16

Relations and Functions

Mathematics
Functions
M-251

Relations and
Functions

Numerical
Q.1 The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3
is ____________.
31st Aug Evening Shift 2021

Q.2

27th Aug Morning Shift 2021

Q.3 Let S = {1, 2, 3, 4, 5, 6, 7}. Then the number of possible functions f : S → S

such that f(m . n) = f(m) . f(n) for every m, n ∈ S and m . n ∈ S is equal to _____________.
27th Jul Morning Shift 2021

Q.4 Let A = {n ∈ N | n2 ≤ n + 10,000}, B = {3k + 1 | k∈ N} an dC = {2k | k∈N}, then the

sum of all the elements of the set A ∩∩(B − C) is equal to _____________.
27th Jul Evening Shift 2021

Q.5 Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions f : A → A

such that f(1) + f(2) = 3 − f(3) is equal to
22th Jul Evening Shift 2021

Q.6 If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x3) + x
g(x3) is divisible by x2 + x + 1, then P(1) is equal to ___________.
18th Mar Evening Shift 2021
Q.7

24th Feb Evening Slot 2021

Q.8

24th Feb Morning Slot 2021

Numerical Answer Key

1. Ans. (5143)
2. Ans. (256)
3. Ans. (490)
4. Ans. (832)
5. Ans. (720)
6. Ans. (0)
7. Ans. (2)
8. Ans. (5)

Numerical Explanation
Ans. 1 A = 4-digit numbers divisible by 3

A = 1002, 1005, ....., 9999.

9999 = 1002 + (n − 1)3

⇒ (n − 1)3 = 8997 ⇒ n = 3000

B = 4-digit numbers divisible by 7

B = 1001, 1008, ......., 9996

⇒ 9996 = 1001 + (n −− 1)7

⇒ n = 1286

A ∩ B = 1008, 1029, ....., 9996

9996 = 1008 + (n − 1)21

⇒ n = 429

So, no divisible by either 3 or 7

= 3000 + 1286 − 429 = 3857

total 4-digits numbers = 9000

required numbers = 9000 − 3857 = 5143

Ans. 2
Ans. 3
Ans. 4
Ans. 5

Ans. 6 Given, p(x) = f(x3) + xg(x3)

Ans. 7
Ans. 8 3 digit number of the form 9K + 2 are {101, 109, .............992}
 MCQ (Single Correct Answer)

Q.1 The range of the function,

1st Sep Evening Shift 2021

Q.2 Let f : N → N be a function such that f(m + n) = f(m) + f(n) for every m, n∈N. If
f(6) = 18, then f(2) . f(3) is equal to :

31st Aug Evening Shift 2021

Q.3 The domain of the function

31st Aug Evening Shift 2021

Q.4 Which of the following is not correct for relation R on the set of real numbers ?

31st Aug Morning Shift 2021

Q.5

26th Aug Evening Shift 2021

Q.6

26th Aug Evening Shift 2021

Q.7 Out of all patients in a hospital 89% are found to be suffering from heart ailment
and 98% are suffering from lungs infection. If K% of them are suffering from both
ailments, then K can not belong to the set :

26th Aug Morning Shift 2021

Q.8 Consider function f : A → B and g : B → C (A, B, C ⊆ R) such that (gof)−−1 exists,

then :
25th Jul Evening Shift 2021

Q.9 If [x] be the greatest integer less than or equal to x,

25th Jul Evening Shift 2021

Q.10 Let N be the set of natural numbers and a relation R on N be defined by

27th Jul Evening Shift 2021

Q.11
27th Jul Evening Shift 2021

Q.12 Let g : N → N be defined as

g(3n + 1) = 3n + 2,

g(3n + 2) = 3n + 3,

g(3n + 3) = 3n + 1, for all n ≥ 0.

Then which of the following statements is true?

25th Jul Morning Shift 2021

Q.13
22th Jul Evening Shift 2021

Q.14 The number of solutions of sin7x + cos7x = 1, x∈ [0, 4ππ] is equal to

22th Jul Evening Shift 2021

Q.15

22th Jul Evening Shift 2021

Q.16
20th Jul Evening Shift 2021

Q.17

20th Jul Morning Shift 2021

Q.18
18th Mar Evening Shift 2021

Q.19

18th Mar Morning Shift 2021

Q.20

18th Mar Morning Shift 2021

Q.21 Consider the function f : R → R defined by

17th Mar Evening Shift 2021

Q.22 In a school, there are three types of games to be played. Some of the students
play two types of games, but none play all the three games. Which Venn diagrams
can justify the above statement?

17th Mar Morning Shift 2021

Q.23
17th Mar Morning Shift 2021

Q.24 Let A = {2, 3, 4, 5, ....., 30} and '≃' be an equivalence relation on A × A, defined
by (a, b) ≃ (c, d), if and only if ad = bc. Then the number of ordered pairs which
satisfy this equivalence relation with ordered pair (4, 3) is equal to :

16th Mar Evening Shift 2021

Q.25 Let f be a real valued function, defined on R − {−1, 1} and given by

Then in which of the following intervals, function f(x) is increasing?

16th Mar Evening Shift 2021

Q.26 The range of a∈R for which the

16th Mar Morning Shift 2021

Q.27 Let [ x ] denote greatest integer less than or equal to x. If for n∈N,

16th Mar Morning Shift 2021

Q.28 The number of elements in the set {x ∈ R : (|x| − 3) |x + 4| = 6} is equal to :

16th Mar Morning Shift 2021

Q.29

26th Feb Evening Shift 2021

Q.30 Let R = {(P, Q) | P and Q are at the same distance from the origin} be a relation,
then the equivalence class of (1, −1) is the set :
26th Feb Morning Shift 2021

Q.31 Let x denote the total number of one-one functions from a set A with 3
elements to a set B with 5 elements and y denote the total number of one-one
functions form the set A to the set A × B. Then :

25th Feb Evening Shift 2021

Q.32

25th Feb Evening Shift 2021

Q.33 Let f, g : N → N such that f(n + 1) = f(n) + f(1) ∀ n∈N and g be any arbitrary
function. Which of the following statements is NOT true?

