1
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.
2
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
$$1$$
3
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
$$1/2$$
4
IIT-JEE 2000 Screening
+2
-0.5
If $${x^2} + {y^2} = 1,$$ then
A
$$yy'' - 2{\left( {y'} \right)^2} + 1 = 0$$
B
$$yy'' + {\left( {y'} \right)^2} + 1 = 0$$
C
$$yy'' + {\left( {y'} \right)^2} - 1 = 0$$
D
$$yy'' + 2{\left( {y'} \right)^2} + 1 = 0$$
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