OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Exercise 2 (C)

Question 1. Let X = {1, 2, 3, 4}. Determine whether or not each relation is a function from X into X.
(i) f = {(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)}
(ii) g = {(3, 1), (4, 2), (1, 1)}
(iii) h = {(2, 1), (3, 4), (1, 4), (4, 4)}
Answer:
(i) The given relation is \( f = \{(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)\} \). For a relation to be a function, each element in the domain (set X) must map to only one unique image (element in set X). In this relation, the element \( 2 \in X \) maps to two different images: 3 and 1. Since an input (2) has multiple outputs (3 and 1), f is not a function.
(ii) The given relation is \( g = \{(3, 1), (4, 2), (1, 1)\} \). Here, each element from set X that is part of the domain (1, 3, 4) maps to a single, distinct image (1, 1, 2). For example, 3 maps only to 1, 4 maps only to 2, and 1 maps only to 1. Since every domain element has a unique image, this relation g is a function. All inputs have a single, predictable output, which is the definition of a function.
(iii) The given relation is \( h = \{(2, 1), (3, 4), (1, 4), (4, 4)\} \). In this relation, each input element from X (1, 2, 3, 4) has only one output image (4, 1, 4, 4). Even though different inputs like 1 and 3 might have the same output (both map to 4), this is allowed in a function as long as each input itself has only one output. Since each element has a unique image, relation h is a function.
In simple words: (i) This relation is not a function because the number 2 points to two different numbers (3 and 1). A function can only point to one number. (ii) This relation is a function because every number from X points to only one specific number. (iii) This relation is a function because each number from X points to only one number. It's okay if different numbers point to the same number, as long as each one only points once.

🎯 Exam Tip: Remember the "one-to-one or many-to-one" rule for functions: each input must have only one output, but multiple inputs can share the same output.

Question 2. State for each of the following relations whether it is function or not. (Write Yes or No)
(i) {(1, 2), (2, 2), (3, 2), (4, 2)}
(ii) {(x, y): x ∈ A, y ∈ B, y is surname of x} where A is the set of people in India and B is the set of surnames.
(iii) {(x, y): x ∈ A, y ∈ B, y is the area of a square of side x} where A is the set of measurements of length.
(iv) {(x, y): x ∈ B, y ∈ P, y is passenger on x} where B is the set of buses of a school and P is the set of pupils of some schools.
(v) {(x, y): x ∈ A, y ∈ B, y is sewn onto y}, where A is the set of buttons and B is the set of shirts.
Answer:
(i) Yes, this relation is a function. Each number in the domain (1, 2, 3, 4) maps to exactly one number in the codomain (2). Even though all inputs map to the same output (2), each input has only one assigned output, which satisfies the function condition. Many-to-one mapping is permitted in a function, but one-to-many is not.
(ii) No, this relation is not a function. A single person (an element 'x' from set A) can have more than one surname (multiple 'y' values in set B, for example, if they have a hyphenated surname or legally changed it). If the relation means one person is mapped to their surname(s), then a person having multiple surnames means 'x' would map to multiple 'y's, making it not a function. The key characteristic of a function is that each input must have exactly one output, which is violated here.
(iii) Yes, this relation is a function. For any given side length 'x' from set A, its area 'y' will always be \( x^2 \). Since \( x^2 \) gives only one specific area for each length 'x', every input 'x' has a unique output 'y'. Mathematical formulas that define a single output for every valid input almost always represent functions.
(iv) No, this relation is not a function. One bus (an element 'x' from set B) can carry many different pupils (multiple 'y' values from set P). If the relation describes which pupils are on a specific bus, then one input (bus) would map to many outputs (pupils), making it not a function. Functions must map each domain element to exactly one codomain element; a bus having many passengers breaks this rule.
(v) Yes, this relation is a function. If a specific button (an element 'x' from set A) is sewn onto one shirt (an element 'y' from set B), it cannot also be sewn onto another different shirt. This means each button 'x' has only one unique shirt 'y' that it is attached to. This scenario ensures that each button has a single, defined location, fulfilling the requirement for a function.
In simple words: (i) Yes, it's a function. Every input number points to only one output number. (ii) No, it's not a function. One person might have more than one surname, meaning one input has many outputs. (iii) Yes, it's a function. For every side length of a square, there's only one exact area. (iv) No, it's not a function. One bus can have many pupils, meaning one input has many outputs. (v) Yes, it's a function. Each button is sewn onto only one shirt, so each input has just one output.