25th Feb Morning Slot 2021

Q.34

24th Feb Morning Slot 2021

MCQ Answer Key

1. Ans. (D)
2. Ans. (B)
3. Ans. (C)
4. Ans. (B)
5. Ans. (D)
6. Ans. (A)
7. Ans. (C)
8. Ans. (C)
9. Ans. (B)
10. Ans. (B)
11. Ans. (C)
12. Ans. (A)
13. Ans. (A)
14. Ans. (C)
15. Ans. (D)
16. Ans. (B)
17. Ans. (C)
18. Ans. (B)
19. Ans. (C)
20. Ans. (B)
21. Ans. (A)
22. Ans. (B)
23. Ans. (B)
24. Ans. (D)
25. Ans. (A)
26. Ans. (C)
27. Ans. (D)
28. Ans. (B)
29. Ans. (B)
30. Ans. (B)
31. Ans. (C)
32. Ans. (A)
33. Ans. (A)
34. Ans. (A)

MCQ Explanation

Ans 1.
Ans 2.
Ans 3.

Ans 4. Note that (a, b) and (b, c) satisfy 0 < |x − y| ≤ 1 but (a, c) does not satisfy it so
0 ≤ |x − y| ≤ 1 is symmetric but not transitive.

For example,

x = 0.2, y = 0.9, z = 1.5

0 ≤ |x – y| = 0.7 ≤ 1

0 ≤ |y – z| = 0.6 ≤ 1
But |x – z| = 1.3 > 1
So, (b) is correct.
Ans 5.

Ans 6. min{x − [x], 1 − x + [x]}

h(x) = min{x − [x], 1 − [x − [x])}

⇒ always continuous in [−2, 2] but not differentiable at 7 points.

Ans 7.
Ans 8.

Ans 9.

Ans 10.
Ans 11.

Ans 12.
Ans 13.

Ans 14.
Ans 15.
Ans 16.
Ans 17. For domain,
Ans 18. Finding inverse of f(x)
Ans 19.
Ans 20.
Ans 21.

Ans 22. As none play all three games the intersection of all three circles must be
zero.

Hence none of P, Q, R justify the given statement

Ans 23.
Ans 24.

Ans 25.
Ans 26.

Ans 27.
Ans 28.
Ans 29.

f(1) = 2

f(2) = 2

f(3) = 4

f(4) = 4
f(5) = 6

f(6) = 6

f(7) = 8

f(8) = 8

f(9) = 10

f(10) = 10

∴ f(1) = f(2) = 2

f(3) = f(4) = 4

f(5) = f(6) = 6

f(7) = f(8) = 8

f(9) = f(10) = 10

Given, g(f(x)) = f(x)

when x = 1, g(f(1)) = f(1) ⇒ g(2) = 2

when, x = 2, g(f(2)) = f(2) ⇒ g(2) = 2

∴ x = 1, 2, g(2) = 2

Similarly, at x = 3, 4, g(4) = 4

at x = 5, 6, g(6) = 6

at x = 7, 8, g(8) = 8

at x = 9, 10, g(10) = 10
Here, you can see for even terms mapping is fixed. But far odd terms 1, 3, 5, 7, 9 we
can map to any one of the 10 elements.

∴ For 1, number of functions = 10

For 3, number of functions = 10

for 9, number of functions = 10

∴ Total number of functions = 10 × 10 × 10 × 10 × 10 = 105

Ans 30. Given R = {(P, Q) | P and Q are at the same distance from the origin}.

Then equivalence class of (1, −1) will contain al such points which lies on
circumference of the circle of centre at origin and passing through point (1, −1).
Ans 31. Number of elements in A = 3

Number of elements in B = 5

Number of elements in A × B = 15

Number of one-one function

x=5×4×3

x = 60

Number of one-one function

y = 15 × 14 × 13
Ans 32.
Ans 33.

Ans 34.
∵ Any horizontal line does not cut the graph at more than one points, so it is one-one
and here, co-domain and range are not equal, so it is into.

Hence, the required function is one-one into.

