Question

If $x^2+\frac{1}{x^2}=A$ and $x-\frac{1}{x}$ = B, where $x\in R$ and $B>0$, then the minimum value of $\frac{A}{B}$ is

• A. $\sqrt{2}$
• B. $2$$\sqrt{2}$
• C. $\sqrt{2}+2$
• D. None of these

Answer 1

Answer: (B)

$2$$\sqrt{2}$

Given $x^2+\frac{1}{x^2}=A$ and $x-\frac{1}{x}=B$

By squaring $B$ , we get $B^2=A-2$

Divide with $B$ on both sides , we get $\frac{A}{B}=B+\frac{2}{B}$

GIven $B>0$

Therefore by $AM\ge GM$ , we get $B+\frac{2}{B}\ge 2\sqrt{2}$

Therefore the minimum value of $\frac{A}{B}$ is $2\sqrt{2}$