If the area bounded by curves \( y=2^{kx}\ ,x=0,\ x=2 \) in first quadrant is \( \displaystyle\frac{3}{\log 2} \), then \( k \) is \( (k\ne 0) \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
The area between the curve and the x-axis from x = 0 to x = 2 is
\( \displaystyle \int_0^2 2^{kx}\,dx=\left[\frac{2^{kx}}{k\log 2}\right]_0^2 \)
\( \displaystyle =\frac{4^k-1}{k\log 2} \)
\( \displaystyle \frac{4^k-1}{k\log 2}=\frac{3}{\log 2} \)
\( \displaystyle 4^k-1=3k \)
k = 1 satisfies this: \( 4-1=3 \). The other values in the options fail: for k = 1/2, 2 − 1 = 1 ≠ 3/2; for k = 2, 16 − 1 = 15 ≠ 6; for k = −1, 1/4 − 1 ≠ −3.