Match the statements/expressions in Column I with the open intervals in Column II :

	Column I		Column II
(A)	Interval contained in the domain of definition of non-zero solutions of the differential equation $${(x - 3)^2}y' + y = 0$$	(P)	$$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$
(B)	Interval containing the value of the integral $$\int\limits_1^5 {(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)dx} $$	(Q)	$$\left( {0,{\pi \over 2}} \right)$$
(C)	Interval in which at least one of the points of local maximum of $${\cos ^2}x + \sin x$$ lies	(R)	$$\left( {{\pi \over 8},{{5\pi } \over 4}} \right)$$
(D)	Interval in which $${\tan ^{ - 1}}(\sin x + \cos x)$$ is increasing	(S)	$$\left( {0,{\pi \over 8}} \right)$$
		(T)	$$( - \pi ,\pi )$$

Let a solution $$y=y(x)$$ of the differential equation,

$$x\sqrt {{x^2} - 1} \,\,dy - y\sqrt {{y^2} - 1} \,dx = 0$$ satify $$y\left( 2 \right) = {2 \over {\sqrt 3 }}.$$

STATEMENT-1 : $$y\left( x \right) = \sec \left( {{{\sec }^{ - 1}}x - {\pi \over 6}} \right)$$ and

STATEMENT-2 : $$y\left( x \right)$$ given by $${1 \over y} = {{2\sqrt 3 } \over x} - \sqrt {1 - {1 \over {{x^2}}}} $$

The differential equation $${{dy} \over {dx}} = {{\sqrt {1 - {y^2}} } \over y}$$ determines a family of circles with

If $$y=y(x)$$ and it follows the relation $$x\cos \,y + y\,cos\,x = \pi $$ then $$y''(0)=$$