Question

If $\sin x+\cos x=\sqrt{y+\frac{1}{y}},x\in (0,\pi )$, then

• A. $x=\frac{\pi }{4},y=1$
• B. $y=0$
• C. $y=2$
• D. $x=\frac{3\pi }{4}$

Answer 1

Answer: (A)

$\sin x+\cos x=\sqrt{y+\frac{1}{y}}$
$-\sqrt{1^2+1^2}\le \sin x+\cos x\le \sqrt{1^2+1^2}$
$-\sqrt{2}\le LHS\le \sqrt{2}$ ....(i)
Also, $y+\frac{1}{y}\ge 2$
$\sqrt{y+\frac{1}{y}}\ge \sqrt{2}$
RHS $\ge \sqrt{2}$ .....(ii)
From (i) and (ii)
$\sin x+\cos x=\sqrt{2}$
$\Rightarrow (\frac{1}{\sqrt{2}}\sin x+\cos x\frac{1}{\sqrt{2}}=1\sin \left(\frac{\pi }{4}+x\right)=1$
$\frac{\pi }{4}+x=n\pi +(-1{)}^n\frac{\pi }{2};\text{ As }x\in (0,\pi )$
$\Rightarrow x=\frac{\pi }{4}$ Also, $y+\frac{1}{y}=2$
$\Rightarrow y=1$