Question

Let $x,y,z$ be elements from the interval $(0,2\pi )$ satisfying the inequality $(4+\sin 4x)(2+{\cot }^{2}y)(1+{\sin }^{4}z)\le 12{\sin }^{2}z.$ Which of the following statements is FALSE?

• A. The number of ordered pairs $(x,y)$ is $8.$
• B. The number of ordered pairs $(y,z)$ is $4.$
• C. The number of ordered pairs $(z,x)$ is $4.$
• D. The number of ordered triplets $(x,y,z)$ is $16.$

Answer 1

The correct option is C The number of ordered pairs $(z,x)$ is $4.$

$(4+\sin 4x)(2+{\cot }^{2}y)(1+{\sin }^{4}z)\le 12{\sin }^{2}z$
Dividing both sides by ${\sin }^{2}z,$ we get
$(4+\sin 4x)(2+{\cot }^{2}y)({\sin }^{2}z+{\text{cosec}}^{2}z)\le 12\cdots \left(1\right)$

$4+\sin 4x\ge 3,$ $2+{\cot }^{2}y\ge 2$ and ${\sin }^{2}z+{\text{cosec}}^{2}z\ge 2$
$\therefore (4+\sin 4x)(2+{\cot }^{2}y)({\sin }^{2}z+{\text{cosec}}^{2}z)\ge 12$
So, eqn. $\left(1\right)$ is true only when
$4+\sin 4x=3,$ $2+{\cot }^{2}y=2$ and ${\sin }^{2}z+{\text{cosec}}^{2}z=2$
$\Rightarrow \sin 4x=-1,$ ${\cot }^{2}y=0$ and ${\sin }^{2}z=1$
$\Rightarrow x\in \{\frac{3\pi }{8},\frac{7\pi }{8},\frac{11\pi }{8},\frac{15\pi }{8}\},$ $y\in \{\frac{\pi }{2},\frac{3\pi }{2}\}$ and $z\in \{\frac{\pi }{2},\frac{3\pi }{2}\}$