Let k be a non-zero real number. If…

Let k be a non-zero real number. If $\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{\left(\mathrm{e}^x-1\right)^2}{\sin \left(\frac{x}{\mathrm{k}}\right) \log \left(1+\frac{x}{4}\right)} & , x \neq 0 \\ 12 & , x=0\end{array}\right.$ is a continuous function, then the value of $k$ is
  1. 1
  2. 2
  3. 4
  4. 3

Solution

$\mathrm{f}(x)$ is continuous at $x=0$. $\begin{aligned} \therefore \quad \mathrm{f}(0) & =\lim _{x \rightarrow 0} \mathrm{f}(x) \\ \Rightarrow 12 & =\lim _{x \rightarrow 0} \frac{\left(\mathrm{e}^x-1\right)^2}{\sin \left(\frac{x}{\mathrm{k}}\right) \cdot \log \left(1+\frac{x}{4}\right)} \\ \Rightarrow 12 & =\lim _{x \rightarrow 0} \frac{\frac{\left(\mathrm{e}^x-1\right)^2}{x^2}}{\frac{\sin \left(\frac{x}{\mathrm{k}}\right)}{x} \cdot \log \frac{\left(1+\frac{x}{4}\right)}{x}} \end{aligned}$ $\begin{aligned} & \Rightarrow 12=\frac{\lim _{x \rightarrow 0}\left(\frac{\mathrm{e}^x-1}{x}\right)^2}{\lim _{x \rightarrow 0} \frac{\sin \frac{x}{\mathrm{k}}}{\frac{x}{\mathrm{k}} \mathrm{k}} \cdot \log \lim _{x \rightarrow 0}\left(1+\frac{x}{4}\right)^{\frac{1}{x}}} \\ & \Rightarrow 12=\frac{\mathrm{k} \lim _{x \rightarrow 0}\left(\frac{\mathrm{e}^x-1}{x}\right)^2}{\lim _{x \rightarrow 0} \frac{\sin \frac{x}{\mathrm{k}}}{\frac{x}{\mathrm{k}}} \cdot \log _{x \rightarrow 0} \lim _{x \rightarrow 0}\left[\left(1+\frac{x}{4}\right)^{\frac{4}{x}}\right]^{\frac{1}{4}}}\end{aligned}$ $\begin{aligned} & \Rightarrow 12=\frac{\mathrm{k}}{1 \times \log ^{\frac{1}{4}}} \quad \cdots\left[\begin{array}{l}\lim _{x \rightarrow 0} \frac{\mathrm{e}^x-1}{x}=1 \\ \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=\mathrm{e}\end{array}\right] \\ & \Rightarrow 12=\frac{\mathrm{k}}{\frac{1}{4}} \\ & \Rightarrow 12=4 \mathrm{k} \\ & \Rightarrow \mathrm{k}=3\end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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