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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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