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

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q13 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #269

For the function

f (x) = e^{\sin | x |} - | x |

x \in R

, consider the following statements : Statement I :

f

is differentiable for all

x \in R

. Statement II :

f

is increasing in

(- π, - \frac{π}{2})

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

Clue 1 of 3: Differentiability and Monotonicity

To check differentiability, analyze points where the function definition changes (e.g., due to absolute value). To check if a function is increasing, determine the sign of its first derivative.

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Step 1: Analyze Statement I (Differentiability)✦ Active

The function is $f (x) = e^{\sin | x |} - | x |$ . The only point where differentiability might be an issue is $x = 0$ . We need to calculate the Left Hand Derivative (LHD) and Right Hand Derivative (RHD) at $x = 0$ . First, $f (0) = e^{\sin 0} - 0 = 1$ .

LHD at x = 0 : lim_{h \to 0^{-}} \frac{f (h) - f (0)}{h} = lim_{h \to 0^{-}} \frac{e^{\sin (- h)} - (- h) - 1}{h} = lim_{h \to 0^{-}} \frac{e^{- \sin h} + h - 1}{h}

Using L'Hopital's rule or Taylor expansion ( $e^{u} \approx 1 + u + \frac{u^{2}}{2}$ and $\sin h \approx h$ ), this limit evaluates to $0$ .

RHD at x = 0 : lim_{h \to 0^{+}} \frac{f (h) - f (0)}{h} = lim_{h \to 0^{+}} \frac{e^{\sin h} - h - 1}{h}

Similarly, using L'Hopital's rule or Taylor expansion, this limit also evaluates to $0$ . Since LHD = RHD = $0$ , $f (x)$ is differentiable at $x = 0$ . For $x \neq 0$ , $| x |$ is differentiable, so $f (x)$ is differentiable everywhere. Thus, Statement I is true.

Step 2: Analyze Statement II (Monotonicity)○ Expand

For $x \in (- π, - \frac{π}{2})$ , $x < 0$ , so $| x | = - x$ . The function becomes $f (x) = e^{\sin (- x)} - (- x) = e^{- \sin x} + x$ . To check if $f (x)$ is increasing, we need to find the sign of its first derivative $f^{'} (x)$ .

f^{'} (x) = \frac{d}{d x} (e^{- \sin x} + x) = e^{- \sin x} (- \cos x) + 1 = 1 - e^{- \sin x} \cos x

In the interval $(- π, - \frac{π}{2})$ , $x$ is in the third quadrant. In this quadrant, both $\sin x$ and $\cos x$ are negative. Specifically, $\sin x \in (- 1, 0)$ and $\cos x \in (- 1, 0)$ . Therefore, $- \sin x \in (0, 1)$ , which implies $e^{- \sin x} \in (e^{0}, e^{1}) = (1, e)$ . Since $\cos x$ is negative, the product $e^{- \sin x} \cos x$ is a negative number. Let $P = e^{- \sin x} \cos x$ . Then $P < 0$ . So, $f^{'} (x) = 1 - P = 1 + (positive value)$ . This means $f^{'} (x) > 1$ for all $x \in (- π, - \frac{π}{2})$ . Since $f^{'} (x) > 0$ , $f (x)$ is increasing in this interval. Thus, Statement II is true.

Step 3: Conclusion○ Expand

Both Statement I and Statement II are true.

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Maths Question 13 – JEE-MAIN 2026

Clue 1 of 3: Differentiability and Monotonicity

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