Question

If \(f(x)\) is continuous at \(x = 0\), where \(f(x) = \frac{8^x-2^x}{k^x-1}\), for \(x\neq 0\) and \(f(0) = 2\), then the value of \(k\) is ...

• A. \(0\)
• B. \(4\)
• C. \(-2\)
• D. \(2\)

Hint:

Continuity needs the limit at 0 to equal 2; factor \(2^x\) from the numerator and use \(\lim\frac{a^x-1}{x}=\ln a\).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
For continuity at \(0\) we need \(\lim_{x\to0}f(x) = f(0) = 2\).

Step 2: Evaluate the limit:
\(8^x - 2^x = 2^x(4^x - 1)\). So \(f(x) = 2^x\cdot\dfrac{4^x-1}{k^x-1}\).
Divide top and bottom by \(x\): \(\dfrac{(4^x-1)/x}{(k^x-1)/x}\to\dfrac{\ln4}{\ln k}\), and \(2^x\to1\).
\[ \lim_{x\to0}f(x) = \frac{\ln 4}{\ln k} \]

Step 3: Solve for k:
\(\frac{\ln4}{\ln k} = 2\), so \(\ln k = \frac{\ln 4}{2} = \ln 2\), and \(k = 2\).
\(k = 0\) is undefined for the logarithm, \(k=-2\) makes \(k^x\) undefined for general \(x\), and \(k = 4\) would give a limit of \(1\).

Final Answer:
The value of \(k\) is \(2\), option (D). \[ \boxed{2} \]