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Let $f : R \to R$ be given by $f (x) = {(x + 1)}^{2} - 1, x \geq - 1$ Then $f^{- 1} (x) =$

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−1+x+1

−1−x+1

does not exist because f is not one - one

does not exist because f is not onto

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detailed solution

Correct option is A

Let y = f(x)

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Let f:R→R be given by f(x)=x+12−1,x≥−1 Then f−1(x)=

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Let $f : R \to R$ be given by $f (x) = {(x + 1)}^{2} - 1, x \geq - 1$ Then $f^{- 1} (x) =$