Question

If G be the GM between x and y, then the value of $\frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}$ is equal to

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

Answer 1

Answer: (C)

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

Answer 2

1. Understanding the Geometric Mean (GM):

If G is the Geometric Mean (GM) between two numbers x and y, then by definition: $G^2 = xy$

2. Substituting into the Expression:

The given expression is $\frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}$. Substitute $G^2 = xy$ into the expression:

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

3. Factoring the Denominators:

Factor out common terms from each denominator:

• $xy-x^2 = x(y-x)$
• $xy-y^2 = y(x-y)$

So the expression becomes:

$\frac{1}{x(y-x)}+\frac{1}{y(x-y)}$

4. Adjusting for Common Denominators:

Notice that $(x-y)$ is the negative of $(y-x)$, i.e., $(x-y) = -(y-x)$. Substitute this into the second term:

$\frac{1}{x(y-x)}+\frac{1}{y(-(y-x))}$

$\frac{1}{x(y-x)}-\frac{1}{y(y-x)}$

5. Combining the Fractions:

Now, find a common denominator, which is $xy(y-x)$:

$\frac{y}{xy(y-x)}-\frac{x}{xy(y-x)}$

Combine the numerators:

$\frac{y-x}{xy(y-x)}$

6. Simplifying the Expression:

Assuming $x \neq y$ (otherwise the original expression would be undefined due to division by zero), we can cancel out the term $(y-x)$ from the numerator and the denominator:

$\frac{1}{xy}$

7. Final Substitution:

Recall from step 1 that $G^2 = xy$. Substitute this back into the simplified expression:

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

Thus, the value of the expression is $\frac{1}{G^2}$.