1
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
$$\sqrt {{{2\left( {{e^2} + 1} \right)} \over 2}}$$
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
$${\pi - 1}$$
D
$$- \pi$$
3
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.
4
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
$$1$$
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