If a function f : R → R is defined by $f(x)=x^2+1$, then the pre images of 17 and -3 respectively are :

$\phi,\{4,-4\}$ $\{3,-3\}, \phi$ $\{4,-4\}, \phi$ $\{4,-4\},\{2,-2\}$

$\{4,-4\}, \phi$

$f(x)=x^2+1 ~~~f: R \rightarrow R$ so  $y=x^2+1 ~~\Rightarrow ~~y-1=x^2$ so  $x= \pm \sqrt{y-1}$   →  image of y  for y = -3 $x = \pm \sqrt{-3-1}$ $= \pm \sqrt{-4}$ X doesn't exist as x ∈ R for y = 17 $x= \pm \sqrt{17-1}$ $x= \pm \sqrt{16}$ $x= \pm 4$ {4, -4}, $\phi$