Consider the curve $C_1$ given by
$$ y=e^{-x} \quad \text { for } x \in[0,10 \pi], $$
and the curve $C_2$ given by
$$ y=e^{-x}(\sin x+\cos x) \quad \text { for } x \in[0,10 \pi] . $$
Let $n$ be the total number of points of intersection of the curves $C_1$ and $C_2$.
Suppose that $\alpha_1, \alpha_2, \ldots, \alpha_n \in[0,10 \pi]$ are the $x$-coordinates of the points of intersection of the curves $C_1$ and $C_2$ such that
$$ \alpha_1<\alpha_2<\cdots<\alpha_n . $$
Let $\beta$ be the area of the region enclosed between the curves $C_1, C_2$, and the lines $x=\alpha_1$ and $x=\alpha_4$. Then the value of
$$ -\frac{1}{\pi} \log _e\left(\beta-2 e^{-\frac{\pi}{2}}\right) $$
is $\_\_\_\_$ .
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 _______.
$$ f(x)= \begin{cases}n(1-2 n x) & \text { if } 0 \leq x \leq \frac{1}{2 n} \\\\ 2 n(2 n x-1) & \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n} \\\\ 4 n(1-n x) & \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\\\ \frac{n}{n-1}(n x-1) & \text { if } \frac{1}{n} \leq x \leq 1\end{cases} $$
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 :
JEE Advanced Subjects
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