Topic Question Set


Q 21 :

Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is:

  • Reflexive and symmetric

     

  • Transitive and symmetric

     

  • Equivalence

     

  • Reflexive, transitive but not symmetric

     

(4)

Ans.      Reflexive, transitive but not symmetric

Explanation:

Since n divides n,∀n∈N,R is reflexive. R is not symmetric since for 3,6∈N, 3R6≠6R3.

R is transitive since for n,m,r, whenever nm and mr⇒nr, i.e., n divides m and m divides r, then n will divide r.



Q 22 :

For real numbers x and y, define xRy if and only if x-y+2 is an irrational number. Then the relation R is:

  • reflexive

     

  • symmetric

     

  • transitive

     

  • equivalence

     

(1)

Ans.      reflexive

Explanation:

Rx=x-x+2=2 is an irrational number.

Thus, R is reflexive.

Also, (2,1)∈R  as  2-1+2=22-1 is an irrational number, but (1,2)∉R  as  1-2+2=1 is a rational number. So R is not symmetric. Since, R21  and  R22  but 1 is not related to 2. So, R is not transitive.



Q 23 :

For the set A = {1, 2, 3}, define a relation R in the set A as follows:

            R = {(1, 1), (2, 2), (3, 3), (1, 3)}

Then, the ordered pair to be added to R to make it the smallest equivalence relation is:

  • (1, 3)

     

  • (3, 1)

     

  • (2, 1)

     

  • (1, 2)

     

(2)

Ans.    (3, 1)

Explanation:

(1,1), (2,2) and (3,3)∈R

⇒R is reflexive.

                            (1,3)∈R

                            (1,3)∈R but (3,1)∉R

⇒R is not symmetric.

To make R an equivalence relation, we can add (3, 1).

Clearly, R is reflexive and transitive. For R to be symmetric, we should add (3, 1) in R.



Q 24 :

Which one of the following is an identity relation?

  • (1, 2), (2, 3), (1, 3)

     

  • (5, 5), (4, 4), (2, 2)

     

  • (1, 3), (3, 1), (2, 3)

     

  • None of the above

     

(2)

Ans.   (5, 5), (4, 4), (2, 2)

Explanation:

A relation is called an identity relation if

A={(a,a):a∈A}

So, (5, 5), (4, 4), (2, 2) is an identity relation.



Q 25 :

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

  • reflexive

     

  • transitive

     

  • symmetric

     

  • none of these

     

(4)

Ans.     none of these

Explanation:

R on the set {1, 2, 3} be defined by  R = {(1, 2)}

It is clear that R is not reflexive, transitive and symmetric.



Q 26 :

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

Then R is:

  • reflexive but not symmetric

     

  • reflexive but not transitive

     

  • symmetric and transitive

     

  • neither symmetric nor transitive

     

(1)

Ans.     reflexive but not symmetric

Explanation:

Given that,

       A = {1, 2, 3}

and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

∵    (1, 1), (2, 2), (3, 3) ∈ R

Hence, R is reflexive.

         (1, 2) ∈ R but (2, 1) ∉ R

Hence, R is not symmetric.

         (1, 2) ∈ R and (2, 3) ∈ R

⇒   (1, 3) ∈ R

Hence, R is transitive.



Q 27 :

Relation R on real numbers is defined as R={(a,b):a≤b}.

  • Reflexive and symmetric but not transitive

     

  • Symmetric and transitive but not reflexive

     

  • Reflexive and transitive but not symmetric

     

  • Equivalence relation

     

(3)

Ans.        Reflexive and transitive but not symmetric

Explanation:

                 R={(a,b):a≤b}

          (a,a)∈R

∵    R is reflexive.

         (a,b), (b,c)∈R

                            a≤b and b≤c

                             a≤c

                      (a,c)∈R

Therefore, R is transitive.

                       (2,4)∈R

                        (4,2)∉R

Therefore R is not symmetric.



