If n is positive integer, then (1 + i)n + (1 - i)n is equal

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 Multiple Choice QuestionsMultiple Choice Questions

241.

If the equations ax2 + 2cx + b = 0 and ax2 + 2bx + c = 0 b  c have a common root, then the value of a + 4b + 4c will be

  • 2

  • 1

  • - 1

  • Non eof these


242.

If one root of ax2 + bx + c = 0 is twice the other root, then

  • b2 = 9ac

  • 2b2 = 9ac

  • 2b2 = ac

  • b2 = ac


243.

The number of solutions of x2 + 3x + 2 = 0 the equation is

  • 0

  • 1

  • 2

  • 4


244.

Which of the following is correct?

  • 2 + 3i > 1 + 4i

  • 6 + 2i > 3 + 3i

  • 5 + 8i > 5 + 7i

  • None of these


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245.

- 1 + - 323n + - 1 - - 323n is equal to

  • 0

  • 1

  • 2

  • 3


246.

Evaluate k = 16sin27 - icos27

  • 2i

  • - i

  • i

  • - 2i


247.

If x2 - 4x + log1/2(a) = 0 does not have two distinct real roots, then maximum value of a is

  • - 14

  • 116

  • 14

  • None of these


248.

The value of (1 + i)3 + (1 - i)6 is

  • i

  • 2(- 1 + 5i)

  • 1 - 5i

  • 2 + 1 - 5i


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249.

If n is positive integer, then (1 + i)n + (1 - i)n is equal to

  • 2n - 2cos4

  • 2n - 2sin4

  • 2n + 2cos4

  • 2n + 2sin4


C.

2n + 2cos4

1 + in + 1 - inUsing polar form,1 + i = 2cosπ4 + isinπ4and 1 - i = 2cosπ4 - isinπ4Now, 1 + in + 1 - in= 2cosπ4 + isinπ4n + 2cosπ4 - isinπ4n= 2ncos4 + isin4 + 2ncos4 - isin4= 22ncos4= 2n + 2cos4


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250.

The square root of 2i is

  • 1 + i

  • 1 - i

  • 2i

  • - 2


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