If a = $\dfrac{4xy}{x+y}$, then $\Big(\dfrac{a+2x}{a-2x} + \dfrac{a+2y}{a-2y}\Big)$ equals :

1. 1

2. 2

3. $\dfrac{1}{2}$

4. none of these

Given,

a = $\dfrac{4xy}{x+y}$

Solving,

$\Rightarrow \dfrac{a+2x}{a-2x} + \dfrac{a+2y}{a-2y} = \dfrac{\dfrac{4xy}{x+y} + 2x}{\dfrac{4xy}{x+y} - 2x} + \dfrac{\dfrac{4xy}{x+y} + 2y}{\dfrac{4xy}{x+y} - 2y} \\[1em] \Rightarrow \text{Multiply numerator and denominator by (x + y)} \\[1em] \Rightarrow \dfrac{\Big(\dfrac{4xy}{x+y} + 2x\Big)\times (x + y)}{\Big(\dfrac{4xy}{x+y} - 2x\Big) \times (x + y)} + \dfrac{\Big(\dfrac{4xy}{x+y} + 2y\Big)\times (x + y)}{\Big(\dfrac{4xy}{x+y} - 2y\Big) \times (x + y)}\\[1em] \Rightarrow \dfrac{4xy + 2x(x+y)}{4xy - 2x(x+y)} + \dfrac{4xy + 2y(x+y)}{4xy - 2y(x+y)} \\[1em] \Rightarrow \dfrac{4xy + 2x^2 + 2xy}{4xy - 2x^2 - 2xy} + \dfrac{4xy + 2xy + 2y^2}{4xy - 2xy - 2y^2} \\[1em] \Rightarrow \dfrac{2x^2 + 6xy}{-2x^2 + 2xy} + \dfrac{6xy + 2y^2}{2xy - 2y^2} \\[1em] \Rightarrow \dfrac{2x(x + 3y)}{2x(y - x)} + \dfrac{2y(3x+y)}{2y(x-y)} \\[1em] \Rightarrow \dfrac{x + 3y}{y - x} + \dfrac{3x + y}{x - y} \\[1em] \Rightarrow \dfrac{x + 3y}{-(x - y)} + \dfrac{3x + y}{x - y} \\[1em] \Rightarrow \dfrac{-(x + 3y)}{x - y} + \dfrac{3x + y}{x - y} \\[1em] \Rightarrow \dfrac{-x - 3y + 3x + y}{x - y} \\[1em] \Rightarrow \dfrac{-x + 3x + y - 3y}{x - y} \\[1em] \Rightarrow \dfrac{2x - 2y}{x - y} \\[1em] \Rightarrow \dfrac{2(x - y)}{x - y} \\[1em] \Rightarrow 2.$

Hence, option 2 is the correct option.