NIMCET 2017 — Mathematics PYQ

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Question

NIMCET 2017 — Mathematics PYQ

NIMCET | Mathematics | 2017

If $α, β$ are the roots of an equation $x^{2} - 2 x cos θ + 1 = 0$ then the equation having $α^{n}$ and $β^{n}$ is ?

Choose the correct answer:

Explanation

Concept:
If $α, β$ are the roots of an equation
$x^{2} - 2 x cos θ + 1 = 0$
then the equation having $α^{n}$ and $β^{n}$ is
$x^{2} - (α^{n} + β^{n}) x + (α β)^{n} = 0$

If $α = cos θ + i sin θ$ , then by De Moivre’s Theorem, we have
$α^{n} = cos n θ + i sin n θ$

Hence, the required equation is
$x^{2} - 2 x cos n θ + 1 = 0$

Calculations:

Consider, the equation $x^{2} - 2 x cos θ + 1 = 0$

$\Rightarrow x = \frac{2 cos θ \pm 4 cos ^{2} θ - 4 ( 1 ) ( 1 )}{2 ( 1 )}$

$\Rightarrow x = cos θ \pm i sin θ$

Given, $α, β$ are the roots of an equation $x^{2} - 2 x cos θ + 1 = 0$

$\Rightarrow α = cos θ + i sin θ$ and $β = cos θ - i sin θ$

$\Rightarrow α^{n} = (cos θ + i sin θ)^{n}$ and $β^{n} = (cos θ - i sin θ)^{n}$

$\Rightarrow α^{n} = cos n θ + i sin n θ$ and $β^{n} = cos n θ - i sin n θ$

$\Rightarrow α^{n} + β^{n} = 2 cos n θ$

and $α^{n} β^{n} = 1$

Required equation is $x^{2} - (α^{n} + β^{n}) x + (α^{n} β^{n}) = 0$

$\Rightarrow x^{2} - 2 cos n θ, x + 1 = 0$