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Question

If the harmonic mean between $$a$$ and $$b$$ be $$H$$, then the value of $$\dfrac {1}{H-a}+\dfrac {1}{H-b}$$


A
a+b
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B
ab
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C
1a+1b
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D
1a1b
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Solution

The correct option is C $$\dfrac {1}{a} +\dfrac {1}{b}$$
$$\because H$$ is harmonic mean between $$a$$ and $$b$$
$$\therefore H=\cfrac{2ab}{a+b}$$
Therefore,  $$\cfrac {1}{H-a}+\cfrac {1}{H-b}=\cfrac {1}{\dfrac{2ab}{a+b}-a}+\cfrac {1}{\dfrac{2ab}{a+b}-b}$$
    $$=\cfrac {a+b}{ab-a^2}+\cfrac {a+b}{ab-b^2}=\cfrac{a+b}{b-a}\left(\cfrac{1}{a}-\cfrac{1}{b}\right)=\cfrac{a+b}{ab}=\cfrac{1}{a}+\cfrac{1}{b}$$
Hence, option 'C' is correct.

Mathematics

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