Question

If $P=x^3-\frac{1}{x^3}$ and $Q=x-\frac{1}{x},x\in \left(0,x\right)$ then minimum value value of $\frac{P}{Q}$ is

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

Answer 1

Answer: (C)

$3$

We have,

If $P=x^3-\frac{1}{x^3}$ and $Q=x-\frac{1}{x}$

Then,

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

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

$\frac{P}{Q}=x^2+\frac{1}{x^2}+1.......\left(1\right)$

Let, $\frac{P}{Q}=y$

Then,

$y=x^2+\frac{1}{x^2}+1$

$\frac{dy}{dx}=2x-\frac{2}{x^3}$

For, maxima and minima

$\frac{dy}{dx}=0$

$2x-\frac{2}{x^3}=0$

$\frac{2}{x^3}=2x$

$x^4=1$

$x=1$

Again differentiation and we get,

$\frac{d^{2}y}{d{x}^2}=2+2\frac{3}{x^4}$

$\frac{d^{2}y}{d{x}^2}=2+\frac{6}{x^4}$

Put $x=1$ and we get,

$\frac{d^{2}y}{d{x}^2}=2+\frac{6}{1}=8$

Since, $\frac{d^{2}y}{d{x}^2}>0$

Hence, at $x=1$, $y=\frac{P}{Q}$ is minimum

Put $x=1$ equation (1) and we get,

$\frac{P}{Q}=1+1+1=3$ is minimum value.

Hence, this is the answer.