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 $x\ne 0$ 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 $x=0$
$\therefore \underset{x\to 0}{\lim }\frac{{\left(e^x-1\right)}^2}{\sin \left(\frac{x}{a}\right)\log \left(1+\frac{x}{4}\right)}=f(0)$
$\Rightarrow \underset{x\to 0}{\lim }\frac{\frac{{\left(e^x-1\right)}^2}{x^2}\cdot x^2}{\frac{\sin \left(\frac{x}{a}\right)}{\left(\frac{x}{a}\right)}\times \left(\frac{x}{a}\right)\cdot \frac{\log \left(1+\frac{x}{4}\right)}{\frac{x}{4}}\times \frac{x}{4}}=12$
$\Rightarrow \underset{x\to 0}{\lim }\frac{\frac{{\left(e^x-1\right)}^2}{x^2}\cdot 4a}{\frac{\sin \left(\frac{x}{3}\right)}{\left(\frac{x}{a}\right)}\cdot \frac{\log \left(1+\frac{x}{4}\right)}{\frac{x}{4}}}=12$
$\Rightarrow 4a=12$
$\Rightarrow a=3$