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If f\left ( x \right )=x^{2}+2x,x\geq 1, then f^{-1}\left ( x \right ) equals

Option: 1

\sqrt{x+1}-1

 

 

 


Option: 2

-\sqrt{x+1}-1


Option: 3

\sqrt{x^{2}+2x}


Option: 4

None of these


Answers (1)

best_answer

\quad y=x^{2}+2 x
\begin{aligned} &\Rightarrow x^{2}+2 x-y=0 \\ &\Rightarrow x=\frac{-2 \pm \sqrt{4+4 y}}{2}=-1 \pm \sqrt{y+1} \\ &\Rightarrow f^{-1}(x)=\pm \sqrt{x+1}-1 \end{aligned}

Now as inverse of a function is unique, we can take only + or - sign for \sqrt{x+1}. 

For this we put x=0 \: in\: f(x) \Rightarrow y=0. So, (0,0) lies in f(x).
and hence (0,0) should also lie in f^{-1}(x).
(As if \left.(a, b) \in f \Rightarrow(b, a) \in f^{-1}\right)


\Rightarrow 0=\pm \sqrt{0+1}-1, clearly this is satisfied for + sign \Rightarrow f^{-1}(x)=\sqrt{x+1}-1.

Posted by

Ritika Jonwal

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