For the function f(x) = e^{sin |x|} - |x|, x | vedclass

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Question

(D) The function is $f(x) = e^{\sin |x|} - |x|$.
First, check the differentiability at $x = 0$. The function $f(x)$ contains the term $|x|$, which is

Vedclass · Accepted Answer

Statement $I$ is false but Statement $II$ is true

Vedclass · Answer

(D) The function is $f(x) = e^{\sin |x|} - |x|$.
First, check the differentiability at $x = 0$. The function $f(x)$ contains the term $|x|$, which is not differentiable at $x = 0$. Therefore, $f(x)$ is not differentiable at $x = 0$. Thus, Statement $I$ is false.
Next, consider Statement $II$ in the interval $(-\pi, -\frac{\pi}{2})$. For $x < 0$, we have $|x| = -x$. Thus, $f(x) = e^{\sin(-x)} - (-x) = e^{-\sin x} + x$.
The derivative is $f'(x) = e^{-\sin x}(-\cos x) + 1 = 1 - e^{-\sin x}\cos x$.
For $x \in (-\pi, -\frac{\pi}{2})$, we have $\sin x \in (-1, 0)$ and $\cos x \in (-1, 0)$.
Since $\sin x$ is negative, $-\sin x$ is positive, so $e^{-\sin x} > e^0 = 1$. Also, $\cos x$ is negative.
Thus, $-e^{-\sin x}\cos x$ is positive. Therefore, $f'(x) = 1 + (	ext{positive value}) > 0$.
Since $f'(x) > 0$, the function $f(x)$ is strictly increasing in the interval $(-\pi, -\frac{\pi}{2})$. Thus, Statement $II$ is true.

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