Question

Let $f(x)=\frac{\left(e^{x}-1\right)^{2}}{\sin \left(\frac{x}{a}\right) \log \left(1+\frac{x}{4}\right)}$ for $x \neq 0$ and $f(0)=12$. If $f$ is continuous at $x=0$, then the value of $a$ is equal to

• A. 1
• B. -1
• C. 2
• D. 3

Answer 1

Answer: (D)

$f(x)=\frac{\left(e^{x}-1\right)^{2}}{\sin \left(\frac{x}{a}\right) \log \left(1+\frac{x}{4}\right)}$
$f(0)=12$
$f(x)$ is continuous at $x=0$
$\therefore \displaystyle\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)^{2}}{\sin \left(\frac{x}{a}\right) \log \left(1+\frac{x}{4}\right)}=f(0)$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{\frac{\left(e^{x}-1\right)^{2}}{x^{2}} \cdot x^{2}}{\frac{\sin \left(\frac{x}{a}\right)}{\left(\frac{x}{a}\right)} \times\left(\frac{x}{a}\right) \cdot \frac{\log \left(1+\frac{x}{4}\right)}{\frac{x}{4}} \times \frac{x}{4}}=12$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{\frac{\left(e^{x}-1\right)^{2}}{x^{2}} \cdot 4 a}{\frac{\sin \left(\frac{x}{3}\right)}{\left(\frac{x}{a}\right)} \cdot \frac{\log \left(1+\frac{x}{4}\right)}{\frac{x}{4}}}=12$
$\Rightarrow 4 a =12$
$\Rightarrow a =3$