Question

For the function \(f(x) = \sin|x| - |x|\), \(x \in \mathbb{R}\), consider the following statements:
Statement I: \(f\) is differentiable for all \(x \in \mathbb{R}\).
Statement II: \(f\) is increasing in \((-\pi, -\pi/2)\).
In the light of the above statements, choose the correct answer from the options given below:

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to assess the differentiability of a function involving absolute values, particularly looking at critical points like \( x=0 \). Then, we must determine the monotonicity of the function in a given interval by inspecting the sign of its first derivative.
Step 2: Key Formula or Approach:
Differentiability at \( x=0 \): Check if left-hand derivative (LHD) equals right-hand derivative (RHD).
\( f'_+(0) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x} \)
\( f'_-(0) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x} \)
Monotonicity: Find \( f'(x) \) for \( x \in (-\pi, -\pi/2) \) and verify if \( f'(x)>0 \).
Step 3: Detailed Explanation:
Statement I: Check differentiability.
The only point of concern for \( |x| \) is \( x=0 \). For \( x \neq 0 \), the function is clearly a composition of differentiable functions.
At \( x=0 \), \( f(0) = e^{\sin 0} - 0 = 1 \).
Right-hand derivative at \( x=0 \):
For \( x>0 \), \( f(x) = e^{\sin x} - x \).
\( f'(x) = e^{\sin x} \cos x - 1 \).
\( f'_+(0) = \lim_{x \to 0^+} (e^{\sin x} \cos x - 1) = e^0(1) - 1 = 1 - 1 = 0 \).
Left-hand derivative at \( x=0 \):
For \( x<0 \), \( f(x) = e^{\sin(-x)} - (-x) = e^{-\sin x} + x \).
\( f'(x) = e^{-\sin x} (-\cos x) + 1 \).
\( f'_-(0) = \lim_{x \to 0^-} (-e^{-\sin x} \cos x + 1) = -e^0(1) + 1 = -1 + 1 = 0 \).
Since \( f'_+(0) = f'_-(0) = 0 \), \( f(x) \) is differentiable at \( x=0 \).
Therefore, \( f(x) \) is differentiable for all \( x \in \mathbb{R} \). Statement I is true.
Statement II: Monotonicity in \( (-\pi, -\pi/2) \).
For \( x \in (-\pi, -\pi/2) \), \( x \) is negative. So \( |x| = -x \).
\( f(x) = e^{\sin(-x)} - (-x) = e^{-\sin x} + x \).
Differentiating with respect to \( x \):
\( f'(x) = e^{-\sin x}(-\cos x) + 1 \).
In the interval \( (-\pi, -\pi/2) \), which is in the third quadrant, \( \cos x \) is negative.
Therefore, \( -\cos x \) is positive.
The exponential function \( e^{-\sin x} \) is always positive.
So, \( e^{-\sin x}(-\cos x)>0 \).
This means \( f'(x) = (\text{positive quantity}) + 1>1>0 \).
Since \( f'(x)>0 \) for all \( x \) in the interval, \( f(x) \) is strictly increasing in \( (-\pi, -\pi/2) \).
Thus, Statement II is true.
Step 4: Final Answer:
Both Statement I and Statement II are true.