Question

Define g : R → R by g(x) = x² + 1. Then g is

(a) neither injective nor surjective 

(b) injective but not surjective 

(c) surjective but not injective 

(d) bijective

Answer 1

Answer: (a) = Neither injective nor surjective.

 g (–1) = 1² + 1 = 2 and g(1) = 1² + 1 = 2 

⇒ Both – 1 and 1 ∈ R have the same image 2 ∈ R 

⇒ g is not one-one or injective 

Let z = g (x) = x² + 1 

⇒x = ± \(\sqrt{z-1}\) 

x is not defined when z – 1 < 0, i.e., z < 1 

∴  for 0 ∈ R, there exists no pre-image in R 

⇒ g(x) is not onto or surjective. 

∴  g is neither injective nor surjective