question_answer

The value of p for which the function \[f(x)=\left\{ \begin{align}   & \frac{{{({{4}^{x}}-1)}^{3}}}{\sin \frac{x}{p}\log \left[ 1+\frac{{{x}^{2}}}{3} \right]},x\ne 0 \\  & 12{{(log\,4)}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x=0 \\ \end{align} \right.\] may be continuous at x = 0, is

A)  1                    

B)  2    

C)  3                    

D)  4

Correct Answer: D

Solution :

[d]: For f(x) to be continuous at x = 0, we have \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)=12{{(log4)}^{3}}\]                ?(i) Now, \[={{(log4)}^{3}}.1.p.\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{p{{x}^{2}}}{\frac{1}{3}{{x}^{2}}-\frac{1}{18}{{x}^{4}}+...} \right)\] \[=3p{{(log\,4)}^{3}}\]                                     ...(ii) \[\therefore \]On comparing (i) and (ii), we get p = 4.