Consider the ellipses given by

$$ x^2+4 y^2=1 \quad \text { and } \quad 4 x^2+y^2=1 . $$

If $\alpha$ is the area of the common region that lies inside both the given ellipses, then the value of $\cot \alpha$ is $\_\_\_\_$ .

Let the function $f:[1, \infty) \rightarrow \mathbb{R}$ be defined by

$$ f(t)=\left\{\begin{array}{cc} (-1)^{n+1} 2, & \text { if } t=2 n-1, n \in \mathbb{N}, \\ \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1), & \text { if } 2 n-1 < t < 2 n+1, n \in \mathbb{N} . \end{array}\right. $$

Define $g(x)=\int_1^x f(t) d t, x \in(1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x)=0$ in the interval $(1,8]$ and $\beta=\lim \limits_{x \rightarrow l+} \frac{g(x)}{x-1}$.

Then the value of $\alpha+\beta$ is equal to _______.

Let $n \geq 2$ be a natural number and $f:[0,1] \rightarrow \mathbb{R}$ be the function defined by

$$ f(x)= \begin{cases}n(1-2 n x) & \text { if } 0 \leq x \leq \frac{1}{2 n} \\\\ 2 n(2 n x-1) & \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n} \\\\ 4 n(1-n x) & \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\\\ \frac{n}{n-1}(n x-1) & \text { if } \frac{1}{n} \leq x \leq 1\end{cases} $$

If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is 4 , then the maximum value of the function $f$ is :

Consider the functions $f, g: \mathbb{R} \rightarrow \mathbb{R}$ defined by

$$ f(x)=x^{2}+\frac{5}{12} \quad \text { and } \quad g(x)= \begin{cases}2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4} \\ 0, & |x|>\frac{3}{4}\end{cases} $$

If $\alpha$ is the area of the region

$$ \left\{(x, y) \in \mathbb{R} \times \mathbb{R}:|x| \leq \frac{3}{4}, 0 \leq y \leq \min \{f(x), g(x)\}\right\}, $$

then the value of $9 \alpha$ is