MathematicsIf the area above the X-axis bounded by the curves y=2kx,x=0 and x=2 is 3log 2 then the value of k is

If the area above the 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

Solution:

$Given area, = \int_{0}^{2} 2^{kx} dx = \frac{3}{\log 2}$

$\begin{array}{l} \Rightarrow \frac{1}{k} {[\frac{2^{kx}}{\log 2}]}_{0}^{2} = \frac{3}{\log 2} \\ \Rightarrow \frac{1}{k} [\frac{2^{2 k}}{\log_{e} 2} - \frac{1}{\log 2}] = \frac{3}{\log 2} \Rightarrow 2^{2 k} - 1 = 3 k \\ ∴ k = 1 \end{array}$

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