Question

The interval for p for which the equation $\sqrt{p}\cos x-2\sin x=\sqrt{2}+\sqrt{2-p}$ has a solution is

• A. $(-\sqrt{5},-1,2)$
• B. $(\frac{\sqrt{5}-1}{2},2)$
• C. $(-1+\sqrt{5},2)$
• D. None of these

Answer 1

Answer: (C)

$(-1+\sqrt{5},2)$

The given equation will have a solution if

$P\ge 0,2-P\ge 0and|\sqrt{P}\cos x-2\sin x|\le \sqrt{P+4}\Rightarrow P\ge 0,2-P\ge 0and|\sqrt{2}+\sqrt{2-p}|\le \sqrt{p+4}\Rightarrow P\ge 0,p\le 2and(\sqrt{2}+\sqrt{2-P}{)}^2\le p+4\Rightarrow P\ge 0,P\le 2and{P}^2+2P-4\ge 0\Rightarrow P\ge 0,P\le 2and(P+1{)}^2\ge 5\Rightarrow P\ge 0,P\le 2and|p+1|\ge 5\Rightarrow P\ge 0,P\le 2andPϵ(-\infty ,-(\sqrt{5}+1\left)\right)U(\sqrt{5}-1,\infty )\Rightarrow Pϵ(-1+\sqrt{5},2).$