Question

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

• A. is $2\sqrt{3}$
• B. is $4\sqrt{3}$
• C. does not exist
• D. None of these

Answer 1

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

Given P $=x^3-\frac{1}{x^3},Q=x-\frac{1}{x}$

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

$Q=x-1/x$

simiplified $P/Q^2$ is $(x-1/x)+3(x-1/x)$

we can apply A.M greater than or equal to G.M

we get the minimum value of

$P/Q^2$ is $2(3{)}^{1/2}=2\sqrt{3}$