Function E

7 inch
TARGET
JEE (MAIN + ADVANCED)
2.5 inch 2.5 inch
XIII
RELATION & FUNCTION
2.5 inch
CONTENT
S.No Pages
1. Theory 01 – 38
2. Exercise-1 (Subjective Questions) 39
3. Exercise-2 (Objective Questions) 40 – 44
4. Exercise-3 (Section-A) 45 – 46
5. Exercise-3 (Section-B) 46
6. Exercise-5 (Rank Booster) 47 – 50
7. Answer Key 51 – 52
TARGET
JEE (MAIN + ADVANCED)
XIII
RELATION & FUNCTION
RELATION & FUNCTION
RELATION & FUNCTION
1. ORDERED PAIRS :
An ordered pair consists of two elements written in a given fixed order. It should not be confused with a
set containing two elements.
Example :- Position of a point in a two dimensional plane is represented by an ordered pair (x, y).
1.1 Equality of ordered pairs :

Two ordered pairs (a1, b1) & (a2, b2) are said to be equal if a1 = a2 & b1 = b2
Eg. Find the values of a and b if (3a – 2, b + 3) = (2a – 1, 3)
3a – 2 = 2a – 1, b + 3 = 3  a = 1, b = 0
1.2 Cartesian Product of two sets :

Let A and B be any two non empty sets. The set of all ordered pairs (a, b) such that a  A and b  B is
called as cartesian product of sets A and B and is denotes by A × B.
A = {1, 2, 3}; B = {p, q, r}
A × B = {(a, b) | a  A and b  B}
= {(1, p), (1, q), (1, r), (2, p), (2, q), (2, r), (3, p), (3, q), (3, r)}
Note : If n(A) = p, n(B) = q then n (A × B) = pq.
2. RELATION:
Every non-zero subset of A × B defines a relation from set A to set B.
Ex. : A = {1, 2, 3} n(A) = 3

B = {5, 6} n(B) = 2
A × B = [(1, 5) (1, 6) (2, 5) (2, 6) (3, 5) (3, 6)]
n (A × B) = 6
R1 = {(1, 5) (2, 5)}
R2 = {(2, 5), (2, 6)}
R3 = {(1, 6), (2, 5), (3, 6)}
Are all examples of relation from A B
2.1 Total number of Relation from A to B

Let number of relations from A to B be x.
LetA contain 'm' elements and B contain 'n' element's.
Number of element's in A × B = m× n
Number of non void subset's = mnC1 + mnC2 + ........ + mnCmn = 2mn – 1
Note : If R1 : A  B such that a R1 b  a + b = 10 means R1 relates those element a  A to

b  B such that sum a & b is 10.
Similarly if R2 :A B
a
Such that aR2b  = 2 means R2 relates those elements a  A to b  B such that ratio of
b
a & b is 2.
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RELATION & FUNCTION
2.2 Domain and Range of Relation:

If R be a relation from a set A to set B. Then set of all first component's or coordinates of ordered pairs
is called the domain of R, while the set of all second component's or coordinates of the ordered pairs is
called as range of relation.
Let R : A  B (R is a relation defined from set A to set B) then
Domain : Set of all the first entries in R
{a | (a, b)  R}
Domain Range
Range : Set of all the second entries in R
{b | (a, b)  R} 3 2
E.g. A = {1, 3, 5, 7}; B = {2, 4, 6, 8}
5 4
Relation is aRb  a > b; a  A, b  B
R = {(3, 2), (5, 2), (5, 4), (7, 2), (7, 4), (7, 6) } 7 7
Domain = {3, 5, 7}
Range = {2, 4, 6}
2.3 Inverse of a Relation: Domain

R
Range
A a+2b=10 B
If R is a relation defined from A  B then R–1 is a relation 2 4
defined from B  A as 4 3
R–1 = {(b, a) | (a, b)  R} 6 2
8 1
i.e. domain is converted in to range element's and range is
10
converted into domain elements.
i.e. Domain of R = Range of R–1
Range of R = Domain of R–1
R–1
Domain Range
Ex. : If A = {2, 4, 6, 8, 10}. A B
B = {1, 2, 3, 4}. 1 2
2 4
& R : A  B such that a R b  a + 2b = 10, a  A, b  B
3 6
then R = {(2, 4) (4, 3) (6, 2), (8, 1)}
4 8
& R–1 = {(4, 2) (3, 4) (2, 6) (1, 8)}. 10
2.4 Types of Relation :
(i) Identity Relation :

Arelation defined on a setAis said to be an Identity relation if every element of Ais related to itself and
only to itself.
Eg.1 A relation defined on the set of natural number's as
with rule aRb  a = b is an identity relation
R = {(1, 1), (2, 2), (3, 3), .........}
Eg: 2 The relation IA = {(1, 1), (2, 2), (3, 3)} is an identity relation on set A = {1, 2, 3} but
{(1, 1), (2, 2), (1, 3)} are not identity relation
Types of Relations
Reflexive Symmetric Transitive
RELATION & FUNCTION
(ii) Reflexive :
A relation defined on a set A is said to be reflexive if every element of A is related to itself but not
necessarily to itself.
i.e. aRb  a R b i.e. if a, b  R
Eg: Let A= {1, 2, 3} be a set then R = {(1, 1), (2, 2), (3, 3), (1, 3), (2, 1)} is a reflexive relation onA.
R1 = {(1, 1), (3, 3), (2, 1), (6, 2)} is not a reflexive relation on A, because 2  A but (2, 2)  R.
Note :
Every identity relation is a reflexive relation but every reflexive relation need not be an identity.
Eg : 2 A relation defined on set of natural number's
aRb  a divides b (a, b)  N
R would always contain (a, a)  a  N because every natural number divides itself and hence it is a
reflexive relation. But it is not an identity relation since elements like (2, 4), (3, 6) …… also appear in R.
(iii) Symmetric Relation :

A relation defined on a set is said to be symmetric
if a R b  bRa i.e. If (a, b)  R then (b, a) must be necessarily there in the same relation.
Eg:
(i) a R b  a is parallel to b
It is a symmetric relation because if a is parallel to b then the line b is parallel to a.
(ii) L1 R L2  L1 is perpendicular to L2 is a symmetric relation.
(iii) a R b  a is brother of b is not a symmetric relation because if a is brother of b then b
is not necessarily brother of a, b can be sister also of a  (b, a)  R
(iv) a R b  a is a cousin of b. This is a symmetric relation.
(iv) Transitive relation :

A relation on set A is said to be transitive if a R b and b R c implies a R c then it is transitive.
(a, b)  R & (b, c)  R  (a, c)  R
On the set of natural numbers
Eg. : 1 a R b  (a – b) is even integer is transitive because if (a, b), (b, c)  R
a – b = 2m b – c = 2m a – c = 2(m + n)  (a, c)  R
Eg. : 2 On the set of natural numbers, the relation R defined by x R y  x < y is transitive because for
any
x , y, z  N if (x, y), (y, z)  N then x < y, y < z  x < z  (x, z)  R
(v) Equivalence Relation :

If a relation is Reflexive, Symmetric and Transitive then it is said to be an equivalence relation.
Eg. : (i) A relation defined on A : x R y  x = y where
A = {1, 2, 3}  R = {(1, 1), (2, 2), (3, 3)} then
(a) Reflexive {(1, 1) (2, 2), (3, 3)}
(b) Symmetric
(c) Transitive  R is equivalence
(ii) A relation defined on a set of | | lines in a plane such that L1 R L2  L1 | | L2 is an
equivalence relation
(iii) Relation defined on the set of integer I
Prove that xRy  (x – y) is an equivalence relation.
RELATION & FUNCTION
Illustration :
Check the following relations for being reflexive, symmetric, transitive and thus choose the equivalence
relations if any.
(i) a R b if a  b; a, b  set of real numbers. (ii) a R b if a < b; a, b  N.
1
(iii) a R b if a  b > ; a, b  R. (iv) a R b if a divides b; a, b  N.
2
(v) a R b if (a – b) is divisible by n; a, b  I, n is a fixed positive integer.
Sol.
(i) Not reflexive, not symmetric but transitive
Since  2 2  –2  (–2, –2)  R hence relation is not Reflexive
Let a = –2 and b = 3; (–2, 3)  R. Since  2  3 is true
Since 3  –2 is wrong hence relation is not symmetric
Now Let a, b, c be three real numbers such that a  b and b  c
a b b 0, so b c  b  c  | a | c  (a, c)  R
the given relation is transitive.
(ii) Not reflexive, not symmetric but transitive.
Since no natural number is less than itself hence not reflexive,
If a < b then b < a is false. Hence not symmetric.
If a < b then b < c clearly a < c. Hence transitive
(iii) Not reflexive, symmetric, not transitive.
1
a  a  0  hence it is not reflexive.
2
1
If (a, b)  R then |a – b| >
2
1 1
Since |a – b| = |b – a|  |a – b| >  |b – a| >  (b, a)  R
2 2
3 1 5 1
Let a = 1, b = – 1 and c = , a  b  2  so (a , b)  R; bc   so (b, c)  R
2 2 2 2
3 1 1
But a  c  1    so (a, c)  R . Hence R is not a transitive relation.
2 2 2
(iv) Reflexive, not symmetric, transitive
a
Since = 1 i.e. every number divides itself, hence R is reflexive.
a
If a divides b then b does not necessarily divide a.
If a divides b and b divides c then it is clear that a will divide c. Hence transitive.
(v) Because (a, a)  R

Reflexive, symmetric as well as transitive, hence it is an equivalence relation.
 0
Since 0 is divisible by n   0  so given relation is reflexive
 n 
If a – b is divisible by n, then (b – a) will also be divisible by n. Hence, symmetric.
If a – b = nI1 and b – c = nI2, where I1, I2 are integer.
Then, a – c = (a – b) + (b – c) = n(I1 + I2) so a – c is also divisible by n, hence transitive.
RELATION & FUNCTION
FUNCTION
3. INTRODUCTION :
A function is like a machine which gives unique output for each input that is fed into it. But every machine
is designed for certain defined inputs for eg. a juicer is designed for fruits & not for wood. Similarly
functions are defined for certain inputs which are called as its "domain and corresponding outputs are
called "Range".
3.1 General Definition :

Definition-1 :
Let A and B be two sets and let there exist a rule or manner or correspondence ‘ f ’which associates to
each element of A, a unique element in B. Then f is called a function or mapping from A to B. It is
denoted by the symbol
f
f : A  B or A  B
which reads ‘ f ’ is a function from A to B’ or ‘f maps A to B,
If an element a  A is associated with an element b  B then b is called ‘the f image of a’or ‘image of
a under f ’or ‘the value of the function f at a’. Also a is called the pre-image of b or argument of b under
the function f. We write it as
b = f (a) or f : a  b or f : (a, b)
f f f f
A B A B A B A B
a p a p a p p
a
b q b q b q q
b
c r c r c r c r
d s d s d s s
d
Function Not a function Function t
Not a function
As a particular type of relation, we can define a function as follows :
Definition-2 :
A relation R from a set A to a set B is called a function if
(i) each element of A is associated with some element of B.
(ii) each element ofAhas unique image in B.
Thus a function ‘ f ’from a set A to a set B is a subset of A × B in which each 'a' belonging to A appears
in one and only one ordered pair belonging to f.
Note : (1) Every function is a relation but every relation is not necessarily a function.
(2) A function is also called a mapping.
3.2 Domain, Co-domain & Range of A Function :

Let f : A  B, then the set A is known as the domain of f & the set B is known as codomain of f.
The set of all f images of elements of A is known as the range of f . Thus :
Domain of f = {a  a  A, (a, f(a))  f}
Range of f = {f(a)  a  A, f(a)  B, (a, f (a) )  f}
RELATION & FUNCTION
(1) It should be noted that range is a subset of codomain .

