Question

Let $f\left(x\right)=x^2+\left(\frac{1}{x^2}\right)$ and $g\left(x\right)=x-\frac{1}{x},x\in R-\left\{-1,0,1\right\}$. If $h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$, then the local minimum value of $h\left(x\right)$ is:

• A. $-2\sqrt{2}$
• B. $2\sqrt{2}$
• C. $3$
• D. $-3$

Answer 1

The correct option is B

$2\sqrt{2}$

Explanation for the correct option:

Given that: $f\left(x\right)=x^2+\left(\frac{1}{x^2}\right)$ , $g\left(x\right)=x-\frac{1}{x},x\in R-\left\{-1,0,1\right\}$, $h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$

Calculating $h\left(x\right)$:

$$\begin{gathered} h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)} \\ =\frac{x^2+{\displaystyle \frac{1}{x^2}}}{x-{\displaystyle \frac{1}{x}}} \\ =\frac{{\left(x-\frac{1}{x}\right)}^2+2}{x-\frac{1}{x}} \\ =x-\frac{1}{x}+\frac{2}{x-\frac{1}{x}} \end{gathered}$$

Let, $$\begin{gathered} x-\frac{1}{x}=t \\ \Rightarrow t+\frac{2}{t}=h\left(x\right) \end{gathered}$$

Differentiating to calculate the maxima or minima:

$\begin{array}{rcl}h\text{'}\left(x\right)& =& 1-\frac{2}{t^2}\\ Puttingh\text{'}\left(x\right)& =& 0\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}1-\frac{2}{t^2}& =& 0\\ & \Rightarrow & t=\pm \sqrt{2}\\ & & \end{array}$

Now,

$\begin{array}{rcl}h\text{'}\text{'}\left(x\right)& =& \frac{4}{t^3}\\ Att& =& \sqrt{2},h\text{'}\text{'}\left(x\right)=2\sqrt{2}\\ Att& =& -\sqrt{2},h\text{'}\text{'}\left(x\right)=-2\sqrt{2}\end{array}$

Therefore, $h\left(x\right)$ is minimum at $t=\sqrt{2}$

So,

$\begin{array}{rcl}h\left(x\right)& =& 1+\frac{2}{t}\\ & =& \sqrt{2}+\frac{2}{\sqrt{2}}\\ & =& 2\sqrt{2}\end{array}$

Hence, the correct option is ( B).