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If $f(x)=\left(p-x^n\right)^{1 / n}, p>0$ and $n$ is a positive integer, then $f[f(x)]$ is equal to
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Verified Answer
The correct answer is:
$x$
Given, $\quad f(x)=\left(p-x^n\right)^{1 / n}, p>0$
Now,
$$
\begin{aligned}
f[f(x)] & =f\left[\left(p-x^n\right)^{1 / n}\right] \\
& =\left\{p-\left(p-x^n\right)^{1 / n \times n}\right\}^{1 / n} \\
& =\left(x^n\right)^{1 / n}=x
\end{aligned}
$$
Now,
$$
\begin{aligned}
f[f(x)] & =f\left[\left(p-x^n\right)^{1 / n}\right] \\
& =\left\{p-\left(p-x^n\right)^{1 / n \times n}\right\}^{1 / n} \\
& =\left(x^n\right)^{1 / n}=x
\end{aligned}
$$
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