Answer : C
f(x,y) = ( x+y , x−y ) . for invertible function you have to find that there should be bijection (one to one correspondence) .
if f(a) = b then a = $f^{-1}$(b)
apply this concept here , f(x,y) = f(x+y , x-y) , so (x,y) = $f^{-1}$(x+y , x-y) -----------(1)
lets assume
p1 = x+y ------------(i)
p2 = x-y ------------(ii)
By Adding (i)+(ii) $\frac{(p1+ p2) }{2}$ = x
By Subtracting (i)-(ii) $\frac{(p1- p2) }{2}$ = y
put value in--(1)
$\left ( \frac{( p1+p2 )}{2} , \frac{( p1-p2 )}{2} \right )$ = $f^{-1}$( p1 , p2 )
$f^{-1}\left ( x,y \right )$ = $\left ( \frac{( x+y )}{2} , \frac{( x-y )}{2} \right )$