 16
TOPIC Ć
Types of Relations, Inverse of a
Relation, Mappings, Mapping of
(a)
(b)
not inective
neither inective nor surective
Functions, Kinds of Mapping of (c) surective but not inective
Functions (d) inective but not surective
1. Let A = {a, b, c} and B = {1, 2, 3, 4}. Then the number of é 1 1ù x
6. The function f : R ® ê - , ú defined as f(x) = , is :
elements in the set C = { f : A ® B | 2 Î f ( A) and f is not ë 2 2 û 1 + x2
one-one} is ___________. [NA Sep. 05, 2020 (II)] [2017]
f : ( 0, ¥ ) ® ( 0, ¥ ) be defined by
(a) neither inective nor surective
2. Let a function
(b) invertible
1 (c) inective but not surective
f ( x) = 1- . Then f is : [Jan. 11, 2019 (II)]
x (d) surective but not inective
(a) not inective but it is surective éx ù
7. The function f : N ® N defined by f ( x ) = x - 5 ê ú , where
(b) inective only ë5û
(c) neither inective nor surective N is set of natural numbers and [x] denotes the greatest
(d) both inective as well as surective integer less than or equal to x, is :
3. The number of functions f from {1, 2, 3, ..., 20} onto [Online April 9, 2017]
{1, 2, 3, ...., 20} such that f (k) is a multiple of 3, whenever k (a) one-one and onto.
is a multiple of 4 is : [Jan. 11, 2019 (II)]
(b) one-one but not onto.
(a) 65 × (15)! (b) 5! × 6!
(c) onto but not one-one.
(c) (15)! × 6! (d) 56 × 15
(d) neither one-one nor onto.
4. Let N be the set of natural numbers and two functions
8. Let A = {x1, x2, ........., x7} and B = {y1, y2, y3} be two sets
f and g be defined as f, g : N ® N such that
containing seven and three distinct elements respectively.
ì n +1 Then the total number of functions f : A ® B that are
ïï if n is odd
onto, if there exist exactly three elements x in A such that
f (n) = í 2
ï n f(x) = y2, is equal to : (Online April 11, 2015)
if n is even
ïî 2
(a) 14.7C3 (b) 16.7C3 (c) 14.7C2 (d) 12.7C2
n
and g(n) = n – (– 1) . Then fog is: [Jan. 10, 2019 (II)] x -1
(a) onto but not one-one. 9. Let f : R ® R be defined by f(x) = x + 1 then f is:
(b) one-one but not onto.
(c) both one-one and onto. [Online April 19, 2014]
(d) neither one-one nor onto. (a) both one-one and onto
5. Let A = {xÎR : x is not a positive integer}. Define a func- (b) one-one but not onto
(c) onto but not one-one
2x
tion f: A ® R as f(x) = , then f is:[Jan. 09, 2019 (II)] (d) neither one-one nor onto.
x -1
 EBD_8344
M-252 Mathematics

10. Let P be the relation defined on the set of all real numbers 17. Consider the following relations:
such that R = {(x, y) | x, y are real numbers and x = wy for some
P = {(a, b) : sec2a – tan2b = 1}. Then P is: æ m pö
[Online April 9, 2014] rational number w}; S = {ç , ÷ | m,n, p and q are
è n qø
(a) reflexive and symmetric but not transitive.
integers such that n, q ¹ 0 and qm = pn}.
(b) reflexive and transitive but not symmetric.
Then [2010]
(c) symmetric and transitive but not reflexive.
(a) Neither R nor S is an equivalence relation
(d) an equivalence relation.
(b) S is an equivalence relation but R is not an equivalence
11. Let R = {(x, y) : x, y Î N and x2 – 4xy + 3y2 = 0}, where N relation
is the set of all natural numbers. Then the relation R is : (c) R and S both are equivalence relations
[Online April 23, 2013] (d) R is an equivalence relation but S is not an equivalence
(a) reflexive but neither symmetric nor transitive. relation
(b) symmetric and transitive. 18. Let R be the real line. Consider the following subsets of
(c) reflexive and symmetric, the plane R × R:
(d) reflexive and transitive. S ={(x, y): y = x + 1 and 0 < x < 2}
12. Let R = {(3, 3) (5, 5), (9, 9), (12, 12), (5, 12), (3, 9), (3, 12), (3, 5)} T ={(x, y): x – y is an integer},
be a relation on the set A = {3, 5, 9, 12}. Then, R is : Which one of the following is true? [2008]
[Online April 22, 2013] (a) Neither S nor T is an equivalence relation on R
(a) reflexive, symmetric but not transitive. (b) Both S and T are equivalence relation on R
(b) symmetric, transitive but not reflexive. (c) S is an equivalence relation on R but T is not
(c) an equivalence relation. (d) T is an equivalence relation on R but S is not
(d) reflexive, transitive but not symmetric. 19. Let W denote the words in the English dictionary. Define
the relation R by R = {(x, y) Î W × W| the words x and y
13. Let A = {1, 2, 3, 4} and R : A ® A be the relation defined have at least one letter in common.} Then R is [2006]
by R = {(l, 1), (2, 3), (3, 4), (4, 2)}. The correct statement is : (a) not reflexive, symmetric and transitive
[Online April 9, 2013] (b) relexive, symmetric and not transitive
(a) R does not have an inverse. (c) reflexive, symmetric and transitive
(b) R is not a one to one function. (d) reflexive, not symmetric and transitive
(c) R is an onto function. 20. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9),
(d) R is not a function. (3, 12), (3, 6)} be a relation on the set
14. If P(S) denotes the set of all subsets of a given set S, then A = {3, 6, 9, 12}. The relation is [2005]
the number of one-to-one functions from the set (a) reflexive and transitive only
S = {1, 2, 3} to the set P(S) is [Online May 19, 2012] (b) reflexive only
(a) 24 (b) 8 (c) 336 (d) 320 (c) an equivalence relation
(d) reflexive and symmetric only
15. If A = {x Î z + : x < 10 and x is a multiple of 3 or 4}, where
21. Let f : (– 1, 1) ® B, be a function defined by
z+ is the set of positive integers, then the total number of
symmetric relations on A is [Online May 12, 2012] -1 2x
f (x) = tan , then f is both one - one and onto when
(a) 25 (b) 215 (c) 210 (d) 220 1 - x2
16. Let R be the set of real numbers. [2011] B is the interval [2005]
Statement-1: A = {(x, y) Î R × R : y – x is an integer} is an æ pö é pö
equivalence relation on R. (a) ç 0, ÷ (b) ê0, ÷
è 2ø ë 2ø
Statement-2: B = {(x, y) Î R × R : x = ay for some rational
number a} is an equivalence relation on R. æ p pö
(c) éê- p , p ùú (d) ç - , ÷
(a) Statement-1 is true, Statement-2 is true; Statement-2 is ë 2 2û è 2 2ø
not a correct explanation for Statement-1. 22. Let R = {(1,3), (4, 2), (2, 4), (2, 3), (3,1)} be a relation on the
(b) Statement-1 is true, Statement-2 is false.
set A = {1, 2,3, 4}. . The relation R is [2004]
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is true; Statement-2 is (a) reflexive (b) transitive
a correct explanation for Statement-1. (c) not symmetric (d) a function
Relations and Functions M-253

