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  • The value of for which the function <img alt="\mathrm{f(x)=\left\{\begin{array}{cc} \frac{\left(4^x-1\right)^3}{\sin \fr

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

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

none of these

 


Answers (1)

best_answer

For f(x)  to be continuous at x = 0, we have

\mathrm{\begin{aligned} \lim _{x \rightarrow 0} f(x)= & f(0)=12(\log 4)^3 \\ \lim _{x \rightarrow 0} f(x)= & \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)} \cdot \frac{p x^2}{\log \left(1+\frac{1}{3} x^2\right)} \\ & (\log 4)^3 \cdot 1 \cdot p \cdot \lim _{x \rightarrow 0}\left(\frac{x^2}{\frac{1}{3} x^2-\frac{1}{18} x^4+\ldots \ldots . . .}\right)=3 p(\log 4)^3 \\ = & \end{aligned}}

Hence, \mathrm{p=4}

Posted by

Deependra Verma

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