NIMCET 2017 Question Paper with solutions

⇒ x = $\frac{2\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\pm \sqrt{4\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }-4}}{2}$ = $\frac{2\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\pm 2\mathit{i}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{2}$
= cosθ ± i sinθ
∴ α = cosθ + i sinθ
and β = cosθ - i sinθ
∴ ${\mathit{\alpha }}^{\mathit{n}}$ = ${\left(\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }+\mathit{i}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\right)}^{\mathit{n}}$ = cosnθ + i sinnθ
and ${\mathit{\beta }}^{\mathit{n}}$ = ${\left(\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }-\mathit{i}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\right)}^{\mathit{n}}$ = cosnθ - i sinnθ [using De-Moivre's theorem]
Now, ${\mathit{\alpha }}^{\mathit{n}}+{\mathit{\beta }}^{\mathit{n}}$ = 2 cos nθ and ${\mathit{\alpha }}^{\mathit{n}}.{\mathit{\beta }}^{\mathit{n}}$ = 1
∴ Equation having ${\mathit{\alpha }}^{\mathit{n}}$ and ${\mathit{\beta }}^{\mathit{n}}$ as roots is
${\mathit{x}}^2-\left({\mathit{\alpha }}^{\mathit{n}}+\mathit{\beta }\mathit{n}\right)\mathit{x}+{\mathit{\alpha }}^{\mathit{n}}{\mathit{\beta }}^{\mathit{n}}$ = 0
⇒ ${\mathit{x}}^2-2\mathit{c}\mathit{o}\mathit{s}\mathit{n}\mathit{\theta }\mathit{x}+1=0$