• # question_answer If the area above the x-axis, bounded by the curves $y={{2}^{kx}}$ and $x=0$ and $x=2$ is $\frac{3}{\ln 2},$ then the value of k is                                                                       [Orissa JEE 2003] A)            $\frac{1}{2}$                       B)            1 C)            $-1$                                        D)            2

$\int_{\,0}^{\,2}{{{2}^{kx}}dx=\frac{3}{\log 2}\Rightarrow {{2}^{2k}}-1=3k}$. Now check from options, only  satisfies the above condition.