Question : If $x=\frac{6pq}{p+q}$, then the value of $\frac{x+3p}{x–3p}+\frac{x+3q}{x–3q}$ is:
Option 1: 6
Option 2: 8
Option 3: 2
Option 4: 3
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Correct Answer: 2
Solution :
Given: $x=\frac{6pq}{p+q}$
⇒ $x=\frac{6pq}{p+q}$ and $x=\frac{6pq}{p+q}$
⇒ $\frac{x}{3p}=\frac{2q}{p+q}$ and $\frac{x}{3q}=\frac{2p}{p+q}$
Applying componendo and dividendo in the above expressions, we get,
⇒ $\frac{x+3p}{x–3p}=\frac{2q+p+q}{2q–(p+q)}$ and $\frac{x+3q}{x–3q}=\frac{2p+p+q}{2p–(p+q)}$
⇒ $\frac{x+3p}{x–3p}=\frac{p+3q}{q–p}$ and $\frac{x+3q}{x–3q}=\frac{3p+q}{p–q}$
The value of the given expression $\frac{x+3p}{x–3p}+\frac{x+3q}{x–3q}$ is as follows,
$\frac{x+3p}{x–3p}+\frac{x+3q}{x–3q}=\frac{p+3q}{q–p}-\frac{3p+q}{q–p}$
$=\frac{p+3q–3p–q}{q–p}$
$=\frac{2(q-p)}{q-p}$
$=2$
Hence, the correct answer is $2$.
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