Using the relationship between the sine function and its reciprocal, which of the following describes the restrictions on the domain of when 0

Question

Using the relationship between the sine function and its reciprocal, which of the following describes the restrictions on the domain of when 0 < x < 2π

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To analyze the domain of f(x) = csc(x) on 0 < x < 2π, recall that csc(x) is 1/sin(x). A reciprocal function is undefined wherever its denominator is zero, so we must exclude x where sin(x) = 0.

Within the interval (0, 2π), the sine function equals zero at x = π. (It also equals zero at x = 0 and x = 2π, but those endpoints are not included since 0 < x < 2π.) Therefore, the only restricted point in this open interval is x = π.

Now evaluate the given options:
- Option a: { x ∈ ℝ, x ≠ π }. This matches the requirement: all real x except x = π are allowed, which aligns with the domain restriction in (0, 2π).
- Option b: { x ∈ ℝ, x ≠ 1 }. Excluding x = 1 has no relation to where sin(x) = 0 in this interval, so this is incorrect.
- Option c: { x ∈ ℝ, x ≠ 3π/2 }. Excluding x = 3π/2 would remove the point where sin(x) = −1, not where sin(x) = 0; thus incorrect for domain of csc.
- Option d: { x ∈ ℝ, x ≠ π/2 }. Excluding x = π/2 removes where sin(x) = 1, not where it vanishes; thus incorrect.

In summary, the correct restriction is that x can be any real number except x = π, which is exactly what option a states.

MHF4U - Advanced Functions 12 (2025-26) - A

Using the relationship between the sine function and its reciprocal, which of the following describes the restrictions on the domain of when 0 < x < 2π

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