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Let $f: R \rightarrow R$ bedefined as $f(x)=x^{2}+1$, find $f^{-1}(-5)$.
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Verified Answer
The correct answer is:
$\phi$
Let $\mathrm{f}^{-1}(-5)=\mathrm{x}$. Then, $\mathrm{f}(\mathrm{x})=-5$
$$
\Rightarrow x^{2}+1=-5 \Rightarrow x^{2}=-6 \Rightarrow x=\pm \sqrt{-6}
$$
which does not belong to $\mathrm{R}$.
$$
\therefore \mathrm{f}^{-1}(-5)=\phi
$$
$$
\Rightarrow x^{2}+1=-5 \Rightarrow x^{2}=-6 \Rightarrow x=\pm \sqrt{-6}
$$
which does not belong to $\mathrm{R}$.
$$
\therefore \mathrm{f}^{-1}(-5)=\phi
$$
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