CBSE Class 11 Maths: Relations and Functions Ex 2.3 - Mastering Functions

What is a Function? The Special Relation

In mathematics, a function is a special type of relation between two non-empty sets, say A and B, where every element of set A is associated with exactly one element of set B. This 'exactly one' condition is crucial and differentiates a function from a general relation. If even a single element in set A is related to more than one element in set B, the relation is not a function.

Think of it like this: for every input you provide (from set A), you must get one and only one output (in set B). You can't put in '2' and sometimes get '4' and sometimes '5' – that's not how a function works! The set A is called the domain of the function, and it contains all possible input values. The set B is called the codomain, which is the set of all possible output values. The actual set of all outputs obtained when you apply the function to every element in the domain is called the range. The range is always a subset of the codomain. A common way to denote a function from set A to set B is f: A → B.

Graphically, a relation is a function if and only if every vertical line intersects its graph at most once. This is known as the Vertical Line Test. If a vertical line intersects the graph at two or more points, it means a single x-value (input) corresponds to multiple y-values (outputs), which violates the definition of a function.

Key Terms: Domain, Codomain, Range

Function

A relation f from a set A to a set B is called a function if every element of set A has one and only one image in set B.

Domain

The set of all first elements (inputs) of the ordered pairs in a function. For f: A → B, the domain is set A.

Codomain

The entire set B, which contains all possible output values for a function f: A → B.

Range

The set of all second elements (outputs or images) of the ordered pairs in a function. The range is a subset of the codomain.

Image

If f(x) = y, then y is called the image of x under f.

Pre-image

If f(x) = y, then x is called the pre-image of y under f.

Worked Examples: Identifying Functions and Finding Properties

• Example 1: Identifying a function from a set of ordered pairs Let R be a relation from A = {1, 2, 3} to B = {4, 5, 6} defined by R = {(1, 4), (2, 5), (3, 6)}. Is R a function? Find its domain and range. Solution: Step 1: Check if every element of set A has an image in set B. Elements of A are 1, 2, 3. Each has an image: 1 has 4, 2 has 5, 3 has 6. Step 2: Check if each element of set A has exactly one image. 1 is related only to 4. 2 is related only to 5. 3 is related only to 6. Since both conditions are met, R is a function. Step 3: Identify the domain and range. Domain = Set of first elements = {1, 2, 3}. Range = Set of second elements = {4, 5, 6}. Final Answer: Yes, R is a function. Domain = {1, 2, 3}, Range = {4, 5, 6}.
• Example 2: Is the given relation a function? Consider the relation f = {(x, y) : y = x^2, x ∈ {-2, -1, 0, 1, 2}}. Is f a function? What are its domain and range? Solution: Step 1: List the ordered pairs for the given x values. If x = -2, y = (-2)^2 = 4. So, (-2, 4). If x = -1, y = (-1)^2 = 1. So, (-1, 1). If x = 0, y = (0)^2 = 0. So, (0, 0). If x = 1, y = (1)^2 = 1. So, (1, 1). If x = 2, y = (2)^2 = 4. So, (2, 4). So, f = {(-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4)}. Step 2: Check if every element in the input set ({-2, -1, 0, 1, 2}) has exactly one image. Each input value yields only one output value (e.g., for x=1, y is only 1, not also some other value). Even though different inputs (-1 and 1) can map to the same output (1), this is allowed in a function. The crucial part is that one input does not map to multiple outputs. Therefore, f is a function. Step 3: Identify the domain and range. Domain = {-2, -1, 0, 1, 2} (the given set of x values). Range = {0, 1, 4} (the set of unique output values). Final Answer: Yes, f is a function. Domain = {-2, -1, 0, 1, 2}, Range = {0, 1, 4}.
• Example 3: Determining domain and range for a function with a square root Find the domain and range of the real function f(x) = √(x - 3). Solution: Step 1: Determine the domain. For f(x) to be a real number, the expression inside the square root must be non-negative. So, x - 3 ≥ 0. This implies x ≥ 3. Therefore, the domain of f is [3, ∞). Step 2: Determine the range. Since x ≥ 3, the minimum value of x - 3 is 0 (when x = 3). As x increases from 3, x - 3 increases, and √(x - 3) also increases. The square root of a non-negative number √(y) is always non-negative, i.e., √(y) ≥ 0. So, f(x) ≥ 0. As x approaches infinity, √(x - 3) also approaches infinity. Therefore, the range of f is [0, ∞). Final Answer: Domain = [3, ∞), Range = [0, ∞).

Exploring Different Types of Functions

In Exercise 2.3, you'll encounter various types of functions, each with unique characteristics regarding their domain and range. Understanding these basic types is essential for solving problems efficiently. Some common functions include:

1. Identity Function: f(x) = x. For any real number x, the output is x. Domain = R, Range = R.
2. Constant Function: f(x) = c, where c is a constant. For any x, the output is always c. Domain = R, Range = {c}.
3. Polynomial Functions: f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0. These are defined for all real numbers. Domain = R. The range depends on the specific polynomial (e.g., for x^2, Range = [0, ∞)).
4. Rational Functions: f(x) = P(x) / Q(x), where P(x) and Q(x) are polynomial functions and Q(x) ≠ 0. The domain excludes any values of x for which Q(x) = 0.
5. Modulus Function (Absolute Value Function): f(x) = |x|. It gives the non-negative value of x. Defined as x if x ≥ 0 and -x if x < 0. Domain = R, Range = [0, ∞).
6. Signum Function: f(x) = 1 if x > 0, 0 if x = 0, and -1 if x < 0. Domain = R, Range = {-1, 0, 1}.
7. Greatest Integer Function (Step Function): f(x) = [x]. It gives the greatest integer less than or equal to x. For example, [3.7] = 3, [-1.2] = -2. Domain = R, Range = Z (set of all integers).

