Explanation

(b)
Area bounded by curves $y = 2^{kx}$ and $x = 0$ and $x = 2$ is given by
$A = \int_{0}^{2}{2^{kx}dx}$
$= \left\lbrack \frac{2^{kx}}{k\log 2} \right\rbrack_{0}^{2} = \left\lbrack \frac{2^{2k} - 1}{k\log 2} \right\rbrack$
But $A = \frac{3}{\log 2}$
$\therefore\ \frac{2^{2k} - 1}{k\log 2} = \frac{3}{\log 2}\ \Rightarrow 2^{2k} - 1 = 3k$
This, relation is satisfied by only option (b)