Question

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

A
G2
B
1G2
C
2G2
D
3G2

Solution

## The correct option is B $$\dfrac {1}{G^2}$$$$G$$ is the geometric mean of $$x$$ and $$y$$.Therefore, $$G^2 = x \times y$$$$\Rightarrow$$ $$\dfrac{1}{G^2- x^2}$$ $$+$$ $$\dfrac{1}{G^2- y^2}$$ $$=$$ $$\dfrac{1}{x \times y- x^2}$$ $$+$$ $$\dfrac{1}{x \times y- y^2}$$ $$=$$ $$\dfrac{1}{x \times y}$$$$=$$ $$\dfrac{1}{G^2}$$Maths

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