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If $f : R \to R$ defined by $f (x) = x^{4} + 2$ then the value of $f^{- 1} (83)$ and $f^{- 1} (- 2)$ respectively are

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$\emptyset, {3, - 3}$

${3, - 3}, \emptyset$

${4, - 4}, {2, - 2}$

${4, - 4}, \emptyset$

answer is B.

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Detailed Solution

$f (x) = x^{4} + 2 = y \Rightarrow x = {(y - 2)}^{1 / 4}$
$\Rightarrow f^{- 1} (x) = {(x - 2)}^{1 / 4}$
$\Rightarrow f^{- 1} (83) = {(81)}^{1 / 4} = \pm 3$
and $f^{- 1} (- 2) = {(- 4)}^{1 / 4} = \emptyset .$

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