Question

If $x=\frac{2ab}{a+b}$ then value of $\frac{x+a}{x-a}+\frac{x+b}{x-b}$

• A. 8
• B. 2
• C. 4
• D. 7

Answer 1

The correct option is B 2

$x=\frac{2ab}{a+b}$
$=>\frac{x}{a}=\frac{2b}{a+b}$
Applying Componendo and Dividendo,we get,
$=>\frac{x+a}{x-a}=\frac{2b+a+b}{2b-a-b}$
$=>\frac{x+a}{x-a}=\frac{3b+a}{b-a}$ (i)
Again,
$x=\frac{2ab}{a+b}$
$=>>\frac{x}{b}=\frac{2a}{a+b}$
Applying Componendo and Dividendo,we get
$=>\frac{x+b}{x-b}=\frac{2a+a+b}{2a-b-a}$
$=>\frac{x+b}{x-b}=\frac{3a+b}{a-b}$ (ii)
Now,from (i) and (ii) we get,
$=>\frac{x+b}{x-b}+\frac{x+b}{x-b}=\frac{3b+a}{b-a}+\frac{3a+b}{a-b}$
$=\frac{3a+b}{a-b}+\frac{3b+a}{a-b}$
$=\frac{3a+b-3b-a}{a-b}$
$=\frac{2(a-b)}{a-b}$
$=2$