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)

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