Question

Let $f\left(x\right)=x^2-\frac{1}{x^2}$ and $g\left(x\right)=x-\frac{1}{x},x\in R--1,0,1$. 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. $2$
• D. $-3$

Answer 1

Answer: (C)

$2$

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

$=\frac{(x^2-\frac{1}{x^2})}{(x-\frac{1}{x})}$

$=\frac{(x-\frac{1}{x})(x+\frac{1}{x})}{(x-\frac{1}{x})}$

$=(x+\frac{1}{x})\therefore h\left(x\right)=(x+\frac{1}{x})$

Now, we know, Arithmetic Mean $\ge$ Geometric Mean

$\therefore$ $\frac{(x+\frac{1}{x})}{2}\ge \sqrt{x-\frac{1}{x}}\ge 1\Rightarrow x+\frac{1}{x}\ge 2$

$\therefore$ The min. value of $h\left(x\right)=2$