23. If f : R ® S , defined by f ( x) = sin x - 3 cos x + 1, is æ 3ö 1- x2

28. For x Î ç 0, ÷ , let f(x) = x , g(x) = tan x and h(x) = .
onto, then the interval of S is [2004] è 2ø 1 + x2
(a) [ –1, 3] (b) [–1, 1] (c) [ 0, 1] (d) [0, 3] æpö
If f (x) = ((hof)og) (x), then f ç ÷ is equal to :
24. A function f from the set of natural numbers to integers è3ø
[April 12, 2019 (I)]
defined by [2003]
p 11p 7p 5p
(a) tan (b) tan (c) tan (d) tan
ìn -1 12 12 12 12
ïï 2 , when n is odd 29. Let f ( x) = x 2 , x Î R . For any A Í R , define g(A) =
f (n) = í is
ï - n , when n is even {x Î R : f ( x) Î A} . If S = [0, 4], then which one of the
ïî 2
following statements is not true ? [April 10, 2019 (I)]
(a) neither one -one nor onto (a) g (f (S)) ¹ S (b) f (g (S)) = S
(b) one-one but not onto (c) g (f (S)) = g (S) (d) f (g (S)) ¹ f (S)
(c) onto but not one-one 1
30. For xÎ R – {0, 1}, let f1 (x) = , f (x) = 1 – x and
(d) one-one and onto both. x 2
1
f3 (x) = be three given functions. If a function, J(x)
Composite Functions & Relations, 1- x
TOPIC n Inverse of a Function, Binary satisfies (f2oJof1) (x) = f3(x) then J(x) is equal to:
Operations
[Jan. 09, 2019 (I)]
1
(a) f3 (x) (b) f (x) (c) f2 (x) (d) f1 (x)
82 x - 8-2 x x 3
25. The inverse function of f ( x) = 2 x -2 x , x Î(-1,1), is
8 +8 31. Let N denote the set of all natural numbers. Define two
________. [Jan. 8, 2020 (I)] binary relations on N as R1 = {(x, y) Î N × N : 2x + y = 10}
and R2 = {(x, y) Î N × N : x + 2y = 10}. Then
1 æ 1+ x ö 1 æ 1- x ö [Online April 16, 2018]
loge ç (log8 e)log e ç
(a)
4 è 1 - x ÷ø (b)
4 è 1 + x ÷ø (a) Both R1 and R2 are transitive relations
(b) Both R1 and R2 are symmetric relations
1 æ 1- x ö 1 æ 1+ x ö (c) Range of R2 is {1, 2, 3, 4}
loge ç (log8 e)log e ç
(c)
4 è 1 + x ÷ø (d)
4 è 1 - x ÷ø (d) Range of R1 is {2, 4, 8}
32. Consider the following two binary relations on the set
A = {a, b, c} : R1 = {(c, a) (b, b) , (a, c), (c, c), (b, c), (a, a)}
æ 5ö and R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c). Then
26. If g(x) = x2 + x – 1 and (gof) (x) = 4x2 – 10x + 5, then f çè ÷ø [Online April 15, 2018]
4
(a) R2 is symmetric but it is not transitive
is equal to: [Jan. 7, 2020 (I)]
(b) Both R1 and R2 are transitive
3 1 1 3 (c) Both R1 and R2 are not symmetric
(a) (b) - (c) (d) - (d) R1 is not symmetric but it is transitive
2 2 2 2
27. For a suitably chosen real constant a, let a function, x -1
33. Let f : A ® B be a function defined as f (x) = , where
x-2
a-x
f : R –{– a}® R be defined by f (x) = . Further sup- A = R – {2} and B = R – {1}. Then f is
a+x
[Online April 15, 2018]
pose that for any real number x ¹ – a and f (x) ¹ – a,
2y +1
(a) invertible and f –1 (y) =
æ 1ö y -1
( fof ) (x) = x. Then f ç - ÷ is equal to:
è 2ø 3y - 1
(b) invertible and f –1 (y) =
[Sep. 06, 2020 (II)] y -1
(c) no invertible
1 1
(a) (b) - (c) – 3 (d) 3 2y -1
3 3 (d) invertible and f –1 (y) =
y -1
 EBD_8344
M-254 Mathematics

34. Let f (x) = 210·x + 1 and g(x) = 310·x – 1. If (fog)(x)=x, then 38. Let f be a function defined by
x is equal to : [Online April 8, 2017]
f ( x) = ( x -1) +1, ( x ³1) .
2
[2011RS]
310 - 1 210 - 1
{ }
(a) (b)
310 - 2 -10 210 - 3-10 Statement - 1 : The set x : f ( x ) = f ( x ) = {1, 2} .
-1