🎯 Exam Tip: When determining if a relation is a function, always ask: "Does each input have exactly one output?" If the answer is no, it's not a function.

Question 3. The ordered pairs are represented by the points shown. For each diagram, state whether it represents a relation or a function. Justify your answer.
Answer:
(i) From the graph, the set of ordered pairs is \( \{(1, 1), (2, 1), (2, 2), (3, 3)\} \). This set of ordered pairs is a relation. However, the element \( x = 2 \) from the domain maps to two different images in the codomain: \( y = 1 \) and \( y = 2 \). Since an element in the domain has more than one image, this relation is not a function. For a function, each input (x-value) must have exactly one output (y-value); here, \( x=2 \) violates this rule.
(ii) From the graph, the set of ordered pairs is \( \{(1, 1), (2, 2), (3, 3)\} \). This is a relation. In this set, no two ordered pairs share the same first component (x-value) with different second components (y-values). Each element in the domain has a unique image. Therefore, this relation represents a function. This is a one-to-one function, where each distinct input maps to a distinct output.
(iii) From the graph, the set of ordered pairs is \( \{(1, 1), (3, 2)\} \). This is a relation. However, if the domain includes integers like 2 (as implied by the typical graph structure), then the element \( x = 2 \) does not have any corresponding image in the codomain (there's no point starting with 2). For a relation to be a function, every element in the domain must be mapped. Since an element from the domain is not mapped, this relation is not a complete function from its intended domain. A complete function requires every element in its specified domain to have an output.
In simple words: (i) This is a relation, but not a function. The number 2 on the 'x' axis points to two different numbers on the 'y' axis (1 and 2). (ii) Yes, this is a function. Each number on the 'x' axis points to only one unique number on the 'y' axis. (iii) This is a relation, but not a function. The number 2 on the 'x' axis doesn't point to any number on the 'y' axis, so it's incomplete for a function.

🎯 Exam Tip: When evaluating a graph for function status, use the vertical line test: if any vertical line intersects the graph at more than one point, it is not a function.

Question 4. The range and domain of a function \( f(x) = \frac{3}{x} + 1 \) are subsets of A and B respectively, where \( A = \{-\frac{1}{2}, 0, \frac{2}{3}, \frac{6}{7}, 1\} \) and \( B = \{-5, 0, 4\frac{1}{2}, 5, 5\frac{1}{2}\} \) as ordered pairs.
Answer:
The function is given as \( f(x) = \frac{3}{x} + 1 \). We need to find the outputs (images) for the elements in set A, which is the domain.
For \( x = -\frac{1}{2} \):
\( f(-\frac{1}{2}) = \frac{3}{-\frac{1}{2}} + 1 = -6 + 1 = -5 \)
For \( x = 0 \):
The term \( \frac{3}{0} \) is undefined, so \( f(0) \) does not exist. This means 0 cannot be in the domain.
For \( x = \frac{2}{3} \):
\( f(\frac{2}{3}) = \frac{3}{\frac{2}{3}} + 1 = \frac{9}{2} + 1 = \frac{11}{2} \)
For \( x = \frac{6}{7} \):
\( f(\frac{6}{7}) = \frac{3}{\frac{6}{7}} + 1 = \frac{21}{6} + 1 = \frac{7}{2} + 1 = \frac{9}{2} \)
For \( x = 1 \):
\( f(1) = \frac{3}{1} + 1 = 3 + 1 = 4 \)
So, the function creates these ordered pairs: \( \{ (-\frac{1}{2}, -5), (\frac{2}{3}, \frac{11}{2}), (\frac{6}{7}, \frac{9}{2}), (1, 4) \} \). A function's domain is restricted by operations like division, where the denominator cannot be zero.
In simple words: We put each number from set A into the function \( f(x) \). We found out that if x is 0, the function cannot work because you cannot divide by zero. For the other numbers, we got the pairs: \( (-\frac{1}{2}, -5) \), \( (\frac{2}{3}, \frac{11}{2}) \), \( (\frac{6}{7}, \frac{9}{2}) \), and \( (1, 4) \).

