Question

The value of a for which the function
$f\left(x\right)=\left\{\begin{array}{ll}\frac{{\left(4^x-1\right)}^3}{\sin \left(x/a\right)\log \left\{\left(1+x^2/3\right)\right\}},& x\ne 0\\ 12{\left(\log 4\right)}^3,& x=0\end{array}\right.$may be continuous at x = 0 is
(a) 1
(b) 2
(c) 3
(d) none of these

Answer 1

(d) none of these

For f(x) to be continuous at $x=0$, we must have
$\underset{\mathrm{x}\to 0}{\lim }f\left(x\right)=f\left(0\right)$

$\underset{x\to 0}{\lim }\left[\frac{{\left(4^x-1\right)}^3}{\sin {\displaystyle \frac{x}{a}}\log \left({\displaystyle 1+\frac{{\displaystyle x^2}}{3}}\right)}\right]=12{\left(\log 4\right)}^3$


$$\begin{gathered} \Rightarrow \underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{{\left(4^x-1\right)}^3}{x^3}}}{{\displaystyle \frac{\sin {\displaystyle \frac{x}{a}}\log \left({\displaystyle 1+\frac{{\displaystyle x^2}}{3}}\right)}{x^3}}}\right]=12{\left(\log 4\right)}^3 \\ \Rightarrow \underset{x\to 0}{\lim }\left[\frac{{\displaystyle a{\left(\frac{4^x-1}{x}\right)}^3}}{{\displaystyle \left(\frac{\sin {\displaystyle \frac{x}{a}}}{{\displaystyle \frac{x}{a}}}\right)\frac{\log \left({\displaystyle 1+\frac{{\displaystyle x^2}}{3}}\right)}{x^2}}}\right]=12{\left(\log 4\right)}^3 \\ \Rightarrow 3a\underset{x\to 0}{\lim }\left[\frac{{\displaystyle {\left(\frac{4^x-1}{x}\right)}^3}}{{\displaystyle \left(\frac{\sin {\displaystyle \frac{x}{a}}}{{\displaystyle \frac{x}{a}}}\right)\frac{\log \left({\displaystyle 1+\frac{{\displaystyle x^2}}{3}}\right)}{\left({\displaystyle \frac{x^2}{3}}\right)}}}\right]=12{\left(\log 4\right)}^3 \\ \Rightarrow 3a\left[\frac{{\displaystyle \underset{x\to 0}{\lim }{\left(\frac{4^x-1}{x}\right)}^3}}{\underset{x\to 0}{\lim }{\displaystyle \left(\frac{\sin {\displaystyle \frac{x}{a}}}{{\displaystyle \frac{x}{a}}}\right)\underset{x\to 0}{\lim }\frac{\log \left({\displaystyle 1+\frac{{\displaystyle x^2}}{3}}\right)}{\left({\displaystyle \frac{x^2}{3}}\right)}}}\right]=12{\left(\log 4\right)}^3 \\ \Rightarrow 3a{\left(\log 4\right)}^3=12{\left(\log 4\right)}^3\left[\because \underset{x\to 0}{\lim }\left(\frac{a^x-1}{x}\right)=\log a,\underset{x\to 0}{\lim }\frac{\log \left({\displaystyle 1+x}\right)}{x}=1\mathrm{and}\underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right] \\ \Rightarrow a=4 \end{gathered}$$