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 __________ .

पर्याय

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Answer 1

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The area bounded by the curves \[y = 2^{kx}, x = 0\], and \[x = 2\] is given by \[\int_0^2 2^{kx} d x\]

It is given that \[\int_0^2 2^{kx} d x = \frac{3}{\log_e \left( 2 \right)}\]

\[\Rightarrow \frac{1}{k} \left[ \frac{2^{kx}}{\log_e \left( 2 \right)} \right]_0^2 = \frac{3}{\log_e \left( 2 \right)}\]

\[ \Rightarrow \frac{1}{k}\left[ \frac{2^{k\left( 2 \right)}}{\log_e \left( 2 \right)} - \frac{2^{k\left( 0 \right)}}{\log_e \left( 2 \right)} \right] = \frac{3}{\log_e \left( 2 \right)}\]

\[ \Rightarrow \frac{1}{k}\left( \frac{2^{2k}}{\log_e \left( 2 \right)} - \frac{1}{\log_e \left( 2 \right)} \right) = \frac{3}{\log_e \left( 2 \right)}\]

\[ \Rightarrow \frac{1}{k}\left( 2^{2k} - 1 \right) = 3\]

\[ \Rightarrow 2^{2k} - 1 = 3k\]

\[ \Rightarrow 2^{2k} - 3k - 1 = 0\]

\[ \Rightarrow k = 1\]

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