Matching Sets of Critical Points and Zeros for Composite Trigonometric Functions

To determine the correct combination, we analyze the sets $X, Y, Z,$ and $W$ defined for $x > 0$.

1. Analysis of the Sets

• Set $X = \{x > 0 : f(x) = 0\}$: $f(x) = \sin(\pi \cos x) = 0 \implies \pi \cos x = k\pi \implies \cos x = k, \quad k \in \mathbb{Z}$ Since $\cos x \in [-1, 1]$, the possible values are $\cos x \in \{-1, 0, 1\}$.

  • $\cos x = 0 \implies x = \frac{\pi}{2} + m\pi = (2m+1)\frac{\pi}{2}$
  • $\cos x = \pm 1 \implies x = m\pi$

  Combining these for $x > 0$, we have: $X = \left\{ \frac{n\pi}{2} : n \in \mathbb{N} \right\} = \left\{ \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \frac{5\pi}{2}, \dots \right\}$ This is an arithmetic progression with common difference $\frac{\pi}{2}$. Thus, $X$ satisfies:

  • (P) since $\left\{\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi\right\} \subseteq X$.
  • (Q) as it is an arithmetic progression.

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• Set $Y = \{x > 0 : f'(x) = 0\}$: $f'(x) = \frac{d}{dx}\big(\sin(\pi \cos x)\big) = -\pi \sin x \cos(\pi \cos x) = 0$ This implies:

  1. $\sin x = 0 \implies x = n\pi, \quad n \in \mathbb{N}$
  2. $\cos(\pi \cos x) = 0 \implies \pi \cos x = (2k+1)\frac{\pi}{2} \implies \cos x = \pm \frac{1}{2}$ $\implies x = 2n\pi \pm \frac{\pi}{3} \quad \text{or} \quad x = 2n\pi \pm \frac{2\pi}{3}$

  Arranging the elements in increasing order: $Y = \left\{ \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi, \dots \right\} = \left\{ \frac{n\pi}{3} : n \in \mathbb{N} \right\}$ Thus, $Y$ satisfies:

  • (Q) as it forms an arithmetic progression with common difference $\frac{\pi}{3}$.
  • (T) since $\left\{\frac{\pi}{3}, \frac{2\pi}{3}, \pi\right\} \subseteq Y$.

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• Set $Z = \{x > 0 : g(x) = 0\}$: $g(x) = \cos(2\pi \sin x) = 0 \implies 2\pi \sin x = (2k+1)\frac{\pi}{2} \implies \sin x = \frac{2k+1}{4}, \quad k \in \mathbb{Z}$ Since $\sin x \in [-1, 1]$, the permissible values are: $\sin x \in \left\{ -\frac{3}{4}, -\frac{1}{4}, \frac{1}{4}, \frac{3}{4} \right\}$ The solutions of these equations do not have a constant difference, so:
  • $Z$ is NOT an arithmetic progression, which satisfies (R).
  • Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \notin \left\{ \pm\frac{1}{4}, \pm\frac{3}{4} \right\}$, $\frac{\pi}{6} \notin Z$. Hence, $Z$ does not contain the sets in (S) or (U).

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• Set $W = \{x > 0 : g'(x) = 0\}$: $g'(x) = \frac{d}{dx}\big(\cos(2\pi \sin x)\big) = -2\pi \cos x \sin(2\pi \sin x) = 0$ This implies:

  1. $\cos x = 0 \implies x = (2n+1)\frac{\pi}{2}$ (for which $\sin x = \pm 1$)
  2. $\sin(2\pi \sin x) = 0 \implies 2\pi \sin x = k\pi \implies \sin x = \frac{k}{2}, \quad k \in \mathbb{Z}$

  Since $\sin x \in [-1, 1]$, the valid values are: $\sin x \in \left\{ -1, -\frac{1}{2}, 0, \frac{1}{2}, 1 \right\}$ The elements in $(0, 2\pi]$ in increasing order are: $\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}, 2\pi$ The consecutive differences in this interval are: $\frac{\pi}{3}, \frac{\pi}{3}, \frac{\pi}{6}, \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{3}, \frac{\pi}{6}$ Since the common difference is not constant:

  • $W$ is NOT an arithmetic progression, so it satisfies (R).

  Checking subsets:

  • For (P):

    • For $x = \frac{\pi}{2}$, $\sin x = 1 \in W$.
    • For $x = \frac{3\pi}{2}$, $\sin x = -1 \in W$.
    • For $x = 4\pi$, $\sin x = 0 \in W$.
    • For $x = 7\pi$, $\sin x = 0 \in W$.

    Hence, $W \supseteq \left\{\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi\right\}$, satisfying (P).

  • For (S):

    • For $x = \frac{\pi}{6}$, $\sin x = \frac{1}{2} \in W$.
    • For $x = \frac{7\pi}{6}$, $\sin x = -\frac{1}{2} \in W$.
    • For $x = \frac{13\pi}{6}$, $\sin x = \frac{1}{2} \in W$.

    Hence, $W \supseteq \left\{\frac{\pi}{6}, \frac{7\pi}{6}, \frac{13\pi}{6}\right\}$, satisfying (S).

Therefore, for set $W$ (IV), the properties (P), (R), and (S) all hold true.

Conclusion

The correct combination is (IV), (P), (R), (S), which corresponds to option B.