Question

$f:R\to R$ is defined by $f\left(x\right)=x^2+4$ then $f^{-1}\left(13\right)=$

• A. $\{-3,3\}$
• B. $\{-2,2\}$
• C. $\{-1,1\}$
• D. Not invertible
  

Answer 1

Answer: (A)

$\{-3,3\}$

$f\left(x\right)=x^2+4=y$
$\Rightarrow x^2=y-4$
$x=\pm \sqrt{y-4}$
$\therefore f^{-1}\left(13\right)=\pm \sqrt{13-4}=\pm 3$
$f\left(3\right)=13\&f(-3)=13$
Thus image $13$ has two pre-images i.e, $3$ and $-3$
$\therefore$ $f$ is not invertible