When finding the domain, always look for values of x that would make the function undefined, such as denominators becoming zero or taking the square root of a negative number. The range requires analyzing the possible output values based on the function's behavior.

Exam Tip: Avoiding Common Mistakes in Functions

Students often confuse a general relation with a function. Remember the golden rule: 'Each input must have exactly one output.'

• Mistake 1: Multiple Outputs for a Single Input: If you see ordered pairs like (2, 5) and (2, 7) in a relation, it is NOT a function because the input '2' has two different outputs ('5' and '7'). This is a very common trap.
• Mistake 2: Missing Domain Elements: In a defined set A for the domain, ensure every element from A has a corresponding image in the relation. If an element from A is left out, it's not a function from A to B.
• Mistake 3: Incorrect Domain Calculation: For functions involving fractions (e.g., 1/x), the denominator cannot be zero. For square roots (e.g., √x), the expression inside the root must be non-negative. Always state these restrictions clearly when defining the domain.
• Mistake 4: Confusing Range with Codomain: The codomain is the possible set of outputs, while the range is the actual set of outputs. The range is always a subset of the codomain. Make sure to identify the actual values obtained by the function.

Practice Questions with Solutions

• Q: Determine which of the following relations are functions. Give reasons. If it is a function, determine its domain and range. (i) R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)} (ii) R = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)} (iii) R = {(1, 3), (1, 5), (2, 5)} A: Step 1: Analyze relation (i). Each first element (2, 5, 8, 11, 14, 17) has exactly one image (which is 1). No first element is repeated with a different second element. So, (i) is a function. Domain = {2, 5, 8, 11, 14, 17}. Range = {1}. Step 2: Analyze relation (ii). Each first element (2, 4, 6, 8, 10, 12, 14) has exactly one image. No first element is repeated with a different second element. So, (ii) is a function. Domain = {2, 4, 6, 8, 10, 12, 14}. Range = {1, 2, 3, 4, 5, 6, 7}. Step 3: Analyze relation (iii). The first element '1' is related to '3' and also to '5'. This violates the 'exactly one image' condition. So, (iii) is not a function. Final answer: (i) Function. Domain = {2, 5, 8, 11, 14, 17}, Range = {1}. (ii) Function. Domain = {2, 4, 6, 8, 10, 12, 14}, Range = {1, 2, 3, 4, 5, 6, 7}. (iii) Not a function, because 1 has two images (3 and 5).
• Q: Find the domain and range of the function f(x) = -|x|. A: Step 1: Determine the domain. The modulus function |x| is defined for all real numbers x. Multiplying by -1 does not change the definability of x. Therefore, the domain of f(x) = -|x| is the set of all real numbers, R or (-∞, ∞). Step 2: Determine the range. We know that |x| ≥ 0 for all real x. Multiplying by -1 reverses the inequality: -|x| ≤ 0. So, f(x) can take any non-positive real value. The maximum value f(x) can take is 0 (when x = 0). Therefore, the range of f(x) = -|x| is (-∞, 0]. Final answer: Domain = R or (-∞, ∞), Range = (-∞, 0].
• Q: Find the domain and range of the function f(x) = √(4 - x^2). A: Step 1: Determine the domain. For f(x) to be a real function, the expression inside the square root must be non-negative. 4 - x^2 ≥ 0 x^2 ≤ 4 Taking the square root of both sides, |x| ≤ 2. This means -2 ≤ x ≤ 2. Therefore, the domain is [-2, 2]. Step 2: Determine the range. We know x^2 ≥ 0 for all x. Since -2 ≤ x ≤ 2, the minimum value of x^2 is 0 (when x = 0), and the maximum value of x^2 is 4 (when x = -2 or x = 2). So, 0 ≤ x^2 ≤ 4. Multiplying by -1 and reversing inequalities: -4 ≤ -x^2 ≤ 0. Adding 4 to all parts: 4 - 4 ≤ 4 - x^2 ≤ 4 - 0 0 ≤ 4 - x^2 ≤ 4. Now, taking the square root: √0 ≤ √(4 - x^2) ≤ √4 0 ≤ f(x) ≤ 2. Therefore, the range is [0, 2]. Final answer: Domain = [-2, 2], Range = [0, 2].
• Q: A function f is defined by f(x) = 2x - 5. Write down the values of: (i) f(0) (ii) f(7) (iii) f(-3) A: Step 1: Calculate f(0). Substitute x = 0 into the function f(x) = 2x - 5. f(0) = 2(0) - 5 = 0 - 5 = -5. Step 2: Calculate f(7). Substitute x = 7 into the function f(x) = 2x - 5. f(7) = 2(7) - 5 = 14 - 5 = 9. Step 3: Calculate f(-3). Substitute x = -3 into the function f(x) = 2x - 5. f(-3) = 2(-3) - 5 = -6 - 5 = -11. Final answer: (i) f(0) = -5 (ii) f(7) = 9 (iii) f(-3) = -11

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