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If _{$f : R \to R$} defined by _{$f (x) = x^{2} - 1$} then the set of all pre-images of _{$15 = f^{- 1} (15)$} the set of all pre-images of _{$- 3 = f^{- 1} (- 3)$} are respectively

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_{$φ, {4, - 4}$}

_{${3, - 3}, ϕ$}

_{${4, - 4}, φ$}

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

answer is C.

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Detailed Solution

$f (x) = x^{2} - 1 \Rightarrow f^{- 1} (x) = \pm \sqrt{x + 1} f^{- 1} (15) = \pm 4 f^{- 1} (- 3) = \emptyset$

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