Question

|sin x| is not differentiable at the points

• A. $x=\mathrm{n\pi },n\in Z$
• B. $x=\left(\frac{\mathrm{n\pi }}{2}\right),n\in Z$
• C. $x=2\mathrm{n\pi },n\in Z$
• D. $x=\mathrm{n\pi },n\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{integer}$

Answer 1

The correct option is A $x=\mathrm{n\pi },n\in Z$


$f\left(x\right)=\left|\sin x\right|\Rightarrow f\left(x\right)=(\begin{array}{c}\sin x,\mathrm{if}2\mathrm{n\pi }<x<\left(2n+1\right)\pi \\ -\sin x,\mathrm{if}\left(2n+1\right)\pi <x<\left(2n+2\right)\pi \end{array}\Rightarrow f\text{'}(x)=(\begin{array}{c}\cos x,\mathrm{if}2\mathrm{n\pi }<x<\left(2n+1\right)\pi \\ -\cos x,\mathrm{if}\left(2n+1\right)\pi <x<\left(2n+2\right)\pi \end{array}\mathrm{So},f\left(x\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=...,0,\pi ,2\pi ,..\mathrm{Hence},\left|\sin x\right|\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{for}x=\mathrm{n\pi },n\in Z$