For the function f(x) = e^ |x| - |x| , x ∈ ℝ , consider the...

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:
$f'(x) = e^{\sin x}\cos x - 1$
The right-hand derivative at $x = 0$ is:
$f'(0^+) = e^{\sin 0}\cos 0 - 1 = 1(1) - 1 = 0$

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

Since $f'(0^+) = f'(0^-) = 0$, the function $f(x)$ is differentiable at $x = 0$. For all other $x \in \mathbb{R}$ ($x \neq 0$), 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, -\dfrac{\pi}{2}\right)$.<br>Since$x < 0$in this interval, the derivative is:<br>$f'(x) = 1 - e^{-\sin x}\cos x$<br>In the interval \left(-\pi, -\dfrac{\pi}{2}\right)$ (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$.<br>Therefore,$f'(x) = 1 - e^{-\sin x}\cos x > 1 > 0$.<br>Since$f'(x) > 0$, the function$f(x)$is strictly increasing in \left(-\pi, -\dfrac{\pi}{2}\right)$. Thus, Statement II is true.

Both Statement I and Statement II are true.

Answer: Both Statement I and Statement II are true