Q 28 :

If A={a,b,c}, B={b,c,d} and C={a,d,c}, then (A-B)×(B∩C)=

  • {(a,c),(a,d)}

     

  • {(a,b),(c,d)}

     

  • {(c,a),(d,a)}

     

  • {(a,c),(a,d),(b,d)}

     

(1)

If A={a,b,c}, B={b,c,d} and C={a,d,c},

A-B={a}, B∩C={c,d}

then  (A-B)×(B∩C)={a}×{c,d}={(a,c),(a,d)}



Q 29 :

Let Y = {1, 2, 3, 4, 5}, A = {1, 2}, B = {3, 4, 5} and ϕ denotes null set. If (A × B) denotes Cartesian product of the sets A and B, then (Y × A) ∩ (Y × B) is

  • Y

     

  • A

     

  • B

     

  • ϕ

     

(4)

Y×A={(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(4,1),(4,2),(5,1),(5,2)}

Y×B={(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),(4,3),(4,4),(4,5),(5,3),(5,4),(5,5)}

∴  (Y×A)∩(Y×B)=ϕ



Q 30 :

Let A={x:x2−5x+6=0}, B = {2, 4}, C = {4, 5} then A×(B∩C)

  • {(2, 4), (3, 4)}

     

  • {(4, 2), (4, 3)}

     

  • {(2, 4), (3, 4), (4, 4)}

     

  • {(2, 2), (3, 3), (4, 4), (5, 5)}

     

(1)

A={x:x2-5x+6=0}

x2-5x+6=0

⇒(x-3)(x-2)=0⇒x=3, 2

A={3,2},  B={2,4},  C={4,5},  B∩C={4}

A×(B∩C)={3,2}×{4}={(3,4),(2,4)}



Q 31 :

The cartesian product A×A has 9 elements among which two elements are found (−1, 0) and (0, 1), then set A = ?

  • {1, 0}

     

  • {1, −1, 0}

     

  • {0, −1}

     

  • {1, −1}

     

(2)

∵  A×A has 9 elements

∴  Set A has exactly 3 elements.

Also (-1,0),(0,1)∈A×A⇒-1,0,1∈A

∴  A={-1,0,1}



Q 32 :

If A and B be two sets such that A×B consists of 6 elements. If three elements of A×B are (1, 4), (2, 6) and (3, 6), find B×A.

  • {(1, 4), (1, 6), (2, 4), (2, 6), (3, 4), (3, 6)}

     

  • {(4, 1), (4, 2), (4, 3), (6, 1), (6, 2), (6, 3)}

     

  • {(4, 4), (6, 6)}

     

  • {(4, 1), (6, 2), (6, 3)}

     

(2)

Since, (1, 4), (2, 6) and (3, 6) are the elements of A × B, therefore 1, 2, 3 are the elements of A and 4, 6 are the elements of B. Also, A × B has 6 elements.

∴  A = {1, 2, 3} and B = {4, 6}

∴  B × A = {(4, 1), (4, 2), (4, 3), (6, 1), (6, 2), (6, 3)}



Q 33 :

If A and B have n elements in common, then the number of elements common to A × B and B × A is

  • 0

     

  • n

     

  • 2n

     

  • n2

     

(4)

 



Q 34 :

A set A has 5 elements. Then the maximum number of relations on A (including empty relation) is

  • 5

     

  • 25

     

  • 225

     

  • 25

     

(3)

 



Q 35 :

The function f satisfies the functional equation 3f(x)+2f(x+59x-1)=10x+30 for all real x≠1. The value of f(7) is

  • 8

     

  • 4

     

  • - 8

     

  • 11

     

(2)

Given functional equation is

3f(x)+2f(x+59x-1)=10x+30                             ...(i)

Put x=x+59x-1, we get

      3f(x+59x-1)+2f[x+59x-1+59x+59x-1-1]=10(x+59x-1)+30

⇒ 3f(x+59x-1)+2f(x)=40x+560x-1                        ...(ii)

Multiply (i) by 3 and (ii) by 2 and then subtract, we get

9f(x)-4f(x)=(30x+90)-(80x+1120x-1)

⇒ 5f(x)=10[(3x+9)-(8x+112x-1)]

⇒ f(x)=2[(3x+9)-(8x+112x-1)]

Putting x=7, we get

f(7)=2[30-56+1126]=2(30-28)=4



Q 36 :

If f(x)=log(1+x1-x), -1<x<1, then f(3x+x31+3x2)-f(2x1+x2) is

  • [f(x)]3

     

  • [f(x)]2

     

  • -f(x)

     

  • f(x)

     

(4)

f(x)=log(1+x1-x)

f(3x+x31+3x2)-f(2x1+x2)

=log(1+3x+x31+3x21-3x+x31+3x2)-log(1+2x1+x21-2x1+x2)

=log(1+3x2+3x+x31+3x2-3x-x3)-log(1+x2+2x1+x2-2x)

=log(1+x1-x)3-log(1+x1-x)2

=log((1+x)3×(1-x)2(1-x)3×(1+x)2)=log(1+x1-x)=f(x)



Q 37 :

Let A = {1, 2, 3} and B = {2, 3, 4}, then which of the following relations is a function from A to B?