(2) If only the rule of function is given then the domain of the function is the set of those real numbers, where
function is defined.
(3) For a continuous function, the interval from minimum to maximum value of a function gives the range.
(4) Let f and g be function with domain D1 and D2 then the functions
f + g, f – g, fg, f g are defined as
(f + g)(x) = f (x) + g (x); Domain D1  D2
(f – g)(x) = f (x) – g (x); Domain D1  D2
(f g)(x) = f (x) · g (x); Domain D1  D2
f  f (x)
  (x) = ; Domain = {x  D1  D2 | g (x)  0}
g g( x )
3.3 Graphical Representation of function :
The equation y = f(x) represents a function if a straight line drawn parallel to y-axis in its domain cuts the
graph of y = f(x) at only one point.
Eg. y = x3
Here all the straight lines parallel to y-axis cut y = x3 only at one print. Hence it is a function.
y
Eg. x2 + y2 = 12
Here line parallel to y-axis is intersecting the circle at two points hence it is not a function.
y
x
(–1,0) (1,0)
Rule for finding Domain :
(i) Expression under even root (i.e. square root, fourth root etc.)  0.
(ii) Denominator  0
(iii) If domain of y = f(x) & y = g(x) are D1 & D2 respectively then the domain of f(x) ± g(x) or f(x) · g(x) is
D1  D2
f (x)
(iv) Domain of is D1  D2 – {g(x) = 0}.
g(x )
RELATION & FUNCTION
3.4 IMPORTANT TYPES OF FUNCTIONS:

(i) Polynomial Function:
If a function f is defined by f (x) = a0 xn + a1 xn1 + a2 xn2 + ... + an1 x + an where n is a non negative
integer and a0, a1, a2, ..., an are real numbers and a0  0, then f is called a polynomial function of degree
n.Apolynomial function is always continuous.
NOTE: (a) A polynomial of degree one with no constant term is called an odd linear function
i.e. f(x) = ax , a  0
(b) There are two polynomial functions , satisfying the relation ;
f(x).f(1/x) = f(x) + f(1/x) . They are :
(i) f(x) = xn + 1 & (ii) f(x) = 1  xn , where n is a positive integer.
(c) A polynomial of degree odd has its range (– , ) but a polynomial of degree even has a range
which is always subset of R.
(ii) Algebraic Function:

A function f is called an algebraic function if it can be constructed using algebraic operations such as
addition, substraction, multiplication, division and taking roots, started with polynomials.
x 4  16x 2
e.g. f (x) = x 1 ;
2 g (x) = + (x – 2) × 3 x  1
x x
Note that all polynomial are algebraic but not the converse. Functions which are not algebraic are
known as transcedental function.
(iii) Fractional Rational Function:

g(x)
A rational function is a function of the form. y = f (x) = , where g (x) & h (x) are polynomials &
h(x)
h(x)  0. The domain of f (x) is set of real x such that h (x)  0.
2x 4  x 2  1
e.g. f (x) = ; D = {x | x  ± 2}
x2  4
(iv) Exponential Function:

A function f(x) = ax = ex ln a (a > 0 , a  1, x  R) is called an 0<a<1 a>1
exponential function. f (x) = ax is called an exponential function
because the variable x is the exponent. It should not be confused y=a
x
with power function. g (x) = x2 in which variable x is the base.

Note : For f (x) = ex domain is R and range is R+.
(v) Logarithmic function:

y = logax, x > 0, a > 0, a  1
a>1
y = loga x
Domain : R  or (0, )
1
Range : R or (, )
0<a<1
RELATION & FUNCTION
(vi) Absolute Value Function:

A function y = f (x) = x is called the absolute value function or Modulus function. It is defined as:
 x if x  0
y = x 
 x if x  0
For f (x) = | x |, domain is R and range is R+  {0}.
1 |x|
For f (x) = or 2 , domain is R – {0} and range is R+.
|x| x
Note : | x|  x,  x  R
(vii) Signum Function:

A function y= f (x) = Sgn (x) is defined as follows :
 1 for x  0 1 1
  , x0
y = f (x) =  0 for x  0 =  x
 1 for x  0 
0, x0
Note that Sgn(Sgn x) = Sgn x;
Domain is R and range is {–1, 0, 1}.
(viii) Greatest Integer Or Step Up Function :
The function y = f (x) = [x] is called the greatest integer function where [x] denotes the greatest integer
less than or equal to x . Note that for :
1  x< 0 ; [x] =  1 0x< 1 ; [x] = 0
1x< 2 ; [x] = 1 2x < 3 ; [x] = 2 and so on.
For f (x) = [x], domain is R and range is I.
Properties of greatest integer function :

(a) [x]  x < [x] + 1 and
x  1 < [x]  x , 0  x  [x] < 1
(b) [x + m] = [x] + m if m is an integer.
(c) [x] + [y]  [x + y]  [x] + [y] + 1
(d) [x] + [ x] = 0 if x is an integer
=  1 otherwise.
(ix) Fractional Part Function :

It is defined as :
g (x) = {x} = x  [x].
e.g. the fractional part of the number 2.1 is
2.1 2 = 0.1 and the fractional part of  3.7 is 0.3. The period
of this function is 1 and graph of this function is as shown. y graph of y={x}
For f (x) = {x}, domain is R and range is [0, 1)
1
1
For f (x) = , domain is R – I, range is (1, )
{x} x
–1 O 1 2 3
Note : {x + n} = {x}
RELATION & FUNCTION
(a) Identity function :

The function f : A  A defined by f(x) = x  x  A is called the y=x
identity of A and is denoted by IA. It is easy to observe that identity
function defined on R is a bijection.
(b) Constant function :

y
A function f : A  B is said to be a constant function if every element
of A has the same f image in B. Thus f : A  B ; f(x) = c ,  x  A , (0, c)
c  B is a constant function. Note that the range of a constant function y=c
is a singleton and a constant function may be oneone or manyone, x
onto or into.
e.g. f (x) = [{x}]; g(x) = sin2x + cos2x; h (x) = sgn(x2 – 3x + 4) etc.
Domain and Range of important trigonometric functions :
Function (f) Domain (Df) Range (Rf)

sin R [–1, 1]
cos R [–1, 1]
cosec R – n (–, –1]  [1, )

sec R – (2n + 1) (–, –1]  (1, )
2

tan R – (2n + 1) R
2
cot R – n R
Note : sin x = 0  x = n
tan x = 0  x = n

cot x = 0  x = (2n + 1)
2

sin x = 1  x = 2n +
2

sin x = –1  x = 2n –
2

cot x = 1  x = (2n + 1)
2
cos x = 1  x = 2n
cos x = –1  x = (2n + 1) , n  I
RELATION & FUNCTION
Brief Recall :
1 y=1/2  5
(i) Solution of sin x  2n + x + 2n
2 /6 /6
6 6
add 2n to both sides
y=1  
(ii) tan x > 1 n + <x< + n
 4 2
4
add n to both sides.
1  
(iii) cos x  –/3 /3
2n –  x  2n +
2 3 3
add 2n to both sides
(iv) ax2 + bx + c > 0  x  R iff a > 0 & D < 0
Illustration :
Find the domain of following functions
1
(i) f (x) = x 2  5x  6 (ii) f (x) = x 2  3x  2 
x 2  3x  4
2 1
(iii) f (x) = 2 + log10 (x3 – x) (iv) f (x) =
x 4 | x | x
1  5x  x 2 
(v) f (x) = (vi) f (x) = log 1  
x  [x]
2
4 
 1
(vii) f(x) = log4 log2 log1/2 (x) (viii) f(x) = log   cos x  
 2
(ix) f(x) = ln (3x2 – 4x + 5) + 2 sin 2 x  5 sin x  2
Sol.
(i) f (x) = x 2  5x  6 + – +
positive negative positive
 x2 – 5x + 6  0
 (x – 2) (x – 3)  0 2 3
 x  (–, 2]  [3, )
RELATION & FUNCTION
1
(ii) f (x) = x 2  3x  2 
2
x  3x  4
x2 – 3x + 2  0 and x2 – 3x – 4 > 0
x2 – 3x + 2  0 + – +
 (x – 2) (x – 1)  0
 x  (–, –1]  [2, ) …(i) 1 2
x2 – 3x – 4 > 0 + – +
(x – 4) (x + 1) > 0
x  (–, –1)  (4, ) …(ii) –1 4
Taking union of (i) & (ii)
x  (–, –1)  (4, )
2
(iii) f (x) = 2 + log10 (x3 – x)
x 4
Following conditions should be followed
x2 – 4  0 & x3 – x > 0
x  ±2
x  R – {–2, 2} …(i)
x3 – x > 0
 x (x2 – 1) > 0
–1 0 1
 x (x – 1) (x + 1) > 0
x  (–1, 0)  (1, ) …(ii)
Taking union of (i) & (ii)
x  (–1, 0)  (1, 2)  (2, )
1
(iv) f (x) =
| x | x
|x|–x>0
|x|>x
This is possible only when x is negative i.e. x < 0, hence
x  (–, 0)
1
(v) f (x) =
[x]  x
x – [x] > 0
x > [x]
but we know that [x]  x and equality holds if x  I.
hence domain is x  R – I
 5x  x 2 
(vi) f (x) = log 1  

2  4 
5x  x 2
0
4
 x (5 – x) > 0  x (x – 5) < 0
x  (0, 5) …(i)
RELATION & FUNCTION
 1 
 5x  x 2   As Base   (0, 1) log is a decreasing function 
Also log 1  0
  2 
 4   sign of inequality changes 
2  
0
5x  x 2  1 
    5x – x2  4
4 2
 x2 – 5x + 4  0
1 4
x  (–, 1]  [4, ) …(ii)
Using (i) and (ii)
x (0, 1] [4, 5)
(vii) f(x) = log4 log2 log1/2 (x)

 log2 log1/2 (x) > 0  log1/2 (x) > (20)
1
1
 log1/2 x > 1 x<  
2
Also x > 0
 1
 x   0, 
 2
 1
(viii) f(x) = log   cos x  
 2
 1
  cos x   > 0 –4/3 –2/3 2/3 4/3
 2
–1
y= 2
1
cos x < –
2
 2 4 
x   2 n  , 2 n  
 3 3 
(ix) f(x) = ln (3x2 – 4x + 5) + 2 sin 2 x  5 sin x  2

3x2 – 4x + 5
Coefficient of x2 = 3
Discriminant (D) = 16 – 4 × 5 × 3 = –44 < 0
hence 3x2 – 4x + 5 > 0, x R
 2 sin2x – 5 sin x + 2 0
 2 sin2 x – sin x – sin x + 2  0
 (2 sin x – 1) (sin x – 2)  0
But sin x – 2  0 .  x  R
 2 sin x – 1  0 y = 1/2
–
1
 sin x  –7/6 /6
2
 7 
 x   2 n  , 2 n  
 6 6
RELATION & FUNCTION
RANGE :
Range of y = f(x) is the collection of all outputs f(x) corresponding to each real number of the domain.
To find the range of function
(i) First of all find the domain of y = f(x).
(ii) If domain is a set having only finite number of points, then range is the set of corresponding f(x) values.
(iii) If domain of y = f(x) is R or R– {Some finite points}, then express x in terms of y. From this find y for
x to be defined or real or form an equation in terms of x & apply the condition for real roots.
Note : For a continuous function interval from least to greatest value gives its range.
Illustration :
Find the range of following functions
(i) f(x) = log3{log1/2(x2 + 4x + 4)} (ii) f(x) = sin2x – 5sin x – 6.
Sol.
(i) f(x) = log3{log1/2(x2 + 4x + 4)}
firstlyfindingthe domain
log1/2(x2 + 4x + 4) > 0
x2 + 4x + 4 < 1  x2 + 4x + 3 < 0  (x + 1) (x +3) < 0  – 3 < x < – 1
Also x2 + 4x + 4 > 0
(x + 2)2 > 0  x  – 2 log1/2x=y
Hence, x  (–3, –1) – {2} and in this domian O 1
x + 4x + 1 (0, 1)  0 < log1/2(x + 4x + 4) < 
2 2
 log3 (log)1/2 (x2 + 4x + 4)  (– , )

