Question

Let f (x) = |sin x|. Then,
(a) f (x) is everywhere differentiable.
(b) f (x) is everywhere continuous but not differentiable at x = n π, n ∈ Z
(c) f (x) is everywhere continuous but not differentiable at $x=\left(2n+1\right)\frac{\pi }{2},n\in Z$.
(d) none of these

Answer 1

(b) f (x) is everywhere continuous but not differentiable at x = n π, n ∈ Z

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=\left|\sin x\right| \\ \Rightarrow f\left(x\right)=\left\{\begin{array}{l}\begin{array}{l}0,x=2n\mathrm{\pi }\\ \sin x,2n\mathrm{\pi }<x<\left(2n+1\right)\mathrm{\pi }\end{array}\\ \begin{array}{l}0,x=\left(2n+1\right)\mathrm{\pi }\\ -\sin x,\left(2n+1\right)\mathrm{\pi }<x<\left(2n+2\right)\mathrm{\pi }\end{array}\end{array}\right. \\ \mathrm{When},x\mathrm{is}\mathrm{in}\mathrm{first}\mathrm{or}\mathrm{second}\mathrm{quadrant},\mathrm{i}.\mathrm{e}.,2n\mathrm{\pi }<x<\left(2\mathrm{n}+1\right)\pi ,\mathrm{we}\mathrm{have} \\ f\left(x\right)=\sin x\mathrm{which}\mathrm{being}\mathrm{a}\mathrm{trigonometrical}\mathrm{function}\mathrm{is}\mathrm{continuous}\mathrm{and}\mathrm{differentiable}\mathrm{in}\left(2n\mathrm{\pi },\left(2n+1\right)\mathrm{\pi }\right) \\ \mathrm{When},x\mathrm{is}\mathrm{in}\mathrm{third}\mathrm{or}\mathrm{fourth}\mathrm{quadrant},\mathrm{i}.\mathrm{e}.,\left(2n+1\right)\mathrm{\pi }<x<\left(2n+2\right)\mathrm{\pi },\mathrm{we}\mathrm{have} \\ f\left(x\right)=-\sin x\mathrm{which}\mathrm{being}\mathrm{a}\mathrm{trigonometrical}\mathrm{function}\mathrm{is}\mathrm{continuous}\mathrm{and}\mathrm{differentiable}\mathrm{in}\left(\left(2n+1\right)\mathrm{\pi },\left(2n+2\right)\mathrm{\pi }\right) \\ \mathrm{Thus}\mathrm{possible}\mathrm{point}\mathrm{of}\mathrm{non}-\mathrm{differentiability}\mathrm{of}f\left(x\right)\mathrm{are}x=2n\mathrm{\pi }\mathrm{and}\left(2n+1\right)\mathrm{\pi } \\ \mathrm{Now},\mathrm{LHD}\left[\mathrm{at}x=2n\mathrm{\pi }\right]=\underset{x\to 2n{\mathrm{\pi }}^{-}}{\lim }\frac{f\left(x\right)-f\left(2n\mathrm{\pi }\right)}{x-2n\mathrm{\pi }} \\ =\underset{x\to 2n{\mathrm{\pi }}^{-}}{\lim }\frac{-\sin x-0}{x-2n\mathrm{\pi }} \\ =\underset{x\to 2n{\mathrm{\pi }}^{-}}{\lim }\frac{-\cos x}{1-0}\left[\mathrm{By}\mathrm{L}\text{'}\mathrm{Hospital}\mathrm{rule}\right] \\ =-1 \\ \mathrm{And}\mathrm{RHD}\left(\mathrm{at}x=2n\mathrm{\pi }\right)=\underset{x\to 2n{\mathrm{\pi }}^{+}}{\lim }\frac{f\left(x\right)-f\left(2n\mathrm{\pi }\right)}{x-2n\mathrm{\pi }} \\ =\underset{x\to 2n{\mathrm{\pi }}^{+}}{\lim }\frac{\sin x-0}{x-2n\mathrm{\pi }} \\ =\underset{x\to 2n{\mathrm{\pi }}^{+}}{\lim }\frac{\cos x}{1-0}\left[\mathrm{By}\mathrm{L}\text{'}\mathrm{Hospital}\mathrm{rule}\right] \\ =1 \\ \therefore \underset{x\to 2n{\mathrm{\pi }}^{-}}{\lim }f\left(x\right)\ne \underset{x\to 2n{\mathrm{\pi }}^{+}}{\lim }f\left(x\right) \\ \mathrm{So}f\left(x\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=2n\mathrm{\pi } \\ \mathrm{Now},\mathrm{LHD}\left[\mathrm{at}x=\left(2n+1\right)\mathrm{\pi }\right]=\underset{x\to \left(2n+1\right){\mathrm{\pi }}^{-}}{\lim }\frac{f\left(x\right)-f\left(\left(2n+1\right)\mathrm{\pi }\right)}{x-\left(2n+1\right)\mathrm{\pi }} \\ =\underset{x\to \left(2n+1\right){\mathrm{\pi }}^{-}}{\lim }\frac{\sin x-0}{x-\left(2n+1\right)\mathrm{\pi }} \\ =\underset{x\to \left(2n+1\right){\mathrm{\pi }}^{-}}{\lim }\frac{\cos x}{1-0}\left[\mathrm{By}\mathrm{L}\text{'}\mathrm{Hospital}\mathrm{rule}\right] \\ =-1 \\ \mathrm{And}\mathrm{RHD}\left(\mathrm{at}x=\left(2n+1\right)\mathrm{\pi }\right)=\underset{x\to \left(2n+1\right){\mathrm{\pi }}^{+}}{\lim }\frac{f\left(x\right)-f\left(\left(2n+1\right)\mathrm{\pi }\right)}{x-\left(2n+1\right)\mathrm{\pi }} \\ =\underset{x\to \left(2n+1\right){\mathrm{\pi }}^{+}}{\lim }\frac{-\sin x-0}{x-\left(2n+1\right)\mathrm{\pi }} \\ =\underset{x\to \left(2n+1\right){\mathrm{\pi }}^{+}}{\lim }\frac{-\cos x}{1-0}\left[\mathrm{By}\mathrm{L}\text{'}\mathrm{Hospital}\mathrm{rule}\right] \\ =1 \\ \therefore \underset{x\to \left(2n+1\right){\mathrm{\pi }}^{-}}{\lim }f\left(x\right)\ne \underset{x\to \left(2n+1\right){\mathrm{\pi }}^{+}}{\lim }f\left(x\right) \\ \mathrm{So}f\left(x\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=\left(2n+1\right)\mathrm{\pi } \\ \mathrm{Therefore},f\left(x\right)\mathrm{is}\mathrm{neither}\mathrm{differentiable}\mathrm{at}2n\mathrm{\pi }\mathrm{nor}\mathrm{at}\left(2n+1\right)\mathrm{\pi } \\ \mathrm{i}.\mathrm{e}.f\left(x\right)\mathrm{is}\mathrm{neither}\mathrm{differentiable}\mathrm{at}\mathrm{even}\mathrm{multiple}\mathrm{of}\pi \mathrm{nor}\mathrm{at}\mathrm{odd}\mathrm{multiple}\mathrm{of}\mathrm{\pi } \\ \mathrm{i}.\mathrm{e}.f\left(x\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}x=n\mathrm{\pi } \\ \mathrm{Therefore},f\left(x\right)\mathrm{is}\mathrm{everywhere}\mathrm{continuous}\mathrm{but}\mathrm{not}\mathrm{differentiable}\mathrm{at}\mathit{}n\pi . \end{gathered}$$