JEE Main 2026 — Mathematics Calculus

For Statement I:

The given function is $f(x) = e^{\sin|x|} - |x|$f(x)=esin∣x∣−∣x∣.

For $x > 0$x>0, $f(x) = e^{\sin x} - x$f(x)=esinx−x.

Differentiating with respect to $x$x, we get:

$$f'(x) = e^{\sin x}\cos x - 1$$f′(x)=esinxcosx−1

The right-hand derivative at $x = 0$x=0 is:

$$f'(0^+) = e^{\sin 0}\cos 0 - 1 = 1(1) - 1 = 0$$f′(0+)=esin0cos0−1=1(1)−1=0

For $x < 0$x<0, $f(x) = e^{\sin(-x)} - (-x) = e^{-\sin x} + x$f(x)=esin(−x)−(−x)=e−sinx+x.

Differentiating with respect to $x$x, we get:

$$f'(x) = -e^{-\sin x}\cos x + 1$$f′(x)=−e−sinxcosx+1

The left-hand derivative at $x = 0$x=0 is:

$$f'(0^-) = -e^{-\sin 0}\cos 0 + 1 = -1(1) + 1 = 0$$f′(0−)=−e−sin0cos0+1=−1(1)+1=0

Since $f'(0^+) = f'(0^-) = 0$f′(0+)=f′(0−)=0, the function $f(x)$f(x) is differentiable at $x = 0$x=0. For all other $x \in \mathbb{R}$x∈R ($x \neq 0$x=0), the function is a composition of differentiable functions and is therefore differentiable. Thus, Statement I is true.

For Statement II:

We need to check the monotonicity of $f(x)$f(x) in the interval $\left(-\pi, -\dfrac{\pi}{2}\right)$(−π,−2π​).

Since $x < 0$x<0 in this interval, the derivative is:

$$f'(x) = 1 - e^{-\sin x}\cos x$$f′(x)=1−e−sinxcosx

In the interval $\left(-\pi, -\dfrac{\pi}{2}\right)$(−π,−2π​) (which lies in the third quadrant), $\cos x < 0$cosx<0.

Since $e^{-\sin x} > 0$e−sinx>0 and $\cos x < 0$cosx<0, the term $-e^{-\sin x}\cos x > 0$−e−sinxcosx>0.

Therefore, $f'(x) = 1 - e^{-\sin x}\cos x > 1 > 0$f′(x)=1−e−sinxcosx>1>0.

Since $f'(x) > 0$f′(x)>0, the function $f(x)$f(x) is strictly increasing in $\left(-\pi, -\dfrac{\pi}{2}\right)$(−π,−2π​). Thus, Statement II is true.

Both Statement I and Statement II are true.

Answer: Both Statement I and Statement II are true