🎯 Exam Tip: Always check for values in the domain that would make the function undefined (e.g., division by zero or square root of a negative number). These values must be excluded from the domain.

Question 5. Which of the four statements given below is different from others ?
(a) f : A\( \rightarrow \) B
(b) f : x \( \rightarrow \) f(x)
(c) f is a mapping of A and B
(d) f A into B
Answer:
All four statements describe the concept of a function or mapping.
(a) \( f: A \rightarrow B \) means that \( f \) is a function from set A to set B.
(b) \( f: x \rightarrow f(x) \) explains how the function maps each element \( x \) to its image \( f(x) \). This describes the rule of the function.
(c) "f is a mapping of A and B" is another way to say \( f \) is a function or relation between sets A and B, often implying a function.
(d) "f A into B" (assuming it means "f maps A into B") is synonymous with statement (a), indicating \( f \) takes elements from A and maps them to elements in B.
Since all statements convey similar information about defining a function from set A to set B or how a function operates, none of them are truly different in their fundamental meaning. Different notations can represent the same mathematical concept, and understanding these equivalences is crucial.
In simple words: All these ways of writing (a), (b), (c), and (d) actually mean the same thing. They all explain what a 'function' is and how it connects one set of numbers (A) to another (B). So, none of them are different from the others.

🎯 Exam Tip: Familiarize yourself with different ways to express function notation and terminology, as they often mean the same thing in context.

Question 6. A = {- 2, - 1, 1, 2} and \( f = \left\{\left(x, \frac{1}{x}\right) : x \in A\right\} \).
(i) List the domain of f
(ii) List the range of f
(iii) Is f a function ?
Answer:
The given function is \( f(x) = \frac{1}{x} \) for \( x \in A = \{-2, -1, 1, 2\} \).
We calculate the output for each element in A:
\( f(-2) = \frac{1}{-2} = -\frac{1}{2} \)
\( f(-1) = \frac{1}{-1} = -1 \)
\( f(1) = \frac{1}{1} = 1 \)
\( f(2) = \frac{1}{2} \)
(i) The **domain of \( f \)** is the set of all possible input values, which is given as set A. So, \( \text{Domain}(f) = \{-2, -1, 1, 2\} \).
(ii) The **range of \( f \)** is the set of all output values we found. So, \( \text{Range}(f) = \{-\frac{1}{2}, -1, 1, \frac{1}{2}\} \).
(iii) Yes, **\( f \) is a function**. For every input value from the domain A, there is only one unique output value. No two ordered pairs have the same first component mapping to different second components. The domain is the set of allowed inputs, and the range is the set of all actual outputs produced by the function.
In simple words: We put each number from A into the function \( f(x) = \frac{1}{x} \). (i) The domain is all the numbers we put in: \( \{-2, -1, 1, 2\} \). (ii) The range is all the answers we got out: \( \{-\frac{1}{2}, -1, 1, \frac{1}{2}\} \). (iii) Yes, it is a function because each number we put in gave us only one specific answer.

🎯 Exam Tip: Be careful with the signs when calculating \( \frac{1}{x} \) for negative values of \( x \).

