Let $X=\left\{(x, y) \in \mathbb{Z} \times \mathbb{Z}: \frac{x^2}{8}+\frac{y^2}{20}<1\right.$ and $\left.y^2<5 x\right\}$. Three distinct points $P, Q$ and $R$ are randomly chosen from $X$. Then the probability that $P, Q$ and $R$ form a triangle whose area is a positive integer, is :

Let $P$ be a point on the parabola $y^2=4 a x$, where $a>0$. The normal to the parabola at $P$ meets the $x$-axis at a point $Q$. The area of the triangle $P F Q$, where $F$ is the focus of the parabola, is 120 . If the slope $m$ of the normal and $a$ are both positive integers, then the pair $(a, m)$ is

Let $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, for $x \in \mathbb{R}$. Then the number of real solutions of the equation $\sqrt{1+\cos (2 x)}=\sqrt{2} \tan ^{-1}(\tan x)$ in the set $\left(-\frac{3 \pi}{2},-\frac{\pi}{2}\right) \cup\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ 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 :