 Rf = R.
5 25 25
(ii) f(x) = sin2x – 5sin x – 6 = sin2x – 2   sin x + –6–
2 4 4
2
 5 49
=  sin x   
 2 4
2 2
9  5 49  5 49
where   sin x     –10   sin x   – 0
4  2 4  2 4
Hence, f(x)  [–10, 0]. Ans.
Illustration :
Find the range of the following functions
x2
(i) f(x) = a sin x + b. a > 0 , b R (ii) f(x) = 4 tan x cos x (iii) y=
1 x2
x2  x x ( x  1)
(iv) y = loge (3x2 – 4x + 5) (v) y = = (vi) y = 3 – 2x
2
x  2x x ( x  2)
Sol.
(i) f(x) = a sin x + b. a > 0 , b R
f(x) = a sin x + b
 –1 sin x 1
–a + b f(x) a + b
Range [b – a, b + a]
RELATION & FUNCTION
(ii) f(x) = 4 tan x cos x

f(x) = 4 sin x for cos x 0
–1 sin x 1

but at x = (2n + 1) , sin x = ±1
2
hence points with sin x = ± 1 will not be included in range.
Range (– 4, 4).
x2
(iii) y=
1 x2
y is defined x R, domain is R
x2 y y
from y =  x2 =  0 (ax x2  0)
1 x2 1 y 1 y
0 y < 1
Range [0, 1)
(iv) y = loge (3x2 – 4x + 5)

y is defined if 3x2 – 4x + 5 > 0
since D = –44
D < 0, coefficient of x2 > 0
hence domain is R and log is increasing function.
D
Minimum value of 3x2 – 4x + 5 is
4a
 D  (44) 11  11 
 = =  y loge  
4a 4(3) 3 3
  11  
Range  log e  ,  
 3 
x2  x x ( x  1)
(v) y= 2 =
x  2x x ( x  2)
Domain is x R – {–2, 0}
x ( x  1)
y=
x ( x  2)
x 1 1  2y
when x  0, y =  x=
x2 1 y
If x is real y – 1 0 y 1
1  2y
Also for x = ; x 0
1 y
1  1 
hence y –  Hence range y R –  , 1
2 2 
(vi) y = 3 – 2x
Domain is x R
0 2x <  – < –2x  0  – < 3 –2x  3
Range (– , 3)
RELATION & FUNCTION
4. EQUAL OR IDENTICAL FUNCTION :
Two functions f & g are said to be equal if

(i) The domain of f = the domain of g. i.e. Df = Dg
(ii) The range of f = the range of g and i.e. Rf = Rg
(iii) f(x) = g(x) , for every x belonging to their common domain.
1 x
e.g. f (x) =
& g(x) = 2 are identical functions.
x x
NOTE : Functions are also equal if their graphs are same
Illustration :
Find the domain of x for which the function f(x) = ln x2 and g(x) = 2ln x are identical.
Sol. f(x) = ln x2 = 2 ln | x |
g(x) = 2ln x
if f(x) = g(x)
2ln | x | = 2ln x Now | x | = x if x  0 but 0  Df , Dg.
function are equal only if x (0, )
Illustration :
Find out which of the following functions are identical.
1 1
(i) f(x) = cosec x, g(x) = (ii) f(x) = tan x, g(x) =
sin x cot x
1 cos 2 x
(iii) f(x) = ln ex, g(x) = eln x (iv) f(x) = , g(x) = sin x
2
1
(v) f (x) = , g(x) = x 2
|x|
Sol.
1
(i) f(x) = cosec x, g(x) =
sin x
Domain of f(x) x  n
Domain of g(x) x  n
Since domain and range are same hence identical function
1
(ii) f(x) = tan x, g(x) =
cot x
f(x) = tan x, 0  Df
1
g(x) = But 0  Dg
cot x
hence f(x) and g(x) are not identical.
(iii) f(x) = ln ex, g(x) = eln x

f(x) = ln ex Domain = R
g(x) = eln x Domain = R+
hence not identical function
RELATION & FUNCTION
1 cos 2 x
(iv) f(x) = , g(x) = sin x
2
f(x) = | sin x | Range [0, 1]
g(x) = sin x Range [– 1, 1]
hence not identical.
1
(v) f (x) = , g(x) = x 2
|x|
1
f(x) =
|x|
1 1
g(x) = 2 =
x |x|
hence identical functions.
5. CLASSIFICATION OF FUNCTIONS :
5.1 One-One Function (Injective Mapping) :
Afunction f :A B is said to be a oneone function or injective mapping if different elements of Ahave
different f images in B. Thus for x1, x2  A & f(x1), f(x2)B,
f(x1) = f(x2)  x1 = x2 or x1  x2  f(x1)  f(x2).
Examples: f : R  R, f (x) = x3 + 1 ; f (x) = e – x ; f (x) = ln x are all one-one functions.
Remember that a linear function is always one-one.

Diagramatically an injective mapping can be shown as
OR
one-one function one-one function

NOTE:
(i) A continuous function which is always increasing or decreasing in whole domain, then f(x) is oneone.
(ii) A function is one to one if and only if a horizontal line intersects its graph at most once.
RELATION & FUNCTION
5.2 Many-One Function (Not Injective) :

Afunction f : A B is said to be a many one function if two or more elements of Ahave the same
f image in B. Thus f : A  B is many one if for;
x1, x2  A , f(x1) = f(x2) but x1  x2.
Examples : R  R f (x) = [ x ] ; f (x) = | x | ; f (x) = ax2 + bx + c ; f (x) = sin x are all examples of many
one function.
Diagramatically a many one mapping can be shown as
OR
many-one function many-one function
NOTE:
(i) Any continuous function which has atleast one local maximum or local minimum, then f(x) is manyone.
In other words, if a line parallel to xaxis cuts the graph of the function atleast at two points, then f
is manyone.
(ii) If a function is oneone, it cannot be manyone and vice versa. if f : A  B

Number of One One mappings + Number of Many One mappings = Total number of mappings from
AB
5.3 Onto Function (Surjective Mapping) :

If the function f : A B is such that each element in B (codomain) is the f image of atleast one element
in A, then we say that f is a function of A 'onto' B. Thus f : A  B is surjective iff  b  B,  some
a  A such that f (a) = b.
f : R  R f (x) = 2x +1; f : R  R+ f (x) = ex; f : R+  R f (x) = ln x are examples of onto functions.
Diagramatically surjective mapping can be shown as
OR
Note that:
(a) if range = codomain, then f (x) is onto.
(b) Any polynomial of degree odd defined on R is onto.
(c) if co-domain of f is not given then it is taken to be R.
RELATION & FUNCTION
5.4 Into Function :

If f : A  B is such that there exists atleast one element in codomain which is not the image of any
element in domain, then f(x) is into.
e.g. f : R  R f (x) = [x], | x |, sgn x, f (x) = ax2 + bx + c are all examples of into functions.
Diagramatically into function can be shown as
OR
Note :
If a function is onto, it cannot be into and vice versa . (b) A polynomial of degree even defined from
R  R will always be into
Thus a function can be one of these four types :
(a) oneone onto (injective & surjective) (I  S)
(b) oneone into (injective but not surjective) (I  S )
(c) manyone onto (surjective but not injective) (S  I )
(d) manyone into (neither surjective nor injective) ( I  S )
Note :
If f is both injective & surjective, then it is called a Bijective mapping. The bijective functions are also
named as invertible, non singular or biuniform functions.
Illustrations :
Classify the following functions as many-one, one-one, onto or into functions.
(i) f(x) = ex + e–x (ii) f(x) = x3
(iii) f(x) = 1 x 2 (iv) f : [–1, 1] [–1, 1], f(x) = sin 2x
Sol.
(i) f(x) = ex + e–x
Domain R
1 1
y = ex + x  y = ex + x 2
e e
Range [2, )  f is into
also f(x) = f(–x)
hence function is many one into
y
(ii) f(x) = x3
Domain R
Range R  f is onto x
Since f(x) = x3 is continuous and strictly increasing on R hence it is one-one.
f(x) is one-one onto y = x3.
RELATION & FUNCTION
(iii) f(x) = 1 x 2
Domain x R
Range y [1, )  f is into
f(x) = f(–x)
y
f(x) is many one-into
–
(iv) f : [–1, 1] [–1, 1], f(x) = sin 2x 4
x
–2  2
From graph we can say that 4
f(x) is many one onto. –1
Illustration :
The function f : [2, ) y defined by f(x) = x2 – 4x + 5 is both one-one and onto if
(A) y = R (B) y = [1, ) (C) y = [4, ] (D) y = [5, ]
Sol. f(x) = x2 – 4x + 5
Minima at x = 2 y
at x = 2, y = 4 – 8 + 5 = 1
For function to be one-one it should be monotonic. x
Hence, for x [2, ), f(x) is increasing. O
at x = 2, y = 1. Hence y [1, )
Illustration :
If f(x) = x2 + bx + 3 is not injective for values of x in the interval, 0 x 1 then b lies in
Sol. If f(x) is not one-one then atleast one horizontal line should intersect it at two points
b
0< <1
2
 0 < –b < 2 b
0 1
 b (2, 0) 2a
Some important points to remember :

If x, y are independent variables, then :
(i) f(xy) = f(x) + f(y)  f(x) = k ln x or f(x) = 0.
(ii) f(xy) = f(x) · f(y)  f(x) = xn , n  R
(iii) f(x + y) = f(x) · f(y)  f(x) = akx.
(iv) f(x + y) = f(x) + f(y)  f(x) = kx, where k is a constant.
6. FUNCTIONAL EQUATIONS :
Illustration :
If f(x) = ax7 + bx3 + cx – 5, where a, b and c are constants. If f(–7) = 7, then find f(7).
Sol. f(–7) = a (–7)7 + b(–7)3 + c(–7) – 5
7 = – (a · 77 + b · 73 + c · 7) – 5
a · 77 + b · 73 + c · 7 = – 12
f(7) = a · 77 + b · 73 + c · 7 – 5 = – 12 – 5 = – 17. Ans.
RELATION & FUNCTION
Illustration :
1
Let f be a real valued function of real and positive argument such that f(x) + 3x f   = 2 (x + 1)  x > 0,
x
find the value of f(10099).
1
Sol. f(x) + 3x f   = 2 (x + 1) ......(1)
x
1
replacing x by
x
1 3 1 
f    f ( x ) = 2  1
x x x 
1
x f   + 3f(x) = 2 (1 + x) ......(2)
x
On solving (1) and (2)
x 1
8f(x) = 4 (1 + x)  f(x) =
2
10100
f(10099) = = 5050. Ans.
2
Illustration :
1
If f(0) = 1, f(1) = 2 & f(x) = [f (x + 1) + f (x + 2)], find the value of f (5).
2
Sol. f(x + 2) = 2f(x) – f(x + 1)
thus f(0 + 2) = f(2) = 2f(0) – f(1) = 2(1) – 2 = 0
f(3) = 2f(1) – f(2) = 2(2) – 0 = 4
f(4) = 2f(2) – f(3) = 0 – 4 = –4
f(5) = 2f(3) – f(4) = 2(4) – (–4) = 12
Illustration :
If f(x) + 2f(1 – x) = x2 + 2  x  R, find f(x).
Sol. f(x) + 2f(1 – x) = x2 …(i)

