Let $f(x) = 2x - 1$. What is the value of $f(3)$?

Correct Option: A. Solution: Substitute $x = 3$ into the function: $f(3) = 2(3) - 1 = 6 - 1 = 5$.

Let $f(x) = x + 1$. What is the value of $f(4)$?

Correct Option: C. Solution: The function $f(x) = x + 1$ implies $f(4) = 4 + 1 = 5$.

If $f(x) = 2x - 3$, what is the value of $f(5)$?

Correct Option: B. Solution: Substitute $x = 5$ into the function: $f(5) = 2(5) - 3 = 10 - 3 = 7$.

What is the domain of the function $f(x) = \sqrt{x-1}$?

Correct Option: A. Solution: The domain of $f(x) = \sqrt{x-1}$ is $x \geq 1$ because the expression under the square root must be non-negative.

If $f(x) = 2x - 1$, find $f(0)$.

Correct Option: A. Solution: Substitute $x = 0$ into the function: $f(0) = 2(0) - 1 = -1$.

Given the function $f(x) = x + 1$, what is $f(2)$?

Correct Option: C. Solution: Substituting $x = 2$ into $f(x) = x + 1$, we get $f(2) = 2 + 1 = 3$.

Relations and Functions

Objectives Notes Exam Tips Practice Qs Ask AI

Ordered Pair

Cartesian Product

Relation

Function

Domain and Range

Real Valued Function

Algebra of Functions

Arrow Diagram

Set-Builder and Roster Method

Modulus and Signum Functions

Learning Objectives

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Learning Objectives

Understand the concept of an Ordered Pair and its significance in distinguishing between different pairings of elements.
Calculate the Cartesian Product of two sets and comprehend how it forms the basis for defining relations.
Define and identify a Relation as a subset of the Cartesian product, and describe how it represents a connection between elements of two sets.
Differentiate between a general relation and a Function, emphasizing the unique mapping of elements in the domain to elements in the codomain.
Identify the Domain and Range of a relation or function, and understand their roles in defining the scope of these mathematical concepts.
Explore Real Valued Functions and recognize their importance when the range is a subset of real numbers.
Perform operations within the Algebra of Functions, including addition, subtraction, multiplication, and division, and understand their application.
Use Arrow Diagrams to visually represent relations and functions, aiding in the comprehension of their structure.
Apply the Set-Builder and Roster Methods to represent relations, understanding the differences and uses of each method.
Analyze the Modulus and Signum Functions, understanding their definitions and graphical representations.

Revision Notes & Summary

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Chapter Notes

Ordered Pair

An ordered pair is a pair of elements grouped together in a specific order, typically represented as $(a, b)$ . The order is crucial, as $(a, b) \neq (b, a)$ .

Cartesian Product

The Cartesian product of two sets $A$ and $B$ , denoted as $A \times B$ , is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$ .
Example: If $A = \{ \text{red, blue} \}$ and $B = \{ \text{b, c, s} \}$ , then $A \times B = \{ (\text{red, b}), (\text{red, c}), (\text{red, s}), (\text{blue, b}), (\text{blue, c}), (\text{blue, s}) \}$ .

Relation

A relation from a set $A$ to a set $B$ is a subset of the Cartesian product $A \times B$ .
Domain: Set of all first elements of the ordered pairs.
Range: Set of all second elements of the ordered pairs.

Function

A function is a specific type of relation where each element in the domain (set $A$ ) is associated with exactly one element in the codomain (set $B$ ).
Notation: $f: A \rightarrow B$ where $f(a) = b$ .

Domain and Range

Domain: The set of all possible inputs (first elements) for a function or relation.
Range: The set of all possible outputs (second elements).

Real Valued Function

A real valued function is a function where the range is a subset of real numbers.

Algebra of Functions

Operations on functions include addition, subtraction, multiplication, and division, performed pointwise on their domains.

Arrow Diagram

An arrow diagram visually represents a relation, showing how elements from the domain are mapped to elements in the range.

Set-Builder and Roster Method

Set-Builder Method: Uses a rule to define a relation, e.g., $R = \{(x, y) : x^2 = y, x \in P, y \in Q\}$ .
Roster Method: Lists all ordered pairs explicitly, e.g., $R = \{(9, 3), (9, -3), (4, 2), (4, -2), (25, 5), (25, -5)\}$ .