-10 -10
1- 3 1- 2 Statement - 2 : f is a biection and
(c) (d)
210 - 3-10 310 - 2 -10
f -1 ( x ) = 1 + x - 1, x ³ 1.
1
35. For x Î R, x ¹ 0 , let f0(x) = and fn+1 (x) = f0(fn(x)), (a) Statement-1 is true, Statement-2 is true; Statement-2 is
1- x a correct explanation for Statement-1.
æ 2ö æ 3ö (b) Statement-1 is true, Statement-2 is true; Statement-2 is
n = 0, 1, 2, .... Then the value of f100(3) + f1 çè ÷ø + f 2 çè ÷ø is NOT a correct explanation for Statement-1.
3 2
equal to : [Online April 9, 2016] (c) Statement-1 is true, Statement-2 is false.
8 4 5 1 (d) Statement-1 is false, Statement-2 is true.
(a) (b) (c) (d)
3 3 3 3 39. Let f(x) = ( x + 1)2 –1, x ³ –1
1
36. If g is the inverse of a function f and f ' ( x ) = , then Statement -1 : The set {x : f(x) = f –1(x) = {0, –1}
1+ x 5
Statement-2 : f is a biection. [2009]
g ¢ ( x ) is equal to: [2014] (a) Statement-1 is true, Statement-2 is true. Statement-2 is
not a correct explanation for Statement-1.
1
(b) 1 + { g ( x )}
5 (b) Statement-1 is true, Statement-2 is false.
(a)
1 + { g ( x )}
5
(c) Statement-1 is false, Statement-2 is true.
(c) 1 + x5 (d) 5x4 (d) Statement-1 is true, Statement-2 is true. Statement-2 is
37. Let A and B be non empty sets in R and f : A ® B is a a correct explanation for Statement-1.
biective function. [Online May 26, 2012] 40. Let f: N®Y be a function defined as f(x) = 4x + 3 where
Statement 1: f is an onto function. Y = {y Î N : y = 4x + 3 for some x Î N}.
Statement 2: There exists a function g : B ® A such that Show that f is invertible and its inverse is [2008]
fog = IB.
3y + 4 y+3
(a) Statement 1 is true, Statement 2 is false. (a) g ( y ) = (b) g ( y ) = 4 +
(b) Statement 1 is true, Statement 2 is true; Statement 2 is 3 4
a correct explanation for Statement 1. y+3 y –3
(c) Statement 1 is false, Statement 2 is true. (c) g ( y ) = (d) g ( y ) =
4 4
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
not the correct explanation for Statement 1.
Relations and Functions M-255

1. (19.00)
ì1, n =1
The desired functions will contain either one element or ï1, n=2
two elements in its codomain of which '2' always belongs ï
to f (A). ï 2, n=3
\ The set B can be {2}, {1, 2}, {2, 3}, {2, 4} ï
ï 2, n=4
Total number of functions = 1 + (23 – 2)3 = 19. ï
í3, n=5
2. (Bonus) f : (0, ฀) ® (0, ฀) ï3, n=6
f(g(n)) = ï
1 ï: :
f (x) = 1 – is not a function ï: :
x ï
ïî: :
Q f (1) = 0 and 1 Î domain but 0 Ï co-domain
Þ fog is onto but not one - one
Hence, f (x) is not a function.
5. (d) As A = {x Î R : x is not a positive integer}
3. (c) Domain and codomain = {1, 2, 3, ¼, 20}.
2x
There are five multiple of 4 as 4, 8, 12, 16 and 20. A function f : A ® R given by f(x) =
x -1
and there are 6 multiple of 3 as 3, 6, 9, 12, 15, 18. f(x1) = f(x2) Û x1 = x2
Since, when ever k is multiple of 4 then f(k) is multiple of 3 So, f is one-one.
then total number of arrangement As f(x) ¹ 2 for any x Î A Þ f is not onto.
Hence f is inective but not surective.
= 6c5 × 5! = 6!
é 1 1ù
Remaining 15 elements can be arranged in 15! ways. 6. (d) We have f : R ® ê - , ú ,
ë 2 2û
Since, for every input, there is an output
x
Þ function f (k) in onto f (x) = "x Î R
1 + x2
\ Total number of arrangement = 15! . 6!
(1 + x 2 ).1 - x.2x -(x + 1)(x - 1)
Þ f ¢(x) = 2 2
=
ì n +1 (1 + x ) (1 + x 2 ) 2
ïï 2 , if n is odd – + –
4. (a) f(n) = íï n
, if n is even
ïî 2 x = –1 x=1
sign of f ¢ (x)
Þ f ¢ (x) changes sign in different intervals.
ì 2, n = 1 \ Not inective
ï1, n = 2
ï x
ï 4, n = 3 Now y =
í 1 + x2
g(n) = ï3, n = 4 Þ y + yx2 = x
ï6, n = 5
ï Þ yx2 – x + y = 0
î5, n = 6 For y ¹ 0, D = 1 – 4y2 ³ 0
Then, é -1 1 ù
Þ y Î ê , ú - {0}
ë 2 2û
ì g ( n) + 1
ïï 2 , if g ( n) is odd For y = 0 Þ x = 0
f(g(n)) = íï g (n) é -1 1 ù
, if g ( n) is even \ Range is ê , ú
ïî 2 ë 2 2û
Þ Surective but not inective
 EBD_8344
M-256 Mathematics

