Let `f(x)=|x|+|sin x|, x in (-pi//2,pi//2).` Then, f is

Options

A. nowhere continuous
B. continuous and differentiable everywhere
C. nowhere differentiable
D. differentiable everywhere except at x=0

Answer

D

Explanation

We have
`f(x)=|x|+|sin x|"for all x"in (-pi//2,pi//2)`
`Rightarrow f(x)={{:(,-x-sin x,"for"-pi//2 lt x lt 0),(,x+sin x,"for "0 le x lt x//2):}`
Clearly, `underset(x to 0^(-))lim f(x)=f(0)=underset(x to 0^(+))lim f(x)`
So, f(x) is continuous at x=0
Also, f(x) is continuous on `(-pi//2,0) uu (0,pi//2)`
Thus, f(x) is continuous on `(-pi//2, pi//2)`
Now,
(LHD at x=0)`={(d)/(dx)(-x-sinx)}_("at x=0")=(-1-cos x)_("at x =0")=-1-1=-2`
and
(RHD at x=0) `={(d)/(dx)(x+sinx)}_("at x=0")=(1+cos x)_("at x =0")=1+1=2`
Clearly, (LHD at x=0) `ne (RHD at x=0)`
So, f(x) is not differentiable at x=0
Clearly f(x) is differentiable on `(-pi//2,0) uu (0,pi//2)`
Hence, f(x) is everywhere differentiable except at x=0.