If $f$ from $R$ into $R$ is defined by $f (x) = x^{3} - 1$ , then $f^{- 1} {- 2, 0, 7} =$

{- 1, 1, 2}

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{0, 1, 2}

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{\pm 1, \pm 2}

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{0, \pm 2}

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Solution

The correct option is A ${- 1, 1, 2}$
$f$ is invertible throughout $R$

$∴ f (x) = x^{3} - 1 = y$

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

$∴ f^{- 1} (- 2) = - 1$

$f^{- 1} (0) = 1$

$f^{- 1} (7) = 2$

$∴ f^{- 1} {- 2, 0, 7} = {- 1, 1, 2}$

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