VITEEE 2014 Paper

Given curve is
y = (1+x)y + sin−1(sin2x)
On differentiating w.r.t x, we get

dy

dx

= (1+x)y [

y

1+x

+log(1+x)

dy

dx

] +

2sinxcosx

√1−sin4x

⇒ (

dy

dx

)at(0,1) = 1 [Since at x = 0 , y = 1]
Slope of normal at (x = 0) = -1
∴ Equation of normal at x = 0 and y = 1 is
y - 1 = - 1 (x - 0)
⇒ y - 1 = - x ⇒ x + y = 1