The function $f(x) = \begin{cases} \frac{x - |x|}{x}, & x \neq 0 \\ 2, & x = 0 \end{cases}$

If $f(x) = \begin{cases} 6 \beta - 3 \alpha x, & \text{if } -4 \leq x < -2 \\ 4x + 1, & \text{if } -2 \leq x \leq 2 \end{cases}$ is continuous on $[-4, 2]$,then $\alpha + \beta = $

If $f(x) = \begin{cases} e^x; & x \le 0 \\ |1 - x|; & x > 0 \end{cases}$,then

If a function $f(x)$ defined by $f(x) = \begin{cases} ax^2 + bx + c, & x \leq -1 \\ 2x^2 + 4x + 1, & -1 < x < 1 \\ cx^2 + bx + a, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$,and $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$,then find $\lim_{x \rightarrow -2} f(x)$.

If $f(x) = \int_{-1}^x |t| \, dt$,$x \ge -1$,then

If $f(x) = \begin{cases} \frac{\tan(2p-7)x + \tan 3x}{x}, & x < 0 \\ p-q, & x=0 \\ q\left(\frac{\sqrt{x^2+x}-\sqrt{x}}{x^{3/2}}\right), & x > 0 \end{cases}$. If $f(x)$ is continuous at $x=0$, then $\frac{q}{p} = $