Let `f(x)=|x|+|sin x|, x in (-pi//2,pi//2).` Then, f isA. nowhere continuousB. continuous and differentiable everywhereC. nowhere differentiableD. differentiable everywhere except at x=0

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

1.	Let `f(x)=\|x\|+\|sin x\|, x in (-pi//2,pi//2).` Then, f isA. nowhere continuousB. continuous and differentiable everywhereC. nowhere differentiableD. differentiable everywhere except at x=0
Answer» Correct Answer - D 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.