  • {(1, 2), (2, 3), (3, 4), (2, 2)}

     

  • {(1, 2), (2, 3), (1, 3)}

     

  • {(1, 3), (2, 3), (3, 3)}

     

  • {(1, 1), (2, 3), (3, 4)}

     

(3)

A relation is a function from set A to set B, if every element in A maps to one and only one element of B. So, relation {(1, 3), (2, 3), (3, 3)} is a function from A to B.



Q 38 :

If a relation R is defined from a set A = {2, 3, 4, 5} to a set B = {3, 6, 7, 10} as follows (x,y)∈R⇔x divides y. Expression of R-1 is represented by

  • {(6, 2), (10, 2), (3, 3), (6, 3)}

     

  • {(6, 2), (3, 3), (10, 5), (10, 2)}

     

  • {(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)}

     

  • None of these

     

(3)

Recall that a∣b stands for a divides b. For the elements of the given sets A and B, we find 2∣6, 2∣10, 3∣3, 3∣6, 5∣10

⇒  R={(2,6),(2,10),(3,3),(3,6),(5,10)}

∴  R-1={(6,2),(10,2),(3,3),(6,3),(10,5)}



Q 39 :

If n(A)=4, n(B)=3, n(A×B×C)=24, then n(C)=

  • 2 

     

  • 1 

     

  • 12 

     

  • 17

     

(1)

n(A×B×C)=n(A)×n(B)×n(C)

∴  n(C)=n(A×B×C)n(A)×n(B)⇒n(C)=244×3=2

∴  n(C)=2



Q 40 :

If n(A) denotes the number of elements in set A and if n(A)=4, n(B)=5 and n(A∩B)=3, then n[(A×B)∩(B×A)]=

  • 8

     

  • 9

     

  • 10

     

  • 11

     

(2)

Given n(A)=4, n(B)=5, n(A∩B)=3

n[(A×B)∩(B×A)]=n[(A∩B)×(B∩A)]

=n(A∩B)×n(B∩A)=3×3=9   [∵ n(A×B)=n(A)×n(B)]



Q 41 :

Let the number of elements of the sets A and B be p and q respectively. Then the number of the relations from the set A to the set B is

  • 2p+q

     

  • 2pq

     

  • p+q

     

  • pq

     

(2)

Number of possible relations from A to B=2pq



Q 42 :

If R is a relation on a finite set having n elements, then the number of relations on A is

  • 2n

     

  • 2n2

     

  • n2

     

  • nn

     

(2)

Number of elements in set A=n

Number of elements in A×A=n×n=n2

∵  Relations R on A are subsets of A×A.

∴  Number of relations =2n2



Q 43 :

Suppose that the number of elements in set A is p, the number of elements in set B is q and the number of elements in A×B is 7 then p2+q2= ________.

  • 50

     

  • 42

     

  • 51

     

  • 49

     

(1)

We have, n(A)=p, n(B)=q

n(A×B)=p×q⇒7=pq

⇒Either p=1, q=7 or q=1, p=7

∴  p2+q2=12+72=50



Q 44 :

If n(A)=5 and n(B)=7, then the number of relations on A×B is 2k, where unit digit of k is ________.



(5)

Total number of relations from A to B is 2mn.

Here, m=5, n=7.

