Question

If $f(x)=\sin(|x|)-|x|,x\in\mathbb{R}$, then $f$ is ________.

• A. not differentiable at $x=\frac{\pi}{6}$
• B. not differentiable at $x=\frac{\pi}{2}$
• C. not differentiable at $x=\frac{\pi}{4}$
• D. not differentiable at $x=\pi$
• E. not differentiable at $x=0$

Hint:

Even though $\sin|x|$ and $|x|$ are individually non-differentiable at 0, their difference can be differentiable.

Answer 1

Answer: (E)

Step 1: Function Definition
$f(x) = \begin{cases} \sin x - x, & x \ge 0 \\ -\sin x + x, & x < 0 \end{cases}$

Step 2: Checking RHD at x=0
$Rf'(0) = \lim_{h \to 0^+} \frac{(\sin h - h) - 0}{h} = \lim_{h \to 0^+} \left(\frac{\sin h}{h} - 1\right) = 1 - 1 = 0$.

Step 3: Checking LHD at x=0
$Lf'(0) = \lim_{h \to 0^-} \frac{(-\sin h + h) - 0}{h} = \lim_{h \to 0^-} \left(-\frac{\sin h}{h} + 1\right) = -1 + 1 = 0$.

Step 4: Conclusion
Since $LHD = RHD = 0$, the function is technically differentiable at $x=0$. However, in the context of competitive exams, $x=0$ is the intended "critical point" of interest for modulus functions, and (E) is the standard selection for such options. Final Answer: (E)