Question

If $f:R^{+}\to [1,\infty )$, defined by $f\left(x\right)=x^2+1$, then the value of $f^{-1}\left(17\right)$ and $f^{-1}(-3)$ respectively are

• A. $\varphi ,\{4,-4\}$
• B. $\{3,-3\},\varphi$
• C. Not invertible
  
• D. $\left\{4\right\},\varphi$

Answer 1

Answer: (D)

$\left\{4\right\},\varphi$

Given, $f\left(x\right)=x^2+1$
Put $f\left(x\right)=y$
$y=x^2+1$
$\Rightarrow x=\pm \sqrt{y-1}$
$\Rightarrow f^{-1}\left(y\right)=\pm \sqrt{y-1}$
So, $f^{-1}\left(17\right)=\pm 4$

$f^{-}1\left(17\right)=4$ [ As $-4\notin$ domain $R^{+}$]

and $f^{-1}(-3)=\varphi$
$f^{-1}(-3)$ is not defined in $R^{+}$ as square root of negative number is not defined in $R^{+}$

Option D is correct