Let $f(x) = \begin{cases} 3x, & x < 0 \\ \min\{1+x+[x],\, x+2[x]\}, & 0 \leq x < 2 \\ 5, & x > 2 \end{cases}$ where $[\cdot]$ denotes the greatest integer function. If $\alpha$ and $\beta$ are the number of points where $f$ is not continuous and not differentiable, respectively, then $\alpha + \beta$ equals ___
Let the function $f(x) = \begin{cases} -3ax^2 - 2, & x < 1 \\ a^2 + bx, & x \geq 1 \end{cases}$ be differentiable for all $x \in \mathbb{R}$, where $a > 1$, $b \in \mathbb{R}$. If the area of the region enclosed by $y = f(x)$ and the line $y = -20$ is $\alpha + \beta\sqrt{3}$, $\alpha, \beta \in \mathbb{Z}$, then the value of $\alpha + \beta$ is ___
Let $f:(0,\pi)\to\mathbb{R}$ be a function given by $f(x)=\begin{cases}\left(\dfrac{8}{7}\right)^{\frac{\tan8x}{\tan7x}}, & 0<x<\dfrac{\pi}{2}\\ a-8, & x=\dfrac{\pi}{2}\\ (1+|\cot x|)^{\frac{b}{a}|\tan x|}, & \dfrac{\pi}{2}<x<\pi\end{cases}$ where $a,b\in\mathbb{Z}$. If $f$ is continuous at $x=\dfrac{\pi}{2}$, then $a^2+b^2$ is equal to:
If $\left(a+\sqrt{2}\,b\cos x\right)\!\left(a-\sqrt{2}\,b\cos y\right)=a^2-b^2$, where $a>b>0$, then $\dfrac{dx}{dy}$ at $\!\left(\dfrac{\pi}{4},\dfrac{\pi}{4}\right)$ is: [If answer expressed as $(a+b)/(a-b)$, find $(a+b)^2$ when $a=5,b=2$]