Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)=|x|+|\sin x|$ for $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then its left hand derivative at $x=0$ is
Options:
Solution:
2834 Upvotes
Verified Answer
The correct answer is:
-2
$\begin{aligned} & f(x)=|x|+|\sin x| \\ & \text { LHD }=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h} \\ & =\lim _{h \rightarrow 0} \frac{|0-h|+|\sin (0-h)|-(0+0)}{0-h} \\ & =\lim _{h \rightarrow 0} \frac{h+\sin h}{-h}=-\lim _{h \rightarrow 0}\left(1+\frac{\sin h}{h}\right) \\ & =-(1+1)=-2\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.