Understanding the Cosine Function and its Properties

The question asks for the specific domain X and codomain Y such that the function \( f(x) = \cos x \) defined from X to Y is both one-one (injective) and onto (surjective).

Conditions for a Function to be One-one and Onto

A function \( f: X \to Y \) is:

• One-one (Injective): If \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \) for all \( x_1, x_2 \in X \). In simple terms, distinct elements in the domain X map to distinct elements in the codomain Y. Graphically, any horizontal line intersects the graph at most once over the domain X.
• Onto (Surjective): If for every element \( y \in Y \), there exists at least one element \( x \in X \) such that \( f(x) = y \). In simple terms, the range of the function (the set of all output values \( f(x) \)) is equal to the codomain Y. Graphically, for every y-value in Y, there is a corresponding point on the graph of \( f(x) \) for some \( x \in X \).

For a function to be both one-one and onto, it must be a bijection. For \( f(x) = \cos x \) to be a bijection from X to Y, it must be one-one on X and its range on X must be exactly Y.

Analyzing the Cosine Function \( f(x) = \cos x \)

Let's consider the graph of \( y = \cos x \).

The cosine function is periodic with period \( 2\pi \). The standard range of \( \cos x \) over its natural domain (all real numbers) is \( [-1, 1] \). It decreases from 1 to -1 on the interval \( [0, \pi] \) and increases from -1 to 1 on the interval \( [\pi, 2\pi] \). It is not one-one over larger intervals like \( [0, 2\pi] \) because, for example, \( \cos(0) = \cos(2\pi) = 1 \) but \( 0 \ne 2\pi \).

Determining the Domain X for \( f(x) = \cos x \) to be One-one

For \( \cos x \) to be one-one, we need to restrict its domain to an interval where it is strictly monotonic (either strictly increasing or strictly decreasing). The standard interval chosen for defining the principal value branch of the inverse cosine function (arccosine) is \( [0, \pi] \). On this interval, \( \cos x \) strictly decreases from 1 to -1. Thus, if \( x_1, x_2 \in [0, \pi] \) and \( x_1 \ne x_2 \), then \( \cos x_1 \ne \cos x_2 \). So, \( f(x) = \cos x \) is one-one on \( [0, \pi] \).

Other intervals where \( \cos x \) is one-one also exist, like \( [-\pi, 0] \) (strictly increasing), \( [\pi, 2\pi] \) (strictly increasing), etc., but \( [0, \pi] \) is a common choice.

Determining the Codomain Y for \( f(x) = \cos x \) to be Onto

For \( f(x) = \cos x \) to be onto the codomain Y, the range of \( f(x) \) on the chosen domain X must be equal to Y.

On the interval \( X = [0, \pi] \), the minimum value of \( \cos x \) is \( \cos(\pi) = -1 \) and the maximum value is \( \cos(0) = 1 \). Since \( \cos x \) is continuous on \( [0, \pi] \), by the Intermediate Value Theorem, it takes every value between -1 and 1. Therefore, the range of \( \cos x \) on \( [0, \pi] \) is the closed interval \( [-1, 1] \).

For \( f(x) = \cos x \) to be onto when its domain is \( [0, \pi] \), the codomain Y must be equal to its range, which is \( [-1, 1] \).

Evaluating the Given Options

Let's examine each option based on our analysis of the cosine function being one-one on X and its range on X being equal to Y:

1. \( X = [0, \pi] \) and \( Y = [-1, 1] \)
2. \( X = \left[ - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right] \) and \( Y = [-1, 1] \)
3. \( X = [0, \pi] \) and \( Y = (-1, 1) \)
4. \( X = [0, \pi] \) and \( Y = [0, 1] \)

Option 1: \( X = [0, \pi] \) and \( Y = [-1, 1] \)

• Domain \( X = [0, \pi] \): \( \cos x \) is strictly decreasing on this interval, so it is one-one.
• Codomain \( Y = [-1, 1] \): The range of \( \cos x \) on \( [0, \pi] \) is \( [-1, 1] \). Since the codomain Y equals the range, the function is onto.

This option satisfies both conditions. For \( X = [0, \pi] \) and \( Y = [-1, 1] \), \( f(x) = \cos x \) is both one-one and onto.

Option 2: \( X = \left[ - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right] \) and \( Y = [-1, 1] \)

• Domain \( X = \left[ - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right] \): On this interval, \( \cos x \) is not one-one (e.g., \( \cos(-\pi/4) = \cos(\pi/4) \)).
• Codomain \( Y = [-1, 1] \): The range of \( \cos x \) on \( \left[ - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right] \) is \( [0, 1] \). The codomain \( [-1, 1] \) does not equal the range \( [0, 1] \), so it is not onto this codomain.

This option is incorrect.

Option 3: \( X = [0, \pi] \) and \( Y = (-1, 1) \)

• Domain \( X = [0, \pi] \): \( \cos x \) is one-one on this interval.
• Codomain \( Y = (-1, 1) \): The range of \( \cos x \) on \( [0, \pi] \) is \( [-1, 1] \). The codomain \( (-1, 1) \) does not equal the range \( [-1, 1] \). The function is not onto \( (-1, 1) \) as the values -1 and 1 are in the range but not in the codomain.

This option is incorrect.

Option 4: \( X = [0, \pi] \) and \( Y = [0, 1] \)

• Domain \( X = [0, \pi] \): \( \cos x \) is one-one on this interval.
• Codomain \( Y = [0, 1] \): The range of \( \cos x \) on \( [0, \pi] \) is \( [-1, 1] \). The codomain \( [0, 1] \) does not equal the range \( [-1, 1] \). The function is not onto \( [0, 1] \) as values between -1 and 0 are in the range but not in the codomain.

This option is incorrect.

Based on the analysis, the function \( f(x) = \cos x \) is one-one and onto if and only if its domain is \( [0, \pi] \) and its codomain is \( [-1, 1] \).

Revision Table: Cosine Function One-one and Onto Properties

Additional Information: Function Mapping and Bijectivity

Understanding function mapping is fundamental in mathematics. A function \( f: X \to Y \) establishes a relationship between elements of set X (the domain) and elements of set Y (the codomain).

• Injective (One-one): Ensures that each element in X maps to a unique element in Y. No two different inputs give the same output.
• Surjective (Onto): Ensures that every element in Y is 'hit' by at least one element from X. The entire codomain is covered by the function's outputs.
• Bijective: When a function is both injective and surjective, it establishes a one-to-one correspondence between the elements of X and Y. This property is essential for the existence of an inverse function \( f^{-1}: Y \to X \).

For trigonometric functions like \( \cos x \), their periodic nature means they are not one-one over their entire natural domain \( (-\infty, \infty) \). To define inverse functions, we restrict the domain to an interval where the function becomes one-one and restrict the codomain to the range on that specific domain, making it onto. For \( \cos x \), the standard interval for the domain restriction to achieve bijectivity is \( [0, \pi] \), which results in the range \( [-1, 1] \).