Question

If the area above the x-axis, bounded by the curves y = 2^kx and x = 0, and x = 2 is $\frac{3}{{\log }_{e}2}$, then the value of k is
(a) 1/2
(b) 1
(c) −1
(d) 2

Answer 1

(b) 1
The area bounded by the curves $y=2^{kx}$, $x=0$, and $x=2$ is given by ${\int }_0^2{2}^{kx}dx$.
It is given that ${\int }_0^2{2}^{kx}dx=\frac{3}{{\log }_e\left(2\right)}$

$$\begin{gathered} \Rightarrow \frac{1}{k}{\left[\frac{2^{kx}}{{\log }_e\left(2\right)}\right]}_0^2=\frac{3}{{\log }_e\left(2\right)} \\ \Rightarrow \frac{1}{k}\left[\frac{2^{k\left(2\right)}}{{\log }_e\left(2\right)}-\frac{2^{k\left(0\right)}}{{\log }_e\left(2\right)}\right]=\frac{3}{{\log }_e\left(2\right)} \\ \Rightarrow \frac{1}{k}\left(\frac{2^{2k}}{{\log }_e\left(2\right)}-\frac{1}{{\log }_e\left(2\right)}\right)=\frac{3}{{\log }_e\left(2\right)} \\ \Rightarrow \frac{1}{k}\left(2^{2k}-1\right)=3 \\ \Rightarrow 2^{2k}-1=3k \\ \Rightarrow 2^{2k}-3k-1=0 \\ \Rightarrow k=1 \end{gathered}$$

Clearly, k = 1 satisfies the equation.Hence, k = 1