Correct Answer (A)
$1.\ f(A_{1}\cup A_{2})=f(A_{1})\cup f(A_{2})$
$f(A_{1}\cup A_{2})=\{y\in B|y=f(x),x\in A_{1}\cup A_{2}\}=\{y\in B|y=f(x),x\in A_{1}\ or\ x\in A_{2}\}=\{y\in B|y=f(x),x\in A_{1}\}\cup \{y\in B|y=f(x),x\in A_{2}\}=f(A_{1})\cup f(A_{2}).$
$2.\ f(A_{1}\cap A_{2})\neq f(A_{1})\cap f(A_{2})$
$A=\{1,2\},B=\{3,4\}$ and $f=\{(1,3),(2,3)\}$
$A_{1}=\{1\},A_{2}=\{2\}$
$f(A_{1}\cap A_{2})=f(\phi)=\phi$ while $f(A_{1})\cap f(A_{2})=\{3\}\cap \{3\}=\{3\}$
Note :
$f(A_{1}\cap A_{2})\subseteq f(A_{1})\cap f(A_{2})$
$f(A_{1}\cap A_{2})=f(A_{1})\cap f(A_{2})$ if f is ONE-ONE