IIT-JEE 2012 Paper 1 Offline | Differential Equations Question 15 | Mathematics

IIT-JEE 2012 Paper 1 Offline

MCQ (More than One Correct Answer)

-1

If $$y(x)$$ satisfies the differential equation $$y' - y\,tan\,x = 2x\,secx$$ and $$y(0)=0,$$ then

$$y\left( {{\pi \over 4}} \right) = {{{\pi ^2}} \over {8\sqrt 2 }}$$

$$y'\left( {{\pi \over 4}} \right) = {{{\pi ^2}} \over {18}}$$

$$y\left( {{\pi \over 3}} \right) = {{{\pi ^2}} \over 9}$$

$$y'\left( {{\pi \over 3}} \right) = {{4\pi } \over 3} + {{2{\pi ^2}} \over {3\sqrt 3 }}$$

IIT-JEE 2006

MCQ (More than One Correct Answer)

-1.25

A curve $$y=f(x)$$ passes through $$(1,1)$$ and at $$P(x,y),$$ tangent cuts the $$x$$-axis and $$y$$-axis at $$A$$ and $$B$$ respectively such that $$BP:AP=3:1,$$ then

equation of curve is $$xy'-3y=0$$

normal at $$(1,1)$$ is $$x+3y=4$$

curve passes through $$(2, 1/8)$$

equation of curve is $$xy'+3y=0$$

IIT-JEE 1999

MCQ (More than One Correct Answer)

-0.75

The differential equation representing the family of curves
$${y^2} = 2c\left( {x + \sqrt c } \right),$$ where $$c$$ is a positive parameter, is of

order $$1$$

order $$2$$

degree $$3$$

degree $$4$$

JEE Advanced Subjects