Replacing x by 1 – x
f(1 – x) + 2f(x) = (1 – x)2 …(ii)
Solving (i) & (ii), we get
3f(x) = 2x2 – (1 – x)2
x 2  2x  1
f(x) =
3
RELATION & FUNCTION
Illustration :
 1 
If 2f(x2) + 3f  2  = x2 – 1 (x  0) then find f(x2).
x 
 1 
Sol. 2f(x2) + 3f  2  = x2 – 1 …(i)
x 
1
Replace x by
x
 1  1
2f  2  + 3f(x2) = 2 – 1 …(ii)
x  x
Solving (i) & (ii) we get
 1 
9f(x)2 – 4f(x2) = 3  2 1 – 2(x2 – 1)
x 
3
5f(x2) = 2 – 2x2 – 1
x
 2x 4  x 2  3 
f(x2) = – 



 5x 2 
Illustration :
Let f(x) & g(x) be functions which take integers as arguments let f(x + y) = f(x) + g(y) + 8 for all integer
x & y. Let f(x) = x for all negative integers x let g(8) = 17, find f(0).
Sol. f(x) = x for integers less than zero
 f(–8) = –8
f(x + y) = f(x) + g(y) + 8
f(–8 + 8) = f(–8) + g(8) + 8
f(0) = –8 + g(8) + 8
f(0) = 17
7. COMPOSITE FUNCTION :
Let f : A  B & g : B  C be two functions. Then the function gof : A  C defined by

(gof) (x) = g (f(x))  x  A
is called the composite of the two functions f & g. Diagramatically
x f (x )

     g (f(x)).
Thus the image of every x Aunder the function gof is
h
the gimage of the fimage of x.
Note that gof is defined only if  x  A, f(x) is an A B C
element of the domain of g so that we can take its f g
gimage. Hence for the product gof of two functions f x f(x) h(x)
& g, the range of f must be a subset of the domain of g.
Note that gof in general is not equal to fog. g{f (x)} = h(x)
RELATION & FUNCTION
Illustration :
x 1 2
Let f(x) = , f (x) = f{f(x)}, f 3(x) = f {f 2(x)} ....... f k+1(x) = f{f k(x)}, for k = 1, 2, 3, .....
x 1
Find f 1998(x).
x 1
Sol. f(x) =
x 1
f (x)  1 1
f 2(x) = f{f(x)} = =
f (x )  1 x
x 1
f 3(x) =
1 x
4
f (x) = x
f 5(x) = f{f 4(x)} = f(x)
1
f 1998(x) = f 2(x) = ..
x
Illustration :
Let f : R  R be the function defined by f(x) = ax2 – 2 for some positive a. If (fof ) 2 = – 2 , then find
the values of a.
Sol. f(x) = ax2 – 2
2
 1  1
(fof )(x) = a(ax2 – 2 )2 – 2  4a  a   0  a = 0,
 2 2
Illustration :
Let f(x) = x ; g(x) = 2  x find the domain of
(i) fog (ii) gof (iii) fof (iv) gog
Sol.
(i) fog

f g ( x )   f 2  x   2x
Domain 2 – x  0
x2  x (–, 2]
(ii) gof
 
gf ( x )   g x  2  x
x  0 & 2 – x  0  x  [0, 4]
(iii) fof
 
f f ( x )   f x  x
Domain  x  0
x  [0, )
RELATION & FUNCTION
(iv) gog
 
gg( x )   g 2  x  2  2  x
2–x0 & 2x  2
x2 & 2–x4
x  [–2, 2]
Illustration :
Let f(x) = x4 & g(x) = x2x , then find f g( x )  .
Sol.    
f g ( x )   f x 2 x  x 2 x
x2x
 x 2 x .x
2x
 x 2x
2 x 1
7.1 Properties of Composite Functions :
(i) The composite of functions is not commutative i.e. gof  fog .

(ii) The composite of functions is associative i.e. if f, g, h are three functions such that
fo (goh) & (fog) oh are defined, then fo (goh) = (fog) oh.
Associativety : f : (N)  I0 f(x) = 2x
1
g : I0  Q g(x) =
x
1
h : Q  R h(x) = ex
(hog)of = ho(gof) = e2x
(iii) The composite of two bijections is a bijection i.e. if f and g are two bijections such that gof is defined,
then gof is also a bijection.
Proof: Let f : A  B and g : B  C be two bijections. Then gof exists such that
gof : A  C
We have to prove that gof is one-one and onto.
One-one : Let a1, a2  A such that (gof)(a1) = (gof)(a2), then

(gof) (a1) = (gof) (a2)  g [f (a1)] = g [ f (a2)]
 f (a1) = f (a2) [ g is one-one]
 a1 = a2 [ f is one-one]
 gof is also one-one function.
Onto : Let c  C, then

cC   b  B s.t. g (b) = c [ g is onto]
and bB   a  A s.t. f (a) = b [ f is onto]
Therefore, we see that
c  C   a  A s.t. (gof) (a) = g[f(a)] = g(b) = c
i.e. every element of C is the gof image of some element of A. As such gof is onto function.
Hence gof befing one-one and onto, is a bijection.
RELATION & FUNCTION
Illustration :
1  x , 0  x  1

Evaluate f {f(x)}, where f ( x )  x  2, 1  x  2
4  x , 2  x  4
1  x , 0  x  1

Sol. f ( x )   x  2, 1  x  2
4
4  x , 2  x  4 x+
2
3 =
graph of f (x) f( x)
2
1  f ( x ), 0  x 1

f f ( x )   f ( x )  2, 1  f (x)  2 f(x) = 4 – x
f(x
1
)=
4  f ( x ), 2x4
1–
x
from the graph we can see that 1 2 3 4
0  f(x)  1 when x  [0, 1]  [3, 4]
1 < f(x)  2 when x  (2, 3]
2 < f(x)  4 when x  [1, 2]
1  (1  x ), f ( x )  1  x, 0  x  1
1  (4  x ), f ( x )  4  x, 3  x  4
f f ( x )   
(4  x )  2, f ( x )  4  x, 2  x  3

4  ( x  2), f ( x )  x  2, 1  x  2
x, 0  x 1 x, 0  x 1
x  3, 3 x  4 2  x , 1 x  2
f f ( x )    = f f ( x )   
6  x, 2x 3 6  x, 2x 3
 
2  x , 1 x  2 x  3, 3 x 4
8.1 Homogeneous Functions :
A function is said to be homogeneous with respect to any set of variables when each of its terms is of
the same degree with respect to those variables.
For example 5x2 + 3y2 – xy is homogeneous in x & y.
x  y cos x
f (x, y) = is not a homogeneous function and
y sin x  x
x y y x
f (x, y) = ln  ln ; x 2  y 2 + x ;
y x x y
y
x + y cos are homogeneous functions of degree one.
x
8.2 Bounded Function :
A function is said to be bounded if f(x)  M , where M is a finite quantity.

e.g. f (x) = sin x is bounded in [–1, 1]
RELATION & FUNCTION
8.3 Implicit & Explicit Function :

A function defined by an equation not solved for the dependent variable is called an IMPLICIT FUNCTION.
For eg. the equation x3 + y3 = 1 defines y as an implicit function. If y has been expressed in terms of
x alone then it is called an EXPLICIT FUNCTION . For ex : y = x3 + 6x2 – 12x + 7 defines y as on explicit
function of x.
8.4 Odd & Even Functions :

Suppose a function f is definedonthe symmetricinterval(–a, a)then
(1) If f (x) = f (x) for all x in the domain of ‘f’then f is said to be an even function.
e.g. f (x) = cos x ; g (x) = x² + 3.
(2) If f (x) = f (x) for all x in the domain of ‘f’then f is said to be an odd function.
e.g. f (x) = sin x; g (x) = x3 + x.
2
y=x y = x3 y=x
y = cos x
O O O O
Odd functions (Symmetric about origin) Even functions (Symmetric about y-axis)
NOTE :
(a) f (x)  f (x) = 0 => f (x) is even & f (x) + f (x) = 0 => f (x) is odd .
(b) A function may neither be odd nor even .
(c) Inverse of an even function is not defined and an even function can not be strictly monotonic
(d) Every even function is symmetric about the yaxis & every odd function is symmetric about the origin.
(e) Every function defined on symmetric interval (– a, a) can be expressed as the sum of an even & an odd
function.
f ( x )  f (  x) f ( x)  f (  x) 2 x  2 x 2x  2x
e.g. f (x)   2x = +
2 2 2 2
EVEN ODD EVEN ODD
(f) The only function which is defined on the entire number line & is even and odd at the same time is f(x) =0.
Any non zero constant is even.
(g) If f and g both are even or both are odd then the function f.g will be even but if any one of them is odd
then f.g will be odd .
f (x) g (x) f (x) + g (x) f (x) - g (x) f (x).g (x) f(x) / g(x) (gof )(x) (f o g) (x)
odd odd odd odd even even odd odd
even even even even even even even even
odd even neither odd nor even neither odd nor even odd odd even even
even odd neither odd nor even neither odd nor even odd odd even even
RELATION & FUNCTION
Illustration :
Identify the functions, as even, odd or neither even nor odd.

 2x  1 
(i) f(x) = ln  x  1 x2  (ii) f(x) = x ·  x  (iii) f(x) = 2x3 – x + 1
   2 1 
(iv) f(x) = 3 (v) f(x) = x2 – |x|
Sol.
(i) f(x) = ln  x  1 x2 
 
f(– x) = ln   x  1  x 2 
 
  1  x 2  x  1  x 2  x  
    
 ln  1  x  x   ln  
2      ln  x  1  x 2   f ( x )
    1  x 2  x    
   
Hence odd function.
 2x  1 
(ii) f(x) = x·  x 
 2 1 
 2 x  1   1  2x   x 
f(–x) = (–x)   x   ( x )   x 2  1   f (x )
x   2x  1 
 2 1  1 2   
hence even function
(iii) f(x) = 2x3 – x + 1

f(–x) = –2x3 + x + 1  f(x) or –f(x)
Hence neither even nor odd function
(iv) f(x) = 3
f(–x) = 3 = f(x)
Hence even function
(v) f(x) = x2 – |x|

f(–x) = x2 – |–x| = f(x)
even function
RELATION & FUNCTION
9. INVERSE OF A FUNCTION :
Let f : A  B be a oneone & onto function, then their exists a unique function
g : B  A such that f(x) = y  g(y) = x,  x  A & y  B. Then g is said to be inverse of f.
Thus g = f1 : B  A = {(f(x), x)  (x, f(x))  f}.
Consider a one-one onto function with domain A = {a, b, c} & range B = {1, 2, 3}
f f –1
A B B A
a 1 1 a
b 2 2 b
c 3 3 c
Domain of f = {a, b, c} = Range of f–1

Range of f = {1, 2, 3} = Domain of f–1
Note : (a) Only one-one onto functions (i.e., Bijections) are invertible.
(b) To find the inverse
Step-1: write y = f (x)
Step-2: solve this equation for x in terms of y (if possible)
Step-3: To express f –1 as a function of x, interchange x and y.
Illustration :
Find the inverse of the function
f (x) = loga  x  x  1  , a > 0, a  1.

