Functions Question 650
Question: The value of $ p $ for which the function $ f(x)= \begin{cases} & \frac{{{(4^{x}-1)}^{3}}}{\sin \frac{x}{p}\log
[ 1+\frac{x^{2}}{3} ]},,x\ne 0 \\ & ,12{{(\log 4)}^{3}},x=0 \\ \end{cases} . $ may be continuous at $ x=0 $ , is [Orissa JEE 2004]
Options:
A) 1
B) 2
C) 3
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
For $ f(x) $ to be continuous at $ x=0, $ we should have $ \underset{x\to 0}{\mathop{\lim }},f(x)=f(0)=12,{{(\log ,4)}^{3}} $ $ \underset{x\to 0}{\mathop{\lim }},f(x)=\underset{x\to 0}{\mathop{\lim }}{{( \frac{4^{x}-1}{x} )}^{3}}\times \frac{( \frac{x}{p} )}{( \sin \frac{x}{p} )}.\frac{px^{2}}{\log ,( 1+\frac{1}{3}x^{2} )} $
$ ={{(\log 4)}^{3}},.,1,.,p,.\underset{x\to 0}{\mathop{\lim }}( \frac{x^{2}}{\frac{1}{3}x^{2}-\frac{1}{18}x^{4}+………} ) $
$ =3p{{(\log 4)}^{3}}. $ Hence $ p=4. $
BETA
