Question

If $G$ be the geometric mean of $x$ and $y,$ where $x,y>0,$ then the value of $\frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}$ is

• A. $G^2$
• B. $-G^2$
• C. $\frac{1}{G^2}$
• D. $-\frac{1}{G^2}$

Answer 1

The correct option is C $\frac{1}{G^2}$

Given, $G=\sqrt{xy}$
$\frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}=\frac{1}{xy-x^2}+\frac{1}{xy-y^2}=\frac{1}{x-y}\{-\frac{1}{x}+\frac{1}{y}\}=\frac{1}{xy}=\frac{1}{G^2}$