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Question

Let y=f(x) is a positive function which satisfies equation y2+2x+y22x=2x2,then dydx is

A
2x3x3y
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B
2x4x2y
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C
2x4x2x6+1
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D
2x3x31+x6
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Solution

The correct option is C 2x4x2x6+1
y2+2x+y22x=2x2...(1)
rationalizing the equation (1) we get
y2+2xy22x=2x ...(2)
adding the equations (1) & (2)
2y2+2x=2x2+2x
(y2+2x)=(x2+1x)2
y2=x4+1x2y=x4+1x2
2y.dydx=4x32x3
dydx=2x3x3y
dydx=2x3x31+x6x2=2x4x21+x6

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