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:
We need to check differentiability of $f(x) = e^{|\sin x|} - |x|$ at all points, paying special attention to points where the inner expressions $|\sin x|$ and $|x|$ lose differentiability.
We also need to check if $f'(x)>0$ on $\left(-\pi, -\dfrac{\pi}{2}\right)$.
Step 2: Checking Statement 1 -- Differentiability:
The function $|x|$ is non-differentiable only at $x = 0$, but $e^{|\sin 0|} = e^0 = 1$ is smooth there, so we check $x = 0$ carefully.
More critically, $|\sin x|$ is non-differentiable at points where $\sin x = 0$, i.e., at $x = n\pi$ for $n \in \mathbb{Z}$.
At $x = \pi$: the left-hand derivative of $e^{|\sin x|}$ uses the expression from the left side $(\sin x>0)$, giving $e^{\sin x}\cos x$, and from the right $(\sin x<0$, so $|\sin x| = -\sin x)$, giving $-e^{-\sin x}\cos x$.
At $x = \pi$, $\cos\pi = -1$, left derivative $= e^0\cdot(-1) = -1$ and right derivative $= -e^0\cdot(-1) = 1$.
These are not equal, so $e^{|\sin x|}$ is not differentiable at $x = \pi$.
Therefore $f(x)$ is not differentiable at $x = \pi$ (and all $x = n\pi,\, n \neq 0$).
Statement 1 is FALSE.
Step 3: Checking Statement 2 -- Monotonicity on $\left(-\pi, -\dfrac{\pi}{2}\right)$:
For $x \in \left(-\pi, -\dfrac{\pi}{2}\right)$: we have $x<0$ so $|x| = -x$, and $\sin x<0$ (since $x$ is in the third quadrant region for sine), so $|\sin x| = -\sin x$.
Therefore on this interval: \[ f(x) = e^{-\sin x} - (-x) = e^{-\sin x} + x \] Differentiating: \[ f'(x) = -e^{-\sin x}\cdot\cos x + 1 \] For $x \in \left(-\pi, -\dfrac{\pi}{2}\right)$: $\cos x<0$ (second quadrant cosine-wise), so $-e^{-\sin x}\cdot\cos x = e^{-\sin x}\cdot(-\cos x)$.
Since $e^{-\sin x}>0$ and $-\cos x>0$ (because $\cos x<0$), we have: \[ -e^{-\sin x}\cos x>0 \] Therefore $f'(x) = (\text{positive}) + 1>0$ on $\left(-\pi, -\dfrac{\pi}{2}\right)$.
Statement 2 is TRUE.
Step 4: Final Answer:
Statement 1 is false and Statement 2 is true.
The answer is Option (3).