Question

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

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5
• F. 6
• G. 7
• H. 8
• I. 9
• J. 0

Answer 1

Answer: (B)

2

Answer 2

The inequality is $(4 + \sin 4x)(2 + \cot^2 y)(1 + \sin^4 z) \leq 12 \sin^2 z$. Let $x, y, z \in [0, \pi]$.

The terms have the following ranges:

• $4 + \sin 4x \in [3, 5]$
• $2 + \cot^2 y \geq 2$ (for $y \in (0, \pi)$)
• $1 + \sin^4 z \in [1, 2]$
• $12 \sin^2 z \in [0, 12]$

The inequality can be written as: $(4 + \sin 4x)(2 + \cot^2 y) \leq \frac{12 \sin^2 z}{1 + \sin^4 z}$.

The left side (LHS) has a minimum value of $3 \times 2 = 6$. Let $u = \sin^2 z$. For $z \in (0, \pi)$, $u \in (0, 1]$. The right side (RHS) becomes $f(u) = \frac{12u}{1+u^2}$. The maximum of $f(u)$ for $u \in (0, 1]$ occurs at $u=1$, giving $f(1) = \frac{12(1)}{1+1^2} = 6$. So, RHS $\leq 6$.

For the inequality LHS $\leq$ RHS to hold, we must have LHS = 6 and RHS = 6.

LHS = 6 implies:

• $4 + \sin 4x = 3 \implies \sin 4x = -1$. For $x \in [0, \pi]$, $4x \in [0, 4\pi]$. Solutions for $4x$ are $\frac{3\pi}{2}, \frac{7\pi}{2}$. Thus, $x = \frac{3\pi}{8}, \frac{7\pi}{8}$ (2 values).
• $2 + \cot^2 y = 2 \implies \cot^2 y = 0 \implies \cot y = 0$. For $y \in [0, \pi]$, $y = \frac{\pi}{2}$ (1 value).

RHS = 6 implies:

• $\frac{12 \sin^2 z}{1 + \sin^4 z} = 6 \implies \sin^2 z = 1$. For $z \in [0, \pi]$, $\sin z = 1$, so $z = \frac{\pi}{2}$ (1 value).

The total number of triplets is the product of the number of solutions for each variable: $2 \times 1 \times 1 = 2$.