11. (d) R = {(x, y) : x, y Î N and x2 – 4xy + 3y2 = 0}

7. (d) f (1) = 1 - 5 [1 5] = 1 üï
ý ® Many one Now, x2 – 4xy + 3y2 = 0
f ( 6 ) = 6 - 5 [6 5] = 1ïþ Þ (x – y) (x – 3y) = 0
f (10) = 10 – 5(2) = 0 which is not in co–domain. \ x = y or x = 3y
Neither one–one nor onto. \ R = {(1, 1), (3, 1), (2, 2), (6, 2), (3, 3),
(9, 3),......}
8. (a) Number of onto function such that exactly three
1 Since (1, 1), (2, 2), (3, 3),...... are present in the relation,
elements in x Î A such that f(x) = is equal to therefore R is reflexive.
2
= 7C3, {24 – 2} = 14. 7C3 Since (3, 1) is an element of R but (1, 3) is not the element of
R, therefore R is not symmetric
x -1
9. (c) f ( x) = Here (3, 1) Î R and (1, 1) Î R Þ (3, 1) Î R
x +1
(6, 2) Î R and (2, 2) Î R Þ (6, 2) Î R
for one-one function if f (x1) = f (x2) then For all such (a, b) Î R and (b, c) Î R
x1 must be equal to x2 Þ (a, c) Î R
Let f (x1) = f (x2) Hence R is transitive.
x1 - 1 x2 - 1 12. (d) Let R = {(3, 3), (5, 5), (9, 9), (12, 12), (5, 12), (3, 9), (3, 12),
=
x1 + 1 x2 + 1 (3, 5)} be a relation on set
A = {3, 5, 9, 12}
x1 x2 + x1 - x2 - 1 = x1 x2 - x1 + x2 - 1 Clearly, every element of A is related to itself.
Þ x1 - x2 = x2 - x1 Therefore, it is a reflexive.
Now, R is not symmetry because 3 is related to 5 but 5 is
2 x1 = 2 x2
not related to 3.
x1 = x2 Also R is transitive relation because it satisfies the property
x1 = x2 , x1 = – x2 that if a R b and b R c then a R c.
here x1 has two values therefore function is many one 13. (c) Domain = {1, 2, 3, 4}
function. Range = {1, 2, 3, 4}
x -1 \ Domain = Range
For onto : f ( x ) = Hence the relation R is onto function.
x +1
14. (c) Let S = {1, 2, 3} Þ n(S) = 3
for every value of f (x) there is a value of x in domain
set. Now, P (S) = set of all subsets of S
If f (x) is negative then x = 0 total no. of subsets = 23 = 8
for all positive value of f (x), domain contain atleast one \ n[P(S)] = 8
element. Hence f (x) is onto function. Now, number of one-to-one functions from S ® P(S) is

10. (d) P = {( a, b) : sec 2 a - tan 2 b = 1} 8!

8P =
3 = 8 × 7 × 6 = 336.
5!
For reflexive :
15. (b) A relation on a set A is said to be symmetric iff
sec 2 a - tan 2 a = 1 (true " a)
(a, b) Î A Þ (b, a) Î A, " a, b Î A
For symmetric :
sec2 b – tan2 a = 1 Here A = {3, 4, 6,8,9}
L.H.S Number of order pairs of A ´ A = 5 ´ 5 = 25
1 + tan 2 b - (sec 2 a - 1) = 1 + tan 2 b - sec 2 a + 1 Divide 25 order pairs of A × A in 3 parts as follows :
= – (sec2 a – tan2b) + 2 Part – A : (3, 3), (4, 4), (6, 6), (8, 8), (9, 9)
= – 1 +2 = 1 Par t – B : (3, 4), (3, 6), (3, 8), (3, 9), (4, 6),
So, Relation is symmetric (4, 8),(4, 9), (6, 8), (6, 9), (8, 9)
For transitive : Part – C : (4, 3), (6, 3), (8, 3),(9, 3), (6, 4), (8, 4), (9, 4),
if sec2 a – tan2 b = 1 and sec2 b – tan2 c = 1 (8, 6), (9, 6), (9, 8)
sec2 a – tan2 c = (1 + tan2 b) – (sec2 b – 1) In part – A, both components of each order pair are same.
= –sec2b + tan2b + 2 In part – B, both components are different but not two
=–1+2=1 such order pairs are present in which first component of
one order pair is the second component of another order
So, Relation is transitive.
pair and vice-versa.
Hence, Relation P is an equivalence relation
Relations and Functions M-257

In part–C, only reverse of the order pairs of part –B are So, S is not reflexive.
present i.e., if (a, b) is present in part – B, then (b, a) will be Hence, S in not an equivalence relation.
present in part –C Given T ={x, y): x – y is an integer}
For example (3, 4) is present in part – B and (4, 3) present in Q x – x = 0 is an integer, " x Î R
part –C. So, T is reflexive.
Number of order pair in A, B and C are 5, 10 and 10 Let (x, y) Î T Þ x – y is an integer then
respectively. y – x is also an integer Þ (y, x) Î R
In any symmetric relation on set A, if any order pair of part \T is symmetric
–B is present then its reverse order pair of
If x – y is an integer and y – z is an integer then
part –C will must be also present.
(x – y) + (y– z) = x – z is also an integer.
Hence number of symmetric relation on set A is equal to
\ T is transitive
the number of all relations on a set D, which contains all
Hence T is an equivalence relation.
the order pairs of part –A and part– B.
Now n(D) = n(A) + n(B) = 5 + 10 = 15 19. (b) Clearly ( x, x) Î R, "x ÎW
Hence number of all relations on set D = (2)15 Q All letter are common in some word. So R is reflexive.
Þ Number of symmetric relations on set D = (2)15 Let ( x, y ) Î R , then ( y , x) Î R as x and y have at least one
16. (a) Q x – x = 0 Î I (\ R is reflexive) letter in common. So, R is symmetric.
Let (x, y) Î R as x – y and y – x Î I (Q R is symmetric) But R is not transitive for example
Now x – y Î I and y – z Î I Þ x – z Î I Let x = BOY, y = TOY and z = THREE
So, R is transative.
then ( x, y ) Î R (O, Y are common) and (y, z) Î R (T is
Hence R is equivalence.
common) but (x, z) Ï R. (as no letter is common)
Similarly as x = ay for a = 1. B is reflexive symmetric and
20. (a) R is reflexive and transitive only.
transative. Hence B is equivalence.
Here (3, 3), (6, 6), (9, 9), (12, 12) Î R [So, reflexive]
Both relations are equivalence but not the correct
explanation. (3, 6), (6, 12), (3, 12) Î R [So, transitive].
17. (b) Let x R y. Q (3, 6) Î R but (6, 3) Ï R [So, non-symmetric]

