Question

If $\mathrm{f}\left(\mathrm{x}+\left(\frac{1}{\mathrm{x}}\right)\right)={\mathrm{x}}^2+\left(\frac{1}{{\mathrm{x}}^2}\right)$,then $f\left(x\right)$ will be

Answer 1

Step 1: Given information

$\mathrm{f}\left(\mathrm{x}+\left(\frac{1}{\mathrm{x}}\right)\right)={\mathrm{x}}^2+\left(\frac{1}{{\mathrm{x}}^2}\right)$

Step 2: Solving the above equation to find $f\left(x\right)$

Let $\left(\mathrm{x}+\left(\frac{1}{\mathrm{x}}\right)\right)=\mathrm{t}$

Taking square on both sides,

$$\begin{gathered} \Rightarrow {\left(\mathrm{x}+\left(\frac{1}{\mathrm{x}}\right)\right)}^2={\mathrm{t}}^2 \\ \Rightarrow {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+2\left(\mathrm{x}\right)\left(\frac{1}{\mathrm{x}}\right)={\mathrm{t}}^2 \end{gathered}$$

$$\begin{gathered} \Rightarrow \left({\mathrm{x}}^2+\left(\frac{1}{{\mathrm{x}}^2}\right)\right)+2={\mathrm{t}}^2 \\ \Rightarrow \left({\mathrm{x}}^2+\left(\frac{1}{{\mathrm{x}}^2}\right)\right)={\mathrm{t}}^2-2 \\ \Rightarrow \mathrm{f}\left(\mathrm{t}\right)={\mathrm{t}}^2-2 \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-2 \end{gathered}$$

Hence, the value of $f\left(x\right)$ is ${\mathrm{x}}^2-2$