Topic Question Set


Q 41 :    

Define a relation R on the interval [0,π2) by xRy if and only if sec2xtan2y=1. Then R is :          [2025]

  • an equivalence relation.

     

  • reflexive but neither symmetric not transitive.

     

  • both reflexive and symmetric but not transitive.

     

  • both reflexive and transitive but not symmetric.

     

(1)

sec2xtan2x=1, x[0,π2)

   R is reflexive

Consider, sec2xtan2y=1

 1+tan2x(sec2y)1)=1

 1+tan2xsec2y+1=1  sec2xtan2x=1

   R is symmetric.

Now, if sec2xtan2y=1 and sec2ytan2z=1

Adding both equation, sec2xtan2y+sec2ytan2z=2

 sec2xtan2z=1        [ sec2ytan2y=1]

   R is transitive

Thus R is an equivalence relation.



Q 42 :    

Let S=N{0}. Define a relation R from S to R by:

R={(x,y):logey=xloge(25), xS, yR}

Then, the sum of all the elements in the range of R is equal to:          [2025]

  • 52

     

  • 53

     

  • 32

     

  • 109

     

(2)

We have, S=N{0}={0,1,2,3,......}

Also, logey=xloge(25)y=(25)x

   Required sum = (25)0+(25)1+(25)2+........

                              =1125=53.



Q 43 :    

The number of relation on the set A = {1, 2, 3}, containing at most 6 elements including (1, 2), which are reflexive and transitive but not symmetric, is __________.          [2025]



5

Given, A = {1, 2, 3}

Let the relation be R on A, which is reflexive and transitive but not symmetric, then

(1, 1), (2, 2), (3, 3), (1, 2)  R

Remaining elements are

(2, 1), (2, 3), (1, 3), (3, 1), (3, 2)

Case I : If relation contains exactly 4 elements  1 way

Case II : If relation contains exactly 5 elements, so we can add (1, 3) or (3, 2)  2 ways

Case III : If relation contains exactly 6 elements, so we can add (2, 3), (1, 3) or (1, 3), (3, 2) or (3, 1), (3, 2)  3 ways

   Total number of relations is 6.



Q 44 :    

Let A = {1, 2, 3}. The number of relations on A, containing (1, 2) and (2, 3), which are reflexive and transitive but not symmetric, is __________.          [2025]



3

For transitive : (1, 2) and (2, 3)  R  (1, 3)  R

For reflexive : (1, 1), (2, 2), (2, 3)  R

Now, for (2, 1), (3, 2), (3, 1); (3, 1) cannot be taken for not symmetric relation.

Case I : (2, 1) taken and (3, 2) not taken

Case II : (3, 2) taken and (2, 1) not taken

Case III : (2, 1) and (3, 2) are not taken

Therefore, 3 relations are possible.