The two curves x3 - 3xy2 + 2 = 0 and 3x2y - y3 - 2 = 0 from Math

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

If 6i3i- 143i- 1203i = x + iy, then

  • x = 3, y = 1

  • x = 1, y = 3

  • x = 0, y = 3

  • x = 0, y = 0


2.

1logbalogab1 is equal to

  • 1

  • 0

  • logab

  • logba


3.

If A = 12- 30 and B = - 1023, then

  • A2 = A

  • B2 = B

  • AB  BA

  • AB = BA


4.

If A = abba and A2 = αββα, then

  • α = a2 + b2, β = ab

  • α = a2 + b2, β = 2ab

  • α = a2 + b2, β = a2 - b2

  • α = 2ab, β= a2 + b2


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5.

If matrix A = 1- 111, then

  • A' = 111- 1

  • A- 1 = 11- 11

  • A . 11- 11 = 2I

  • λA = λ- λ- 11, where λ is a non- zero scalar


6.

If for AX = B, B = 9520 and A- 1 = 3- 1/2- 1/2- 43/45/42- 1/4- 3/4, then X is equal to

  • 135

  • - 1/2- 1/22

  • - 423

  • 33/4- 3/4


7.

If cos-1xa + cos-1yb = αx2a2 - 2xyabcosα + y2b2, then x2a2 - 2xyabcosα + y2b2 is equal to

  • sin2α

  • acos2α

  • atan2α

  • acot2α


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8.

The two curves x3 - 3xy2 + 2 = 0 and 3x2y - y3 - 2 = 0

  • cut at right angle

  • touch each other

  • cut at an angle π3

  • cut at an angle π4


A.

cut at right angle

The equation of given curves are       x3 - 3xy2 + 2 = 0    ...iand 3x2y - y3 - 2 = 0    ...iiOn differentiating Eq. (i) w.r.t. x, we get3x2 - 6xydydx - 3y2 = 0 dydx = 3x2 - 3y26xy = x2 - y22xy m1 = dydxc1 = x2 - y22xyNow, on differentiating Eq. (ii) w.r.t. x, we get3x2dydx + 6xy - 3y2dydx = 0 dydx = - 6xy3x2 - 3y2 = - 2xyx2 - y2  m2 = dydxc2 = - 2xyx2 - y2 m1 × m2 = x2 - y22xy × - 2xyx2 - y2                      = - 1

Thus, Given curves cut each other at right angle


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9.

The domain of the function sin-1log2x22 is

  • - 2, 2 ~ - 1, 1

  • - 1, 2 ~ 0

  • [1,2]

  • - 2, 2 ~ 0


10.

If the function f(x) = 1 + sinπx2, for -  < x  1ax +b,           for 1 < x <36tanπx12,      for 3  x < 6 is continuous in the interval (- , 6), then the values of a and b are respectively

  • 0, 2

  • 1, 1

  • 2, 0

  • 2, 1


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