For the function $$f(x) = e^{\sin|x|} - |x|$$, $$x \in \mathbf{R}$$, consider the following statements :
Statement I : $$f$$ is differentiable for all $$x \in \mathbf{R}$$.
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 :

This problem requires checking the differentiability and monotonicity of a composite function involving absolute values. Let's break it down statement by statement.

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

Statement I: Differentiability at $$x = 0$$

Since $$e^{\sin |x|}$$ and $$|x|$$ are differentiable everywhere except potentially at $$x = 0$$ (due to the modulus), we need to check the derivative at the origin.

• For $$x > 0$$: $$f(x) = e^{\sin x} - x$$
• For $$x < 0$$: $$f(x) = e^{\sin(-x)} - (-x) = e^{-\sin x} + x$$
• $$\cos x$$: In the third quadrant, $$\cos x$$ is negative.
• $$-\sin x$$: In the third quadrant, $$\sin x$$ is negative, so $$-\sin x$$ is positive. This makes $$e^{-\sin x} > 1$$.
• The product $$-e^{-\sin x} \cos x$$:
• Statement I: True
• Statement II: True

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

Right Hand Derivative (RHD) at $$0$$: $$f'(0^+) = e^0 \cdot \cos(0) - 1 = 1(1) - 1 = 0$$.

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

Left Hand Derivative (LHD) at $$0$$: $$f'(0^-) = e^0 \cdot (-1) + 1 = -1 + 1 = 0$$.

Since LHD = RHD = 0, the function is differentiable at $$x = 0$$. Since it is also differentiable for all other $$x \in \mathbb{R}$$, Statement I is True.

Statement II: Increasing in $$(-\pi, -\pi/2)$$

To check if the function is increasing, we look at the sign of $$f'(x)$$ in the given interval.

In the interval $$(-\pi, -\pi/2)$$, $$x$$ is negative, so we use the derivative formula for $$x < 0$$:

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

Let's analyze the components in $$(-\pi, -\pi/2)$$:

$$\text{(negative)} \cdot \text{(positive)} \cdot \text{(negative)} = \text{Positive}$$

Since $$f'(x)$$ is the sum of a positive term and $$1$$, $$f'(x) > 0$$ for all $$x \in (-\pi, -\pi/2)$$.

Thus, the function is strictly increasing in this interval. Statement II is True.

Final Conclusion

Correct Option: A