A curve passes through (2,0) and the slope of tangent at a point P(x,y) is equal to $(x + 1)^{2} + y - 3 / (x + 1) .$ Then equation of the curve is: [IIT 2004]

y = x^{2} + 2 x

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y = x^{2} - 2 x

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y = 2 x^{2} - x

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None of these

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Solution

The correct option is B $y = x^{2} - 2 x$
$\frac{d y}{d x} = \frac{(x + 1)^{2} + y - 3}{x + 1} = (x + 1) + \frac{y - 3}{x + 1}$

Putting x+1=X, y-3=Y, $\frac{d y}{d x} = \frac{d Y}{d X},$ the equation becomes $\frac{d Y}{d X} = X + \frac{Y}{X} o r \frac{d Y}{d X} - \frac{1}{X} .$ Y = X [L.D.E]

I.F. = $e^{\int (- 1 / X) d X} = e^{- l o g X} = X^{- 1} = \frac{1}{X}$

∴ the solution is $Y . (\frac{1}{X}) = c + \int X . (\frac{1}{X}) d x = c + X o r \frac{(y - 3)}{(x + 1)} = c + x + 1$

x = 2, y = 0 $\Rightarrow \frac{0 - 3}{2 + 1} c + 2 + 1 c = - 4.$

∴ the equation of the curve is $\frac{y - 3}{x + 1} = x - 3 o r y = x^{2} - 2 x .$

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Which of the following is only correct combination?

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A curve passes through (2,0) and the slope of tangent at a point P(x,y) is equal to (x+1)2+y−3/(x+1). Then equation of the curve is: [IIT 2004]

A curve passes through (2,0) and the slope of tangent at a point P(x,y) is equal to $(x + 1)^{2} + y - 3 / (x + 1) .$ Then equation of the curve is: [IIT 2004]