Question

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

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

Answer 1

Answer: (B)

$1$

Area bounded by the curves $2^{kx}=y,$ $x=0$ and $x=2$ is given by
$A={\int }_0^2{2}^{kx}dx$
$=\frac{1}{k}{\left[\frac{2^{kx}}{\log 2}\right]}_0^2=\left[\frac{2^{2k}-1}{k\log 2}\right]$
But $A=\frac{3}{\log 2}$
$\therefore \frac{2^{2k}-1}{k\log 2}=\frac{3}{\log 2}$
$\Rightarrow 2^{2k}-1=3k$

$\Rightarrow 4^k=3k+1$
This relation is satisfied by only option (b)

Hence, $k=1$.