Question

Let k be a non-zero real number. If $f(x) = \begin{cases} \frac{\left(e^x-1\right)^2}{sin\left(\frac{x}{k}\right)log\left(1+\frac{x}{4}\right)}, & \text{x $\ne$ 0} \\[2ex] 12, & \text{x = 0} \end{cases}$ is a continuous function, then the value of $k$ is :

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is C

To determine the value of $k$ for which the given function $f(x)$ is continuous at $x = 0$, we need to ensure that the left-hand limit, the right-hand limit, and the value of the function at this point are all equal. Let's analyze the problem step-by-step:

1. Given the piecewise function:
   f(x) = \begin{cases} \frac{\left(e^x-1\right)^2}{\sin\left(\frac{x}{k}\right)\log\left(1+\frac{x}{4}\right)}, & x \neq 0 \\[2ex] 12, & x = 0 \end{cases}
2. For the function to be continuous at $x = 0$, the limit of $f(x)$ as $x$ approaches 0 must be equal to 12. That is, \lim_{{x \to 0}} f(x) = f(0) = 12.
3. Calculate the limit:
   \lim_{{x \to 0}} \frac{(e^x - 1)^2}{\sin\left(\frac{x}{k}\right)\log\left(1 + \frac{x}{4}\right)}
4. Using the Taylor series expansions around $x = 0$:
   • e^x - 1 \approx x \, \text{as} \, x \to 0
   • \sin\left(\frac{x}{k}\right) \approx \frac{x}{k} \, \text{as} \, x \to 0
   • \log\left(1 + \frac{x}{4}\right) \approx \frac{x}{4} \, \text{as} \, x \to 0
5. Substitute these approximations into the limit:
   \lim_{{x \to 0}} \frac{x^2}{\frac{x}{k} \cdot \frac{x}{4}} = \lim_{{x \to 0}} \frac{x^2}{\frac{x^2}{4k}}
   = \lim_{{x \to 0}} \frac{4k x^2}{x^2} = 4k
6. Set this limit equal to the function value at $x = 0$:
   4k = 12
7. Solve for $k$:
   k = \frac{12}{4} = 3
8. Therefore, the value of $k$ is $3$, which makes the function continuous at $x = 0$.

Hence, the correct answer is 3.