Modulus and Signum Functions

Modulus Function: $f(x) = |x|$ $f (x) = ∣ x ∣$ .
- $f(x) = x$ if $x \geq 0$ .
- $f(x) = -x$ if $x < 0$ .
Signum Function: $f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases}$ .

Exam Tips & Common Mistakes

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Common Mistakes & Exam Tips

Ordered Pair

Mistake: Confusing the order of elements in an ordered pair.
- Tip: Remember that $(a, b)$ is not the same as $(b, a)$ . The order is crucial.

Cartesian Product

Mistake: Assuming $A \times B$ $A \times B$ is the same as $B \times A$ $B \times A$ .
- Tip: The Cartesian product $A \times B$ consists of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$ . Always consider the order of sets.

Relation

Mistake: Misunderstanding that a relation is not necessarily a function.
- Tip: A relation from $A$ to $B$ is a subset of $A \times B$ . Verify if each element in $A$ is related to one or more elements in $B$ .

Function

Mistake: Confusing functions with general relations.
- Tip: A function requires every element in the domain to map to exactly one element in the codomain. Check for unique mappings.

Domain and Range

Mistake: Incorrectly identifying the domain and range of a function.
- Tip: The domain is the set of all first elements, and the range is the set of all second elements in the ordered pairs.

Real Valued Function

Mistake: Forgetting that the range must be real numbers.
- Tip: Ensure that the function outputs are real numbers and check the domain for any restrictions.

Algebra of Functions

Mistake: Incorrectly performing operations on functions.
- Tip: When adding, subtracting, multiplying, or dividing functions, perform operations pointwise and be mindful of domains.

Arrow Diagram

Mistake: Misinterpreting the mapping in an arrow diagram.
- Tip: Use arrow diagrams to visualize how each element in the domain maps to elements in the range. Ensure clarity in one-to-one or many-to-one mappings.

Set-Builder and Roster Method

Mistake: Confusing the two methods of representation.
- Tip: Use the set-builder method for rules and the roster method for listing elements explicitly.

Modulus and Signum Functions

Mistake: Misapplying the definitions of modulus and signum functions.
- Tip: The modulus function returns $|x|$ , and the signum function returns $1$ , $0$ , or $-1$ based on whether $x$ is positive, zero, or negative, respectively.

AI Learning Assistant

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Practice Test – MCQs, True/False

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Multiple Choice Questions

Correct Answer: C

Solution:

The function

f(x) = x + 1

means for any input

x

, we add 1 to it. Thus,

f(3) = 3 + 1 = 4

All real numbers except

x = 1

and

x = 4

All real numbers

All real numbers except

x = -1

and

x = -4

All integers

Correct Answer: B

Solution:

The function

f(x) = x^2 + 5x + 4

is a polynomial and is defined for all real numbers.

Correct Answer: A

Solution:

Substitute

x = 3

into the function:

f(3) = 2(3) - 1 = 6 - 1 = 5

x = 1

and

x = 4

x = -1

and

x = 4

x = 0

and

x = 5

x = 2

and

x = 3

Correct Answer: A

Solution:

The function

f(x) = x^2 - 5x + 4

can be factored as

(x-1)(x-4)

. Setting

f(x) = 0

, we get

x = 1

and

x = 4

2.7

3.7

Correct Answer: A

Solution:

The greatest integer less than or equal to 2.7 is 2. Therefore,

f(2.7) = 2

(a, a) \in R

for all

a \in N

(a, b) \in R

implies

(b, a) \in R

(a, b) \in R

and

(b, c) \in R

implies

(a, c) \in R

None of the above

Correct Answer: D

Solution:

For the relation

R = \{(a, b) : a = b^2\}

, none of the given statements are true for all natural numbers.

x = 1, 4

x = -2, -3

x = 0, 3

f(x)

is defined for all real

x

Correct Answer: D

Solution:

The function

f(x) = x^2 + 5x + 6

is a polynomial, and polynomials are defined for all real numbers. Therefore,

f(x)

is defined for all real

x

x = -\frac{5}{2}

x = -\frac{5}{3}

x = -\frac{5}{4}

x = -\frac{5}{6}

Correct Answer: A

Solution:

The quadratic function

f(x) = x^2 + 5x + 6

is in the form

ax^2 + bx + c

, where

a > 0

. Its minimum value occurs at

x = -\frac{b}{2a} = -\frac{5}{2}

(a, a) \in R

for all

a \in N

(a, b) \in R

implies

(b, a) \in R

(a, b) \in R

and

(b, c) \in R

implies

(a, c) \in R

None of the above

Correct Answer: D

Solution:

For the relation

R = \{(a, b) : a = b^2\}

, it is not true that

(a, a) \in R

for all

a \in N

because

a

must be a perfect square. Also,

(a, b) \in R

does not imply

(b, a) \in R

unless

b

is also a perfect square, and the transitive property does not hold in general. Hence, none of the statements are true.

