If the area above \( x \)-axis, bounded by the curves \( y=2^{kx} \) and \( x=0 \) and \( x=2 \) is \( \displaystyle\frac{3}{\log_e 2} \), then the value of \( k \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1
Since \( 2^{kx}\gt 0 \) for all x, the curve lies above the x-axis and the area is the integral. For \( k\ne 0 \), using \( \int a^{kx}dx=\frac{a^{kx}}{k\log_e a} \):
\( \displaystyle \int_{0}^{2}\displaystyle 2^{kx}\,dx=\left[\frac{2^{kx}}{k\log_e 2}\right]_{0}^{2} \)
\( \displaystyle =\frac{2^{2k}-1}{k\log_e 2} \)
\( \displaystyle =\frac{4^k-1}{k\log_e 2} \)
Equating it to the given area:
\( \displaystyle \frac{4^k-1}{k\log_e 2}=\frac{3}{\log_e 2} \)
\( \displaystyle \frac{4^k-1}{k}=3 \)
This equation is not solved by algebra, so test the values offered:
\( \displaystyle k=1:\ \frac{4-1}{1}=3 \)
\( \displaystyle k=2:\ \frac{16-1}{2}=\frac{15}{2} \)
\( \displaystyle k=-3:\ \frac{\frac{1}{64}-1}{-3}=\frac{21}{64} \)
For k = 0 the curve is y = 1 and the area is 2, not \( \frac{3}{\log_e 2}\approx 4.33 \). So k = 1.