Question

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

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

Answer 1

Answer: (B)

$\frac{1}{G^2}$

As given $G=\sqrt{xy}$

$\therefore \frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}=\frac{1}{xy-x^2}+\frac{1}{xy-y^2}$

$=\frac{1}{x-y}[-\frac{1}{x}+\frac{1}{y}]=\frac{1}{xy}=\frac{1}{G^2}$