\mathbb{R}

[0, \infty)

(-\infty, \infty)

(-\infty, 0]

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

outputs non-negative values for any real number

x

. Therefore, the range of

f(x)

[0, \infty)

2.5

3.5

Correct Answer: A

Solution:

The greatest integer function

f(x) = [x]

returns the greatest integer less than or equal to

x

. For

x = 2.5

, the greatest integer less than or equal to 2.5 is 2.

x \geq 1

x > 1

x \leq 1

x < 1

Correct Answer: A

Solution:

The function

f(x) = \sqrt{x - 1}

is defined for

x - 1 \geq 0

, which simplifies to

x \geq 1

. Therefore, the domain is

x \geq 1

The range is

(-\infty, \infty)

The range is

[0, \infty)

The range is

(-\infty, 0]

The range is

[0, 1]

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

outputs the absolute value of

x

, which is always non-negative. Therefore, the range of

f(x)

[0, \infty)

, as it includes all non-negative real numbers.

Correct Answer: A

Solution:

Since 3 is greater than or equal to 2, we use the second definition of the function:

f(x) = 2x + 1

. Substituting

x = 3

, we get

f(3) = 2(3) + 1 = 7

The graph is a straight line passing through the origin.

The graph is a V-shaped curve with the vertex at the origin.

The graph is a parabola opening upwards.

The graph is a circle centered at the origin.

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

has a V-shaped graph with the vertex at the origin, reflecting the property that it is linear for both positive and negative values of

x

All real numbers

y \geq -\frac{1}{4}

y \geq -\frac{1}{2}

y \geq -\frac{1}{3}

Correct Answer: B

Solution:

The function

f(x) = x^2 + 5x + 6

is a quadratic function opening upwards. The vertex form is

f(x) = (x + \frac{5}{2})^2 - \frac{1}{4}

, indicating the minimum value is

-\frac{1}{4}

x = -\frac{5}{2}

. Thus, the range is

y \geq -\frac{1}{4}

Correct Answer: D

Solution:

The number of relations from set

A

to set

B

is given by the number of subsets of the Cartesian product

A \times B

. Since

|A \times B| = 3 \times 2 = 6

, there are

2^6 = 64

possible relations.

2.7

3.7

Correct Answer: A

Solution:

The greatest integer function

\lfloor x \rfloor

returns the greatest integer less than or equal to

x

. Therefore,

f(2.7) = 2

Correct Answer: D

Solution:

The number of relations from set

A

to set

B

2^{m \times n}

, where

m

and

n

are the number of elements in

A

and

B

, respectively. Here,

m = 3

and

n = 2

, so the number of relations is

2^{3 \times 2} = 64

\mathbb{R}

[0, \infty)

(-\infty, 0]

(-\infty, \infty)

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

outputs non-negative values for any real input

x

. Thus, the range of

f(x)

[0, \infty)

Correct Answer: C

Solution:

The function

f(x) = x + 1

implies

f(4) = 4 + 1 = 5

-3

-2

-1

Correct Answer: A

Solution:

The greatest integer function

[x]

returns the greatest integer less than or equal to

x

. For

x = -2.5

, the greatest integer less than or equal to

-2.5

-3

. Therefore,

f(-2.5) = -3

[0, \infty)

(-\infty, 0]

(-\infty, \infty)

[1, \infty)

Correct Answer: A

Solution:

The function

f(x) = \sqrt{x - 1}

is defined for

x \geq 1

. The smallest value of

\sqrt{x - 1}

is 0, which occurs when

x = 1

. As

x

increases,

\sqrt{x - 1}

can take any non-negative real value. Hence, the range is

[0, \infty)

For all

a \in \mathbb{N}

(a, a) \in R

(a, b) \in R

, then

(b, a) \in R

(a, b) \in R

and

(b, c) \in R

, then

(a, c) \in R

None of the above

Correct Answer: D

Solution:

For the relation

R = \{(a, b) : a = b^2\}

(a, a) \in R

is false because

a

is not necessarily a perfect square.

