Solve this

Solution:

Given: $f(x)=x^{2}+1$

To find: (i) $f^{-1}\{10\}$

We know that, if $f: X \rightarrow Y$ such that $y \in Y$. Then $f^{-1}(y)=\{x \in X: f(x)=y\}$.

In other words, $\mathrm{f}^{-1}(\mathrm{y})$ is the set of pre - images of $\mathrm{y}$

Let $f^{-1}\{10\}=x$. Then, $f(x)=10$...(i)

and it is given that $f(x)=x^{2}+1 \ldots$ (ii)

So, from (i) and (ii), we get

$x^{2}+1=10$

$\Rightarrow x^{2}=10-1$

$\Rightarrow x^{2}=9$

$\Rightarrow x=\sqrt{9}$

$\Rightarrow x=\pm 3$

$\therefore f^{-1}\{10\}=\{-3,3\}$

To find: (ii) $f^{-1}\{-3\}$

Let $f^{-1}\{-3\}=x$. Then, $f(x)=-3 \ldots$ (iii)

and it is given that $f(x)=x^{2}+1 \ldots$ (iv)

So, from (iii) and (iv), we get

$x^{2}+1=-3$

$\Rightarrow x^{2}=-3-1$

$\Rightarrow x^{2}=-4$

Clearly, this equation is not solvable in R

$\therefore f^{-1}\{-3\}=\phi$