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If $f(x)=x^2+1$, then $f^{-1}(17)$ and $f^{-1}(-3)$ will be
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Verified Answer
The correct answer is:
None of these
$\begin{aligned}
& \text { Let } y=x^2+1 \Rightarrow x= \pm \sqrt{y-1} \\
& \Rightarrow f^{-1}(y)= \pm \sqrt{y-1} \Rightarrow f^{-1}(x)= \pm \sqrt{x-1} \\
& \Rightarrow f^{-1}(17)= \pm \sqrt{17-1}= \pm 4
\end{aligned}$
and $f^{-1}(-3)= \pm \sqrt{-3-1}= \pm \sqrt{-4}$, which is not possible.
& \text { Let } y=x^2+1 \Rightarrow x= \pm \sqrt{y-1} \\
& \Rightarrow f^{-1}(y)= \pm \sqrt{y-1} \Rightarrow f^{-1}(x)= \pm \sqrt{x-1} \\
& \Rightarrow f^{-1}(17)= \pm \sqrt{17-1}= \pm 4
\end{aligned}$
and $f^{-1}(-3)= \pm \sqrt{-3-1}= \pm \sqrt{-4}$, which is not possible.
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