2
 
Sol. Since x2 1  x2  x  R
Hence x + x 2  1 > 0  x  R
f(x) is one-one onto hence invertible
y  log a  x  x 2  1 
 
a y  x  x2 1 …(i)
1
a y   x  x 2  1 …(ii)
2
x  x 1
(i) – (ii)
ay – a–y = 2x  x=
2

1 y y
a a 
Hence f 1 ( x ) 
1 x
2

a  a x 
RELATION & FUNCTION
Illustration :
Find the inverse of the following bijective function

(i) f : R  R+ , f(x) = 10x+1 (ii) f(x) = 3x – 5
1 2x
(iii) f : [1, )  [2, ), f(x) = x + (iv) f : R  (0, 1), f(x) =
x 1  2x
Sol.
(i) y = 10x+1
x + 1 = log10 y
x = –1 + log10 y
 f–1 = y = –1 + log10 x, f–1 : R+  R
(ii) f(x) = 3x – 5
y = 3x – 5
y5
x=
3
x5
 f–1 (x) = y =
3
(iii) f : [1, )  [2, )

1
y = f(x) = x +
x
y  y2  4
 x2 – xy + 1 = 0  x=
2
x  x2  4
 f –1 (x) =
2
Since range is [1, ), hence
x  x2  4
f –1 (x) =
2
2x
(iv) f : R  (0, 1), f(x) =
1  2x
2x
y=
1  2x
y
 y + 2x y = 2x  2x =
1 y
 y 
 x = log2  
1 y 
 x 
 f–1 (x) = y = log2  
1 x 
RELATION & FUNCTION
9.1 Properties of Inverse of a Function :
(i) The inverse of Bijection is unique.

(ii) The inverse of Bijection is also bijection.
(iii) If f : A  B is Bijection & g : B  A is inverse of f, then
fog = IB & gof = IA
(iv) If f : A  B and g : B  C are two bijections, then gof : A  C is bijections and
(gof)–1 = f–1og–1
(v) fog  gof but if fog = gof then either f–1 = g or g–1 = f also
(fog) (x) = (gof) (x) = x
(vi) The graphs of f & g are the mirror images of each other in the line y = x. As shown in the figure given
below a point (x , y ) corresponding to y = x2(x>0) changes to (y, x) corresponding to y   x , the
changed form of x = y .
Illustration :
If f : R  R is defined by f(x) = x3 + 1, then find value of f–1 (28).
Sol. f–1 (28) = x  f(x) = 28  x3 + 1 = 28  x=3
Illustration :
If the function f & g be defined as f(x) = ex and g(x) = 3x – 2 where f : R  R and g : R  R then find
the function fog and gof. Also find the domain of (fog)–1 and (gof )–1 .
Sol. (fog) (x) = f{g(x)}
f{g(x)} = f (3x – 2) = e3x–2
(gof) (x) = g{f(x)} = g(ex) = 3ex – 2
To find (fog)–1 & (gof )–1
(fog) (x) = y = e3x–2
 3x – 2 = log y
log y  2 log x  2
 x=  (fog)–1 x =
3 3
Domian of (fog)–1 is x > 0 i.e. x  (0, )
Again (gof ) x = y = 3ex – 2
y2 y2
 ex =  x = log  
3  3 
 x2
 (gof)–1 x = log  
 3 
x2
Domain of (gof )–1 is >0
3
x > –2
 x  (–2, )
RELATION & FUNCTION
Illustration :
ex  e x
If f: [0, )  [1, ), f (x) = . Find f–1(x).
2
e x  e x
Sol. f (x) =
2
1
 2y = ex +  e2x – 2ex y + 1 = 0
ex
 e2x – 2ex y + y2 = y2 – 1  (ex – y)2 = y2 – 1
 2 
 ex = y ± y2  1  x = log  y  y  1 
 
 2 
 f–1 (x) = y = log  x  x  1 
 
Since range is [0, ) hence
 2 
 f–1 (x) = y = log  x  x  1 
 
Illustration :
Find the inverse of the function f : N  N, f (x) = x + (–1)x–1 .
Sol. f (x) = x + (–1)x–1 , x  N
Then we have f (1) = 1 + 1 = 2, f (2) = 1
f (3) = 4, f (4) = 3
f (5) = 6, f (6) = 5
The points on graph are (1, 2), (2, 1), (3, 4) (4, 3), (5, 6), (6, 5) ....... etc. Thus if (a, b) is a point on the
graph then (b, a) is also a point on the graph. Hence f is the inverse of itself.
i.e. f–1(x) = x + (–1)x–1 , x  N
10. PERIODIC FUNCTION:

A function f(x) is called periodic if there exists a positive number T(T > 0) called the period of the
function such that f (x + T) = f(x) , for all values of x within the domain of x.
e.g. The function sin x & cos x both are periodic over 2 & tan x is periodic over .
Graphically :
If the graph repeats at fixed interval then function is said , 1
to be periodic and its period is the width of that interval. 2
For example graph of sin x repeats itself at an interval
of 2 0  2
f (x) = sin x
RELATION & FUNCTION
Illustration :
Show f(x) satisfying the following conditions is periodic
(i) f(x–1) + f(x + 3) = f(x + 1) + f(x + 5)
(ii) f(x) = {x} + cos x
where {·} denotes fraction part function.
Sol.
(i) f(x – 1) + f(x + 3) = f(x + 1) + f(x + 5) .......(1)
Replacing x by x + 2
f(x + 1) + f(x + 5) = f(x + 3) + f(x + 7) .......(2)
Adding (1) and (2), we get
f(x – 1) = f(x + 7) i.e. f(x) = f(x + 8)
Hence, f(x) is periodic
(ii) f(x) = {x} + cos x

Period of {x} = 1
2
Period of cos x = =2

Hence f(x) is periodic.
10.1 Properties of periodic function :
(a) f (T) = f (0) = f (T) , where ‘T’ is the period.
(b) Inverse of a periodic function does not exist.
(c) Every constant function is always periodic, with no fundamental period .
(d) If f (x) has a period T & g (x) also has a period T then it does not mean that f (x) + g (x)
must have a period T. e.g. f (x) =| sin x | + | cos x |; sin4x + cos4x
1
(e) If f(x) has a period p, then and f (x) also has a period p.
f (x )
(f) If f (x) and g (x) are periodic then f (x) + g (x) need not be periodic.
e.g. f (x) = cos x and g (x) = { x }
(g) if f(x) has a period T then f(ax + b) has a period T/a (a > 0).
(h) if period of f(x) is T1 and g(x) is T2 then period of f ± g , f · g, f/g is T = L.C.M. (T1, T2) provided
there does not exist any positive real less than T after which value repeats.
 a c  L.C.M (a , c)
Note : (1) L.C.M.  ,  = H.C.F ( b, d ) .
b d
(2) L.C.M. of rational with irrational is not possible.
RELATION & FUNCTION
Illustration :
Find the period of the following functions.
2x 4x
(i) f (x) = cos – sin ; (ii) f (x) = cos (sin x)
3 5
(iii) f (x) = sin (cos x); (iv) f (x) = [x] + [2x] + [3x] + …… + [nx]
where n  N & [ ] denotes greatest integer function
cos 2 x sin 4 x
Sol. f (x )  
3 5
cos 2 x 2(3)
Period of  = 3
3 2
4x 2 5
Period of sin = ×5= 
5 4 2
5
L.C.M. of 3 &  = 15
2
(ii) f (x) = cos (sin x)
Since cos is even functions f ( + x) = cos (sin ( + x)) = cos (–sin x) = cos (sin x) = f(x)
Hence  is period.
(iii) f (x) = sin (cos x)
Period is 2
(iv) f (x) = [x] + [2x] + [3x] + …… + [nx]
Period of [x] = 1
1
period of [2x] =
2
1
period of [3x] =
3
……………………
……………………
1 1 1
L.C.M. of 1, , …… = 1
2 3 n
Illustration :
sin nx
If f (x) = x has its period as 4, then find the integral values of n.
sin
n
2
Sol. Period of sin nx =
n
x
Period of sin = 2n
n
2
L.C.M. of , 2n = 2n
n
2n = |4|
n = 2, – 2
RELATION & FUNCTION
Illustration :
Find the period of f (x) = | sin x | + | cos x | .
Sol. | sin x | has period 

| cos x | has period 
f (x) is an even function & sin x, cos x are complementary then period of
1 
f (x) = {LCM of  & } =
2 2
Illustration :
Prove that if f (x) = sin x + cos ax is a periodic function then a must be rational.
Sol. f (x) = sin x + cos ax

Period of sin x = 2
2
Period of cos ax =
a
2
LCM of 2 & is possible only when a is rational, hence a must be rational.
a
11. SOME GRAPHICAL TRANSFORMATION :

Y
y = f (x)
Consider the graph y = f(x) shown alongside. X

O
Y
y –  = f(x – )
(i) Graph of y –  = f (x – ) is drawn by shifting the
origin to (, ) & then translating the graph of
y = f(x) w.r.t. new axes (,)
O X
Y
y = – f(x)
(ii) The graph of y = – f(x) is the mirror image of O X

f(x) in X-axis.
RELATION & FUNCTION
y = |f (x)|
(iii) y = |f (x)| is mirror image of negative portion of

y = f(x) in X-axis. O X
y = f (| x |)
(iv) y = f (| x |) is drawn by taking the mirror image of
positive x-axis graph in y-axis.
O X
(v) The graph of |y| = f(x) is drawn by deleting |y| = f(x)

those portions of the graph y = f(x) which lie
below the X-axis and then taking the mirror O X
image of the remaining portion in the X-axis, as
shown alongside.
Y
x = f (y)
y=x
y = f (x)
(vi) x = f(y) is drawn by taking mirror image of

X
y = f(x) in the line y = x. O
y = f(–x)
(vii) y = f(–x) is drawn by taking the mirror image

of y = f(x) in Y-axis. O X
RELATION & FUNCTION
Definition :
A binary composition (or operation) in a set S is a mapping from S × S to S.
The word binary is used since for two elements of a set S there exists a unique element of S, i.e. (a, b) is
mapped on some element of S.
Examples :
(1) Ordinary addition of numbers, i.e. +, is a binary operation in any set of numbers.
Addition is a binary operation in the set of natural numbers since the sum of any two natural numbers is
also a natural number. Similar is the case for integers, rationals, reals and complex numbers.
(2) Ordinary multiplication, i.e. ×, is a binary operation in the set N, Z, Q, R and C.
(3) Ordinary subtraction of numbers, i.e. –, is a binary operation in Z, Q and R. But it is not binary in N,
since 5 – 6  N.
(4) Division of numbers, i.e. ÷, is a binary operation in Q0 (non-zero rationals), R0 (non-zero reals) and
C0 (non zero complex numbers). But it is not binary in N and Z.
Laws of Binary Operation :

Let * be a binary operation defined in S, then
(i) * is said to be commutative if  a, b,  S,
(ii) * is said to be associative if  a, b, c  S,
a * (b * c) = (a * b) * c
Example :
(1) (a) Ordinary addition and multiplication of numbers over the set of real numbers are commutative and
associative but subtraction is neither commutative nor associative.
(b) Ordinary multiplication of numbers is distributive w.r. to addition of numbers.
(c) Additive identity is 0 (zero) and multiplicative identity is 1 (one).
1
(d) Additiveinverseof ais–aand multiplicativeinverseof ais .
a
(2) Division of numbers is neither commutative nor associative.
Illustration :
Let * be an operation on the set of real numbers defined by a * b = a + b + a2b, where a, b  R.
(i) is * commutative ? (ii) is * associative ?
Sol.
(i) We have a * b = a + b + a2b
and b * a = b + a + b2a
 a * b  b * a.
Thus * is not commutative.
(ii) a * (b * c) = a * (b + c + b2c)
= a + b + c + b2c + a2(b + c + b2c)
= a + b + c + b2c + a2b + a2c + a2b2c
and (a * b) * c = (a + b + a2b) * c = a + b + a2b + c + (a + b + a2b)2c
 a * (b * c) = (a * b) * c
Thus * is not associative.
RELATION & FUNCTION
Illustration :
A binary operation * one the set {0, 1, 2, 3, 4, 5} is defined as :
a  b, if ab6
a * b = a  b  6, if ab6

