Q. 21 B

# 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.

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