Let $f:[0, 4\pi] \rightarrow [0, \pi]$ be defined by $f(x) = \cos^{-1}(\cos x)$. The number of points $x \in [0, 4\pi]$ satisfying the equation $f(x) = \frac{10-x}{10}$ is

Let $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,for $x \in R$. 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

The number of solutions of $\tan^{-1}\left(x+\frac{2}{x}\right) - \tan^{-1}\left(\frac{4}{x}\right) - \tan^{-1}\left(x-\frac{2}{x}\right) = 0$ is:

Let $x = \frac{m}{n}$ ($m, n$ are co-prime natural numbers) be a solution of the equation $\cos(2 \sin^{-1} x) = \frac{1}{9}$ and let $\alpha, \beta$ $(\alpha > \beta)$ be the roots of the equation $mx^2 - nx - m + n = 0$. Then the point $(\alpha, \beta)$ lies on the line

Let $E_1 = \{x \in R : x \neq 1 \text{ and } \frac{x}{x-1} > 0\}$ and $E_2 = \{x \in E_1 : \sin^{-1}(\log_e(\frac{x}{x-1})) \text{ is a real number}\}$. (Here,the inverse trigonometric function $\sin^{-1} x$ assumes values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$). Let $f : E_1 \rightarrow R$ be the function defined by $f(x) = \log_e(\frac{x}{x-1})$ and $g : E_2 \rightarrow R$ be the function defined by $g(x) = \sin^{-1}(\log_e(\frac{x}{x-1}))$. Match the items in $LIST I$ with $LIST II$.

$LIST I$	$LIST II$
$P$. The range of $f$ is	$1$. $(-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty)$
$Q$. The range of $g$ contains	$2$. $(0, 1)$
$R$. The domain of $f$ contains	$3$. $[-\frac{1}{2}, \frac{1}{2}]$
$S$. The domain of $g$ is	$4$. $(-\infty, 0) \cup (0, \infty)$
	$5$. $(-\infty, \frac{e}{e-1}]$
	$6$. $(-\infty, 0) \cup (\frac{1}{2}, \frac{e}{e-1}]$

For $x \in (-1, 1]$,the number of solutions of the equation $\sin^{-1} x = 2 \tan^{-1} x$ is equal to