For the function \(\mathrm{f}(\mathrm{x})=\mathrm{e}^{\sin |\mathrm{x}|}-|\mathrm{x}|, \mathrm{x} \in \mathbf{R}\), consider the following statements: Statement I : f is differentiable for all \(\mathrm{x} \in \mathbf{R}\). Statement II : f is increasing in \(\left(-\pi,-\frac{\pi}{2}\right)\). 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

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Continuity and Differentiability

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 :

Answer: (A)

Write

f (x) = e^{s i n ∣ x ∣} - ∣ x ∣

. For

x > 0

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

, giving

f^{'} (0^{+}) = 1 \cdot 1 - 1 = 0

; for

x < 0

f (x) = e^{- s i n x} + x

, so

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

and

f^{'} (0^{-}) = - 1 \cdot 1 + 1 = 0

. The two one-sided derivatives agree, so

f

is differentiable at

x = 0

and hence everywhere; Statement I holds. On

(- π, - \frac{π}{2})

we have

x < 0

with

∣ x ∣ \in (\frac{π}{2}, π)

, where

cos ∣ x ∣ < 0

, so

f^{'} (x) = 1 - e^{s i n ∣ x ∣} cos ∣ x ∣ > 0

and

f

is increasing there; Statement II holds. Therefore both Statement I and Statement II are true.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option