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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) (ii) (Q)

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B

(II) (iv) (P)

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C

(III) (i) (R)

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D

(IV) (iv) (Q)

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Solution

The correct option is B

(II) (iv) (P)


y=y2x22xy

Let y=vx

y=v+xdvdx

Putting the value of y in original equation, we get,

v+xdvdx=v212v

xdvdx=v212v

2vv2+1dv=dxx

Let v2+1=t

2vdv=dt

Hence, integral becomes dtt=dxx

lnt=lnx+C

ln(x2+y2x2)=lnx+C

ln(x2+y2x)=C

x2+y2x=eC

Given, y(1)=1

eC=2

C=ln2

So, the function is x2+y22x=0

It is a circle with centre (1, 0) and radius =1

So, area =π

To find the maximum value, differentiate the function:

x+yy1=0

y=1x2xx2

For maximum or minimum value, y=0x=1

Differentiating again,

yy′′+(y)2=1

Since, y′′ is negative, so x=1 is the maxima.

So, maximum value of y at x=1 is 1.

Thus, correct combination is (II) (iv) (P)


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