NIMCET 2018 Question Paper with solutions

Given that roots are $\mathit{s}\mathit{i}{\mathit{n}}^{2}18°$ and $\mathit{c}\mathit{o}{\mathit{s}}^{2}36°$
∴ Sum of roots = $\mathit{s}\mathit{i}{\mathit{n}}^{2}18°$ + $\mathit{c}\mathit{o}{\mathit{s}}^{2}36°$
= ${\left(\frac{\sqrt{5}-1}{4}\right)}^2+{\left(\frac{\sqrt{5}+1}{4}\right)}^2$
= $\frac{5+1-2\sqrt{5}+5+1+2\sqrt{5}}{16}$ = $\frac{12}{16}$ = $\frac{3}{4}$
Product of roots = ($\mathit{s}\mathit{i}{\mathit{n}}^2$ 18°) × ($\mathit{c}\mathit{o}{\mathit{s}}^2$ 36°)
= ${\left(\frac{\sqrt{5}-1}{4}\right)}^2$ × ${\left(\frac{\sqrt{5}+1}{4}\right)}^2$ = $\frac{1}{16}$
Hence, required equation will be
${\mathit{x}}^2$ - (Sum of roots) x+ (Product of roots) = 0
⇒ ${\mathit{x}}^2-\frac{3}{4}\mathit{x}+\frac{1}{16}$ = 0
⇒ $16{\mathit{x}}^2-12\mathit{x}+1$ = 0