For the function …, …, consider the…

Single correct+4 / -1Answer matches NTA’s final key

For the function $f(x) = e^{\sin|x|} - |x|$ , $x \in \mathbb{R}$ , consider the following statements:

Statement I: $f$ is differentiable for all $x \in \mathbb{R}$ .

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:

Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true

Solution

$f$ is even, so only $x = 0$ needs checking. For $x > 0$ , $f(x) = e^{\sin x} - x$ and $f'(x) = \cos x\, e^{\sin x} - 1 \to 0$ as $x \to 0^+$ . By symmetry the left derivative is $-0 = 0$ . Both one-sided derivatives are $0$ , so $f$ is differentiable everywhere. Statement I is true.

For $x < 0$ , $f(x) = e^{-\sin x} + x$ , so $f'(x) = 1 - \cos x\, e^{-\sin x}$ . On $\left(-\pi, -\frac{\pi}{2}\right)$ , $\cos x < 0$ , so $-\cos x\,e^{-\sin x} > 0$ and $f'(x) > 1 > 0$ . $f$ is increasing there. Statement II is true.

Answer: both statements are true.

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Limit, Continuity and Differentiability

Solution

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