Question

Which of the following is differentiable at \(x=0\)?

• A. \(f(x)=\cos|x|+|x|\)
• B. \(f(x)=\sin|x|+|x|\)
• C. \(f(x)=\cos|x|-|x|\)
• D. \(f(x)=\sin|x|-|x|\)

Hint:

For functions involving \(|x|\), always replace \(|x|\) by \(x\) for \(x\gt 0\) and by \(-x\) for \(x\lt 0\), then compare the left and right derivatives.

Answer 1

The Correct Option is D

Step 1: Examine each function near \(x=0\).
Since \[ |x|= \begin{cases} x,&x\gt 0,\\ -x,&x\lt 0, \end{cases} \] we compare the left-hand and right-hand derivatives at \(x=0\).

Step 2: Check Option (1).
\[ f(x)=\cos|x|+|x|. \] For \(x\gt 0\), \[ f(x)=\cos x+x. \] Hence, \[ f'_+(0)=-\sin0+1=1. \] For \(x\lt 0\), \[ f(x)=\cos x-x. \] Hence, \[ f'_-(0)=-\sin0-1=-1. \] Since \[ f'_+(0)\neq f'_-(0), \] the function is not differentiable at \(0\).

Step 3: Check Option (2).
\[ f(x)=\sin|x|+|x|. \] For \(x\gt 0\), \[ f(x)=\sin x+x, \] so \[ f'_+(0)=2. \] For \(x\lt 0\), \[ f(x)=-\sin x-x, \] so \[ f'_-(0)=-2. \] Hence, not differentiable.

Step 4: Check Option (3).
\[ f(x)=\cos|x|-|x|. \] For \(x\gt 0\), \[ f'_+(0)=-1. \] For \(x\lt 0\), \[ f'_-(0)=1. \] Therefore, not differentiable.

Step 5: Check Option (4).
\[ f(x)=\sin|x|-|x|. \] For \(x\gt 0\), \[ f(x)=\sin x-x. \] Thus, \[ f'_+(0)=\cos0-1=0. \] For \(x\lt 0\), \[ f(x)=-\sin x+x. \] Thus, \[ f'_-(0)=-\cos0+1=0. \] Since \[ f'_+(0)=f'_-(0)=0, \] the function is differentiable at \(x=0\).

Step 6: Final conclusion.
Therefore, \[ \boxed{f(x)=\sin|x|-|x|} \]