If $x \in [0, 1]$,then the number of solution$(s)$ of the equation $2[\cos^{-1}x] + 6[\text{sgn}(\sin x)] = 3$ is (where $[.]$ denotes the greatest integer function and $\text{sgn}(x)$ denotes the signum function of $x$)-

Which pair$(s)$ of function$(s)$ is/are equal? (where ${x}$ and $[x]$ denote the fractional part and integral part functions respectively.)

If $\sin^{-1}(\tan \frac{\pi}{4}) - \sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{6}$, then $x$ is a root of the equation:

If $u = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)$ and $v = 2 \tan^{-1} x$,then $\frac{du}{dv}$ is equal to

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	$4$. $(-\infty, 0) \cup (0, \infty)$
	$5$. $(-\infty, \frac{e}{e-1}]$
	$6$. $(-\infty, 0) \cup (\frac{1}{2}, \frac{e}{e-1}]$

The value of $\lim_{x \to 1/\sqrt{2}} \frac{x - \cos(\sin^{-1} x)}{1 - \tan(\sin^{-1} x)}$ is