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,nN,R is reflexive. R is not symmetric since for 3,6N, 3R66R3.

R is transitive since for n,m,r, whenever nm and mrnr, 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):aA}

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):ab}.

  • 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):ab}

          (a,a)R

    R is reflexive.

         (a,b), (b,c)R

                            ab and bc

                             ac

                      (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)×(BC)=

  • {(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}, BC={c,d}

then  (A-B)×(BC)={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:x25x+6=0}, B = {2, 4}, C = {4, 5} then A×(BC)

  • {(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)=0x=3, 2

A={3,2},  B={2,4},  C={4,5},  BC={4}

A×(BC)={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-1,0,1A

  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 x1. 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)Rx 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 ab stands for a divides b. For the elements of the given sets A and B, we find 26, 210, 33, 36, 510

  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)=

  •  

  •  

  • 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(AB)=3, then n[(A×B)(B×A)]=

  • 8

     

  • 9

     

  • 10

     

  • 11

     

(2)

Given n(A)=4, n(B)=5, n(AB)=3

n[(A×B)(B×A)]=n[(AB)×(BA)]

=n(AB)×n(BA)=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×q7=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