A function f: R→R defined by f(x) = 1 + x2 is-

We have f: R→R defined by f(x) =  1 + x²

Let x₁, x₂ ∈R such that f(x₁) = f(x₂)

⇒ 1+ (x1)2 = 3 -4 (x₂)2

⇒  (x₁)² =  (x₂)²

⇒ x₁ =  ±x₂

So f(x₁) = f(x₂) does not imply that  x₁ =  x₂

so f is not one-one.

Consider an element -2 in co-domain in R.

It is seen that f(x) =  1 + x² 

is positive for all x ∈ R.

Thus, there does not exist any x in domain R such that f(x) = -2

so f is not onto.

hence f is neither one-one nor onto.