Question

If \[ f(x) = \begin{cases} \frac{\log(1 + ax) + \log(1 - bx)}{x}, & \text{for } x \neq 0 \\ k, & \text{for } x = 0 \end{cases} \] is continuous at $x = 0$, then the value of $k$ is:

• A. $a$
• B. $a + b$
• C. $a - b$
• D. $b$

Hint:

For continuity at $x = 0$, ensure the limit of the function as $x$ approaches 0 equals the function's value at $x = 0$.

Answer 1

The Correct Option is B

For $f(x)$ to be continuous at $x = 0$, the limit of $f(x)$ as $x$ approaches 0 must equal $f(0)$, which is $k$. For $x eq 0$, we simplify the expression: \[ \lim_{x \to 0} \frac{\log(1 + ax) + \log(1 - bx)}{x}. \] Applying the logarithm property $\log A + \log B = \log(AB)$: \[ \lim_{x \to 0} \frac{\log\left((1 + ax)(1 - bx)\right)}{x}. \] Expanding the product inside the logarithm: \[ (1 + ax)(1 - bx) = 1 + ax - bx - abx^2. \] For small $x$, the $x^2$ term is negligible. Thus, we have: \[ \lim_{x \to 0} \frac{\log(1 + (a - b)x)}{x}. \] Using the approximation $\log(1 + u) \approx u$ for small $u$: \[ \lim_{x \to 0} \frac{(a - b)x}{x} = a - b. \] Consequently, for continuity, $k = a - b$.