Mark the correct alternative in each of the following:
The function f : R → R, f(x) = x2 is
Given that f : R → R, f(x) = x2
Let f(x)=y(x)
⇒ x2 = y2
⇒ x = ±y
So, it is not one-one.
f(x) = y
⇒ x2 = y
⇒ x = ±√y
But co domain is R.
Hence, f is neither injective nor surjective.