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1

### IIT-JEE 2005 Screening

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 2005 Screening

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
$$\sqrt {{{2\left( {{e^2} + 1} \right)} \over 2}}$$
3

### IIT-JEE 2005 Screening

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

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
$$- \pi$$

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