∴  Number of relations=25×7=235



Q 45 :

Consider the relation R on the set {–2,–1,0,1,2} defined by (a,b)∈R if and only if 1 + ab > 0. Then, among the statements:

I. The number of elements in R is 17

II. R is an equivalence relation          [2026]

  • Only I is true

     

  • Only II is true

     

  • Both I and II are true

     

  • Neither I nor II is true

     

(1)

R on set {–2,–1,0,1,2}, if 1 + ab > 0

R = {(–2, –2), (–1, –1), (0, 0), (1, 1), (2, 2), (–2, –1), (1, 2), (–2, 0), (–1, 0), (1, 0), (2, 0), (–1, –2), (2, 1), (0, –2), (0, –1), (0, 1), (0, 2)}

Number of elements = 17

Clearly, 'R' is reflexive and symmetric but not transitive,

Hence, not an equivalence relation.

∴  Only statement I is true.



Q 46 :

Let A = {2, 3, 4, 5, 6}. Let R be a relation on the set A x A given by (x, y) R (z, w) if and only if x divides z and y ≤ w. Then the number of elements in R is ________.          [2026]



(120)

We have, A = {2, 3, 4, 5, 6}

We have, y ≤ w

∙  If y = 2, then w ∈ {2, 3, 4, 5, 6}, Number of pairs = 5

∙  If y = 3, then w ∈ {3, 4, 5, 6}, Number of pairs = 4

∙  If y = 5, then w ∈ {4, 5, 6}, Number of pairs = 3

∙  If y = 5, then w ∈ {5, 6}, Number of pairs = 2

∙  If y = 6, then w ∈ {6}, Number of pairs = 1

    Total pairs 5 + 4 + 3 + 2 + 1 = 15

    Now, x divides z

∴  If x = 2, then z ∈ {2, 4, 6}, Number of pairs = 3

    If x = 3, then z ∈ {3, 6}, Number of pairs = 2

    If x = 4, then z ∈ {4}, Number of pairs = 1

    If x = 5, then z ∈ {5}, Number of pairs = 1

    If x = 6, then z ∈ {6}, Number of pairs = 1

∴    Total pairs = 3 + 2 + 1 + 1 + 1 = 8

  Total elements = 15 x 8 = 120.



Q 47 :

Let A = {1,4, 7} and B = {2, 3, 8}. Then the number of elements, in the relation R={((a1,b1),(a2b2))∈((A×B)×(A×B)) : a1+b2 divides a2+b1} is ________.          [2026]



(18)

Given, A = {1, 4,  7} and B = {2, 3,8}

All possible sums:

A/B 2 3 8
1 3 4 9
4 6 7 12
7 9 10 15

 

a1+b2 a2+b1 Number of pairs
15 15 1
12 12 1
10 10 1
9 9 4
7 7 1
6 6, 12 2
4 4, 12 2
3 3, 6, 9 12, 15 6

∴  Number of relations = 18



Q 48 :

Let R={(x,y)∈N×N:loge(x+y)≤2}. Then the minimum number of elements, required to be added in R to make it a transitive relation, is ________.          [2026]



(15)

Given : loge(x+y)≤2

⇒ x+y≤e2

Since, e2≈7.29, so x+y≤7

∴ R = {(1, 1), (1, 2), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1) (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (6, 1)}

For R to be transitive, we have to add (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (5, 3), (5, 4), (5, 5), (5, 6), (4, 4), (4,5), (4, 6), (3, 5),(3, 6), (2, 3) i.e., 15 elements



Q 49 :

Let R = {(a,b)|a−b is irrational; a, b are real numbers}, then relation R is

  • reflexive and symmetric relation

     

  • transitive and symmetric

     

  • symmetric relation

     

  • equivalence relation

     

(3)

(a,a)∉R

If (a,b)∈R⇒(b,a)∈R

If (a,b)∈R, (b,c)∈R⇒(a,c)∉R



Q 50 :

Let R be a relation on real numbers given by R={(a,b):3a-3b+7 is an irrational number}. Then R is

  • Reflexive but neither symmetric nor transitive

     

  • Reflexive and transitive but not symmetric

     

  • Reflexive and symmetric but not transitive

     

  • An equivalence relation

     

(1)

Check for reflexivity:    As 3(a-a)+7=7 which belongs to relation

So relation is reflexive.

Check for symmetric

Take a=73, b=0 Now (a,b)∈R but (b,a)∉R

As 3(b-a)+7=0 which is rational so relation is not symmetric

Check for Transitivity,    Take (a,b) as (73,1) & (b,c) as (1,273)

So now (a,b)∈R & (b,c)∈R but (a,c)∉R,

Which means relation is not transitive