Question

If $x=\frac{6pq}{p+q}$ then value of $\frac{x+3p}{x-3p}+\frac{x+3q}{x-3q}$

• A. 2
• B. 9
• C. 5
• D. 8

Answer 1

The correct option is A 2

$x=\frac{6pq}{p+q}andx=\frac{6pq}{p+q}$
$\therefore x=\frac{3p\times 2q}{p+q}$ $\therefore x=\frac{2p\times 3q}{p+q}$
$\therefore \frac{x}{3p}=\frac{2q}{p+q}$ $\therefore \frac{x}{3q}=\frac{2p}{p+q}$
By componendo- dividendo
$\frac{x+3p}{x-3p}=\frac{2q+p+q}{2q-(p+q)}and\frac{x+3q}{x-3q}=\frac{2p+p+q}{2p-(p+q)}$
$\therefore \frac{x+3p}{x-3p}=\frac{p+3q}{q-p}and\frac{x+3q}{x-3q}=\frac{3p+q}{p-q}$
$\therefore \frac{x+3p}{x-3p}+\frac{x+3q}{x-3q}=\frac{p+3q}{q-p}-\frac{3p+q}{q-p}$
$=\frac{p+3q}{q-p}-\frac{3p+q}{q-p}=\frac{p+3q-3p-q}{q-p}$
$\therefore \frac{x+3p}{x-3p}+\frac{x+3q}{x-3q}=\frac{2q-2p}{q-p}=2$