(b, a) \in R

is false because

b^2 = a

does not imply

a^2 = b

. The transitive property does not hold because if

a = b^2

and

b = c^2

, it does not imply

a = c^4

. Therefore, none of the statements are true.

All real numbers

All positive real numbers

All negative real numbers

All integers

Correct Answer: A

Solution:

The function

g(x) = 2x - 3

is a linear function with a non-zero slope. Therefore, it can take any real value as

x

varies over the real numbers, making the range all real numbers.

Correct Answer: B

Solution:

Substitute

x = 5

into the function:

f(5) = 2(5) - 3 = 10 - 3 = 7

(2, 8)

(3, 9)

(5, 25)

(7, 343)

Correct Answer: D

Solution:

The prime numbers less than 10 are 2, 3, 5, and 7. The cube of 7 is 343, making (7, 343) a correct element of the relation.

x \geq 1

x > 1

x \leq 1

x < 1

Correct Answer: A

Solution:

The domain of

f(x) = \sqrt{x-1}

x \geq 1

because the expression under the square root must be non-negative.

(1, 6)

(2, 7)

(3, 8)

(4, 9)

Correct Answer: D

Solution:

The relation is defined for natural numbers less than 4. Thus, the valid pairs are (1, 6), (2, 7), and (3, 8). The pair (4, 9) is not valid because 4 is not less than 4.

Correct Answer: B

Solution:

The Cartesian product

A \times B

has

3 \times 2 = 6

elements.

Undefined

Correct Answer: C

Solution:

The expression

\frac{x^2 - 1}{x - 1}

simplifies to

x + 1

for

x \neq 1

. Therefore, as

x

approaches 1, the limit of

f(x)

1 + 1 = 2

All real numbers except

x = 1

and

x = 4

All real numbers

All integers

All real numbers except

x = 0

Correct Answer: B

Solution:

The function

f(x) = x^2 + 5x + 4

is a polynomial, and polynomials are defined for all real numbers. Therefore, the domain is all real numbers.

-3

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

returns the absolute value of

x

. Therefore,

f(-3) = 3

Correct Answer: B

Solution:

Since

x = 2

falls in the case

x \geq 2

, we use

f(x) = 2x + 1

. Thus,

f(2) = 2(2) + 1 = 5

All real numbers except

x = 1

and

x = 4

All real numbers

All real numbers except

x = 0

All real numbers except

x = -5

Correct Answer: A

Solution:

The function

f(x) = x^2 + 5x + 4

is defined for all real numbers except where the denominator is zero, which occurs at

x = 1

and

x = 4

Correct Answer: A

Solution:

For

x = -2

f(-2) = 1 - (-2) = 3

. For

x = 3

f(3) = 3 + 1 = 4

. Thus,

f(-2) + f(3) = 3 + 4 = 7

. Therefore, the correct answer is 0.

x = 1

and

x = 4

x = -1

and

x = -4

x = 0

f(x)

is defined for all real

x

Correct Answer: D

Solution:

The function

f(x) = x^2 + 5x + 4

is a polynomial, which is defined for all real numbers.

All real numbers except 2

All real numbers

All real numbers except 0

All real numbers except -2

Correct Answer: A

Solution:

The function

f(x) = \frac{1}{x-2}

is undefined when

x - 2 = 0

, i.e., when

x = 2

. Therefore, the domain is all real numbers except 2.

The graph is a straight line for

x < 0

and

x > 0

The graph is a parabola for

x < 0

and a straight line for

x > 0

The graph is a V-shape with vertex at the origin.

The graph is a U-shape with vertex at the origin.

Correct Answer: C

Solution:

The modulus function

f(x) = |x|

has a V-shaped graph with the vertex at the origin, as it is linear for both

x < 0

and

x > 0

Correct Answer: C

Solution:

The function

f(x) = |x - 2| + |x + 3|

reaches its minimum value when

x

is the midpoint of the points 2 and -3, which is

x = \frac{2 + (-3)}{2} = -0.5

. At this point,

f(x) = |(-0.5) - 2| + |(-0.5) + 3| = 2.5 + 2.5 = 5

\{5, 6, 7, 8\}

\{6, 7, 8, 9\}

\{7, 8, 9, 10\}

\{4, 5, 6, 7\}

Correct Answer: A

Solution:

For

x = 1, 2, 3

, the values of

y

are

6, 7, 8

respectively. Thus, the range is

\{6, 7, 8\}

. However, since

x < 4

y = 5

is also included when

x = 0

. Therefore, the range is

\{5, 6, 7, 8\}

2x - 2

2x - 1

2x + 1

2x + 3

Correct Answer: A

Solution:

