Question

If the area above the x- axis bounded by the curves $y=2^{kx}$ and $x=0$ and $x=2$ is $\frac{3}{ln2}$, then the value of k is

• A. $\frac{1}{2}$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (B)

$1$

Equation of curve is given as;

$y=2^{kx}$

Let $2^{kx}=e^{\left(\log 2\right)kx}⟶\left(1\right)$

area bounded by the curve above x-axis is given as;

${\int }_0^{2}ydx=\frac{3}{\log 2}$

on solving the above euation we get;

$\Rightarrow {\int }_0^2{e}^{\left(\log 2\right)kx}dx=\frac{3}{\log 2}\Rightarrow \frac{1}{\left(\log 2\right)k}{\left[e^{\left(\log 2\right)kx}\right]}_0^2=\frac{3}{\log 2}\Rightarrow \frac{1}{k}[e^{\left(\log 2\right)2k}-e^0]=3\Rightarrow 2^{2k}-1=3k\Rightarrow 2^{2k}-3k-1=0\Rightarrow k=1$

Clearly for $k=1$ saties the above equation.