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If $\mathbf{f}: \mathbf{R} \rightarrow \mathbf{R}$ be given by $\mathbf{f}(\mathbf{x})=\left(3-x^3\right)^{\frac{1}{3}}$, then fof $(x)$ is
(a) $x^{\frac{1}{3}}$
(b) $\mathbf{x}^3$
(c) $\mathbf{x}$
(d) $\left(3-x^3\right)$
(a) $x^{\frac{1}{3}}$
(b) $\mathbf{x}^3$
(c) $\mathbf{x}$
(d) $\left(3-x^3\right)$
Solution:
2773 Upvotes
Verified Answer
(c) $\mathrm{f}: \mathrm{R} \rightarrow$ R defined by $\mathrm{f}(\mathrm{x})=\left(3-\mathrm{x}^3\right)^{1 / 3}$ fof $(x)=f[f(x)]=f\left(\left(3-x^3\right)^{1 / 3}\right)$
$$
\begin{aligned}
&=\left[3-\left\{\left(3-x^3\right)^{1 / 3}\right\}^3\right]^{1 / 3} \\
&=\left[3-\left(3-x^3\right)\right]^{1 / 3}=\left(x^3\right)^{1 / 3}=x
\end{aligned}
$$
$$
\begin{aligned}
&=\left[3-\left\{\left(3-x^3\right)^{1 / 3}\right\}^3\right]^{1 / 3} \\
&=\left[3-\left(3-x^3\right)\right]^{1 / 3}=\left(x^3\right)^{1 / 3}=x
\end{aligned}
$$
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