The equation of normal of x2 + y2 - 2x + 4y - 5 = 0 at (2, 1

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471.

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The minimum value of f(x) = ex4 - x3 + x2

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  • a < 0, b > 0

  • a  0, b > 0

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475.

If the normal to the curve y = f(x) at the point (3, 4) make an angle 3π/4 with the positive x-axis, then f'(3) is

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476.

The equation of normal of x+ y2 - 2x + 4y - 5 = 0 at (2, 1) is

  • y = 3x - 5

  • 2y = 3x - 4

  • y = 3x + 4

  • y = x + 1


A.

y = 3x - 5

Given equation is,

x+ y2 - 2x + 4y - 5 = 0

On differentiating, we get

2x +2ydydx - 2 +4dydx = 0 y + 2dydx = 1 - x dydx2, 1 = 1 - 21 + 2 = - 13 - dxdy2, 1 = 3Now, equation of nrmal is(y - 1) = 3(x - 2)  y - 1 = 3x - 6     y = 3x - 5


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Angle between y2 = x and x2 = y at the origin is

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The distance covered by a particle in t seconds is given by x = 3 + 8t - 4t2. After 1 s its velocity will be

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If the tangent to the parabola y = x(2 - x) at the point (1, 1) intersects the parabola at P. Find the coordinate of P.


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The maximum value of xy subject to x + y = 8

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