Define $f: R \rightarrow R$ by $f(x) = \cos(\tan^{-1}(\sin(\tan^{-1} x)))$. Then $\lim_{x \rightarrow \infty} (f \circ f)(x)$ is equal to

If $p, q, r$ are in $G.P.$ and $\tan^{-1} p, \tan^{-1} q, \tan^{-1} r$ are in $A.P.$,then $p, q, r$ satisfy the relation:

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

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$,then $x(y+z)+y(z+x)+z(x+y)$ equals to

Considering only the principal values of the inverse trigonometric functions,the set $\{x \geq 0 : \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\}$

Let $(x, y)$ be such that $\sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2}$. Match the statements in Column $I$ with the statements in Column $II$.

Column $I$	Column $II$
$(A)$ If $a=1$ and $b=0$,then $(x, y)$	$(p)$ lies on the circle $x^2+y^2=1$
$(B)$ If $a=1$ and $b=1$,then $(x, y)$	$(q)$ lies on $(x^2-1)(y^2-1)=0$
$(C)$ If $a=1$ and $b=2$,then $(x, y)$	$(r)$ lies on $y=x$
$(D)$ If $a=2$ and $b=2$,then $(x, y)$	$(s)$ lies on $(4x^2-1)(y^2-1)=0$