1
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
$$- \pi$$
2
IIT-JEE 2005 Screening
+2
-0.5
The differential equation $${{dy} \over {dx}} = {{\sqrt {1 - {y^2}} } \over y}$$ determines a family of circles with
A
variable radii and a fixed centre at $$(0,1)$$
B
variable radii and a fixed centre at $$(0,-1)$$
C
fixed radius $$1$$ and variable centres along the $$x$$-axis.
D
fixed radius $$1$$ and variable centrs along the $$y$$-axis.
3
IIT-JEE 2004 Screening
+2
-0.5
If $$y=y(x)$$ and $${{2 + \sin x} \over {y + 1}}\left( {{{dy} \over {dx}}} \right) = - \cos x,y\left( 0 \right) = 1,$$
then $$y\left( {{\pi \over 2}} \right)$$ equals
A
$$1/3$$
B
$$2/3$$
C
$$-1/3$$
D
$$1$$
4
IIT-JEE 2003 Screening
+2
-0.5
If $$y(t)$$ is a solution of $$\left( {1 + t} \right){{dy} \over {dt}} - ty = 1$$ and $$y\left( 0 \right) = - 1,$$ then $$y(1)$$ is equal to
A
$$- 1/2$$
B
$$e+1/2$$
C
$$e-1/2$$
D
$$1/2$$
EXAM MAP
Medical
NEET