JEE Advanced 2016 Paper 1

(x2+xy+4x+2y+4)

dy

dx

−y2=0
((x+2)2+y(x+2))

dy

dx

=y2
Let x+2=X,y=Y
(X)(X+Y)

dY

dX

=Y2
−X2dY=XYdY−Y2dX
−X2dY=Y(XdY−YdX)
−

dY

Y

=

XdY−YdX

X2

−ln|Y|=(

Y

X

)+C
−ln|y|=

y

x+2

+C
∵ it is passing through(1,3)−ln3=1+C
C+−1−ln3
∵ curve

y

x+2

=ln|y|−1−ln3=0,x>0......(i)
put y=x+2 in equation (i)
then

x+2

x+2

+ln|x+2|−1−ln3=0
x=1,−5(reject)
∴ curve intersect y=x=2 at point (1,3)
for option (C),put y=(x+2)2,we will get
x+2+2ln(x+2)=1=ln3
Clearly left hand side is an increasing function
Hence, it is always greater than 2+2ln2
therefore no solution
for option (C) put y=(x+3)2 in equation (i)

(x+3)2

x+2

+ln((x+3)2−1−ln3=0

(x+3)2

x+2

+ln

(x+3)2

3

−1=0
∵x>0⇒x=3>x+2
                     and x+3>3
So

(x+3)2

x+2

+ln

(x+3)2

3

>1
∴

(x+3)2

x+2

+ln

(x+3)2

3

−1=0
⇒curve y=(x+3)2 does not intersect