Consider the function $f : \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \rightarrow (-\infty, \infty)$ defined by

$f(x) = (|x| + |x - 1|) \sin x + [x \sin x]$,

where $[x \sin x]$ is the greatest integer less than or equal to $x \sin x$.

Let $\alpha$ be the total number of points in the interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ at which $f$ is NOT continuous, and let $\beta$ be the total number of points in the interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ at which $f$ is NOT differentiable.

Then the value of $\alpha + \beta$ is ___________.

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 $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
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.
At $x = 0$, $x \sin x = 0$.
As $x \to \dfrac{\pi}{2}$, $x \sin x \to \dfrac{\pi}{2} \approx 1.57$.
Thus, $x \sin x \in [0, 1.57)$ for $x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
The value of $[x \sin x]$ will be $0$ when $0 \le x \sin x < 1$, and $1$ when $1 \le x \sin x < \dfrac{\pi}{2}$.
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: $5$