Show that zero is the identity for this operation and each element 'a' of the set is invertible with 6 – a,
being the inverse of 'a'. [NCERT, CBSE (AI) 2011]
Sol. For identity element :
Let a be an arbitrary element of set {0, 1, 2, 3, 4, 5}
Now, a * 0 = a + 0 = a ........(i)
0*a=0+a ........(ii) [ a + 0 = 0 + a < 6  a  {0, 1, 2, 3, 4, 5}]
Equation (i) and (ii)  a * 0 = 0 * a = a  a  {0, 1, 2, 3, 4, 5}
Now, a * (6 – a) = a + (6 – a) – 6 [ a + (6 – a)  6]
=a+6–a–6
= 0 (identity) .......(iii)
Also, (6 – a) * a = (6 – a) + a – 6 [ a + (6 – a)  6]
=6–a+a–6
= 0 (identity) .......(iv)
Equation (iii) and (iv)  a * (6 – a) = (6 – a) * a = 0 (identity)  a  {0, 1, 2, 3, 4, 5}
Hence, each element 'a' of given set is invertible with inverse 6 – a.
Illustration :
For each binary operation * defined below, determine whether * is commutative or associative.
(i) on Z , defined as a * b = a – b
ab
(ii) on Q, defined as a * b =
2
a
(iii) on R – {– 1}, defined as a * b = [NCERT]
b 1
Sol.
(i) Given a, b   Z
 a*b=a–b  Z
b * a = b – a = – (a – b)   Z
but a*bb*a
So, * is not commutative.
Now a * (b * c) = a * (b – c) = a – (b – c)
= a – b + c, a, b, c  z
(a * b) * c = (a – b) * c = a – b – c
Here a * (b * c)  (a * b) * c
So, * is not associative.
Hence, * is neither commutative nor associative.
ab
(ii) Given a * b = a, b  Q
2
ba ab
 b*a= =
2 2
a*b=b*a
RELATION & FUNCTION
So, it is commutative.
 bc  abc
Again a * (b * c) = a *   =
 2 4
ab abc
(a * b) * c = *c=
2 4
a * (b * c) = (a * b) * c  It is associative.
a
(iii) Given a * b = , a, b  R – {– 1}
b 1
b
Now, b * a = . Here a * b  b * a, so it is not commutative.
a 1
 b  a a (c  1)
Again a * *b * c) = a *   = =
 c 1  b
1 b  c 1
c 1
a
a a
(a * b) * c = * c = b  1 = ( b  1) (c  1)
b 1 c 1
Here a * (b * c)  (a * b) * c, so it is not associative.
Hence * is neither commutative nor associative.
Illustration :
If the operation * on Q – {1}, defined by a * b = a + b – ab for all a, b  Q – {1}, then find

(i) the identity element and (ii) the inverse of a for each a  Q – {1}.
Sol. We have, a * b = a + b – ab, for all a, b  Q {– 1}.

(i) Let e be the identity element.
Then, for all a  Q – {1}, we have
a * e = a  a + e – ae = a
 (1 – a) e = 0
 a  1 
 e = 0  Q – {1}  1  a  0
Now, a * 0 = a + 0 – a × 0 = a
and, 0 * a = 0 + a – 0 × a = a
Thus, 0 is the inverse element of a, for each a  Q – {1}
(ii) Let b be the inverse element of a, for each a  Q – {1}.

Then, a * b = e = 0  a + b – ab = 0
a
 a = ab – b  a = b (a – 1)  b =  Q – {1}.
a 1
a
Therefore, for each a the corresponding inverse element is  Q – {1}.
a 1
RELATION & FUNCTION
Illustration :
Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = min. {a, b}. Write the
operation table of the operation *. [CBSE (Delhi) 2011]
Sol. Required operation table of the operation * is given as
* 1 2 3 4 5
1 1 1 1 1 1
2 1 2 2 2 2
3 1 2 3 3 3 .
4 1 2 3 4 4
5 1 2 3 4 5
Illustration :
Determine which of the followingbinary operation on the set N are associative and which are commutative.
ab
(i) a * b = 1 for all a, b  N (ii) a*b= , for all a, b  N
2
Sol.
(i) We have,
a * b = 1,  a, b  N
Now, a * b = 1 = b * a  a, b  N
 a*b=b*a
So, * is commutative on N.
Again, let a, b, c  N
Therefore, a * (b * c) = a * 1 = 1  a, b, c  N
and, a * b * c) = (a * b) * c
So, * is associative on N.
Hence, * is both associative and commutative.
(ii) We have,
ab
a*b= ,  a, b  N
2
ba
= =b*a
2
Thus, a * b = b * a
So, * is commutative on N.
Now, let a, b, c  N
bc
bc a  2 2a  b  c
Then, a * (b * c) = a *  = =
 2  2 4
ab
c
ab  2 a  b  2c
and (a * b) * c =  *c = = .
 2  2 4
Obviously, a * (b * c)  (a * b) * c
So, * is not associative on N.
So, * is commutative but not associative.
RELATION & FUNCTION
EXERCISE-1 (Subjective Questions)

Q.1 Find the domains of definitions of the following functions :
(Read the symbols [*] and {*} as greatest integers and fractional part functions respectively.)
(i) f (x) = cos2x  16  x 2 (ii) f (x) = log7 log5 log3 log2 (2x3 + 5x2  14x)
 2 log10 x  1 
(iii) y = log10 sin (x  3)  16  x 2 (iv) f (x) = log100 x  
 x 
1 1
(v) f (x) =  ln x(x 2  1) (vi) f (x) = x2  | x | 
4x 2  1 9  x2
Q.2 Find the domain & range of the following functions.
2x x 2  3x  2
(i) y = log 5  2 (sin x  cos x)  3  (ii) y =
1 x2
(iii) f (x) =
x2  x  6
x x  4 3
(iv) f (x) = 1 | x | (v) y = 2  x  1  x (vi) f (x) =
x 5
Q.3 If f (x) = max x, 1 x  for x > 0 where max (a, b) denotes the greater of the two real numbers a and b.
Define the function g(x) = f (x) · f 1 x  and plot its graph.
Q.4 Identify the pair(s) of functions which are identical?

(where [x] denotes greatest integer and {x} denotes fractional part function)
1  cos 2 x
(i) f (x) = sgn (x2 – 3x + 4) and g (x) = e[{x}] (ii) f (x) = and g (x) = tan x
1  cos 2 x
cos x 1 sin x
(iii) f (x) = ln(1 + x) + ln(1 – x) and g (x) = ln(1 – x2) (iv) f (x) = and g (x) =
1  sin x cos x
Q.5 Let f be a oneone function with domain {x, y, z} and range {1,2,3}. It is given that exactly one of the
following statements is true and the remaining two are false.
f(x) = 1 ; f(y)  1 ; f(z)  2. Find the function f (x).
    5
Q.6 If f(x) = sin²x+ sin²  x    cos x cos x   and g    1 , then find (gof)(x).
 3  3 4
 1 x 
Q.7 A function f : R  R is such that f   = x for all x  – 1. Prove the following.
1 x 
(a) f  f ( x )  = x (b) f 1 x  = – f (x), x  0 (c) f (– x – 2) = – f (x) – 2.
x
Q.8(i) Find the formula for the function fogoh, given f (x) = ; g (x) = x10 and h (x) = x + 3. Find also the
x 1
domain of this function.Also compute (fogoh)(–1).
(ii) Given F (x) = cos2(x + 9). Find the function f, g, h such that F = fogoh.
Q.9 Polynomial P(x) contains only terms of odd degree. When P(x) is divided by (x – 3), the remainder is 6.
If P(x) is divided by (x2 – 9) then remainder is g (x). Find the value of g (2).
RELATION & FUNCTION
EXERCISE-2 (Objective Questions)
[SINGLE CORRECT CHOICE TYPE]

x2  a
Q.1 The set of all real values of a so that the range of the function y = is R, is
x 1
(A) [1, ) (B) (– , – 1) (C) (1, ) (D) (– , – 1]
| sin x |  | cos x |
Q.2 The period of the function f (x) = is
| sin x  cos x |
(A) /2 (B) /4 (C)  (D) 2
x lnx
Q.3 f (x) = and g (x) = . Then identify the CORRECT statement
lnx x
1 1
(A) and f (x) are identical functions (B) and g(x) are identical functions
g( x ) f (x )
1
(C) f (x) . g (x) = 1  x  0 (D)  1  x 0
f (x ) . g(x )
Q.4 A function f (x) = 1 2x + x is defined from D1  D2 and is onto. If the set D1 is its complete
domain then the set D2 is
 1
(A)   ,  (B) (– , 2) (C) (– , 1) (D) (– , 1]
 2
Q.5 Which of the following function is surjective but not injective

(A) f : R  R f (x) = x4 + 2x3 – x2 + 1 (B) f : R  R f (x) = x3 + x + 1
(C) f : R  R+ f (x) = 1 x 2 (D) f : R  R f (x) = x3 + 2x2 – x + 1
Q.6 Range of the function f (x) = tan–1 1 is

[ x]  [  x ]  2 |x| 
x2
where [*] is the greatest integer function.
LM 1 , IJ RS 1 UV  g RS 1 , 2UV LM 1 , 2OP
(A)
N4 K (B)
T4 W 2, (C)
T4 W (D)
N4 Q
Q.7 Which of the following statements are incorrect?
I If f (x) and g (x) are one to one then f (x) + g(x) is also one to one.
II If f (x) and g (x) are one-one then f (x) · g(x) is also one-one.
III If f (x) is odd then it is necessarily one to one.
(A) I and II only (B) II and III only (C) III and I only (D) I, II and III
8
Q.8 The range of the function y = is
9  x2
8   8 8 
(A) (– , ) – {± 3} (B)  ,   (C)  0,  (D) (– , 0)  ,  
9   9 9 
RELATION & FUNCTION
{x}
Q.9 Range of the function f (x) = where {x} denotes the fractional part function is
1  {x}
 1  1
(B) 0, 
1
(A) [0 , 1) (C) 0,  (D)  0, 
 2  2  2
ex  1
Q.10 For the function f (x) = , if n(d) denotes the number of integers which are not in its domain and
ex 1
n(r) denotes the number of integers which are not in its range, then n(d) + n(r) is equal to
(A) 2 (B) 3 (C) 4 (D) Infinite
 x7 4
Q.11 If f(x) =    , then the range of function y = sin (2f(x)) is

 x  9 
 1   1   1 
(A) [0, 1] (B)  0, (C)  0,  , 1 (D) (0, 1]
 2   2   2 
 1 
 
  log e 
e  x 
2x
Q.12 Let f(x) = (x + 2) eln (x + 2) and g(x) =  . If h(x) = f(x) + g(x),
x e ln 2
then the smallest positive integer in the range of h(x) is
(A) 2 (B) 3 (C) 6 (D) 7
Q.13 If the functions f(x) = (k2 – 3k + 2) x2 + (k2 – 1)  x  R and
g(x) = (k2 – 6k + 5) x 3 + (k2 – 2k + 1) x + (k2 – k)  x  R have the same graph then the number of
real values of k, is
(A) 0 (B) 1 (C) 2 (D) 3
Q.14 Let f (x) = x2 – 4x + 3. If | f (x) | > – f (x), then the true set of values of x, is
(A) (1, 3) (B) (1, 3]
(C) (– , 1)  (3, ) (D) (– , 1]  (3, )
Q.15 Let f : [0, 2]  [2, 5] be defined as f(x) = 3x2 – 6x + 5, then f(x) is

(A) injective but not surjective (B) surjective but not injective
(C) injective as well as surjective (D) neither injective nor surjective
Q.16 Let f (x) = sin2x + cos4x + 2 and g (x) = cos (cos x) + cos (sin x). Also let period of f (x) and g (x) be
T1 and T2 respectively then
(A) T1 = 2T2 (B) 2T1 = T2 (C) T1 = T2 (D) T1 = 4T2
4 4x
Q.17 Let f : R     R    be a function defined as f(x) =
4
. The inverse of f is the map
 3  3 3x  4
4  4
g : R–  R–   is given by
3  3 
3y 4y 4y 3y
(A) g(y) = (B) g(y) = (C) g(y) = (D) g(y) =
3  4y 4  3y 3  4y 4  3y
RELATION & FUNCTION
[MULTIPLE CORRECT CHOICE TYPE]

Q.18 Let f(x) = [x]2 + [x + 1] – 3, where [x] denotes greatest integer less than or equal to x, then which of the
following statement(s) is/are CORRECT?
(A) f(x) is many one function. (B) f(x) vanishes for atleast three values of x.
(C) f(x) is neither even nor odd function. (D) f(x) is aperiodic.
Q.19 Consider the function f (x) = x  1  x  , then which of the following is/are CORRECT?
(A) Range of f (x) is  1, 2 .  

