Question

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

• A. $G^2$
• B. $\frac{1}{G^2}$
• C. $\frac{2}{G^2}$
• D. $3G^2$

Answer 1

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

$G$ is the geometric mean of $x$ and $y$.

Therefore, $G^2=x\times y$
$\Rightarrow$ $\frac{1}{G^2-x^2}$ $+$ $\frac{1}{G^2-y^2}$ $=$ $\frac{1}{x\times y-x^2}$ $+$ $\frac{1}{x\times y-y^2}$
$=$ $\frac{1}{x\times y}$
$=$ $\frac{1}{G^2}$