Understanding the Function f(x) = sin x on [0, 2π]

The given function is f : [0, 2π] → R, defined by f(x) = sin x. Here, the domain is the interval [0, 2π], and the codomain is the set of all real numbers, R.

We need to determine if this function f(x) = sin x is one-to-one or onto on the given domain and codomain.

Is f(x) = sin x One-to-One on [0, 2π]?

A function is called one-to-one (or injective) if every unique element in the domain maps to a unique element in the codomain. Mathematically, f(x₁) = f(x₂) implies x₁ = x₂ for all x₁, x₂ in the domain.

Let's consider the function f(x) = sin x on the domain [0, 2π].

• We know that sin(0) = 0 and sin(π) = 0. Here, we have two different values in the domain, 0 and π, that map to the same value in the codomain (0). Since 0 ≠ π but f(0) = f(π), the condition for one-to-one is not met.
• Another example: sin(π/6) = 1/2 and sin(5π/6) = 1/2. Again, π/6 ≠ 5π/6, but f(π/6) = f(5π/6).

Because there are distinct values of x in the domain [0, 2π] for which sin x has the same value, the function f(x) = sin x is not a one-to-one function on this domain.

Is f(x) = sin x Onto R?

A function is called onto (or surjective) if every element in the codomain is the image of at least one element in the domain. Mathematically, for every y in the codomain, there exists at least one x in the domain such that f(x) = y.

The domain of the function is [0, 2π]. We need to find the range of f(x) = sin x for x ∈ [0, 2π].

• As x varies from 0 to π/2, sin x increases from sin(0) = 0 to sin(π/2) = 1.
• As x varies from π/2 to 3π/2, sin x decreases from sin(π/2) = 1 to sin(3π/2) = -1.
• As x varies from 3π/2 to 2π, sin x increases from sin(3π/2) = -1 to sin(2π) = 0.

Therefore, the set of all possible values that f(x) = sin x can take when x is in the domain [0, 2π] is the interval [-1, 1]. This is the range of the function.

The codomain of the function is given as R (the set of all real numbers). For the function to be onto, the range must be equal to the codomain.

Since the range [-1, 1] is a proper subset of the codomain R (e.g., the number 2 is in R but not in [-1, 1]), there are many elements in the codomain R that are not the image of any x in the domain [0, 2π]. For example, there is no x in [0, 2π] such that sin x = 2.

Thus, the function f(x) = sin x is not an onto function with codomain R.

Conclusion for f(x) = sin x on [0, 2π]

Based on our analysis, the function f(x) = sin x defined on the domain [0, 2π] with codomain R is neither one-to-one nor onto.

Let's check the given options:

1. one-to-one function: Incorrect, as shown above.
2. one-to-one and onto function: Incorrect, as it is neither.
3. onto function: Incorrect, as shown above.
4. More than one of the above: Incorrect, since none of the above options are true.
5. None of the above: This option aligns with our conclusion that the function is neither one-to-one nor onto.

Therefore, the function f(x) = sin x on [0, 2π] is none of the properties listed in options 1, 2, or 3.