JEE Main 2026 Apr 8 Shift 2, Mathematics Q13: Differentiability Of A Function

JEE Main2026Apr 8, Shift 2Mathematics

For the function $f (x) = e^{s i n ∣ x ∣} - ∣ x ∣$ , $x \in R$ , 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 :

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

Solution

Differentiability: $∣ x ∣$ is non-differentiable at $x = 0$ . While $e^{s i n ∣ x ∣}$ is differentiable at $x = 0$ (since $sin ∣ x ∣$ has matching one-sided derivatives there), the term $- ∣ x ∣$ ensures $f$ is not differentiable at $x = 0$ . Statement I is false.

Monotonicity on $(- π, - π /2)$ : For $x < 0$ , $∣ x ∣ = - x$ , so $f (x) = e^{s i n (- x)} - (- x) = e^{- s i n x} + x$ .

$f^{'} (x) = - cos x \cdot e^{- s i n x} + 1$

On $(- π, - \frac{π}{2})$ , $cos x < 0$ , so $- cos x \cdot e^{- s i n x} > 0$ , and $f^{'} (x) > 1 > 0$ .

Hence $f$ is strictly increasing on that interval. Statement II is true.

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q13: Differentiability Of A Function

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