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If _{$f : R \to R$} defined by _{$f (x) = x^{2} + 1$} , then the set of all pre-images of _{$17 = f^{- 1} (17)$} , the set of all pre-images of _{$- 3 = f^{- 1} (- 3)$} are respectively

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_$φ$ , {4, –4}

_{${4, - 4}, φ$}

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

_{${3, - 3}, φ$}

answer is C.

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Detailed Solution

$f^{- 1} (17) = x$ $f^{- 1} (- 3) = x$

$\Rightarrow f (x) = 17$ $f (x) = - 3$

$\Rightarrow x^{2} + 1 = 17$ $x^{2} + 1 = - 3$

$∴ x = \pm 4$ $x^{2} = - 2$ $\Rightarrow$ $x \notin R$

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