Practicing Success
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 |
1 2 3 none of these |
none of these |
For continuity lim = value = 12 (log4)3 (given) $f(x)=\frac{(4^x-1)^3}{\sin\left(\frac{x}{p}\right)\log\sin\left(1+\frac{x^2}{3}\right)},x≠0$ $\underset{x→0}{\lim}\frac{\left(\frac{4^x-1}{x}\right).x^3}{\frac{\sin (x/p)}{(x/p)}.\left(\frac{x}{p}\right).\frac{x^3}{3}\ln\begin{Bmatrix}1+\frac{x^2}{3}\end{Bmatrix}^{3/x^2}}=3p(\log 4)^3$ $\left[∵\underset{x→0}{\lim}\begin{Bmatrix}1+\frac{x^2}{3}\end{Bmatrix}^{3/x^2}=e\,and\,\underset{x→0}{\lim}\frac{\sin (x/p)}{(x/p)}=1\right]$ $∴3p(\log 4)^3=12(\log 4)^3⇒p=4$ |