DM-Sets,Relations and Functions-Q6

Question

Let be one to one function.

S1:

S2:

Which of the following is TRUE?

(A). Only S1

(B). Only S2

(C), Both S1 and S2

(D). Neither S1 nor S2

Answer 1

Both S1 and S2 are true. Please note that this statements are true for any function. Function need not be one-one. 

Comments

sir please give me one example in S2 in which LHS would be a proper subset of RHS and a function need to one to one

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We can not get such example for one-one function as  for one to one function f(S1 intersection S2) = f(S1) intersection f(S2).

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so the equation becomes wrong na at every time it has to be equal but in ques the equation says it can be proper subset of RHS

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Equation can not be wrong. Try to understand the definition of subset OR equal.
For A subset or equal to B to be true A =B is sufficient.

Answer 2

A should be the answer S2 should be true when it is not a one to one mapping but there is a one to one mapping so S2 equations would like LHS=RHS

Comments

provide a counter example for S2

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for which case when it onto or one to one

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when function is one to one then
suppose 1->a
2-> b
3-> c
then two subsets 12 and 23 then  f(12) give a,b and for f(23) give b,c intersection of both gives b only now when f(12 intersection 23) it gives f(2) it will gives b only it is true .
now when function is not one to one
a-> 1
b->2
c-> 2
now consider two subsets ab and c
then do ab intersection c it is phie f(phie)=phie
and if we do f(ab)intersection f(c) then 12 intersection 2 it would come 2 now equation be like f(A interection B)<= f(A)intersection f(B)
correct me if i am wrong

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so phi ⊆ {2}
this satisfies

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but the function which i had taken that was not onto i was saying if it was not a one to one function then S2 would be true

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S and T mean what here... nothing is given about them. And i think it is similar to a gate question in which S2 also had equal to operator not the subset operator

Answer 3

I think C is correct.

Answer 4

Let's assume  but doesn't exist in 

and also f(x) = y

So if y   then x  

but we are also stating that 

this is only possible when x 

So this is a contradiction

Hence we can say that S2 is true.

Note: Notice that y can only exist because of x because f is one to one

 

Comments

now see my answer

Answer 5

see this i have provided a counter example clearly functions is not one to one 

Comments

This is not a counter example

Provide an example where function is one to one and S2 condition is not satisfied
Then only we can say say S2 is false

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only i wanted to say in S2 LHS=RHS
can you provide me any counter example in which function is one to one and
LHS is a proper subset of RHS as it is written as LHS  can bv proper as well improper suset in the question.