To find

(f \circ g)(x)

, substitute

g(x)

into

f(x)

f(g(x)) = f(2x - 3) = (2x - 3) + 1 = 2x - 2

-1

Correct Answer: A

Solution:

Substitute

x = 0

into the function:

f(0) = 2(0) - 1 = -1

\{1, 2, 3, 4, 6\}

\{1, 2, 3, 4\}

\{2, 4, 6\}

\{1, 2, 4, 6\}

Correct Answer: D

Solution:

The range of the relation

R

consists of elements in

A

that can be expressed as

b

where

b

is divisible by some

a \in A

. Thus, the range is

\{1, 2, 4, 6\}

x \geq 1

x > 1

x \geq 0

x > 0

Correct Answer: A

Solution:

The function

f(x) = \sqrt{x-1}

is defined for all

x

such that the expression under the square root is non-negative. Therefore,

x-1 \geq 0

, which implies

x \geq 1

. Hence, the domain is

x \geq 1

All real numbers

Non-negative real numbers

Negative real numbers

Integers only

Correct Answer: B

Solution:

The range of the modulus function

f(x) = |x|

is non-negative real numbers because

|x|

is always zero or positive.

All real numbers

All non-negative real numbers

All integers

All positive real numbers

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

outputs the absolute value of

x

. Therefore, it can only take non-negative values, meaning the range is all non-negative real numbers.

(-\infty, \infty)

[0, \infty)

(-\infty, 0]

[0, 1]

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

outputs non-negative values for all

x

. Thus, the range is

[0, \infty)

-5

Correct Answer: A

Solution:

The modulus function

f(x) = |x|

returns the non-negative value of

x

. Therefore,

f(-5) = 5

Correct Answer: C

Solution:

Substituting

x = 2

into

f(x) = x + 1

, we get

f(2) = 2 + 1 = 3

(2, 7)

(3, 8)

(4, 9)

(5, 10)

Correct Answer: A

Solution:

For the pair

(x, y)

to be in

R

y

must equal

x + 5

and

x

must be less than 4. For

(2, 7)

7 = 2 + 5

and

2 < 4

, so this pair is in

R

The domain is

x \geq 1

The domain is

x > 1

The domain is

x \leq 1

The domain is all real numbers.

Correct Answer: A

Solution:

The function

f(x) = \sqrt{x-1}

is defined only when the expression under the square root is non-negative. Therefore,

x-1 \geq 0

, which implies

x \geq 1

. Thus, the domain of the function is

x \geq 1

\{1, 2, 3, 4, 6\}

\{2, 4, 6\}

\{1, 2, 4, 6\}

\{1, 2, 3, 4\}

Correct Answer: A

Solution:

The range of

R

includes all elements

b

A

that are exactly divisible by some

a

A

. Since every element in

A

is divisible by itself, the range is

\{1, 2, 3, 4, 6\}

-3

Correct Answer: A

Solution:

Substituting

x = 0

into

f(x) = 2x - 3

gives

f(0) = 2(0) - 3 = -3

\{1, 2, 3\}

\{0, 1, 2, 3\}

\{0, 1, 2\}

\{0, 1, 2, 3, 4\}

Correct Answer: A

Solution:

The domain of

R

consists of natural numbers

x

such that

x < 4

. Therefore, the domain is

\{1, 2, 3\}

x \geq 1

x > 1

x \leq 1

x < 1

Correct Answer: A

Solution:

The function

f(x) = \sqrt{x - 1}

is defined for values of

x

such that the expression under the square root is non-negative. Therefore,

x - 1 \geq 0

, which simplifies to

x \geq 1

Correct Answer: B

Solution:

For

x = -2

f(-2) = 1 - (-2) = 3

. For

x = 0

f(0) = 0

. For

x = 3

f(3) = 3 + 1 = 4

. Therefore,

f(-2) + f(0) + f(3) = 3 + 0 + 4 = 7

Correct Answer: B

Solution:

Substituting

x = 4

into

f(x) = 2x - 3

, we get

f(4) = 2(4) - 3 = 8 - 3 = 5

It is only defined for

x \geq 0

It is defined for all real numbers.

It is only defined for

x < 0

It is not defined for

x = 0

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

is defined for all real numbers

x

. It returns

x

for

x \geq 0

and

-x

for

x < 0

. Therefore, it is defined for all real numbers.