x æ 2x ö
Þx = wy Þy= 21. (d) Given f (x) = tan -1 ç = 2tan–1x
w è 1 - x 2 ÷ø
Þ (y, x) Ï R
R is not symmetric for x Î (-1, 1)
m p æ -p p ö
Let S : S If x Î( -1, 1) Þ tan -1 x Î ç
è 4 4 ÷ø
,
n q
p m
Þ qm = pn Þ =
æ -p p ö
q n Þ 2 tan -1 x Î ç
è 2 2 ÷ø
,
m m
Q = \ reflexive.
n n æ p pö
Clearly, range of f (x) = ç - , ÷
m p p m è 2 2ø
= Þ q = n \ symmetric
n q
For f to be onto, codomain = range
m p p r
Let S , S æ p pö
n q q s \ Co-domain of function = B = ç - , ÷ .
è 2 2ø
Þ qm = pn, ps = rq
22. (c) Q (1, 1) Ï R Þ R is not reflexive
p m r
Þ = = Q (2, 3) Î R but (3, 2) Ï R
q n s \ R is not symmetric
Þ ms = rn transitive. Q (4, 2) and (2, 4) Î R but (4, 4) Ï R
S is an equivalence relation. Þ R is not transitive
18. (d) Given that
23. (a) Given that f ( x) is onto
S = {(x , y) : y = x + 1 and 0 < x < 2}
Q x ¹ x + 1 for any x Î(0, 2) \ range of f (x) = codomain = S
Þ (x, x) Ï S
 EBD_8344
M-258 Mathematics

26. (b) (gof) (x) = g(f(x)) = f 2(x) + f(x) – 1

Now, f (x) = sin x - 3cos x + 1
2
æ pö æ æ 5 öö æ5ö 5 5
= 2sin ç x - ÷ + 1 g ç f ç ÷ ÷ = 4 ç ÷ - 10. + 5 = -
è 3ø è è øø
4 è ø
4 4 4
[Q g(f(x)) = 4x2 – 10x +5]
æ pö
we know that -1 £ sin ç x - ÷ £ 1
è 3ø æ æ 5 öö æ5ö æ5ö
g ç f ç ÷ ÷ = f 2 ç ÷ + f ç ÷ -1
è è øø
4 è ø
4 è4ø
æ pö
-1 £ 2sin ç x - ÷ + 1 £ 3 \ f ( x) Î[ -1, 3] = S
è 3ø 5 æ5ö æ5ö
- = f 2ç ÷+ f ç ÷ -1
4 è4ø è4ø
24. (d) We have f : N ® I
Let x and y are two even natural numbers, æ5ö æ5ö 1
f 2ç ÷+ f ç ÷+ =0
-x - y è ø
4 è4ø 4
and f ( x) = f ( y ) Þ = Þx=y
2 2 2
æ æ5ö 1ö
\ f (n) is one-one for even natural number. ç f ç ÷+ ÷ =0
Let x and y are two odd natural numbers and è è4ø 2ø

x -1 y - 1 æ5ö
f ( x) = f ( y) Þ
1
= Þx=y tç ÷ =-
2 2 è ø
4 2
\ f (n) is one-one for odd natural number.
Hence f is one-one. æ a - xö
a-ç
è a + x ÷ø
n –1 27. (d) f ( f ( x )) = =x
Let y = Þ 2 y +1 = n æ a - xö
a+ç
è a + x ÷ø
2
This shows that n is always odd number for y Î I.
..........(i) a - ax a(1 - x ) a - x
Þ = f ( x) Þ = Þ a =1
–n 1+ x 1+ x a+ x
and y = Þ –2 y = n
2 1- x æ 1ö
This shows that n is always even number for y Î I. \ f ( x) = Þ f ç- ÷ = 3
1+ x è 2ø
..........(ii)
28. (b) Q f (x) = (( hof ) og)(x)
From (i) and (ii)
Range of f = I = codomain æpö æ æ æ p öö ö
\ f is onto. Q f ç 3 ÷ = h ç f ç g ç 3 ÷ ÷ ÷ = h( f ( 3)) = h(31/4 )
è ø è è è øøø
Hence f is one one and onto both.
1- 3 1
= = - (1 + 3 - 2 3) = 3 - 2 = -(- 3 + 2)
82 x - 8-2 x 1+ 3 2
25. (a) y = 2x
8 + 8-2 x
æ pö 11p
= – tan 15 = tan (180 – 15 ) = tan ç p - ÷ = tan
1+ y 8 2x 1+ y è 12 ø 12
= Þ 84 x =
1 - y 8-2 x 1- y 29. (c) f (x) = x2 ; x Î R
g (A) = {x Î R : f (x) Î A} S = [0, 4]
æ1+ y ö g (S) = {x Î R : f (x) Î S}
Þ 4 x = log8 ç ÷
è1- y ø = {x Î R : 0 £ x2 £ 4} = {x Î R : – 2 £ x £ 2}
\ g (S) ¹ S \ f (g (S)) ¹ f (S)
1 æ 1+ y ö g (f (S)) = {x Î R : f (x) Î f (S)}
Þ x = log 8 ç ÷
4 è1- y ø = {x Î R : x2 Î S2} = {x Î R : 0 £ x2 £ 16}
= {x Î R : – 4 £ x £ 4}
1 æ1+ x ö
\ f -1( x) = log8 ç ÷ \ g (f (S)) ¹ g (S)
4 è 1- x ø
\ g (f (S)) = g (S) is incorrect.
Relations and Functions M-259

