Matching Properties of Zero Sets for Trigonometric Functions

To find the correct combination, let us analyze the given sets $X$ and $Y$ derived from the function $f(x) = \sin(\pi \cos x)$ for $x > 0$:

1. Analysis of Set $X$: The set $X$ consists of the roots of: $f(x) = 0 \iff \sin(\pi \cos x) = 0$ This implies: $\pi \cos x = n\pi \quad (n \in \mathbb{Z}) \implies \cos x = n$ Since $-1 \le \cos x \le 1$, the possible values of $n$ are $-1, 0, 1$.

   • For $\cos x = 0$: $x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$ (odd multiples of $\frac{\pi}{2}$)
   • For $\cos x = 1$: $x = 2\pi, 4\pi, 6\pi, \dots$ (even multiples of $\pi$)
   • For $\cos x = -1$: $x = \pi, 3\pi, 5\pi, \dots$ (odd multiples of $\pi$)

   Combining all these solutions in increasing order for $x > 0$: $X = \left\{ \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \frac{5\pi}{2}, 3\pi, \dots \right\} = \left\{ \frac{k\pi}{2} : k \in \mathbb{N} \right\}$

   • Is $X$ an arithmetic progression? Yes, the successive elements of $X$ have a common difference of $\frac{\pi}{2}$. Thus, $X$ forms an arithmetic progression, corresponding to property (Q).
   • Does $X$ satisfy property (P)? $X$ contains $\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi$, so $X \supseteq \left\{\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi\right\}$. Thus, property (P) also holds for $X$.
   • Does $X$ satisfy property (U)? Property (U) requires $X \supseteq \left\{\frac{\pi}{6}, \frac{3\pi}{4}\right\}$. Since neither $\frac{\pi}{6}$ nor $\frac{3\pi}{4}$ belongs to $X$, property (U) does NOT hold for $X$.

   Therefore:

   • Option (A) is (I), (P), (R): Incorrect because $X$ is an AP, so it cannot satisfy (R).
   • Option (C) is (I), (Q), (U): Incorrect because $X$ does not satisfy (U).

2. Analysis of Set $Y$: The set $Y$ consists of the roots of $f'(x) = 0$ for $x > 0$: $f'(x) = \cos(\pi \cos x) \cdot (-\pi \sin x) = 0$ This gives two cases:

   • Case 1: $\sin x = 0 \implies x = m\pi$ for $m \in \mathbb{N}$.
   • Case 2: $\cos(\pi \cos x) = 0 \implies \pi \cos x = \left(k + \frac{1}{2}\right)\pi \implies \cos x = k + \frac{1}{2}$ for $k \in \mathbb{Z}$. Since $-1 \le \cos x \le 1$, the only possible values for $\cos x$ are: $\cos x = \frac{1}{2} \quad \text{or} \quad \cos x = -\frac{1}{2}$
     • $\cos x = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \dots$
     • $\cos x = -\frac{1}{2} \implies x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \dots$

   Combining all roots of $f'(x) = 0$ in the interval $(0, 2\pi]$: $x \in \left\{ \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi \right\}$ Arranging all positive elements in increasing order: $Y = \left\{ \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi, \frac{7\pi}{3}, \frac{8\pi}{3}, 3\pi, \dots \right\} = \left\{ \frac{n\pi}{3} : n \in \mathbb{N} \right\}$

   • Is $Y$ an arithmetic progression? Yes, the successive elements of $Y$ have a common difference of $\frac{\pi}{3}$. Hence, $Y$ forms an arithmetic progression, which corresponds to property (Q).
   • Does $Y$ satisfy property (T)? Property (T) states: $\supseteq \left\{\frac{\pi}{3}, \frac{2\pi}{3}, \pi\right\}$ Since $\frac{\pi}{3}, \frac{2\pi}{3}, \pi \in Y$, $Y$ clearly satisfies (T).

   Therefore, set $Y$ satisfies both (Q) and (T), which means the combination (II), (Q), (T) is completely correct.

Let us briefly verify Option (D):

• Option (D) proposes (II), (R), (S). Since $Y$ is an arithmetic progression, (R) is false for $Y$. Furthermore, $\frac{\pi}{6} \notin Y$, so (S) is also false for $Y$. Thus, (D) is incorrect.

Hence, the only correct combination is (B).