1
JEE Advanced 2013 Paper 1 Offline
MCQ (Single Correct Answer)
+4
-1
A curve passes through the point $$\left( {1,{\pi \over 6}} \right)$$. Let the slope of
the curve at each point $$(x,y)$$ be $${y \over x} + \sec \left( {{y \over x}} \right),x > 0.$$
Then the equation of the curve is
the curve at each point $$(x,y)$$ be $${y \over x} + \sec \left( {{y \over x}} \right),x > 0.$$
Then the equation of the curve is
2
IIT-JEE 2009 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0
Match the statements/expressions in Column I with the values given in Column II:
Column I | Column II | ||
---|---|---|---|
(A) | The number of solutions of the equation $$x{e^{\sin x}} - \cos x = 0$$ in the interval $$\left( {0,{\pi \over 2}} \right)$$ | (P) | 1 |
(B) | Value(s) of $$k$$ for which the planes $$kx + 4y + z = 0,4x + ky + 2z = 0$$ and $$2x + 2y + z = 0$$ intersect in a straight line | (Q) | 2 |
(C) | Value(s) of $$k$$ for which $$|x - 1| + |x - 2| + |x + 1| + |x + 2| = 4k$$ has integer solution(s) | (R) | 3 |
(D) | If $$y' = y + 1$$ and $$y(0) = 1$$ then value(s) of $$y(\ln 2)$$ | (S) | 4 |
(T) | 5 |
3
IIT-JEE 2009 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-0
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) | $$( - \pi ,\pi )$$ |
4
IIT-JEE 2008 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Let a solution $$y=y(x)$$ of the differential equation,
$$x\sqrt {{x^2} - 1} \,\,dy - y\sqrt {{y^2} - 1} \,dx = 0$$ satify $$y\left( 2 \right) = {2 \over {\sqrt 3 }}.$$
$$x\sqrt {{x^2} - 1} \,\,dy - y\sqrt {{y^2} - 1} \,dx = 0$$ 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 $${1 \over y} = {{2\sqrt 3 } \over x} - \sqrt {1 - {1 \over {{x^2}}}} $$
Questions Asked from Differential Equations (MCQ (Single Correct Answer))
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