(B) f is many one.
(C) f is either even or odd.
 
(D) Range of f (x) is identical to range of g (x) = 2 cos x   .
 4
Q.20 Let f : [– 1, 1] onto [3, 5] be a linear polynomial. Which of the following can be true?
 1  7  15  1  1   1
(A) f   = (B) f–1   = (C) f(0)  4 (D) f    f   = 8
 2  2 4 4  2  2 
Q.21 Which of the following pair of functions have the same graph?
[Note : [k], {k} and sgn k denote the largest integer less than or equal to k, fractional part of k
and signum function of k respectively.]
(A) f (x) = ln 1  {x}  and g (x) = ln  1  {x}  ,
1 1
(B) f (x) =  and g (x) = 2 sin2x + cos 2x
1  tan x 1  cot 2 x
2
2
 3 x  3)
(C) f (x) = 2sgn( x and g(x) = 2 sgn (x2 + 3x + 3)
sgn x |x|
(D) f (x) = and g (x) =
sgn x |x|
 
Q.22 Let f(x) = sgn(arc cot x) + tan  [ x ]  , where [x] is the greatest integer function less than or equal to x.
2 
Then which of the following alternatives is/are true?
(A) f (x) is many one but not even function (B) f(x) is periodic function
(C) f(x) is bounded function (D) Graph of f(x) remains above x-axis
1  sin x
Q.23 The values of x in [–2, 2], for which the graph of the function y = – secx and
1  sin x
1  sin x
y=– + secx, coincide are
1  sin x
 3   3   3     3 
(A) 2,     , 2  (B)   ,     , 
 2 2  2 2 2 2
     3 
(C)   ,  (D) [–2, 2] –  ,  
2 2  2 2
RELATION & FUNCTION
Q.24 Which of the following function(s) would represent a non singular mapping.
(A) f : R  R f (x) = | x | Sgn x where Sgn denotes Signum function
(B) g : R  R g (x) = x3/5
(C) h : R  R h (x) = x4 + 3x2 + 1
3x 2  7 x  6
(D) k : R  R k (x) =
x  x2  2
Q.25 Which of the following are identical functions?

(A) f (x) = sgn  x  1  (B) g (x) = sin2 (ln x) + cos2 (ln x)
(C) h (x) =
2


sin 1x  cos 1x  (D) k (x) = sec2 x  tan2  x 
(where [ x ] denotes greatest integer less than or equal x, {x} denotes fractional part of x and sgn x
denotes signum function of x respectively.)
[MATRIX TYPE]
Q.26 [Note : [k], {k} and sgn k denote the largest integer less than or equal to k, fractional part of k
and signum function of k respectively.]
COLUMN-I COLUMN-II
x 1 x 1
(A) f(x) =  . (P) Both many one and odd function
x 1 x 1
1 x4 
(B) f(x) = sgn  2  . (Q) Even and periodic function
 x 1
(C) f(x) = log1. 3 cos x . (R) Bounded function

(S) Range contains atleast one integer and
atmost three integers
Q.27 Column-I Column-II

2
x | x |
(A) Let f : [– 1, )  (0, ) defined by f (x) = e , then f (x) is (P) one-one
(B) Let f : (1, )  [3, ) defined by f (x) = 10  2 x  x 2 , then f (x) is (Q) into
(C) Let f : R  I defined by f (x) = tan5[x2 + 2x +3] (R) many one
where [ ] denotes greatest integer function, then f (x) is
(D) Let f : [3, 4]  [4, 6] defined by f (x) = x  1  x  2  x  3  x  4 (S) onto
then f (x) (T) periodic
RELATION & FUNCTION
Q.28 Column I contains functions and column II contains their natural domains. Exactly one entry of column II
matches with exactly one entry of column I.
Column I Column II
 x 1
(A) f (x) = sin 1   (P) (1, 3)  (3, )
 x 
 x 2  3x  2 
(B) g (x) = ln   (Q) (– , 2)
 x  1 
1  1
(C) h (x) = (R)   ,  
 1
x  2
ln  
 2 
(D) (x) = ln  x 2  12  2x  (S) [–3, –1)  [1, )

 
Q.29 The graph of the function y = f (x) is as follows.
Match the function mentioned in Column-I with the respective graph given in Column-II.
Column-I Column-II
y
1
–2 –1 1 2 x
(A) y = | f (x) | (P) O
–1
y
1
(B) y = f ( | x | ) (Q) –2 –1 O 1 2
x
–1
y
1
(C) y = f (– | x | ) (R) –2 –1 O 1 2
x
–1
y
1 1
(D) y = ( | f (x) | – f (x) ) (S) x

2 –2 –1 O 1 2
–1
RELATION & FUNCTION
EXERCISE-3
SECTION-A
(IIT JEE Previous Year's Questions)
Q.1(a) Suppose f(x) = (x + 1)2 for x  –1. If g(x) is the function whose graph is the reflection of the graph of f (x)
with respect to the line y = x, then g(x) equals
1
(A) – x – 1, x > 0 (B) , x > –1 (C) x  1 , x > –1 (D) x – 1, x > 0
(x  1) 2
(b) Let function f : R R be defined by f (x) = 2x + sinx for x  R. Then f is
(A) one to one and onto (B) one to one but NOT onto
(C) onto but NOT one to one (D) neither one to one nor onto
[JEE 2002 (Screening), 3 + 3]
x2  x  2
Q.2(a) Range of the function f (x) = is
x2  x 1
 7  7
(A) [1, 2] (B) [1,  ) (C) 2 ,  (D) 1, 
 3  3
x
(b) Let f (x) = defined from (0, )  [ 0, ) then by f (x) is
1 x
(A) one- one but not onto (B) one- one and onto
(C) Many one but not onto (D) Many one and onto [JEE 2003 (Scr),3+3]
Q.3 Let f (x) = sin x + cos x, g (x) = x2 – 1. Thus g ( f (x) ) is invertible for x 
          
(A)  , 0 (B)  ,  (C)  ,  (D) 0, 
 2   2   4 4  2
[JEE 2004 (Screening)]
Q.4 If the functions f (x) and g (x) are defined on R  R such that
0, x  rational 0, x  irrational
 
f (x) =  , g (x) = 
 x, x  irrational  x, x  rational
then (f – g)(x) is
(A) one-one and onto (B) neither one-one nor onto
(C) one-one but not onto (D) onto but not one-one [JEE 2005 (Scr.)]
Q.5 Let f (x) = x2 and g(x) = sin x for all x  R. Then the set of all x satisfying
(f o g o g o f ) (x) = (g o g o f) (x), where (f o g) (x) = f (g(x)), is [JEE 2011, 3]
(A)  n , n  {0, 1, 2, .......} (B)  n , n  {1, 2, .......}

(C) + 2n, n  {....... – 2, – 1, 0, 1, 2, .......} (D) 2n, n  {......., – 2, – 1, 0, 1, 2, .......}
2
Q.6 The function f : [0, 3]  [1, 29], defined by f(x) = 2x 3 – 15x2 + 36x + 1, is
(A) one-one and onto. (B) onto but not one-one.
(C) one-one but not onto. (D) neither one-one nor onto. [JEE 2012, 3]
RELATION & FUNCTION
2    
Q.7 Let f : (–1, 1)  R be such that f (cos 4) = 2 for    0,    ,  .
2  sec   4 4 2
1
Then the value(s) of f   is(are)
3
3 3 2 2
(A) 1  (B) 1  (C) 1  (D) 1 
2 2 3 3
[JEE 2012, 4]
SECTION-B
(AIEEE Previous Year's Questions)
Q.1 Let f : N  Y be a function defined as f(x) = 4x + 3 where Y = |y  N : y = 4x + 3 for some x  N|.

Show that f is invertible and its inverse is [AIEEE 2008]
y3 y3 y3 3y  4
(A) g(y) = 4 + (B) g(y) = (C) g(y) = (D) g(y) =
4 4 4 3
Q.2 For real x, let f(x) = x3 + 5x + 1, then - [AIEEE 2009]

(A) f is one – one but not onto R (B) f is onto R but not one – one
(C) f is one – one and onto R (D) f is neither one – one nor onto R
Q.3 Let f(x) = (x + 1)2 –1, x > –1 [AIEEE 2009]

Statement – 1 : The set {x : f(x) = f–1(x)} = {0, –1}.
Statement – 2 : f is a bijection.
(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.
Q.4 Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational
 m p 
number w} ; S =  ,  m, n, p and q are integers such that n, q  0 and qm = pm}. Then
 n q 
(A) R is an equivalence relation but S is not an equivalence relation [AIEEE 2010]
(B) Neither R nor S is an equivalence relation
(C) S is an equivalence relation but Ris not an equivalence relation
(D) R and S both are equivalence relations
Q.5 Let R be the set of real numbers. [AIEEE 2011]

Statement-1 : A = {(x, y)  R × R : y – x is an integer} is an equivalence relation on R.
Statement-2 : B = {(x, y)  R × R : x = y for some rational number } is an equivalence relation
on R.
(A) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of Statement-1.
(B) Statement-1 is true, Statement-2 is false and Statement-2 is not the correct explanation of Statement-1
(C) Statement-1 is true, Statement-2 is false
(D) Statement-1 is false, Statement-2 is true.
RELATION & FUNCTION
EXERCISE-4 (Rank Booster)
Q.1 The graph of the function y = g (x) is shown.

