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.You need to login to perform this action.
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