Question

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

Answer 1

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