1
IIT-JEE 2007 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1

The differential equation $$\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{y}$$ 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 centres along the $$y$$-axis
2
IIT-JEE 2005 Screening
MCQ (Single Correct Answer)
+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 2005 Screening
MCQ (Single Correct Answer)
+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
MCQ (Single Correct Answer)
+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
$$\sqrt {{{2\left( {{e^2} + 1} \right)} \over 2}} $$

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