Question

The harmonic mean of $\dfrac{a}{1-ab}$ and $\dfrac{a}{1+ab}$ is
(A) $a$
(B) $\dfrac{a}{1-{{a}^{2}}{{b}^{2}}}$
(C) $\dfrac{1}{1-{{a}^{2}}{{b}^{2}}}$
(D) $\dfrac{a}{1+{{a}^{2}}{{b}^{2}}}$

Answer 1

We solve this question by first considering the formula, harmonic mean of two numbers $a$ and $b$ is $\dfrac{2ab}{a+b}$. Then we use this formula and substitute the given numbers in it to find the harmonic mean. Then we simplify the values in numerator and denominator using the formula $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. Then we substitute the numerator and denominator in the harmonic mean and simplify it to find the harmonic mean of the given numbers.

Complete step by step answer:
The numbers we are given are $\dfrac{a}{1-ab}$ and $\dfrac{a}{1+ab}$.
We need to find the harmonic mean of given two numbers.
First let us consider the formula for the harmonic mean of two numbers. Harmonic Mean of two numbers $a$ and $b$ is $\dfrac{2ab}{a+b}$
Now, using this formula we can write the harmonic mean of $\dfrac{a}{1-ab}$ and $\dfrac{a}{1+ab}$ as,
$\Rightarrow Harmonic\ Mean=\dfrac{2\times \dfrac{a}{1-ab}\times \dfrac{a}{1+ab}}{\dfrac{a}{1-ab}+\dfrac{a}{1+ab}}............\left( 1 \right)$
Now let us consider the numerator of the harmonic mean obtained above.
$\Rightarrow 2\times \dfrac{a}{1-ab}\times \dfrac{a}{1+ab}=\dfrac{2{{a}^{2}}}{\left( 1-ab \right)\left( 1+ab \right)}$
Now let us consider the formula, $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$.
Using this we can write the numerator obtained above as,
$\Rightarrow 2\times \dfrac{a}{1-ab}\times \dfrac{a}{1+ab}=\dfrac{2{{a}^{2}}}{1-{{a}^{2}}{{b}^{2}}}..........\left( 2 \right)$
Now let us consider the denominator in the equation (1).

& \Rightarrow \dfrac{a}{1-ab}+\dfrac{a}{1+ab}=\dfrac{a}{1-ab}\times \dfrac{1+ab}{1+ab}+\dfrac{a}{1+ab}\times \dfrac{1-ab}{1-ab} \\\ & \Rightarrow \dfrac{a}{1-ab}+\dfrac{a}{1+ab}=\dfrac{a\left( 1+ab \right)}{\left( 1-ab \right)\left( 1+ab \right)}+\dfrac{a\left( 1-ab \right)}{\left( 1-ab \right)\left( 1+ab \right)} \\\ & \Rightarrow \dfrac{a}{1-ab}+\dfrac{a}{1+ab}=\dfrac{a\left( 1+ab \right)+a\left( 1-ab \right)}{\left( 1-ab \right)\left( 1+ab \right)} \\\ \end{aligned}$$ Now let us consider the formula, $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. Using this we can write the above equation as, $$\begin{aligned} & \Rightarrow \dfrac{a}{1-ab}+\dfrac{a}{1+ab}=\dfrac{a\left( 1+ab+1-ab \right)}{1-{{a}^{2}}{{b}^{2}}} \\\ & \Rightarrow \dfrac{a}{1-ab}+\dfrac{a}{1+ab}=\dfrac{a\left( 2 \right)}{1-{{a}^{2}}{{b}^{2}}} \\\ & \Rightarrow \dfrac{a}{1-ab}+\dfrac{a}{1+ab}=\dfrac{2a}{1-{{a}^{2}}{{b}^{2}}}.............\left( 3 \right) \\\ \end{aligned}$$ Substituting the values of numerator and denominator from equations (2) and (3) in equation (1) we get, $\Rightarrow Harmonic\ Mean=\dfrac{2\times \dfrac{a}{1-ab}\times \dfrac{a}{1+ab}}{\dfrac{a}{1-ab}+\dfrac{a}{1+ab}}$ $\Rightarrow Harmonic\ Mean=\dfrac{\dfrac{2{{a}^{2}}}{1-{{a}^{2}}{{b}^{2}}}}{\dfrac{2a}{1-{{a}^{2}}{{b}^{2}}}}$ $\begin{aligned} & \Rightarrow Harmonic\ Mean=\dfrac{2{{a}^{2}}}{2a} \\\ & \therefore Harmonic\ Mean=a \\\ \end{aligned}$ So, we get the harmonic mean of $\dfrac{a}{1-ab}$ and $\dfrac{a}{1+ab}$ equal to $a$. **Hence the answer is Option A.** **Note:** We can also solve this question in an alternate process by considering the property that Harmonic mean of two numbers is equal to the inverse of the arithmetic mean of their inverses, that is, harmonic mean of $a$ and $b$ is equal to inverse of arithmetic mean of $\dfrac{1}{a}$ and $\dfrac{1}{b}$. So, we need to find the arithmetic mean of their inverses, that is $$\dfrac{1-ab}{a}$$ and $\dfrac{1+ab}{a}$. Then its inverse is our required harmonic mean.