(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.
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