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If $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ is defined by $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$, then $\mathrm{f}^{-1}(8)$ is equal to
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The correct answer is:
$\{2\}$
Let
$$
y=f(x)=x^{3}
$$
$\therefore$
$\mathrm{x}=\mathrm{y}^{1 / 3}$
$\Rightarrow \quad \mathrm{f}^{-1}(\mathrm{x})=\mathrm{x}^{1 / 3}$
$\Rightarrow \quad \mathrm{f}^{-1}(8)=(8)^{1 / 3}$
$=2$
$$
y=f(x)=x^{3}
$$
$\therefore$
$\mathrm{x}=\mathrm{y}^{1 / 3}$
$\Rightarrow \quad \mathrm{f}^{-1}(\mathrm{x})=\mathrm{x}^{1 / 3}$
$\Rightarrow \quad \mathrm{f}^{-1}(8)=(8)^{1 / 3}$
$=2$
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