Question

If ${\sin }^{4}x+{\cos }^{4}y+2=4\sin x\cos y$ and $0\le x,y\le \frac{\pi }{2},$ then $\sin x+\cos y$ is equal to

• A. $-2$
• B. $0$
• C. $2$
• D. $5$

Answer 1

Answer: (C)

$2$

Put $a=sinx$ and $b=cosy$
$a^4+b^4+2=4ab$

Let, $b=ka$

Then we have
$a^4+(ka{)}^4+2=4ka²$
$\Rightarrow (1+k^4)a^4+2=4ka²$

now, Put $c=a²$

$\Rightarrow (1+k^4)c²+2=4kc$

This is a quadratic equation in $c$ :

$\Rightarrow (1+k^4)c²-4kc+2=0$

discriminant $D=16k^2-8k^4-8$

Now we investigate the sign of the discriminant

Put, $r=k²$, then we have

$-8(r²-2r+1)=-8(r-1)²\le 0$

The discriminant is always negative, so we have only real solutions for c if the discriminant is zero.
Then $r=1\Rightarrow k=\pm 1c=\left(4k\right)/\left(2\right(1+k^4\left)\right)=a²$

$k=1\Rightarrow c=1\Rightarrow a=\pm 1\Rightarrow b=\pm 1$

$k=-1\Rightarrow$ no solution as $c=a²>0$

So we have two solutions : $(a,b)=(-1,-1)or(1,1)$

This is $(x,y)=(2n\pi +\frac{3}{2}\pi ,(2n+1\left)\pi \right)$

or $(x,y)=(2n\pi +\frac{\pi }{2},2n\pi )$

as $0\le x,y\le \frac{\pi }{2}$

$\therefore (x,y)=(2n\pi +\frac{\pi }{2},2n\pi )$

and $sinx+cosy=1+1=2$

Therefore, Answer is $2$