Let `A}x :-1lt=xlt=1}a n df: Avec` such that `f(x)=x|x|,` then `f` is a bijection (b) injective but not surjective Surjective but not injective (d) neither injective nor surjective

Explanation

Injectivity:

⇒ Let `x_1`​ and `x_2`​ be any two elements in the domain A.

CaseI:

Let `x_1`​ and `x_2`​ be two positive numbers,
such that

`⇒ f(x_1​)=f(x_2​)`

`⇒ x_1​∣x_1​∣=x_2​∣x_2​∣`

`⇒ x_1​(x_1​)=x_2​(x_2​)`

`⇒ (x_1​)^2​=(x_2)^2​`

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