IIT-JEE 2009 Paper 1 Offline | JEE Advanced Year Wise Previous Years Questions

IIT-JEE 2009 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Let $$f$$ be a non-negative function defined on the interval $$[0,1]$$.

If $$\int\limits_0^x {\sqrt {1 - {{(f'(t))}^2}dt} = \int\limits_0^x {f(t)dt,0 \le x \le 1} } $$, and $$f(0) = 0$$, then

	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)	$$( - \pi ,\pi )$$