1
IIT-JEE 2008 Paper 2 Offline
+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 $$y\left( 2 \right) = {2 \over {\sqrt 3 }}.$$

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

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

A
STATEMENT-1 is True, STATEMENT-2 is True;STATEMENT-2 is a correct explanation for STATEMENT-1
B
STATEMENT-1 is True, STATEMENT-2 is True;STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C
STATEMENT-1 is True, STATEMENT-2 is False
D
STATEMENT-1 is False , STATEMENT-2 is True
2
IIT-JEE 2005 Screening
+2
-0.5
If $$y=y(x)$$ and it follows the relation $$x\cos \,y + y\,cos\,x = \pi$$ then $$y''(0)=$$
A
$$1$$
B
$$-1$$
C
$${\pi}$$
D
$$- \pi$$
3
IIT-JEE 2005 Screening
+2
-0.5
For the primitive integral equation $$ydx + {y^2}dy = x\,dy;$$
$$x \in R,\,\,y > 0,y = y\left( x \right),\,y\left( 1 \right) = 1,$$ then $$y(-3)$$ is
A
$$3$$
B
$$2$$
C
$$1$$
D
$$5$$
4
IIT-JEE 2005 Screening
+2
-0.5
The solution of primitive integral equation $$\left( {{x^2} + {y^2}} \right)dy = xy$$
$$dx$$ is $$y=y(x),$$ If $$y(1)=1$$ and $$\left( {{x_0}} \right) = e$$, then $${{x_0}}$$ is equal to
A
$$\sqrt {2\left( {{e^2} - 1} \right)}$$
B
$$\sqrt {2\left( {{e^2} + 1} \right)}$$
C
$$\sqrt 3 \,e$$
D
$$\sqrt {{{2\left( {{e^2} + 1} \right)} \over 2}}$$
EXAM MAP
Medical
NEET