JEE Main 2026 — Continuity and Differentiability Question with Solution

For Statement I:
The given function is $f(x)=e^{\sin |x|}-|x|$.
For $x>0$, $f(x)=e^{\sin x}-x$.
Differentiating with respect to $x$, we get:
$f^{\prime }(x)=e^{\sin x}\cos x-1$
The right-hand derivative at $x=0$ is:
$f^{\prime }(0^{+})=e^{\sin 0}\cos 0-1=1(1)-1=0$

For $x<0$, $f(x)=e^{\sin (-x)}-(-x)=e^{-\sin x}+x$.
Differentiating with respect to $x$, we get:
$f^{\prime }(x)=-e^{-\sin x}\cos x+1$
The left-hand derivative at $x=0$ is:
$f^{\prime }(0^{-})=-e^{-\sin 0}\cos 0+1=-1(1)+1=0$

Since $f^{\prime }(0^{+})=f^{\prime }(0^{-})=0$, the function $f(x)$ is differentiable at $x=0$. For all other $x\in \mathbb{R}$ ($x\ne 0$), 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 $f(x)$ in the interval $\left(-\pi ,-\frac{\pi }{2}\right)$.
Since $x<0$ in this interval, the derivative is:
$f^{\prime }(x)=1-e^{-\sin x}\cos x$
In the interval $\left(-\pi ,-\frac{\pi }{2}\right)$ (which lies in the third quadrant), $\cos x<0$.
Since $e^{-\sin x}>0$ and $\cos x<0$, the term $-e^{-\sin x}\cos x>0$.
Therefore, $f^{\prime }(x)=1-e^{-\sin x}\cos x>1>0$.
Since $f^{\prime }(x)>0$, the function $f(x)$ is strictly increasing in $\left(-\pi ,-\frac{\pi }{2}\right)$. Thus, Statement II is true.

Both Statement I and Statement II are true.

Answer: Both Statement I and Statement II are true