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$S u p p o s e f (x) = {(x + 1)}^{2} f o r x \geq - 1. I f g (x)$ is the function whose graph is the reflection of the graph of f(x ) w.r.t the line y = x , then g ( x) is equal to

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a

$- \sqrt{x} - 1, x \geq 0$

b

$\frac{1}{{(x + 1)}^{2}}, x > - 1$

c

$\sqrt{x + 1}, x \geq - 1$

d

$\sqrt{x} - 1, x \geq 0$

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = {(x + 1)}^{2} f o r x \geq - 1 \\ ∴ g (x) w i l l b e t h e i n v e r s e o f f (x) \\ ∴ L e t f^{- 1} (x) = y \\ \Rightarrow x = f (y) \\ \Rightarrow x = {(y + 1)}^{2} \\ \Rightarrow \sqrt{x} = y + 1 \\ \Rightarrow y = \sqrt{x} - 1 \\ ∴ f^{- 1} (x) = \sqrt{x} - 1 \Rightarrow g (x) = \sqrt{x} - 1, x \geq 0 \end{array}$

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