NIMCET 2012 Question Paper with solutions

Given equation is
(cos p - 1) ${\mathit{x}}^2$ + (cos p) x + sin p = 0
Since, the equation has real roots
so, Δ = ${\mathit{B}}^2-4\mathit{A}\mathit{C}$ ≥ 0
⇒ $\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{p}$ - 4 (cos p - 1) sin p ≥ 0
⇒ $\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{p}$ 4 sin p . cos p + 4 sin p ≥ 0
⇒ For real value of p
${\left(-4\mathit{s}\mathit{i}\mathit{n}\mathit{p}\right)}^2$ - 4 . 1 . (4 sin p) > 0
⇒ $16\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{p}$ - 16 sin p > 0
⇒ sin p (sin p - 1) > 0
⇒ sin p > sin 0 or sin p > sin $\frac{\mathit{\pi }}{2}$
⇒ p > nπ + ${\left(-1\right)}^{\mathit{n}}$ . 0 or p > nπ + ${\left(-1\right)}^{\mathit{n}}\frac{\mathit{\pi }}{2}$
⇒ p ∊ (0 , π) or (not possible)