Question 7. f : x \( \rightarrow \) highest prime factor of x.
(i) Find the range of f when the domain is {12, 13, 14, 15, 16 and 17}
(ii) State a domain of five integers for which the range is (3).
(iii) A set of positive integers is called S. What can be said about these integers if f(S) = S ?
Answer:
The function \( f(x) \) gives the highest prime factor of \( x \).
(i) For the domain \( \{12, 13, 14, 15, 16, 17\} \):
Highest prime factor of 12 is 3 (since \( 12 = 2^2 \times 3 \)).
Highest prime factor of 13 is 13 (since 13 is prime).
Highest prime factor of 14 is 7 (since \( 14 = 2 \times 7 \)).
Highest prime factor of 15 is 5 (since \( 15 = 3 \times 5 \)).
Highest prime factor of 16 is 2 (since \( 16 = 2^4 \)).
Highest prime factor of 17 is 17 (since 17 is prime).
The **range of \( f \)** is the set of these highest prime factors: \( \{2, 3, 5, 7, 13, 17\} \).
(ii) We need a domain of five integers where the highest prime factor for each is 3. Examples of numbers whose highest prime factor is 3 include:
3 (highest prime factor is 3)
6 (highest prime factor is 3, from \( 2 \times 3 \))
9 (highest prime factor is 3, from \( 3^2 \))
12 (highest prime factor is 3, from \( 2^2 \times 3 \))
18 (highest prime factor is 3, from \( 2 \times 3^2 \))
A possible **domain of five integers** is \( \{3, 6, 9, 12, 18\} \).
(iii) If \( S \) is a set of positive integers and \( f(S) = S \), it means that for every number \( x \) in \( S \), its highest prime factor \( f(x) \) is also in \( S \), and all elements in \( S \) are themselves highest prime factors of some numbers in \( S \). This condition is only met if \( S \) is the **set of all prime numbers**. For any prime number \( p \), its highest prime factor is \( p \) itself. If \( S \) contained a composite number (e.g., 4), then \( f(4) = 2 \). If 2 is not in \( S \), then \( f(S) \neq S \). If 2 is in \( S \), but 4 is in \( S \), then \( f(S) \) would contain 2 but S would contain 4, still making \( f(S) \neq S \) unless S *only* contained primes. Prime factorization helps us understand the fundamental building blocks of composite numbers.
In simple words: The function finds the biggest prime number that divides into a given number. (i) For the numbers \( \{12, 13, 14, 15, 16, 17\} \), the biggest prime factors are \( \{2, 3, 5, 7, 13, 17\} \). (ii) To get '3' as the biggest prime factor, we can pick numbers like \( \{3, 6, 9, 12, 18\} \). (iii) If a set \( S \) gives back itself when we find the highest prime factor of its numbers, then \( S \) must be a set of only prime numbers. This is because a prime number's biggest prime factor is itself.

🎯 Exam Tip: To find the highest prime factor, first list all prime factors of the number, then identify the largest one. For prime numbers, the highest prime factor is the number itself.

Question 8.
(i) For \( x > 3, f(x) = 3x - 2 \) and for \( -2 < x < 2, f(x) = x^2 - 2 \), find \( f(0) + f(4) \).
(ii) If \( f : \mathbb{R} \rightarrow \mathbb{R} \) defined by \( f(x) = \begin{cases} 4x-1 & \text{ for } x>4 \\ x^2-2 & \text{ for } -2 \leq x<4 \\ 3x+4 & \text{ for } x<-2 \end{cases} \) find \( f(5) + f(0) + f(-5) \).
Answer:
(i) We need to find the value of \( f(0) + f(4) \) using the given piecewise definition.
For \( x = 0 \): The condition \( -2 < x < 2 \) applies. So, \( f(x) = x^2 - 2 \).
\( f(0) = (0)^2 - 2 = 0 - 2 = -2 \).
For \( x = 4 \): The condition \( x > 3 \) applies. So, \( f(x) = 3x - 2 \).
\( f(4) = 3(4) - 2 = 12 - 2 = 10 \).
Therefore, \( f(0) + f(4) = -2 + 10 = 8 \). Piecewise functions use different rules for different parts of their domain, making careful interval checking essential.
(ii) We need to find the value of \( f(5) + f(0) + f(-5) \) using the given piecewise function definition.
For \( x = 5 \): The condition \( x > 4 \) applies. So, \( f(x) = 4x - 1 \).
\( f(5) = 4(5) - 1 = 20 - 1 = 19 \).
For \( x = 0 \): The condition \( -2 \leq x < 4 \) applies. So, \( f(x) = x^2 - 2 \).
\( f(0) = (0)^2 - 2 = 0 - 2 = -2 \).
For \( x = -5 \): The condition \( x < -2 \) applies. So, \( f(x) = 3x + 4 \).
\( f(-5) = 3(-5) + 4 = -15 + 4 = -11 \).
Therefore, \( f(5) + f(0) + f(-5) = 19 + (-2) + (-11) = 19 - 2 - 11 = 6 \). Always check which interval the input value belongs to before applying the corresponding function rule.
In simple words: (i) We use the right rule for \( f(x) \) based on the 'x' value. For \( x=0 \), we use \( x^2-2 \), which gives -2. For \( x=4 \), we use \( 3x-2 \), which gives 10. Adding them up gives \( -2 + 10 = 8 \). (ii) We look at the 'x' value and pick the correct rule for \( f(x) \). For \( x=5 \), the answer is 19. For \( x=0 \), it's -2. For \( x=-5 \), it's -11. Adding them up gives \( 19 - 2 - 11 = 6 \).

