Question

If $\sin x+\cos x=\sqrt{y+\frac{1}{y}},xϵ[0,\pi ]$, then:

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

Answer 1

The correct option is A $x=\frac{\pi }{4},y=1$

$cosx+sinx=\sqrt{y+\frac{1}{y}}$

we know $cosx+sinx\le \sqrt{2}$ and $y+1/y\le 2$, assuming $y>0$

and in $x\in [0,\pi ]$ equality only exit if $x=\frac{\pi }{4}$ and $y=1$ as

$sin\frac{\pi }{4}+cos\frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$ and $1+\frac{1}{1}=2$ $\dots \left(1\right)$

when substituted in given equation

$cosx+sinx=\sqrt{y+\frac{1}{y}}$

$\Rightarrow \sqrt{2}=\sqrt{2}$ for these values it satisfies the equation

thus solution is, $x=\frac{\pi }{4},y=1$

Hence, option 'A' is correct.