Question

The number of triplets $(x,y,z)$, where $x,y,z\in [0,2\pi ]$, satisfying the inequality
$(1+{\cos }^{4}x)(2+{\cot }^{2}y)(4+\sin 4z)\le 12{\cos }^{2}x$ are

• A. $16$
• B. $18$
• C. $20$
• D. $24$

Answer 1

The correct option is D $24$

Rearrange the expression as,
$\left(\frac{1+{\cos }^{4}x}{{\cos }^{2}x}\right)(2+{\cot }^{2}y)(4+\sin 4z)\le 12$

$\therefore ({\cos }^{2}x+{\sec }^{2}x)(2+{\cot }^{2}y)(4+\sin 4z)\le 12$..................$\left(1\right)$

$({\cos }^{2}x+{\sec }^{2}x)\ge 2$......................$(A.M\ge G.M)$

$(2+{\cot }^{2}y)\ge 2$......................$({\cot }^{2}y\ge 0)$

$(4+\sin 4z)\ge 3$......................$(\sin 4z\ge -1)$

$\therefore ({\cos }^{2}x+{\sec }^{2}x)(2+{\cot }^{2}y)(4+\sin 4z)\ge 12$...................$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ it implies that only limiting case is possible.

Therefore,
${\cos }^{2}x={\sec }^{2}x=1\to x=0$ or $x=\pi$ or $x=2\pi$ gives $3$ values.

and ${\cot }^{2}y=0\to x=\frac{\pi }{2}$ or $x=\frac{3\pi }{2}$ gives $2$ values.

and $\sin 4z=-1\to z=\frac{3\pi }{8}$ or $z=\frac{7\pi }{8}$ or $z=\frac{11\pi }{8}$ or $z=\frac{15\pi }{8}$ gives $4$ values.

$\therefore$ Total number of triplets is $3\times 2\times 4$.
Hence, answer is $24$.