Consider the function f : (-(π)/(2), (π)/(2)) arrow (-∞, ∞) defined...

Let $f(x) = g(x) + h(x)$, where $g(x) = (|x| + |x - 1|) \sin x$ and $h(x) = [x \sin x]$.

First, we analyze $h(x) = [x \sin x]$ on the interval \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.<br>The function$y = x \sin x$is an even function. For$x \in \left[0, \dfrac{\pi}{2}\right)$,$x \sin x$is strictly increasing. <br>At$x = 0$,$x \sin x = 0$.<br>As$x \to \dfrac{\pi}{2}$,$x \sin x \to \dfrac{\pi}{2} \approx 1.57$.<br>Thus,$x \sin x \in [0, 1.57)$for$x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.<br>The value of$[x \sin x]$will be$0$when$0 \le x \sin x < 1$, and$1$when$1 \le x \sin x < \dfrac{\pi}{2}$.<br>Let$x_1 \in \left(0, \dfrac{\pi}{2}\right)$be the unique point where$x_1 \sin x_1 = 1$. By symmetry,$-x_1$is the unique point in \left(-\dfrac{\pi}{2}, 0\right)$ where $(-x_1) \sin(-x_1) = 1$.
Therefore, $h(x)$ has jump discontinuities at exactly two points: $x = x_1$ and $x = -x_1$.
At $x = 0$, $x \sin x = 0$ and is non-negative in its neighborhood, so $[x \sin x] = 0$ around $x = 0$, making $h(x)$ continuous and differentiable at $x = 0$.

Next, we analyze $g(x) = (|x| + |x - 1|) \sin x$.
The absolute value functions have critical points at $x = 0$ and $x = 1$.
For $x$ in a small neighborhood of $0$ (specifically $x \in (-1, 1)$):
If $x \in [0, 1)$, $g(x) = (x - (x - 1)) \sin x = \sin x \Rightarrow g'(0^+) = \cos 0 = 1$.
If $x \in (-1, 0)$, $g(x) = (-x - (x - 1)) \sin x = (1 - 2x) \sin x \Rightarrow g'(0^-) = -2 \sin 0 + 1 \cdot \cos 0 = 1$.
Since $g'(0^+) = g'(0^-)$, $g(x)$ is differentiable at $x = 0$.

For $x$ in a small neighborhood of $1$:
If $x \in (0, 1]$, $g(x) = \sin x \Rightarrow g'(1^-) = \cos 1$.
If $x \in \left(1, \dfrac{\pi}{2}\right)$, $g(x) = (x + (x - 1)) \sin x = (2x - 1) \sin x \Rightarrow g'(1^+) = 2 \sin 1 + 1 \cdot \cos 1$.
Since $g'(1^-) \neq g'(1^+)$, $g(x)$ is not differentiable at $x = 1$.

Now, combining the results for $f(x) = g(x) + h(x)$:
1. Discontinuities: $g(x)$ is continuous everywhere. $h(x)$ is discontinuous at $x_1$ and $-x_1$. Thus, $f(x)$ is discontinuous at $2$ points. $\Rightarrow \alpha = 2$.
2. Non-differentiability: $f(x)$ is not differentiable where it is discontinuous ($x_1$ and $-x_1$). Additionally, at $x = 1$, $h(x)$ is constant ($0$) because $1 \cdot \sin 1 \approx 0.84 < 1$, meaning $h(x)$ is differentiable at $x = 1$. However, $g(x)$ is not differentiable at $x = 1$. Thus, $f(x)$ is not differentiable at $x = 1$.
This gives $3$ points of non-differentiability: $-x_1, 1, x_1$. $\Rightarrow \beta = 3$.

Finally, $\alpha + \beta = 2 + 3 = 5$.

Answer: $5$