Q74 / 125 · BITSAT 2021 · Mathematics
If the harmonic mean between $a$ and $b$ be $H$, then the value of $\frac{1}{\mathrm{H}-a}+\frac{1}{\mathrm{H}-b}$ is
- A$a+b$
- B$a b$
- Correct answerC$\frac{1}{a}+\frac{1}{b}$
- D$\frac{1}{a}-\frac{1}{b}$
Solution
Putting $\mathrm{H}=\frac{2 a b}{a+b}$, we have
$\begin{array}{l}
\frac{1}{\mathrm{H}-a}+\frac{1}{\mathrm{H}-b} \\
=\frac{1}{\left(\frac{2 a b}{a+b}-a\right)}+\frac{1}{\left(\frac{2 a b}{a+b}-b\right)}=\frac{a+b}{a b-a^{2}}+\frac{a+b}{a b-b^{2}}
\end{array}$
$=\left(\frac{a+b}{b-a}\right)\left(\frac{1}{a}-\frac{1}{b}\right)=\left(\frac{a+b}{b-a}\right)\left(\frac{b-a}{a b}\right)=\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$
Hence, the correct answer is option: 'C'.