The harmonic conjugate of (2, 3, 4)

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 Multiple Choice QuestionsMultiple Choice Questions

271.

The number of real roots of the equation x5 +3x3 + 4x + 30 = 0 is

  • 1

  • 2

  • 3

  • 5


272.

If the coefficients of the equation whose roots are k times the roots of the equation x3 + 14x2 - 116x + 1144 = 0, are integers, then a possible value of k is

  • 3

  • 12

  • 9

  • 4


273.

If the coordinate axes are rotated through an angle π6 about the origin, then the transformed equation of 3x2 - 4xy + 3y2 = 0 is

  • 3y2 + xy = 0

  • x2 - y2 = 0

  • 3y2 - xy = 0


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274.

The harmonic conjugate of (2, 3, 4) with respect to the points (3, - 2, 2), (6, - 17, - 4) is

  • 12, 13, 14

  • 185, - 5, 45

  • - 185, 54, 45

  • 185, - 5, - 45


B.

185, - 5, 45

Let P (2, 3, 4),  A(3, - 2, 2) and B(6, - 17, - 4). Let P divides AB in the ratio k : 1, then2 = 6k +3k + 1 2k + 2 = 6k +3 - 4k = 1 k = - 14Harmonic conjugate Q  divides in the ratio - k : 1, i.e. 14 :1Q = 146 + 314 +1, 14 - 17 - 214 +1, 14 - 4 + 214 +1= 6 + 125,  - 17 - 85, - 4 + 85= 185, - 254,  - 4 + 85= 185, - 255, 45= 185, - 5, 45


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275.

The harmonic mean of two numbers is - 85 and their geometric mean is 2. The quadratic equation whose roots are twice those numbers is

  • x2 + 5x + 4 = 0

  • x2 + 10x + 16 = 0

  • x2 - 10x + 16 = 0

  • x2 - 5x + 4 = 0


276.

If z is a complex number with z  5. Then the least value of z + 2z is

  • 245

  • 265

  • 235

  • 295


277.

If α  is  a  non-real  root  of  x7 = 1,  then α(1 + α) (1 + α2 + α4) =

  • 1

  • 2

  • - 1

  • - 2


278.

If ω is a complex root of unity, then for anyn > 1, r = 1n - 1rr  + 1 - ωr +1 - ω2 = 

  • n2n + 124

  • nn +12n +16

  • nn - 14n2 + 3n +4

  • nn +12n + 14


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279.

If α, β, γ are the roots of x3 + px2 + qx +r = 0then the value of 1 + α21 + β21 + γ2 is

  • r - p2 + r - q2

  • 1 + p2 + 1 + q2

  • r+ p2 + q + 12

  • r - p2 + q - 12


280.

Let α and β be the roots of the equation, 5x2 + 6x  2 = 0. If Sn = αn+ βn, n = 1, 2, 3,..., then :

  • 6S6 +5S5 = 2S4

  • 5S6 +6S5 = 2S4

  • 5S6 +6S5 +2S4 = 0 

  • 6S6 + 5S5 +2S4 = 0


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