Question

If $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{{(4^x-1)}^3}{\sin \left(\frac{x}{a}\right)\log [1+\left(\frac{x^2}{3}\right)]},& x\ne 0\end{array}\\ \begin{array}{cc}9{\left(\log 4\right)}^3,& x=0\end{array}\end{array}$ is a continuous function at $x=0$, then the value of $a$ is equal to

• A. $3$
• B. $1$
• C. $2$
• D. $0$

Answer 1

Answer: (A)

$3$

We have $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{{(4^x-1)}^3}{\sin \left(\frac{x}{a}\right)\log [1+\left(\frac{x^2}{3}\right)]},& x\ne 0\end{array}\\ \begin{array}{cc}9{\left(\log 4\right)}^3,& x=0\end{array}\end{array}$

For $f\left(x\right)$ to be continuous at $x=0$, we should have

$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)=9{\left(\log 4\right)}^3$

$\therefore \underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}{\left(\frac{4^x-1}{x}\right)}^3\frac{\frac{x}{a}}{\sin \left(\frac{x}{a}\right)}.\frac{{ax}^2}{\log (1+\frac{x^2}{3})}$

$={\left(\log 4\right)}^{3}a\underset{x\to 0}{lim}\left(\frac{x^2}{\frac{1}{3}{x}^2-\frac{1}{18}{x}^4+...}\right)$

$\Rightarrow 9{\left(\log 4\right)}^3=3a{\left(\log 4\right)}^3$

$\Rightarrow a=3$