The correct option is C surjective but not injective
We have, f(x)=x1+x2
∴ f(1x)=1x1+1x2=x1+x2=f(x)∴ f(12)=f(2) or f(13)=f(3) and so on
So, f(x) is many – one function.
Again, let y=f(x)⇒y=x1+x2
⇒y+x2y=x⇒yx2−x+y=0As xϵR∴ (−1)2−4(y)(y)≥0⇒ 1−4y2≥0⇒ yϵ[−12,12]
∴ Range = Codomain =[−12, 12]