Question

Let $x,y\in [0,2\pi ]$ and satisfying the equation ${\sin }^{3}x+{\cos }^{3}y+6\sin x\cos y=8.$ If $a=x+y$ and $b=x-y,$ then

• A. minimum value of $a$ is $\frac{\pi }{2}$
• B. maximum value of $a$ is $\frac{5\pi }{2}$
• C. minimum value of $b$ is $-\frac{3\pi }{2}$
• D. maximum value of $b$ is $\pi$

Answer 1

Answer: (A), (B), (C)

The correct options are
A minimum value of $a$ is $\frac{\pi }{2}$
B maximum value of $a$ is $\frac{5\pi }{2}$
C minimum value of $b$ is $-\frac{3\pi }{2}$

We know that,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
If $a+b+c=0,$ then
$a^3+b^3+c^3=3abc$

Now, ${\sin }^{3}x+{\cos }^{3}y+(-2{)}^3=3\sin x\cdot \cos y\cdot (-2)$
$\Rightarrow \sin x+\cos y-2=0$
$\Rightarrow \sin x+\cos y=2$
$\Rightarrow \sin x=1,\cos y=1$
$\Rightarrow x=\frac{\pi }{2},y=0,2\pi$
$\therefore a=x+y=\frac{\pi }{2},\frac{5\pi }{2}$
$\therefore b=x-y=\frac{\pi }{2},-\frac{3\pi }{2}$