If $f : R \to R$ define by $f (x) = x^{3} - 1$ , then $f^{- 1} {- 2, 0, 7}$ is

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$\{\pm 1, \pm 2\}$

$\{- 1, 1, 2\}$

$\{0, 1, 2\}$

$\{0, \pm 2\}$

answer is A.

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Detailed Solution

$f (x) = x^{3} - 1 = y \Rightarrow x = f^{- 1} (y)$

$\Rightarrow x^{3} = y + 1 \Rightarrow x = (y + 1)^{\frac{1}{3}}$

$\Rightarrow x = \sqrt[3]{(y + 1)} = f^{- 1} (y)$

$\Rightarrow f^{- 1} (x) = \sqrt[3]{x + 1}$

$∴ f^{- 1} (- 2) = \sqrt[3]{- 2 + 1} = \sqrt[3]{- 1} = - 1$

$∴ f^{- 1} (0) = \sqrt[3]{1} = 1$

$∴ f^{- 1} (7) = \sqrt[3]{7 + 1} = 2$

$∴ f^{- 1} (- 2, 0, 7) = {- 1, 1, 2}$

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