If the mean and variance of a binomial distribution are 4 and 2,

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

Which of the following options is not the asymptote of the curve 3x3 + 2x2y- 7xy2 + 2y- 14xy + 7y2 + 4x + 5y = 0?

  • y = - 12x - 56

  • y = x - 76

  • y = 2x + 37

  • y = 3x - 32


2.

The two lines x = my + n, z = py + q and x = m'y + n', z =p'y + q' are perpendicular to each other, if

  • mm' + pp' = 1

  • mm' + pp' = - 1

  • mm' + pp' = 1

  • mm' + pp' = - 1


3.

If p, q, r are simple propositions with truth values T, F, T, then the truth value of ~ p  q  ~ r  p is

  • true

  • false

  • true, if r is false

  • true, if q is true


4.

If z  3, then the least value of z + 14

  • 112

  • 114

  • 3

  • 14


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5.

The normal at the point (at12, 2at1) on the parabola meets the parabola again in the point (at22, 2at2), then

  • t2 = - t1 + 2t1

  • t2 = - t1 - 2t1

  • t2 = t1 - 2t1

  • t2 = t1 + 2t1


6.

If a = i^ - j^ + 2k^ and b = 2i^ - j^ +k^, then the angle θ between a and b is given by

  • tan-11

  • sin-112

  • sec-11

  • tan-113


7.

If the rectangular hyperbola is x2 - y2 = 64. Then, which of the following is not correct?

  • The length of latusrectum is 16

  • The eccentricity is 2

  • The asymptotes are parallel to each other

  • The directrices are x = ± 42


8.

The equation of tangents to the hyperbola 3x2 - 2y2 = 6, which is perpendicular to the line x - 3y = 3, are

  • y = - 3x ± 15

  • y = 3x ± 6

  • y = - 3x ± 6

  • y = 2x ± 15


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9.

limxπ4tanx - 1cos2x is equal to

  • 1

  • 0

  • - 2

  • - 1


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10.

If the mean and variance of a binomial distribution are 4 and 2, respectively. Then, the probability of atleast 7 successes is

  • 3214

  • 4173

  • 9256

  • 7231


C.

9256

Given, mean = 4 and variance = 2

 np = 4 and npq = 2

 npqnp = 24  q = 12

Then, p = 1 - q = 1 - 12 = 12

   Mean = np = 4

      n × 12 = 4               n = 8   PX = r = Crnprqn - r                      = Cr8128             p = q = 12

The required probability of atleast 7 successes is

PX  7 = PX = 7 + PX = 8= C78 + C88128= 8!7!1! + 8!8!0!128= 8 + 1128 = 9256


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