If $f : R \to R$ defined by $f (x) = x^{2} + 1$ , then the values of $f^{- 1} (17)$ are a and b, then a+b =?

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Solution

Given, $f (x) = x^{2} + 1$
$f^{- 1} (x^{2} + 1) = x$
Clearly for $x^{2} + 1 = 17 \Rightarrow x = 4, - 4$
$\Rightarrow f^{- 1} (17) = 4, - 4$
That is, let $a = 4$ and $b = - 4,$
$\Rightarrow a + b = 0$
So ,the correct answer is $0$

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