JEE Challenger
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MathMedMain 2026 • 8th Apr - Shift 2

Differentiability and Monotonicity Analysis of Exponential Absolute Sine Function

For the function f(x)=esin⁡∣x∣−∣x∣,x∈Rf(x) = e^{\sin|x|} - |x|, x \in \mathbf{R}, consider the following statements : Statement I : ff is differentiable for all x∈Rx \in \mathbf{R}. Statement II : ff is increasing in (−π,−π2)\left(-\pi, -\frac{\pi}{2}\right).

In the light of the above statements, choose the correct answer from the options given below :

Options

A

Both Statement I and Statement II are true

Correct
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

Topics & Concepts

Continuity and DifferentiabilityDifferentiabilityMonotonicityFunctions

Step-by-Step Solution

To determine the correctness of the given statements, let us analyze the function f(x)=esin⁡∣x∣−∣x∣f(x) = e^{\sin|x|} - |x| for x∈Rx \in \mathbf{R}.

Analysis of Statement I:

We need to check the differentiability of f(x)f(x) for all x∈Rx \in \mathbf{R}.

  1. For x>0x > 0: f(x)=esin⁡x−xf(x) = e^{\sin x} - x Since esin⁡xe^{\sin x} and xx are differentiable functions, f(x)f(x) is differentiable for all x>0x > 0. f′(x)=esin⁡xcos⁡x−1f'(x) = e^{\sin x} \cos x - 1

  2. For x<0x < 0: f(x)=esin⁡(−x)−(−x)=e−sin⁡x+xf(x) = e^{\sin(-x)} - (-x) = e^{-\sin x} + x Similarly, f(x)f(x) is differentiable for all x<0x < 0. f′(x)=−e−sin⁡xcos⁡x+1f'(x) = -e^{-\sin x} \cos x + 1

  3. At x=0x = 0: Let us find the Right-Hand Derivative (RHD) and Left-Hand Derivative (LHD) at x=0x = 0:

    • RHD at x=0x = 0: RHD=lim⁡h→0+f(0+h)−f(0)h=lim⁡h→0+esin⁡h−h−1h\text{RHD} = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{e^{\sin h} - h - 1}{h} RHD=lim⁡h→0+(esin⁡h−1sin⁡h⋅sin⁡hh−1)=(1⋅1)−1=0\text{RHD} = \lim_{h \to 0^+} \left( \frac{e^{\sin h} - 1}{\sin h} \cdot \frac{\sin h}{h} - 1 \right) = (1 \cdot 1) - 1 = 0

    • LHD at x=0x = 0: LHD=lim⁡h→0+f(0−h)−f(0)−h=lim⁡h→0+esin⁡h−h−1−h=−0=0\text{LHD} = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{e^{\sin h} - h - 1}{-h} = -0 = 0

    Since RHD=LHD=0\text{RHD} = \text{LHD} = 0, f′(0)f'(0) exists and is equal to 00.

Hence, f(x)f(x) is differentiable for all x∈Rx \in \mathbf{R}. Statement I is TRUE.

Analysis of Statement II:

We need to determine if f(x)f(x) is increasing in the interval (−π,−π2)\left(-\pi, -\frac{\pi}{2}\right).

For x∈(−π,−π2)x \in \left(-\pi, -\frac{\pi}{2}\right), we have x<0x < 0, so ∣x∣=−x|x| = -x. f(x)=e−sin⁡x+xf(x) = e^{-\sin x} + x

Differentiating f(x)f(x) with respect to xx: f′(x)=ddx(e−sin⁡x+x)=1−e−sin⁡xcos⁡xf'(x) = \frac{d}{dx}\left(e^{-\sin x} + x\right) = 1 - e^{-\sin x}\cos x

In the third quadrant, i.e., x∈(−π,−π2)x \in \left(-\pi, -\frac{\pi}{2}\right):

  • sin⁡x<0  ⟹  −sin⁡x>0  ⟹  e−sin⁡x>1>0\sin x < 0 \implies -\sin x > 0 \implies e^{-\sin x} > 1 > 0
  • cos⁡x<0  ⟹  −cos⁡x>0\cos x < 0 \implies -\cos x > 0

Thus, the term −e−sin⁡xcos⁡x=e−sin⁡x(−cos⁡x)-e^{-\sin x}\cos x = e^{-\sin x}(-\cos x) is strictly positive. f′(x)=1+e−sin⁡x(−cos⁡x)>1>0f'(x) = 1 + e^{-\sin x}(-\cos x) > 1 > 0

Since f′(x)>0f'(x) > 0 for all x∈(−π,−π2)x \in \left(-\pi, -\frac{\pi}{2}\right), f(x)f(x) is strictly increasing in (−π,−π2)\left(-\pi, -\frac{\pi}{2}\right). Statement II is TRUE.

Conclusion:

Both Statement I and Statement II are true.

Correct Answer: Option A

Differentiability and Monotonicity Analysis of Exponential Absolute Sine Function | Mathematics PYQ Solution - JEE Challenger