@Chirag_Hegde bro we r sure that d is correct but for c we have to check if its one one , as such elements r different so we r sure that some of X or some of Y may not be image or coimage properly. Function f is defined properly but only one info is given for its inverse which is not saying for all elements its one one
Yeah bro, but y has a preimage in X right?
And b is inside y
And b has a preimage inside c
Yeah @Chirag_Hegde bro but for only b belongs to 4,2,3 it's giving the result , but for 5, ist not giving although b has 5,6 but f(inverse) don't exist so its not necessarily true for every element in y
By y I mean the subset of Y which is defined in the question as f(c)=y
And every element in y has a preimage under c in X
Yeah but for inverse , in question it's taking subset from Y not from y @Chirag_Hegde bro
Option C me small y hi to hai
Oops yeah my apologies , didn't read that b belongs to subset y in c option , its should be both c and d @Chirag_Hegde bro
A is not a function it will violate definition of function
2 g(x) nhi hona chaiye ek x ke liye.
Are ha right
Why it will be d ?? @Chirag_Hegde
@Chirag_Hegde its C
My bad didn't see options
For the very first question D will always be true however C may or may not be always true
Think about the case when codomain is not equal to the range
But dor domain the function is always defined
But here we have defined a subset y in Y which definitely had a preimage in X
And b is again the subset of this subset, so naturally even b should have a preimage in X?