MCQ
The value of $p$ for which the function
$f(x)=\left\{\begin{array}{ll}\frac{\left(4^{\prime}-1\right)^3}{\sin \left(\frac{x}{p}\right) \log \left(1+\frac{x^2}{3}\right)} & , x \neq 0 \$12)(\log 4)^3 & , x=0\end{array}\right. $
may be continuous at $x=0$, is
  • A
    1
  • B
    2
  • C
    3
  • 4

Answer

Correct option: D.
4
(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow 12(\log 4)^3=\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)^3}{\sin \frac{x}{p} \log \left(1+\frac{x^2}{3}\right)}$
$\Rightarrow 12(\log 4)^3$ $=\lim _{x \rightarrow 0}\left(\frac{4^x-1}{x}\right)^3 \times \frac{\left(\frac{x}{ p }\right)}{\left(\sin \frac{x}{ p }\right)} \times \frac{ p }{\frac{\log \left(1+\frac{1}{3} x^2\right)}{\frac{x^2}{3} \times 3}}$
$\Rightarrow 12(\log 4)^3=(\log 4)^3(1)\left(\frac{3 p }{1}\right)$
$\Rightarrow p =4$

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