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If $\mathrm{f}(x)=x^2+1$ and $\mathrm{g}(x)=\frac{1}{x}$, then the value of $\mathrm{f}(\mathrm{g}(\mathrm{g}(\mathrm{f}(x))))$ at $x=1$ is
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Verified Answer
The correct answer is:
$5$
$$
\begin{aligned}
\mathrm{f}(x)=x^2+1, & \mathrm{~g}(x)=\frac{1}{x} \\
\therefore \quad \mathrm{f}(\mathrm{g}(\mathrm{g}(\mathrm{f}(x)))) & =\mathrm{f}\left(\mathrm{g}\left(\mathrm{g}\left(x^2+1\right)\right)\right) \\
& =\mathrm{f}\left(\mathrm{g}\left(\frac{1}{x^2+1}\right)\right) \\
& =\mathrm{f}\left(x^2+1\right) \\
& =\left(x^2+1\right)^2+1
\end{aligned}
$$
$\therefore \quad$ At $x=1$, we get the value of above function
$$
\begin{aligned}
& =\left[(1)^2+1\right]^2+1 \\
& =5
\end{aligned}
$$
\begin{aligned}
\mathrm{f}(x)=x^2+1, & \mathrm{~g}(x)=\frac{1}{x} \\
\therefore \quad \mathrm{f}(\mathrm{g}(\mathrm{g}(\mathrm{f}(x)))) & =\mathrm{f}\left(\mathrm{g}\left(\mathrm{g}\left(x^2+1\right)\right)\right) \\
& =\mathrm{f}\left(\mathrm{g}\left(\frac{1}{x^2+1}\right)\right) \\
& =\mathrm{f}\left(x^2+1\right) \\
& =\left(x^2+1\right)^2+1
\end{aligned}
$$
$\therefore \quad$ At $x=1$, we get the value of above function
$$
\begin{aligned}
& =\left[(1)^2+1\right]^2+1 \\
& =5
\end{aligned}
$$
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