If the curves x2 + py2 = l and qx2 + y2 = l are orthogonal t

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371.

If the curves x2 + py= l and qx2 + y2 = l are orthogonal to each other, then

  • p - q = 2

  • 1p - 1q = 2

  • 1p + 1q = - 2

  • 1p + 1q = 2


D.

1p + 1q = 2

Given curves arex2 + py2 = 1           ...iand qx2 +y2 = 1  ...iiOn differentiating Eq (i), w r t , x we get2x + 2ypdydx = 0 dydx = m1 = - xpyOn differentiating Eq (ii), w r t , x we get2qx + 2ydydx = 0 dydx = m2 = - qxySince, both the curves are orthogonal to each other,Then, m1m2 = - 1 - xpy . - qxy = - 1 qx2 = - py2       iii q1 - pqy2 = - py2  from eq i q - pqy2 = - py2 q = pq - p = y2 y2 = qpq - pOn putting x2 and y2 in eq ii we get- pqpq - p + qpq - p = 1 - pq + q = pq - p  p + q = 2pq 1p + 1q = 2


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372.

The focal length of a mirror is given by 2f = 1v - 1u. In finding the values of u and v, the errors are equal to ' 'p'. Then, the relative error in f is

  • p21u + 1v

  • p1u + 1v

  • p21u - 1v

  • p1u - 1v


373.

If u = logx3 +y3 + z3 - 3xyz, then x + y + zux + uy +uz = ?

  • 0

  • x - y + z

  • 2

  • 3


374.

An integrating factor of the equation 1 + y +x2ydx + x + x3dy = 0 is

  • ex

  • x2

  • 1x

  • x


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375.

The solution of the differential equationdydx - 2ytan2x = exsec2x is

  • ysin2x = ex + C

  • ycos2x = ex . 1dx + C

  • y = excos2x + C

  • ycos2x + ex = C


376.

If fx = x1 + x and gx = ffx, then g'x = 

  • 12x + 32

  • 1x + 12

  • 1x2

  • 12x + 12


377.

The differential equation of the family of parabolas with vertex at (0, - 1) and having axis along the Y-axis is

  • yy' + 2xy + 1 = 0

  • xy' + y + 1 = 0

  • xy' - 2y - 2

  • xy' - y - 1 = 0


378.

The solution of xdydx = y + eyx with y1 = 0 is

  • 1 = logx + e yx

  • logx = e - yx

  • 1 = 2logx + e - yx

  • logx + e - yx = 1


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379.

The solution of cos(y) + (xsin(y) - 1)dy/dx = 0 is, 

  • xsecy = tany + C

  • tany - secy = Cx

  • tany + secy = Cx

  • xsecy + tany = C


380.

If x is real, then the minimum value of y = x2 - x + 1x2 + x + 1 is

  • 3

  • 13

  • 12

  • 2


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