -1   (6) = =    4, 6, 
4   5  
-1 -1
x=  f (y)   f (x) = .
y y x g g(6) 5 g 

15. If f = {(1,2) (2,-3) (3,-1)} then (iv) {(4, 1), (6, 1) (v) {4, - 6), (6, 12)}
find (i) 2f (ii) f2 (iii) 2 +f (iv) f (vi) {(4, 5), (5, 6), (6, 4)} (vii)  4, 5 5, 6 
A: Given f = {(1, 2)(2, -3)(3, -1)}
(viii) {(4, 25), (5, 36), (6, 16)}
i) take 2f(1) = 2[f(1)] = 2(2) = 4
2f(2) = 2[f(2)] = 2(-3) = - 6 17.On what domain the function f(x) = x2 - 2x and
2f(3) = 2[f(3)] = 2(-1) = -2 g(x) = -x + 6 are equal?
 2f = {(1, 4)(2, -6) (3, -2)} A: Take f(x) = g(x)  x2 - 2x = -x + 6
ii) take f2(1) = [f(1)]2 = (2)2 = 4  x2 - x - 6 = 0  (x - 3) (x + 2) = 0 x = 3, -2
f2(2) = [f(2)]2 = (-3)2 = 9 f(x) and g(x) are equal on the domain {-2,3}
f2(3) = [f(3)]2 = (-1)2 = 1
 f2 = {(1, 4)(2, 9) (3, 1)} 18.Find the domain of definition of the function
iii) take (2 + f) (1) = 2 + f(1) = 2 + 2 = 4 y(x), given by the equation 2x + 2y = 2.
(2 + f) (2) = 2 + f(2) = 2 - 3 = - 1 A: Given equation is 2x + 2y = 2.
(2 + f) (3) = 2 + f(3) = 2 - 1 = 1 2x = 2 - 2y
 2 + f = {(1, 4)(2, -1) (3, 1)} 2x < 2
log 2x < log 2
f (1)  f(1)  2 xlog 2 < log 2
f (2)  f(2)  3 (not valid) x < 1
 x  (- ,1)
f (3)  f(3)  1 (not valid)
iv) take  Domain = (- ,1)
 f  1, 2 
2 + x + 2 -x
19.Find the domain of .
16.If f = {(4, 5), (5, 6), (6,- 4)} g = {(4, -4), (6, 5), x
(8, 5) then find (i) f + 4 (ii) fg (iii) f/g
2 + x + 2 -x
(iv) f + g (v) 2f + 4g (vi) |f| (vii) f (viii) f
2 A: Let f(x) =
x
A: Given f = {(4,5), (5, 6), (6, -4)}, g = {(4, -4), (6, The function f(x) is defined for
2 + x > 0  x > - 2  (1) and
5), (8, 5)
2 - x > 0  x < 2  (2) and
Here Domain of f  g = {4, 6} x  0  (3)
from (1) and (2) and (3)
i) take (f + 4) (4) = f(4) + 4 = 5 + 4 = 9
(f + 4) (5) = f(5) + 4 = 6 + 4 = 10 x  [-2,2] - {0} (or) x  [-2,0)  (0,2]
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1 x x
20.Find the domain of x+2 + 24. Determine the function f(x) = + + 1 is
log10 (1-x) .
x
e -1 2
even or odd.
A: The function is defined for
x + 2 > 0  x > - 2  (1) and x x
A: take f(-x)   1
1 - x > 0 and 1 - x  1 x
e 1 2
x - 1 < 0 and x  0.
x  [-2,1) - {0} x x  xe x x
  1   1
or) x  [-2, 0)  (0,1). 1 2 1  ex 2
 1
ex
21. Find the domain of the function
 xe x  x  x x
1   1
(i) f (x) = (ii) f (x) = |x |- x 1  ex 2
|x|-x
x
x(1  e )  x x
1   1
A: (i) f (x) = 1  ex 2
| x | -x  R
x x x x
 |x| - x > 0. x  x
 1  x   1 f(x)
1 e 2 e 1 2
 |x| > x Hence, f(x) is an even function.
 x  (-  , 0)
 Domain of f = (-  , 0) LEVEL - II (LAQ)
(ii) f (x) = | x | -x
 |x| - x  0 1. If A = {1, 2, 3}, B = {a, b, c}, C = {p, q, r} and
 |x|  x f : A  B, g : B  C are defined by
 x R f  1, a  2, c  3, b  ,
 Domain of f = R or (-  ,  )
22. Find the domain of the function
g   a,q ,  b, r  ,  c,p  then
(i) f (x) = x - [x] show that f-1 o g-1 = (gof)-1.
(ii) f (x) = [x] - x A: f = { (1, a), (2, c), (3, b)} g = {(a, q), (b, r), (c,
A: (i) f (x) = x - [x]  R p)} then gof = {(1, q) (2, p) (3, r)}
  gof    q,1 p, 2  r,3   
1
 x - [x]  0
 x  [x] g = {(q, a) (r, b), (p, c)}
-1

 x R f-1 = {(a, 1) (c, 2) (b, 3)}

 Domain of f = R or ( -  ,  )  f 1o g1   q,1 r,3  p,2   
from  and 
(ii) f (x) = [x] - x  R
(gof)-1 = f-1 o g-1
 [x] - x  0
 [x]  x 2. I f t h e f u n c t i o n f : R  R d e f i n e d b y
 x  [x]
3x + 3-x
xZ f(x) = , then show that
2
 Domain of f = Z. f(x + y) + f(x - y) = 2f(x) f(y).
sin [x]
23.Find the range of . 3 x + 3-x
1+ [x]2 A: Given that f : R  R and f (x ) =
2
A: The function is defined for 1 + [x]2  0
Which is true  x  R Hence, Domain = R 3x+y + 3-(x+y) 3x-y + 3-(x-y)
 f(x  y)  , f(x  y) 
If x  R then [x]  Z  sin [x] = 0 2 2
sin [x]
 = 0  x  R Hence, Range = {0}.
1+ [x]2
12
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