Question

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

• A. 2
• B. 4
• C. 3
• D. 1

Answer 1

Answer: (C)

3

We know that a function is said to be continuous if

${lim}_{x\to a^{+}}f\left(x\right)=f\left(a\right)={lim}_{x\to a^{-}}f\left(x\right)$

$\therefore {lim}_{x\to 0}\frac{{(e^x-1)}^2}{\sin \left(\frac{x}{k}\right)\log (1+\frac{x}{4})}=12$

We know that ${lim}_{x\to 0}\frac{e^x-1}{x}=1,{lim}_{x\to 0}\frac{\sin x}{x}=1,{lim}_{x\to 0}\frac{\log (1+x)}{x}=1$

Using the above formulae,we have

$\Rightarrow {lim}_{x\to 0}\frac{\frac{{(e^x-1)}^2}{x^2}\times x^2}{\frac{\sin \left(\frac{x}{k}\right)}{\frac{x}{k}}\times \frac{x}{k}\frac{\log (1+\frac{x}{4})}{\frac{x}{4}}\times \frac{x}{4}}=12$

$\Rightarrow {lim}_{x\to 0}\frac{{\left(\frac{e^x-1}{x}\right)}^2\times x^2}{\frac{\sin \frac{x}{k}}{\frac{x}{k}}\times \frac{1}{4k}\times x^2\times \frac{\log (1+\frac{x}{4})}{\frac{x}{4}}}=12$

$\Rightarrow \frac{1\times 4k}{1\times 1\times 1}=12$

$\Rightarrow 4k=12$

$\therefore k=\frac{12}{4}=3$