Question

lf $x^2+\frac{1}{x^2}=A$ and $x-\frac{1}{x}=B$, then least value of $\frac{A}{B}$ is

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

Answer 1

The correct option is D $2\sqrt{2}$

$\frac{A}{B}=\frac{x^2+\frac{1}{x^2}}{x-\frac{1}{x}}=\frac{{(x-\frac{1}{x})}^2+2}{x-\frac{1}{x}}=(x-\frac{1}{x})+\frac{2}{(x-\frac{1}{x})}$
If we take $x-\frac{1}{x}$ as $t$, then the given expression is
$t+\frac{2}{t}$
Using $A.M.\ge G.M.$, we get
$\frac{t+\frac{2}{t}}{2}\ge \sqrt{t.\frac{2}{t}}\Rightarrow t+\frac{2}{t}\ge 2\sqrt{2}$
Thus, the least value of $\frac{A}{B}=2\sqrt{2}$
Hence, option 'D' is correct.