Question

Let f(x) =$x^2+\frac{1}{x^2}$ and g(x) = x - $\frac{1}{x},x\in R-\{-1,0,1\}.IFh\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)},thenthelocalminimumvalueofh\left(x\right)is:$

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

Answer 1

The correct option is C $2\sqrt{2}$

Given:

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

$g\left(x\right)=x-\frac{1}{x}$

$h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$

To find:

Minimum value of h(x)

Solution:

$h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$

$\Rightarrow h\left(x\right)=\frac{x^2+\frac{1}{x^2}}{x-\frac{1}{x}}$

$\Rightarrow h\left(x\right)=\frac{{(x-\frac{1}{x})}^2+2}{x-\frac{1}{x}}$

$\Rightarrow h\left(x\right)=(x-\frac{1}{x})+\frac{2}{x-\frac{1}{x}}$

Differentiate both the sides

$\Rightarrow h^{\prime }\left(x\right)=(1+\frac{1}{x^2})-\frac{2}{{(x-\frac{1}{x})}^2}(1+\frac{1}{x^2})$

$\Rightarrow h^{\prime }\left(x\right)=(1+\frac{1}{x^2})(1-\frac{2}{{(x-\frac{1}{x})}^2})$

$\Rightarrow h^{\prime }\left(x\right)=(1+\frac{1}{x^2})\left(\frac{{(x-\frac{1}{x})}^2-2}{{(x-\frac{1}{x})}^2}\right)$

$\Rightarrow h^{\prime }\left(x\right)=\frac{(1+\frac{1}{x^2})}{{(x-\frac{1}{x})}^2}({(x-\frac{1}{x})}^2-2)$

$\Rightarrow h^{\prime }\left(x\right)=\frac{x^2+1}{{(x-\frac{1}{x})}^2\cdot x^2}\left(\frac{(x^2-1{)}^2-2x^2}{x^2}\right)$

Equating h'(x)=0, we get

$\Rightarrow h^{\prime }\left(x\right)=\frac{x^2+1}{{(x-\frac{1}{x})}^2\cdot x^4}\left(\right(x^2-1{)}^2-2x^2)=0$

$\Rightarrow \left(\right(x^2-1{)}^2-2x^2)=0$

$\Rightarrow (x^4+1-2x^2-2x^2)=0$

$\Rightarrow (x^4+1-4x^2)=0$

$\Rightarrow (x^2-2{)}^2-3=0$

$\Rightarrow (x^2-2{)}^2=3$

$\Rightarrow (x^2-2)=\pm \sqrt{3}$

$\Rightarrow x^2=2\pm \sqrt{3}$

$\Rightarrow x^2=\frac{(\sqrt{3}\pm 1{)}^2}{2}$

$\Rightarrow x=\pm \frac{\sqrt{3}\pm 1}{\sqrt{2}}$

$Asx\in R-\{-1,0,1\}$

$\therefore x=\frac{\sqrt{3}+1}{\sqrt{2}}$

and

$\Rightarrow \frac{1}{x}=\frac{\sqrt{2}}{\sqrt{3}+1}$

$\Rightarrow \frac{1}{x}=\frac{\sqrt{2}}{\sqrt{3}+1}\cdot \frac{\sqrt{3}-1}{\sqrt{3}-1}$

$\Rightarrow \frac{1}{x}=\frac{\sqrt{3}-1}{\sqrt{2}}$

Now,

$\Rightarrow x-\frac{1}{x}=\frac{\sqrt{3}+1}{\sqrt{2}}-\frac{\sqrt{3}-1}{\sqrt{2}}$

$\Rightarrow x-\frac{1}{x}=\frac{2}{\sqrt{2}}$

$\Rightarrow x-\frac{1}{x}=\sqrt{2}$

Also,

$\Rightarrow h\left(x\right)=(x-\frac{1}{x})+\frac{2}{x-\frac{1}{x}}$

putting the value,

$\Rightarrow h\left(x\right)=\sqrt{2}+\frac{2}{\sqrt{2}}$

$\Rightarrow h\left(x\right)=2\sqrt{2}$