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 the value of $f(x)$ at $x = 0$, which is $k$. We first simplify the expression for $x \neq 0$: \[ \lim_{x \to 0} \frac{\log(1 + ax) + \log(1 - bx)}{x}. \] Using the logarithmic property, $\log A + \log B = \log(AB)$, we get: \[ \lim_{x \to 0} \frac{\log\left((1 + ax)(1 - bx)\right)}{x}. \] Now, expand the product inside the logarithm: \[ (1 + ax)(1 - bx) = 1 + ax - bx - abx^2. \] For small values of $x$, we can ignore the $x^2$ term. Thus, we have: \[ \lim_{x \to 0} \frac{\log(1 + (a - b)x)}{x}. \] Using the approximation $\log(1 + u) \approx u$ for small $u$, we get: \[ \lim_{x \to 0} \frac{(a - b)x}{x} = a - b. \] Therefore, for continuity, $k = a + b$.