Question

The area of the region bounded by the curve \(y = 2^{kx}\) and \(x = 0,x = 2\), in first quadrant is \(\frac{2}{log2}\), then the value of \(k\) is

• A. \(1\)
• B. \(2\)
• C. \(-\frac{1}{2}\)
• D. \(\frac{1}{2}\)

Hint:

Integrate \(2^{kx}\) and test the options in \(2^{2k}-1=2k\).

Answer 1

The Correct Option is D

Step 1: Plan:
Think of the area as a function of $k$ and check the sizes.

Step 2: Steps:
If $k = \frac12$, the curve is $y = 2^{x/2} = \sqrt2^{\,x}$, and the area is $\frac{2^{1}-1}{\frac12\ln2} = \frac{1}{\frac12\ln2} = \frac{2}{\ln2}$. That matches.
For $k = 1$, $y = 2^x$, area $= \frac{3}{\ln2}$, too big. For $k=2$ the area is even bigger. For negative $k$ the area is below $2$, but $\frac{2}{\ln 2}\approx2.89 > 2$. So only $k=\frac12$ works.

Final Answer:
The value is $k=\frac12$, option (D). \[ \boxed{\frac12} \]