If `X and Y` are two non-empty sets where `f: X->Y`,is function is defined such that`f(c) = {f (x): x in C}` for `C sube X and f^-1 (D) = {x: f(x) in D}` for `D sube Y`,for any `A sube Y and B sube Y`, thenA. `f^(-1)(f(A))=A`B. `f^(-1)(f(A))=A" only if f "(X)=Y`C. `f(f^(-1)(B))=B" only if B "subseteqf(X)`D. `f(f^(-1)(B))=B`

If `X and Y` are two non-empty sets where `f: X->Y`,is function is def

1.	If `X and Y` are two non-empty sets where `f: X->Y`,is function is defined such that`f(c) = {f (x): x in C}` for `C sube X and f^-1 (D) = {x: f(x) in D}` for `D sube Y`,for any `A sube Y and B sube Y`, thenA. `f^(-1)(f(A))=A`B. `f^(-1)(f(A))=A" only if f "(X)=Y`C. `f(f^(-1)(B))=B" only if B "subseteqf(X)`D. `f(f^(-1)(B))=B`
Answer» Correct Answer - C We have `A subseteq f^(-1)(f(A)) and A=f^(-1)(f(A))` iff f is a bijection. So, options (a) and (b) are not correct `f(f^(-1)(B)) subseteq B and f(f^(-1)(B))=B" iff "B subseteqf(X)` So, options ( c) is correct but options (d) is not correct.