The common roots of the equations z3 + 2z2 

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 Multiple Choice QuestionsMultiple Choice Questions

261.

If ZR = cosπ2r + isinπ2r for r = 1, 2, 3, ...,then Z1Z2Z3 ...  = ?

  • - 2

  • 1

  • 2

  • - 1


262.

If x1 and xare the real roots of the equation x2 - kx + c = 0, then the distance between the points A(x1, 0) and B(x2, 0) is

  • k2 + 4c

  • k2 - c

  • c - k2

  • k2 - 4c


263.

If p and q are distinct prime numbers and if the equation x2 - px + q = 0 has positive integers as its roots, then the roots of the equation are

  • 1, - 1

  • 2, 3

  • 1, 2

  • 3, 1


264.

The cubic equation whose roots are the squares of the roots of x3 - 2x2 + 10x - 8 = 0, is

  • x3 + 16x2 + 68x - 64 = 0

  • x3 + 8x2 + 68x - 64 = 0

  • x3 + 16x2 - 68x - 64 = 0


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265.

If ω is a complex cube root of unity, thenω13 + 29 + 427 +     + ω12 + 38 + 932 +    = ?

  • 1

  • - 1

  • ω

  • i


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266.

The common roots of the equations z3 +2z2 + 2z + 1 = 0,  z2014 + z2015 + 1 = 0 are

  • ω, ω2

  • 1, ω, ω2

  • - 1, ω, ω2

  • - ω, - ω2


A.

ω, ω2

The given equationz3 +2z2 + 2z + 1 = 0,  z2014 + z2015 + 1 = 0 can be written asz + 1z2 + z + 1 = 0Since, its roots are - 1, ω, ω2Let fz = z2014 + z2015 + 1 = 0Put z = - 1, ω, ω2 respectively, we getf - 1 = - 12014 +  - 12015 + 1 = 0= 1  0Therefore, - 1 is not a root of the equation f(z) = 0Again, f(ω) = ω2014 + ω2015 + 1 = 0= ω3671ω + ω3671ω2 + 1 = 0 ω + ω2 + 1 ω2 + ω + 1 = 0 0 = 0Therefore, ω is a root of the equation f(z) = 0Similarly,f(ω3) =ω22014 + ω32015 + 1 = 0 ω31342 . ω2 + ω31343ω 1 = 0 ω2 + ω +1 = 0 0 = 0Hence, ω and ω2 are the common roots


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267.

If the harmonic mean of the roots of 2x2 - bx + 8 - 2d = 0 is 4, then the value of b is

  • 2

  • 3

  • 4 - 5

  • 4 + 5


268.

For real value of x, the rane of x2 + 2x +1x2 + 2x - 1 is

  • - , 0  1, 

  • 12, 2

  • - , - 29  1, 

  • - , - 6 - 2, 


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269.

The locus of the point representing the complex number z for which z + 32 - z - 32 = 15 is

  • a circle

  • a parabola

  • a straight line

  • an ellipse


270.

1 + i20161 - i2014 = ?

  • - 2i

  • 2i

  • 2

  •  - 2


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