Question

If $P=x^3-\frac{1}{x^3}$ and $Q=x-\frac{1}{x}$, $x\ne (0,\infty )$ , then local minimum value of $\frac{P}{Q^2}$ is

• A. $2\sqrt{3}$
• B. $-2\sqrt{3}$
• C. Does not exist
• D. None of these

Answer 1

The correct option is A $2\sqrt{3}$

$P=x^3-\frac{1}{x^3}=(x-\frac{1}{x})(x^2+\frac{1}{x^2}+1)$

We have $Q=(x-\frac{1}{x})$

Now $\frac{P}{Q^2}=P\times \frac{1}{Q^2}$

$=(x^3-\frac{1}{x^3})\frac{1}{{(x-\frac{1}{x})}^2}$

$=[{(x-\frac{1}{x})}^3+3x\times \frac{1}{x}(x-\frac{1}{x})]\frac{1}{{(x-\frac{1}{x})}^2}$

$=(x-\frac{1}{x})+\frac{3}{(x-\frac{1}{x})}$

As A.M$\ge$G.M

Minimum value of $\frac{P}{Q^2}=2\sqrt{(x-\frac{1}{x})\times \frac{3}{(x-\frac{1}{x})}}=2\sqrt{3}$