Question

$f\left(x\right)=\frac{(e^x-1{)}^2}{\sin \left(\frac{x}{a}\right)log(1+\frac{x}{4})}$ when $x\ne 0$ and $f\left(0\right)=12$. If $f\left(x\right)$ is continuous at $x=0$, then $a=$

• A. 3
• B. -1
• C. 2
• D. -2

Answer 1

Answer: (A)

3

$f\left(x\right)=\frac{{(e^x-1)}^2}{\sin \left(\frac{x}{a}\right)\log (1+\frac{x}{4})}$
$f\left(0\right)=12$
$f\left(x\right)$ is continuous at $x=0$
$\therefore \underset{x\to 0}{lim}\frac{{(e^x-1)}^2}{\sin \left(\frac{x}{a}\right)\log (1+\frac{x}{4})}=f\left(0\right)$

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

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

$\Rightarrow 4a=12$
$\Rightarrow a=3$