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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
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B
1G2
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C
2G2
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D
3G2
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Solution

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

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