Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$is defined by $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3$, then ${\mathrm{f}}^{-1}\left(8\right)$ is equal to

• A. $\left\{2\right\}$
• B. $\left\{2,2\mathrm{\omega },2{\mathrm{\omega }}^2\right\}$
• C. $\left\{2,-2\right\}$
• D. $\left\{2,2\right\}$

Answer 1

The correct option is B

$\left\{2,2\mathrm{\omega },2{\mathrm{\omega }}^2\right\}$

Explanation for the correct answer:

The inverse function of the given function is given by interchanging the $x\mathrm{and}y$ and finding the function (the inverse function which is now y) in terms of $x$.

For, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3$

Let, $\begin{array}{rcl}\mathrm{f}\left(\mathrm{x}\right)& =& \mathrm{y}\end{array}$

$$\begin{gathered} \Rightarrow \mathrm{x}={\mathrm{f}}^{-1}\left(\mathrm{y}\right) \\ \Rightarrow \mathrm{y}={\mathrm{x}}^3 \\ \Rightarrow \mathrm{x}={\mathrm{y}}^{\frac{1}{3}} \end{gathered}$$

Now,

$\begin{array}{rcl}\therefore {\mathrm{f}}^{-1}\left(\mathrm{y}\right)& =& {\mathrm{y}}^{\frac{1}{3}}\\ \Rightarrow {\mathrm{f}}^{-1}\left(8\right)& =& 8^{\frac{1}{3}}\\ & =& 2\times {\left(1\right)}^{\frac{1}{3}}\end{array}$

We know that the cube roots of unity (1) are $1,\mathrm{\omega }\mathrm{and}{\mathrm{\omega }}^2$

$\begin{array}{rcl}\therefore {\mathrm{f}}^{-1}\left(8\right)& =& 2\times \left\{1,\mathrm{\omega },{\mathrm{\omega }}^2\right\}\\ {\mathrm{f}}^{-1}\left(8\right)& =& \left\{2,\mathrm{\omega },{\mathrm{\omega }}^2\right\}\end{array}$

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