Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

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

Options:

1

2

3

none of these

Correct Answer:

none of these

Explanation:

For continuity lim = value = 12 (log4)3 (given)

$f(x)=\frac{(4^x-1)^3}{\sin\left(\frac{x}{p}\right)\log\sin\left(1+\frac{x^2}{3}\right)},x≠0$

$\underset{x→0}{\lim}\frac{\left(\frac{4^x-1}{x}\right).x^3}{\frac{\sin (x/p)}{(x/p)}.\left(\frac{x}{p}\right).\frac{x^3}{3}\ln\begin{Bmatrix}1+\frac{x^2}{3}\end{Bmatrix}^{3/x^2}}=3p(\log 4)^3$

$\left[∵\underset{x→0}{\lim}\begin{Bmatrix}1+\frac{x^2}{3}\end{Bmatrix}^{3/x^2}=e\,and\,\underset{x→0}{\lim}\frac{\sin (x/p)}{(x/p)}=1\right]$

$∴3p(\log 4)^3=12(\log 4)^3⇒p=4$