We need to check which function is differentiable at \( x = 0 \). Step 1: Differentiability Condition A function \( f(x) \) is differentiable at \( x = 0 \) if: 1. It is continuous at \( x = 0 \): \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \] 2. Left-hand derivative (LHD) equals right-hand derivative (RHD): \[ \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} \] We check these conditions for each function. --- Step 2: Analyze Each Option \[(A) f(x) = \cos(|x|) + |x| \] - \( \cos(|x|) \) is differentiable everywhere. - \( |x| \) is not differentiable at \( x = 0 \) (since LHD and RHD differ). - Thus, not differentiable at \( x = 0 \). \[(B) f(x) = \cos(|x|) - |x| \] - \( \cos(|x|) \) is differentiable everywhere. - \( -|x| \) is not differentiable at \( x = 0 \). - Thus, not differentiable at \( x = 0 \). \[(C) f(x) = \sin(|x|) + |x| \] - \( \sin(|x|) \) is differentiable everywhere. - \( |x| \) is not differentiable at \( x = 0 \). - Thus, not differentiable at \( x = 0 \). \[(D) f(x) = \sin(|x|) - |x| \] - \( \sin(|x|) \) is differentiable everywhere. - \( -|x| \) is not differentiable at \( x = 0 \), but we check derivatives. For \( x > 0 \): \[ f(x) = \sin(x) - x \] \[ f'(x) = \cos(x) - 1 \] For \( x < 0 \): \[ f(x) = \sin(-x) - (-x) = -\sin(x) + x \] \[ f'(x) = -\cos(x) + 1 \] At \( x = 0 \): \[ \lim_{h \to 0^-} f'(h) = -\cos(0) + 1 = -1 + 1 = 0 \] \[ \lim_{h \to 0^+} f'(h) = \cos(0) - 1 = 1 - 1 = 0 \] Since both limits are equal, \( f(x) \) is differentiable at \( x = 0 \). Final Answer \[ \boxed{D} \]