Suppose $f: [-2, 2] \rightarrow R$ is defined by $f(x) = \begin{cases} -1, & -2 \leq x \leq 0 \\ x - 1, & 0 < x \leq 2 \end{cases}$. Then the set $\{x \in [-2, 2] : x \leq 0 \text{ and } f(|x|) = x\}$ is equal to

If $f(x) = \frac{2^{2x}}{2^{2x} + 2}$,$x \in R$,then $f\left(\frac{1}{2023}\right) + f\left(\frac{2}{2023}\right) + \dots + f\left(\frac{2022}{2023}\right)$ is equal to

The function $f(x) = |px - q| + r|x|$,$x \in (-\infty, \infty)$,where $p > 0, q > 0, r > 0$ assumes its minimum value only at one point,if

If $f(x) = x \left( \frac{1}{x-1} + \frac{1}{x} + \frac{1}{x+1} \right)$ for $x > 1$, then:

Consider the function $f: [\frac{1}{2}, 1] \rightarrow \mathbb{R}$ defined by $f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$. Consider the following statements:
$(I)$ The curve $y = f(x)$ intersects the $x$-axis exactly at one point.
$(II)$ The curve $y = f(x)$ intersects the $x$-axis at $x = \cos \frac{\pi}{12}$.
Then:

Let $f(x) = Ax^3 - Bx - \tan x \cdot \text{sgn}(x)$ be an even function $\forall x \in R - \left\{ (2n + 1)\frac{\pi}{2}, n \in I \right\}$,where $A = \sin^2 \alpha - \sin \alpha + \frac{1}{4}$ and $B = \tan^2 \alpha + \frac{2}{\sqrt{3}} \tan \alpha + \frac{1}{3}$. Then the number of values of $\alpha$ in $\left[ -\frac{3\pi}{2}, 2\pi \right]$ is (where $\text{sgn}(x)$ denotes the signum function of $x$).