The quadratic equations x2 – 6x + a = 0 and x2 – cx + 6 = 0

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11.

If z is a complex number of unit modulus and argument θ, then arg open parentheses fraction numerator 1 plus straight z over denominator 1 plus straight z with bar on top end fraction close parentheses is equal to

  • π/2-θ

  • θ

  • θ

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12.

The equation esinx-e-sinx -4 = 0 has

  • infinite number of real roots

  • No real root

  • exactly one real root

  • exactly one real root

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13.

Let α,  β be real and z be a complex number. If z2 + αz + β = 0 has two distinct roots on the line Re z = 1, then it is necessary that

  • β ∈(0, 1)

  • β ∈(-1, 0)

  • |β| = 1

  • |β| = 1

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14.

If ω(≠1) is a cube root of unity, and (1 + ω)7 = A + Bω.Then (A, B) equals

  • (0,1)

  • (1,1)

  • (1,0)

  • (1,0)

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15.

The Number of complex numbers z such that |z– 1| = |z + 1| = |z – i| equals 

  • 0

  • 1

  • 2

  • 2

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16.

If α and β are the roots of the equation x2 – x +1 =0, then α2009 + β2009 =

  • -2

  • -1

  • 1

  • 1

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17.

If, for a positive integer n, the quadratic equation,
x(x + 1) + (x + 1) (x + 2) + .....
+ (x + n -1 ) (x + n) = 10n
has two consecutive integral solutions, then n is equal to :

  • 11

  • 12

  • 9

  • 9

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18.

If the roots of the equation bx2+ cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2 + 6bcx + 2c2 is

  • greater than 4ab

  • less than 4ab

  • greater than -4ab

  • greater than -4ab

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19.

The conjugate of a complex number is fraction numerator 1 over denominator straight i minus 1 end fraction. Then the complex number is

  • fraction numerator negative 1 over denominator straight i minus 1 end fraction
  • fraction numerator 1 over denominator straight i plus 1 end fraction
  • fraction numerator negative 1 over denominator 1 plus straight i end fraction
  • fraction numerator negative 1 over denominator 1 plus straight i end fraction
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20.

The quadratic equations x2 – 6x + a = 0 and x2 – cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is

  • 1

  • 4

  • 3

  • 3


D.

3

Let α and 4β be roots of x2– 6x + a = 0 and
α, 3β be the roots of x2– cx + 6 = 0, then
α + 4β = 6 and 4αβ = a
α + 3β = c and 3αβ = 6.
We get αβ = 2 ⇒ a = 8
So the first equation is x2 – 6x + 8 = 0 ⇒ x = 2, 4
If α = 2 and 4β = 4 then 3β = 3
If α = 4 and 4β = 2, then 3β = 3/2 (non-integer)
∴ common root is x = 2.

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