🎯 Exam Tip: When working with piecewise functions, precisely identify the interval for each input value to apply the correct function rule. Pay close attention to inclusive vs. exclusive boundaries ( \( \leq, < \) ).

Question 9. What is the fundamental difference between a function and a relation? Let \( X = \{1,2, 3, 4\} \) and \( Y = \{1, 5, 9, 11, 15, 16\} \). Determine which of the following sets are :
(i) relation
(ii) function
(iii) neither
(a) \( f_1 = \{(x, y) : y = x^2, x \in X, y \in Y\} \)
(b) \( f_2 = \{(1, 1), (2, 11), (3, 1), (4, 15)\} \)
(c) \( f_3 = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\} \)
(d) \( f_4 = \{(1, 1), (2, 7), (3,5)\} \)
Answer:
A **relation** is simply any set of ordered pairs, usually connecting elements from one set (domain) to elements in another set (codomain). It can be thought of as any subset of the Cartesian product \( X \times Y \). A **function** is a special type of relation where every element in the domain (the first set, X) is mapped to *exactly one* element in the codomain (the second set, Y). This means no element in the domain can have more than one image. All functions are relations, but not all relations are functions; the "one output per input" rule is what makes a relation a function.

(a) \( f_1 = \{(x, y) : y = x^2, x \in X, y \in Y\} \)
Let's find the ordered pairs for \( f_1 \).
For \( x = 1 \): \( y = 1^2 = 1 \). So, \( (1, 1) \). This is in \( X \times Y \).
For \( x = 2 \): \( y = 2^2 = 4 \). This is **not** in Y, so \( (2, 4) \) is not part of \( f_1 \).
For \( x = 3 \): \( y = 3^2 = 9 \). So, \( (3, 9) \). This is in \( X \times Y \).
For \( x = 4 \): \( y = 4^2 = 16 \). So, \( (4, 16) \). This is in \( X \times Y \).
So, \( f_1 = \{(1, 1), (3, 9), (4, 16)\} \).
(i) **\( f_1 \) is a relation** because it is a subset of \( X \times Y \).
(ii) **\( f_1 \) is not a function**. This is because the element \( x = 2 \) from set X does not have an image in set Y. For a function, every element in the domain must be mapped.
(iii) This means \( f_1 \) is **neither** a function from X to Y nor a complete relation from X to Y, though it is a relation itself. The codomain Y defines the possible output values, and if a calculation results in a value outside Y, that pair cannot be part of the function/relation *into* Y.

