Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

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