Answer:
Option C
Explanation:
Given , p(x)$=\begin{cases}\frac{log(1+2x)\sin x^{0}}{x^{2}} &for x\neq0\\k & for x=0\end{cases}$
= $f(x)=\begin{cases}\frac{log(1+2x)\sin \frac{\pi x}{180}}{x^{2}} &for x\neq0\\k & for x=0\end{cases}$
Since f(x) is continuous at x=0
$\therefore$ LHL=f(0)
Now, LHL= $\lim_{x \rightarrow 0-}f(x)=\lim_{h \rightarrow 0}f(0-h)$
=$\lim_{h \rightarrow 0}\frac{\log(1+2(0-h)\sin\frac{\pi}{180^{0}}(0-h))}{(0-h)^{2}}$
=$\lim_{h \rightarrow 0}\frac{\log(1-2h)\left\{-\sin\frac{\pi h}{180}\right\}}{h^{2}}$
= $\lim_{h \rightarrow 0}(-2)\frac{\log(1-2h)}{-2h}\times(-)\lim_{h \rightarrow 0}\frac{\sin \frac{\pi h}{180}}{\frac{\pi h}{180}}\times\frac{\pi}{180}$
= $(-2)\times(-1)\times 1\times \frac{\pi}{180}$
$\left[\because \lim_{x \rightarrow 0}\log\frac{1+x}{x}=1 and \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\right]$
=$\frac{\pi}{90}$
