Question

Statement 1: \( f(x) = e^{|\sin x| - |x|} \) is differentiable for all \( x \in \mathbb{R} \).
Statement 2: \( f(x) \) is increasing in \( x \in \left( -\pi, -\frac{\pi}{2} \right) \).

• A. Statement 1 and 2 are false
• B. Statement 1 is true and statement 2 is false
• C. Statement 1 is false and statement 2 is true
• D. Statement 1 and 2 are true

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Differentiability of \(|g(x)|\) fails where \(g(x) = 0\) and \(g'(x) \neq 0\). Monotonicity is checked by the sign of the derivative \(f'(x)\).

Step 2: Key Formula or Approach:
For \(x \in (-\pi, -\pi/2)\), \(\sin x\) is negative and \(x\) is negative. Thus, \( |\sin x| = -\sin x \) and \( |x| = -x \). The function becomes: \( f(x) = e^{-\sin x} + x \).

Step 3: Detailed Explanation:
1. **Differentiability (Statement 1):** - \(|x|\) is not differentiable at \(x=0\). - \(e^{|\sin x|}\) is not differentiable where \(\sin x = 0\) (like \(x=0, \pi, \dots\)). - At \(x=0\), LHD of \(f(x)\) is \(-1 - (-1) = 0\) and RHD is \(1 - 1 = 0\). (Actually, check carefully: \(f'(0)\) exists, but \(f(x)\) fails differentiability at other points where sine crosses zero). **Statement 1 is generally false.** 2. **Increasing Nature (Statement 2):** - In \((-\pi, -\pi/2)\), \( f(x) = e^{-\sin x} + x \). - \( f'(x) = e^{-\sin x}(-\cos x) + 1 \). - In the 3rd quadrant \((-\pi, -\pi/2)\), \(\cos x\) is negative. - So, \(-\cos x\) is positive. Since \(e^{-\sin x} > 0\), the whole term \(e^{-\sin x}(-\cos x) + 1 > 0\). - Since \(f'(x) > 0\), the function is increasing. **Statement 2 is true.**

Step 4: Final Answer:
Statement 1 is false and Statement 2 is true.