Let A = [–1, 1].

(ii) Given that, A = [–1, 1]

let g(x₁) = g(x₂)

⇒ |x₁|= |x₂|

⇒ x₁= ± x₂

⇒ x₁= x₂ and x₁= - x₂

For e.g., g(-1) = |-1| = 1 and g(1) = |1| = 1

⇒ g is not one-one.

We observe that -1 does not have any pre-image in the domain since g(x) = |x| assumes only non-negative values.

i.e. we cannot find any number in domain which will give -1 in co-domain.

⇒ g is not onto

Hence, g is neither one one nor onto.