Ans: A
$\begin{array}{rcl}
\dfrac{2+a}{a} & = & \dfrac{2-x}{x} \\
x(2+a) & = & a(2-x) \\
x(2+a) & = & 2a-ax \\
x(2+a) +ax & = & 2a \\
x(2+2a) & = & 2a \\
x & = & \dfrac{2}{2+2a} \\
x & = & \dfrac{1}{1+a}
\end{array}$
$\begin{array}{rcl}
\dfrac{2+a}{a} & = & \dfrac{2-x}{x} \\
x(2+a) & = & a(2-x) \\
x(2+a) & = & 2a-ax \\
x(2+a) +ax & = & 2a \\
x(2+2a) & = & 2a \\
x & = & \dfrac{2}{2+2a} \\
x & = & \dfrac{1}{1+a}
\end{array}$