(b) \( f_2 = \{(1, 1), (2, 11), (3, 1), (4, 15)\} \)
(i) **\( f_2 \) is a relation** as all its ordered pairs are members of \( X \times Y \).
(ii) **\( f_2 \) is a function**. Each element in set X (1, 2, 3, 4) has a unique image in set Y, and no element in X is repeated as a first component with different second components. Therefore, it is not "neither". Functions are often characterized by passing the vertical line test on a graph, where any vertical line intersects the graph at most once.

(c) \( f_3 = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\} \)
(i) **\( f_3 \) is a relation** because all its ordered pairs are members of \( X \times Y \).
(ii) **\( f_3 \) is not a function**. The element \( x = 2 \) from set X maps to two different images in set Y: \( y = 9 \) and \( y = 11 \). This violates the definition of a function. Therefore, it is not "neither". A single input leading to multiple outputs is the most common reason a relation fails to be a function.

(d) \( f_4 = \{(1, 1), (2, 7), (3,5)\} \)
Let's check the ordered pairs for \( f_4 \).
\( (1, 1) \) is in \( X \times Y \).
\( (2, 7) \): Here, \( 2 \in X \) but \( 7 \notin Y \). So, \( (2, 7) \) is **not** a member of \( X \times Y \).
\( (3, 5) \) is in \( X \times Y \).
(i) **\( f_4 \) is neither a relation nor a function from X to Y**. Since one of its ordered pairs \( (2, 7) \) is not part of the Cartesian product \( X \times Y \) (because \( 7 \notin Y \)), it cannot be considered a relation *from X to Y*. Consequently, it cannot be a function either. For a relation to exist between two sets, all its ordered pairs must be formed by taking the first element from the first set and the second element from the second set.
(iii) Thus, \( f_4 \) is **neither**.
In simple words: A relation just connects numbers from one group to numbers in another group. A function is a special kind of connection where each number from the first group only points to one number in the second group.
(a) \( f_1 = \{(1, 1), (3, 9), (4, 16)\} \). This is a relation. It is not a function because the number 2 from set X doesn't have an output in set Y. So, it is neither a complete function nor a complete relation from X to Y.
(b) \( f_2 = \{(1, 1), (2, 11), (3, 1), (4, 15)\} \). This is a relation and also a function because each number from X points to only one number in Y.
(c) \( f_3 = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\} \). This is a relation. It is not a function because the number 2 from set X points to two different numbers in set Y (9 and 11).
(d) \( f_4 = \{(1, 1), (2, 7), (3,5)\} \). This is neither a relation nor a function because the number 7 is not in set Y, so the pair \( (2, 7) \) doesn't fit the rules for X and Y.

🎯 Exam Tip: Always verify two conditions: 1) all elements of the domain are used, and 2) each element of the domain maps to exactly one element in the codomain. If either fails, it's not a function.

Question 10. A certain jet plane has an average speed of 500 km per hour. It can carry sufficient fuel for a 5 hour flight.
(i) Define the relation, as a set, between the distance d (in km) and time t (in hours) for this plane.
(ii) State the domain of the relation.
(iii) State the range of this relation.
(iv) Is this relation a function ?
Answer:
The jet plane travels at 500 km/hr and can fly for up to 5 hours.
(i) **Define the relation:** We know that speed = distance / time. So, \( 500 = \frac{d}{t} \), which means \( d = 500t \). The time \( t \) can be any real number from just above 0 hours up to 5 hours.
The relation \( R \) can be written as: \( R = \{(t, d) : d = 500t, t \in \mathbb{R} \text{ and } 0 < t \leq 5\} \).
(ii) **State the domain:** The domain is the set of all possible time values \( t \). Since the plane flies for up to 5 hours, the time is positive and includes 5 hours.
\( \text{Domain}(R) = \{t : 0 < t \leq 5, t \in \mathbb{R}\} \).
(iii) **State the range:** The range is the set of all possible distances \( d \). When \( t \) is just above 0, \( d \) is just above 0. When \( t = 5 \), \( d = 500 \times 5 = 2500 \) km.
\( \text{Range}(R) = \{d : 0 < d \leq 2500, d \in \mathbb{R}\} \).
(iv) **Is this relation a function?** Yes, this relation is a function. For every single value of time \( t \) in the domain, there is only one unique distance \( d \) calculated by \( d = 500t \). This means each input \( t \) has only one output \( d \). Real-world problems often impose natural constraints on domain and range, such as time always being positive.
In simple words: A jet flies at 500 km/hr for up to 5 hours. (i) The relation is \( d = 500t \), where \( t \) is time and \( d \) is distance. (ii) The domain (possible times) is from just above 0 hours to 5 hours. (iii) The range (possible distances) is from just above 0 km to 2500 km. (iv) Yes, it is a function. For every flying time, there's only one possible distance covered.

