IIT-JEE 2008 Paper 2 Offline | Differential Equations Question 25 | Mathematics

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Let a solution $$y=y(x)$$ of the differential equation,

$$x\sqrt {{x^2} - 1} \,\,dy - y\sqrt {{y^2} - 1} \,dx = 0$$ satify $$y\left( 2 \right) = {2 \over {\sqrt 3 }}.$$

STATEMENT-1 : $$y\left( x \right) = \sec \left( {{{\sec }^{ - 1}}x - {\pi \over 6}} \right)$$ and

STATEMENT-2 : $$y\left( x \right)$$ given by $${1 \over y} = {{2\sqrt 3 } \over x} - \sqrt {1 - {1 \over {{x^2}}}} $$

STATEMENT-1 is True, STATEMENT-2 is True;STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True;STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False , STATEMENT-2 is True

IIT-JEE 2007 Paper 2 Offline

MCQ (Single Correct Answer)

-1

The differential equation $$\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{y}$$ determines a family of circles with :

variable radii and a fixed centre at $$(0,1)$$

variable radii and a fixed centre at $$(0,-1)$$

fixed radius 1 and variable centres along the $$x$$-axis

fixed radius 1 and variable centres along the $$y$$-axis

IIT-JEE 2005 Screening

MCQ (Single Correct Answer)

-0.5

The differential equation $${{dy} \over {dx}} = {{\sqrt {1 - {y^2}} } \over y}$$ determines a family of circles with

variable radii and a fixed centre at $$(0,1)$$

variable radii and a fixed centre at $$(0,-1)$$

fixed radius $$1$$ and variable centres along the $$x$$-axis.

fixed radius $$1$$ and variable centrs along the $$y$$-axis.

IIT-JEE 2005 Screening

MCQ (Single Correct Answer)

-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

$$3$$

$$2$$

$$1$$

$$5$$

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