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: Set up the area:
\[ A = \int_0^2 2^{kx}dx = \left[\frac{2^{kx}}{k\ln2}\right]_0^2 = \frac{2^{2k}-1}{k\ln 2} \]

Step 2: Use the given area:
\(\frac{2^{2k}-1}{k\ln2} = \frac{2}{\ln2}\), so \(2^{2k} - 1 = 2k\).

Step 3: Solve:
Test \(k = \frac12\): \(2^{1} - 1 = 1\), and \(2k = 1\). It satisfies the equation.
Test the others: \(k=1\) gives \(3 \ne 2\). \(k=2\) gives \(15\ne4\). \(k=-\frac12\) gives \(-\frac12 \ne -1\).
So \(k = \frac12\).

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