🎯 Exam Tip: When formulating real-world relations, carefully consider the physical constraints to correctly define the domain and range (e.g., time cannot be negative).

Question 11. The domain of a function is the set of positive integers less than 12. Given \( y = f(x) = |X -4| \), find all ordered pairs satisfying the function. Graph the function.
Answer:
The domain of the function is all positive integers less than 12, so \( D_f = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} \). The function is defined as \( y = f(x) = |x - 4| \).
We find the output \( y \) for each \( x \) in the domain:
\( f(1) = |1 - 4| = |-3| = 3 \implies (1, 3) \)
\( f(2) = |2 - 4| = |-2| = 2 \implies (2, 2) \)
\( f(3) = |3 - 4| = |-1| = 1 \implies (3, 1) \)
\( f(4) = |4 - 4| = |0| = 0 \implies (4, 0) \)
\( f(5) = |5 - 4| = |1| = 1 \implies (5, 1) \)
\( f(6) = |6 - 4| = |2| = 2 \implies (6, 2) \)
\( f(7) = |7 - 4| = |3| = 3 \implies (7, 3) \)
\( f(8) = |8 - 4| = |4| = 4 \implies (8, 4) \)
\( f(9) = |9 - 4| = |5| = 5 \implies (9, 5) \)
\( f(10) = |10 - 4| = |6| = 6 \implies (10, 6) \)
\( f(11) = |11 - 4| = |7| = 7 \implies (11, 7) \)
The set of ordered pairs satisfying the function is \( \{(1, 3), (2, 2), (3, 1), (4, 0), (5, 1), (6, 2), (7, 3), (8, 4), (9, 5), (10, 6), (11, 7)\} \). The absolute value function \( |x-a| \) creates a symmetrical graph with its vertex at \( x=a \).
In simple words: We use the numbers from 1 to 11 for \( x \). For each \( x \), we find \( |x - 4| \) to get \( y \). This gives us a list of pairs like \( (1, 3), (2, 2) \), and so on. If you draw these points, they will form a 'V' shape on a graph.

🎯 Exam Tip: When graphing points for a discrete domain, plot distinct points rather than drawing a continuous line, as the function is only defined for specific integer values.

Question 12. Let \( X = \{2, 3\} \) and \( Y = \{1, 3, 5\} \). How many different functions are there from X into Y ?
Answer:
We are given set \( X = \{2, 3\} \) and set \( Y = \{1, 3, 5\} \). We need to find the number of different functions that can map elements from \( X \) to \( Y \).
The number of elements in set \( X \) is \( n(X) = 2 \).
The number of elements in set \( Y \) is \( n(Y) = 3 \).
The total number of functions from set \( X \) to set \( Y \) is given by the formula \( n(Y)^{n(X)} \). This happens because each element in X can independently map to any element in Y.
So, the number of functions \( = 3^2 = 9 \). Each element in the domain has an independent choice of output from the codomain.
In simple words: Set X has 2 numbers, and Set Y has 3 numbers. To find how many different ways we can make a function from X to Y, we take the number of elements in Y and raise it to the power of the number of elements in X. So, \( 3^2 = 9 \). There are 9 different functions.

🎯 Exam Tip: Remember the formula for the number of functions from set A to set B is \( n(B)^{n(A)} \), where \( n(A) \) is the number of elements in set A and \( n(B) \) is the number of elements in set B.