30. (a) The given relation is Now, for R2 , (b, a) Î R2, (a, c) Î R2 but (b, c) Ï R2.
1 Similarly, for R1 , (b, c) Î R1, (c, a) Î R1 but (b, a) Ï R1.
(f2o Jo f1) (x) = f3(x) = Therefore, neither R1 nor R2 is transitive.
1- x
33. (d) Suppose y = f (x)
1
Þ (f2oJ) (f1(x)) = x -1
1- x Þ y=
x-2
1
Þ yx – 2y = x – 1
1 x
= Þ (y – 1) x = 2y – 1
æ 1ö 1 1 é 1ù
Þ (f2o J) çè ÷ø = 1 - 1 -1 êëQ f1 ( x ) = x úû 2y -1
x x
Þ x = f –1 (y) =
x y -1
As the function is invertible on the given domain and its
x é1 ù
Þ (f2o J)(x) = êë x is replaced by x úû inverse can be obtained as above.
x -1 34. (d) f (g(x)) = x
x Þ f (310x – 1) = 210 (310 . x – 1) + 1 = x
Þ f2 (J(x)) =
x -1 Þ 210 (310x – 1) + 1 = x
Þ x (610 – 1) = 210 – 1
x
Þ 1 – J(x) = [Q f2(x) = 1 – x]
x -1 210 - 1 1- 2-10
Þ x= =
x 1 610 - 1 310 - 2-10
\ J(x) = 1 - = = f 3 ( x)
x -1 1- x 1 x -1
31. (c) Here, 35. (c) f1 (x) = f0 + 1 (x) = f0 (f0 (x)) = =
1 x
R1 = {(x, y) Î N × N : 2x + y = 10} and 1-
1- x
R2 = {(x, y) Î N × N : x + 2y = 10}
For R1; 2x + y = 10 and x, y Î N 1
f2 (x) = f1 + 1 (x) = f0 (f1 (x)) = =x
So, possible values for x and y are: x -1
1-
x = 1, y = 8 i.e. (1, 8); x
x = 2, y = 6 i.e. (2, 6);
1
x = 3, y = 4 i.e. (3, 4) and f3 (x) = f2 + 1 (x) = f0 (f2 (x)) = f0 (x) =
1- x
x = 4, y = 2 i.e. (4, 2).
R1 = {(1, 8), (2, 6), (3, 4), (4, 2)} x -1
f4 (x) = f3 + 1 (x) = f0 (f3 (x)) =
Therefore, Range of R1 is {2, 4, 6, 8} x
R1 is not symmetric
1
Also, R1 is not transitive because (3, 4), (4, 2) Î R1 but \ f0 = f3 = f6 = .......... =
(3, 2) Ï R1 1- x
Thus, options A, B and D are incorrect. x -1
For R2; x + 2y = 10 and x, y Î N f1 = f4 = f7 = .......... =
x
So, possible values for x and y are: f2 = f5 = f8 = .......... = x
x = 8, y = 1 i.e. (8, 1);
x = 6, y = 2 i.e. (6, 2); 2
3 -1 2 æ 2 ö 3 - 1 1
x = 4, y = 3 i.e. (4, 3) and f100 (3) = = f1 ç ÷ = =-
x = 2, y = 4 i.e. (2, 4) 3 3 è 3ø 2 2
R2 = {(8, 1), (6, 2), (4, 3), (2, 4)} 3
Therefore, Range of R2 is {1, 2, 3, 4}
æ 3ö 3
R2 is not symmetric and transitive. f2 ç ÷ =
è 2ø 2
32. (a) Both R1 and R2 are symmetric as
For any (x, y) Î R1, we have æ 3ö
æ 2ö 5
(y, x) Î R1 and similarly for R2 \ f100 (3) + f1 ç ÷ + f2 ç ÷=
è 3ø è ø
2 3
 EBD_8344
M-260 Mathematics

36. (b) Since f (x) and g(x) are inverse of each other Now f ( x ) = f -1 ( x )

Þ f ( x ) = x Þ ( x - 1) + 1 = x
1 2
\ g'( f (x)) =
f '( x )
Þ x 2 - 3 x + 2 = 0 Þ x = 1, 2
æ 1 ö Hence statement-1 is correct
Þ g '( f ( x)) = 1 + x5 çèQ f ¢ ( x) = ÷
1 + x5 ø 39. (d) Given that f (x) = (x + 1)2 –1, x ³ –1
Here x = g(y) Clearly Df = [–1, ¥ ) but co-domain is not given. Therefore
f (x) is onto.
g ¢( y ) = 1 + [ g ( y )]
5
\ Let f (x1) = f (x2)
Þ (x1 + 1)2 – 1 = (x2 + 1)2 – 1
Þ g ¢ ( x ) = 1 + ( g ( x) ) 5 Þ x1 = x2
37. (d) Let A and B be non-empty sets in R. \ f (x) is one-one, hence f (x) is biection
Let f : A ® B is biective function. Q (x + 1) being something +ve, " x > –1
\ f –1(x) will exist.
Clearly statement - 1 is true which says that f is an onto
Let (x + 1)2 –1 = y
function.
Statement - 2 is also true statement but it is not the Þ x +1 = y +1 (+ve square root as x +1 ³ 0 )
correct explanation for statement-1
Þ x =–1+ y +1
38. (a) Given f is a biective function
Þ f –1 (x) = x +1 –1
\ f :[1, ¥) ® [1, ¥)
Then f (x) = f –1 (x)
f ( x ) = ( x - 1) + 1, x ³ 1
2
Þ (x + 1)2 – 1 = x +1 –1

Let y = ( x - 1) + 1 Þ ( x - 1) = y - 1
2 2 Þ (x + 1)2 = x + 1 Þ (x + 1)4 = (x + 1)
Þ (x + 1) [ (x + 1)3 – 1] = 0 Þ x = – 1, 0
Þ x = 1± y - 1 Þ f -1 ( y ) = 1 ± y - 1 \ The statement-1 and statement-2 both are true.
40. (d) Clearly f (x) = 4x + 3 is one one and onto, so it is
Þ f -1 ( x ) = 1 + x - 1 {\ x ³ 1} invertible.
Hence statement-2 is correct Let f (x) = 4x + 3 = y
y –3 y –3
Þx= \ g ( y) =
4 4