If $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$,then $x \in$

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 $(4x^2-1)(y^2-1)=0$

If $\alpha$ and $\beta$ are the roots of the equation $6x^2 - 5x + 1 = 0$,then the value of $\tan^{-1}\alpha + \tan^{-1}\beta$ is:

The derivative of ${\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ with respect to ${\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ is

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:

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: