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

For continuity at $x=0$
$ \displaystyle\lim _{x \rightarrow \theta}\left\{\frac{\left(e^{x}-1\right)^{2}}{\sin \left(\frac{x}{k}\right) \cdot \ln \left(1+\frac{x}{4}\right)}\right\}=12$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0}\left[\frac{\left(\frac{e^{x}-1}{x}\right)^{2}}{\frac{\sin \left(\frac{x}{k}\right)}{k\left(\frac{x}{k}\right)} \cdot \frac{\ln \left(1+\frac{x}{4}\right)}{4 \cdot\left(\frac{x}{4}\right)}}\right]$
$\Rightarrow \left\{\frac{(1)^{2}}{\left(\frac{1}{k}\right)} \cdot \frac{1}{\frac{1}{4}(1)}\right\}=12$
$\Rightarrow 4 k=12 \Rightarrow k=3$