1
The number of solutions of the equation g ( x )  1  , is
2
(A) 4
(B) 5
(C) 6
(D) 8
Q.2 Let f : A  B and g : B  C be two functions and gof : A  C is defined. Then which of the following
statement(s) is true?
(A) If gof is onto then f must be onto.
(B) If f is into and g is onto then gof must be onto function.
(C) If gof is one-one then g is not necessarily one-one.
(D) If f is injective and g is surjective then gof must be bijective mapping.
Q.3 If f (x) = x2 + bx + c and f (2 + t) = f (2 – t) for all real numbers t, then which of the following is true?
(A) f (1) < f (2) < f (4) (B) f (2) < f (1) < f (4)
(C) f (2) < f (4) < f (1) (D) f (4) < f (2) < f (1)
1
 1 7
Q.4 If g(x) =  4 cos 4 x  2 cos 2 x  cos 4 x  x 7  , then the value of gg(100)  is equal to
 2 
(A) – 1 (B) 0 (C) 1 (D) 100
x 2  2x  a
Q.5 Let f : D  R be defined as f(x) = where D and R denote the domain of f and
x 2  4x  3a
the set of all real numbers respectively. If f is surjective mapping then the range of a is
(A) 0  a  1 (B) 0  a  1 (C) 0  a  1 (D) 0  a  1
Paragraph for question nos. 6 to 9

Let f (x) = x2 – 2x – 1  x  R. Let f : (– , a]  [b, ), where 'a' is the largest real number for which
f (x) is bijective.
Q.6 The value of (a + b) is equal to

(A) – 2 (B) – 1 (C) 0 (D) 1
Q.7 Let f : R  R, g (x) = f (x) + 3x – 1, then the least value of function y = g(| x |) is
(A) – 9/4 (B) – 5/4 (C) – 2 (D) – 1
Q.8 Let f : [a, )  [b, ), then f –1(x) is given by
(A) 1 + x2 (B) 1 – x3 (C) 1 – x2 (D) 1 + x3
Q.9 Let f : R  R, then range of values of k for which equation f (| x |) = k has 4 distinct real roots is
(A) (– 2, – 1) (B) (– 2, 0) (C) ( – 1, 0) (D) (0, 1)
RELATION & FUNCTION
Paragraph for question nos. 10 to 12

Consider the function y = f(x) = x2 – 6x + 5
Q.10 The domain of definition of the function g (x) = ln f ( x ) 

(A) (4, ) (B) (– , 1) (C) R – {x | 1  x  5} (D) R
Q.11 Range of the function y = ln f ( x )  is

(A) R (B) [1, ) (C) [0, ) (D) [– 4, )
Q.12 Number of points of intersection of a line y = c (where c is a fixed constant and 0 < c < 4)
with graphs y = f | x | is
(A) 2 (B) 8 (C) 6 (D) 4
Q.13 Which of the following statement(s) is(are) correct?

(A) If f is a one-one mapping from set A to A, then f is onto.
(B) If f is an onto mapping from set A to A, then f is one-one.
(C) Let f and g be two functions defined from R  R such that gof is injective, then f must be
injective.
(D) If set Acontains 3 elements while set B contains 2 elements, then total number of functions from
A to B is 8.
Q.14 Let f : R  R and g : R  R be mappings such that f (g(x)) is an injective mapping,

then which of the following statement(s) is(are) correct?
(A) g(x) must be an injective mapping.
(B) If g(x) is surjective then f(x) may not be injective mapping.
(C) If g(x) is not surjective, then f(x) may be injective or many-one mapping.
(D) f(x) may not be an injective mapping.
Q.15 If the function f (x) = ax + b has its own inverse then the ordered pair (a, b) can be
(A) (1, 0) (B) (–1, 0) (C) (–1, 1) (D) (1, 1)
RELATION & FUNCTION
Q.16 Column-I Column-II

(A) Let f : R R defined as (P) Odd
2
f (x) = esgn x  e x , where sgn x
denotes signum function of x, then f (x) is
(B) Let f : (–1, 1)  R defined as (Q) Even
4 1
f (x) = x [ x ] 
1 x2
where [ x ] denotes greatest integer less than
or equal to x, then f (x) is
(C) Let f : R R defined as (R) Neither odd nor even
x ( x  1)(x 4  1)  2x 4  x 2  2
f (x) =
x2  x 1
then f (x) is
(D) Let f : R R defined as (S) One-One
f (x) = x + 3x3 + 5x5 + ....... + 101x101
then f (x) is (T) Many-One
Q.17 Find the domains of definitions of the following functions :

(i) f (x) = ( x 2  3x  10) . ln 2 ( x  3)
(ii) f (x) = 
log1 / 3 log 4  [x ]
2
5 
1 1 1
(iii) f (x) = + log1 – {x}(x2 – 3x + 10) + +
[x ] 2| x| sec(sin x)
1
 7 
(iv) f (x) = (5x  6  x ) lnx +
2
(7 x  5  2x ) +  ln
2   x  
 2 
Q.18 Read the symbols [ ] and { } as greatest integer function less than or equal to x and fractional part function..
(i) Find the number of real values of x, satisfying the equation (x – 2)[x] = {x} – 1.
(ii) Find the number of solutions of the equation, x2  3x + [x] = 0 in the interval [0, 3].
(iii) If [x]2 + 3[x] – 10  0, then find the range of x.
(iv) If [x]2 – 5[x] + 6 = 0, then find the range of x.
RELATION & FUNCTION
Q.19 Consider the graph of a real-valued continuous function f(x) defined on R (the set of all real numbers)
as shown below.
y
5
(–2,4)
4 (4,4)
2
(–1,2) (2,2)
1
x
–5 –4 –3 –2 –1 O 1 2 3 4 5
–1
–2
Find the number of real solutions of the equation f (f(x)) = 4.
 x 2  3, x  5

x   5  x  1
Q.20 Let f(x) = (  7) | 1  x |  | 1  x |  1  x  1


x  6 1 x  5
 2
3  x x 5
f(x) is an odd function then find the value of ( + ).
Q.21 If f : R  R be an injective mapping and p, q, r are non-zero distinct real quantities satisfying
p  pq q r
f    f   and f    f   .
r  qr  r p
2
If the graph of g(x) = px + qx + r passes through M (1, 6) then find the value of q.
RELATION & FUNCTION
EXERCISE-1
 5   3     3 5    1
Q.1 (i)  4 , 4    4 , 4    4 , 4  (ii)   4 ,   (2, )(iii) (–  , – 3]
 2
     
 1   1 1 
(iii) (3  2 < x < 3  )  (3 < x  4) (iv)  0,  , 
 100   100 10 
(v) (1 < x < 1/2)  (x > 1) (vi) (3, 1]  {0}  [ 1,3 )
Q.2 (i) D : x R R : [0 , 2] (ii) D = R ; range [ –1 , 1 ]

(iii) D : {xx R ; x  3 ; x  2} R : {f(x)f(x)R , f(x)  1/5 ; f(x)  1}
(iv) D : R ; R : (–1, 1) (v) D : 1  x  2 R :  3, 6 
 1   1 1
(vi) D : [– 4, ) – {5}; R :  0,    , 
 6   6 3
1
if 0x 1
 x2
Q.3 g( x )   Q.4 (i), (iii) are identical Q.5 f (x) = {(x, 2)(y, 1)(z, 3)}
 2
 x if x 1
( x  3)10 1024
Q.6 1 Q.8 (i) , domain is R, (ii) f(x) = x2; g (x) = cos x; h (x) = x + 9
( x  3)10  1 1025
Q.9 4
EXERCISE-2
Q.1 B Q.2 C Q.3 A Q.4 D Q.5 D
Q.6 C Q.7 D Q.8 D Q.9 C Q.10 C
Q.11 C Q.12 D Q.13 B Q.14 C Q.15 B
Q.16 C Q.17 B Q.18 A, B, C, D Q.19 A, B
Q.20 A, B, D Q.21 A, C, D Q.22 A, B, C, D Q.23 A, C
Q.24 A, B Q.25 A, C, D
Q.26 (A) P, R, S ;(B) R, S ;(C) P, Q, R, S Q.27 (A) Q, R; (B) P, Q; (C) Q, R, T; (D) P, S
Q.28 (A) R; (B) S; (C) P; (D) Q Q.29 (A) S; (B) R; (C) P; (D) Q
EXERCISE-3
SECTION-A
Q.1 (a) D; (b) A Q.2 (a) D , (b) A Q.3 C Q.4 A Q.5 A
Q.6 B Q.7 Marks to all
SECTION-B
Q.1 C Q.2 C Q.3 B Q.4 C Q.5 C
RELATION & FUNCTION
EXERCISE-4
Q.1 D Q.2 C Q.3 B Q.4 D
Q.5 D Q.6 B Q.7 C Q.8 A
Q.9 A Q.10 C Q.11 A Q.12 B
Q.13 C, D Q.14 A, C, D Q.15 A, B, C
Q.16 (A) R, T ; (B) Q, T ; (C) R, T ; (D) P, S
Q.17 (i) { 4 }  [ 5, ) , (ii) [– 3,– 2)  [ 3,4), (iii) (–2, –1)  (–1, 0)  (1, 2) , (iv) (1, 2)  (2, 5/2)
Q.18 (i) infinite; (ii) 2; (iii) x  (– , –4) [2, ); (iv) x  [2, 4)
Q.19 7 Q.20 1 Q.21 8
RELATION & FUNCTION
Bansal Quick Review Table
Instruction to fill
(A) Write down the Question Number you are unable to solve in column A below, by Pen.
(B) After discussing the Questions written in column A with faculties, striks off them in the
manner so that you can see at the time of Revision also, to solve these questions again.
(C) Write down the Question Number you feel are important or good in the column B.
COLUMN : A COLUMN : B
EXERCISE NO. Question I am unable to Good / Important

solve in first attempt questions
Exercise # 1 (Subjective)
Exercise # 2 (Objective)
Exercise # 3 (Section-A)
Exercise # 3 (Section-B)
Exercise # 4 (Rank Booster)
Advantages
1. It is advised to the students that they should prepare a question bank for the revision as it is very
difficult to solve all the questions at the time of revision.
2. Using above index you can prepare and maintain the questions for your revision.

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7 inch

2. Exercise-1 (Subjective Questions) 39

3. Exercise-2 (Objective Questions) 40 – 44

6. Exercise-5 (Rank Booster) 47 – 50

RELATION & FUNCTION

1.1 Equality of ordered pairs :

1.2 Cartesian Product of two sets :

Ex. : A = {1, 2, 3} n(A) = 3

2.1 Total number of Relation from A to B

Note : If R1 : A  B such that a R1 b  a + b = 10 means R1 relates those element a  A to

2.2 Domain and Range of Relation:

2.3 Inverse of a Relation: Domain

2.4 Types of Relation :

(i) Identity Relation :

Reflexive Symmetric Transitive

(iii) Symmetric Relation :

(iv) Transitive relation :

(v) Equivalence Relation :

(v) Because (a, a)  R

3.1 General Definition :

3.2 Domain, Co-domain & Range of A Function :

(1) It should be noted that range is a subset of codomain .

3.3 Graphical Representation of function :

Rule for finding Domain :

3.4 IMPORTANT TYPES OF FUNCTIONS:

(ii) Algebraic Function:

(iii) Fractional Rational Function:

(iv) Exponential Function:

with power function. g (x) = x2 in which variable x is the base.

(v) Logarithmic function:

(vi) Absolute Value Function:

(vii) Signum Function:

(viii) Greatest Integer Or Step Up Function :

Properties of greatest integer function :

(ix) Fractional Part Function :

(a) Identity function :

(b) Constant function :

e.g. f (x) = [{x}]; g(x) = sin2x + cos2x; h (x) = sgn(x2 – 3x + 4) etc.

Domain and Range of important trigonometric functions :

Function (f) Domain (Df) Range (Rf)

add 2n to both sides

add n to both sides.

add 2n to both sides

(iv) ax2 + bx + c > 0  x  R iff a > 0 & D < 0

(ix) f(x) = ln (3x2 – 4x + 5) + 2 sin 2 x  5 sin x  2

(vii) f(x) = log4 log2 log1/2 (x)

(ix) f(x) = ln (3x2 – 4x + 5) + 2 sin 2 x  5 sin x  2

 log3 (log)1/2 (x2 + 4x + 4)  (– , )

(ii) f(x) = 4 tan x cos x

(iv) y = loge (3x2 – 4x + 5)

4. EQUAL OR IDENTICAL FUNCTION :

Two functions f & g are said to be equal if

(iii) f(x) = ln ex, g(x) = eln x

5.1 One-One Function (Injective Mapping) :

Examples: f : R  R, f (x) = x3 + 1 ; f (x) = e – x ; f (x) = ln x are all one-one functions.

Remember that a linear function is always one-one.

one-one function one-one function

5.2 Many-One Function (Not Injective) :