Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ defined by $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4+2$ then the value of ${\mathrm{f}}^{-1}\left(83\right)$ and ${\mathrm{f}}^{-1}(-2)$ respectively are:

• A. $\mathrm{\varphi },\left\{3,-3\right\}$
• B. $\left\{3,-3\right\},\mathrm{\varphi }$
• C. $\left\{4,-4\right\},\mathrm{\varphi }$
• D. $\left\{4,-4\right\},\left\{2,-2\right\}$

Answer 1

The correct option is B

$\left\{3,-3\right\},\mathrm{\varphi }$

Explanation for the correct answer:

Step 1: Finding the value of $x$

Given, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4+2$

Let, $\begin{array}{rcl}y& =& x^4+2\end{array}$

$$\begin{gathered} \Rightarrow x^4=y-2 \\ \Rightarrow x={\left(y-2\right)}^{\frac{1}{4}} \end{gathered}$$

Therefore, $f^{-1}\left(x\right)={(x-2)}^{\frac{1}{4}}$

Step 2: Finding the value of ${\mathrm{f}}^{-1}\left(83\right)$

$$\begin{gathered} \therefore f^{-1}\left(\mathrm{x}\right)={\left(\mathrm{x}-2\right)}^{\frac{1}{4}} \\ \Rightarrow {\mathrm{f}}^{-1}\left(83\right)={\left(83-2\right)}^{\frac{1}{4}}\left[\because x=83\right] \\ \Rightarrow {\mathrm{f}}^{-1}\left(83\right)={\left(81\right)}^{\frac{1}{4}} \\ \Rightarrow {\mathrm{f}}^{-1}\left(83\right)=\pm 3 \end{gathered}$$

Step 2: Finding the value of ${\mathrm{f}}^{-1}(-2)$

$$\begin{gathered} \Rightarrow {\mathrm{f}}^{-1}(-2)={\left(-2-2\right)}^{\frac{1}{4}}\left[\because x=-2\right] \\ \Rightarrow {\mathrm{f}}^{-1}(-2)={\left(-4\right)}^{\frac{1}{4}} \\ \Rightarrow {\mathrm{f}}^{-1}=\mathrm{\varphi } \end{gathered}$$

Therefore, ${\mathrm{f}}^{-1}\left(83\right)\mathrm{and}{\mathrm{f}}^{-1}(-2)\mathrm{are}\left\{3,-3\right\},\mathrm{\varphi }$

Hence, the correct answer is an option (B).