A) \[\frac{1}{2}\]
B) 1
C) \[-1\]
D) 2
Correct Answer: B
Solution :
Area bounded by curves \[{{2}^{kx}}=y\]and \[x=0\]and \[x=2\]is given by \[A=\int\limits_{0}^{2}{{{2}^{kx}}dx}\] \[=\frac{1}{k}\left[ \frac{{{2}^{kx}}}{\log 2} \right]_{0}^{2}=\left[ \frac{{{2}^{2k}}-1}{k\log 2} \right]\] But \[A=\frac{3}{\log 2}\] \[\therefore \] \[\frac{{{2}^{2k}}-1}{k\,\log 2}=\frac{3}{\log 2}\] \[{{2}^{2k}}-1=3k\] This relation is satisfied by only option (b).
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