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Column - IColumn - IIColumn - III(I)y.(y)2xy(1+y)+x2=(i)[y]=1,where [.] is greatest(p)Curve is bounded with area, π0, y(3)=2integer function(II)y=y2x22xy, y(1)=1(ii)Maximum value of y is 3(Q)Area bounded by curve in first quadrant withco-ordinate axes is 3π4(III)y=9xy, y(1)=0(iii)Maximum value of y is not defined(R)Curve is conic with eccentricty, 12(IV)y=xy, y(2)=0(iv)Maximum value of y is 1(S)Curve is conic with eccentricity, 2

The correct combination is


A

(I) (iii) (S)

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B

(II) (ii) (R)

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C

(III) (iv) (Q)

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D

(IV) (iv) (S)

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Solution

The correct option is A

(I) (iii) (S)


(I) y(y)2=xy(1+y)+x2=0
(yyx)(yx)=0
yy=x or y=x
y2=x2+C y=x22+C
y(3)=2
y2=x2+1

Max. value of y is not defined.
Eccentricity =2


(II) y=y2x22xy; y(1)=1
dydx=12(yxxy)
It is a homogeneous differential equation

y2+x2=cx
y(1)=11+1=cc=2
x2+y22x=0
Maximum value of y is 1
Area =πr2=π

(III)

y=9xy, y(1)=0
y dy+9x dx=0
y22+9x2=8x2+y29=1
Maximum value of y is 3
Area =3π

(IV) y=xy; y(2)=0
y dy=x dx
y2=x24
x2y2=4
Maximum value of y is not defined and eccentricity =2

Thus, the correct combination is (I) (iii) (S)


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