All real numbers

Non-negative real numbers

Positive real numbers

Non-positive real numbers

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

outputs the absolute value of

x

, which is always non-negative. Therefore, the range is non-negative real numbers.

Correct Answer: C

Solution:

Substitute

x = 2

into the function:

f(2) = 2(2) - 1 = 4 - 1 = 3

All real numbers except

x = -1

All real numbers

All real numbers except

x = 1

All real numbers except

x = 0

Correct Answer: A

Solution:

The function

f(x) = \frac{1}{x+1}

is undefined when the denominator is zero. Therefore,

x \neq -1

, and the domain is all real numbers except

x = -1

{6, 7, 8}

{5, 6, 7}

{4, 5, 6}

{7, 8, 9}

Correct Answer: A

Solution:

The domain is

\{1, 2, 3\}

. For

x = 1

y = 6

; for

x = 2

y = 7

; for

x = 3

y = 8

. Thus, the range is

\{6, 7, 8\}

(1, 6)

(2, 7)

(3, 8)

(4, 9)

Correct Answer: D

Solution:

The relation

R

includes pairs where

x

is a natural number less than 4. Thus, (4, 9) is not included.

Correct Answer: B

Solution:

Substitute

x = 3

into the function:

f(3) = 2(3) - 1 = 6 - 1 = 5

. Therefore,

x = 3

x = 1

x = 4

x = 0

x = 5

Correct Answer: A

Solution:

The function

f(x) = \frac{x^2 - 5x + 4}{x - 1}

is undefined when the denominator is zero. Setting

x - 1 = 0

, we find

x = 1

-5

-10

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

returns the absolute value of

x

. Therefore,

f(-5) = |-5| = 5

{5, 6, 7, 8}

{0, 1, 2, 3}

{1, 2, 3, 4}

{6, 7, 8, 9}

Correct Answer: A

Solution:

For each

x

in the set {0, 1, 2, 3}, compute

y = x + 5

y = 5, 6, 7, 8

. Thus, the range is {5, 6, 7, 8}.

Reflexive only

Symmetric only

Transitive only

Reflexive, Symmetric, and Transitive

Correct Answer: D

Solution:

(i) Reflexive: For any

a \in \mathbb{Q}

a - a = 0 \in \mathbb{Z}

, so

(a, a) \in R

. (ii) Symmetric: If

(a, b) \in R

, then

a - b \in \mathbb{Z}

implies

b - a \in \mathbb{Z}

, so

(b, a) \in R

. (iii) Transitive: If

(a, b) \in R

and

(b, c) \in R

, then

a - b \in \mathbb{Z}

and

b - c \in \mathbb{Z}

imply

a - c = (a - b) + (b - c) \in \mathbb{Z}

, so

(a, c) \in R

(-\infty, \infty)

[0, \infty)

(-\infty, 0]

\{0\}

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

outputs non-negative values, so the range is

[0, \infty)

Reflexive only

Symmetric only

Transitive only

Reflexive, Symmetric, and Transitive

Correct Answer: D

Solution:

The relation is reflexive because for any integer

a

a - a = 0

, which is even. It is symmetric because if

a - b

is even, then

b - a

is also even. It is transitive because if

a - b

is even and

b - c

is even, then

a - c = (a - b) + (b - c)

is also even.

Correct Answer: C

Solution:

First, calculate

f(2) = 2 + 1 = 3

. Then, substitute into

g(x)

g(3) = 2(3) - 3 = 6 - 3 = 3

. Therefore,

g(f(2)) = 3

2x - 2

2x + 1

2x - 3

2x + 3

Correct Answer: A

Solution:

The composition

f(g(x))

is found by substituting

g(x)

into

f(x)

. So,

f(g(x)) = f(2x - 3) = (2x - 3) + 1 = 2x - 2

6x - 2

6x + 2

3x^2 - 2

3x^2 + 2x

Correct Answer: A

Solution:

The derivative of

g(x) = 3x^2 - 2x + 1

g'(x) = \frac{d}{dx}(3x^2) - \frac{d}{dx}(2x) + \frac{d}{dx}(1) = 6x - 2

Yes, because each element in

A

is related to exactly one element in

B

No, because not all elements in

A

are related to elements in

B

Yes, because

f

is a bijection.

No, because

f

is not injective.

Correct Answer: A

Solution:

For

f

to be a function, each element in the domain

A

must be related to exactly one element in the codomain

B

. Here, each element of

A

(1, 2, 3, 4) is related to exactly one element in

B

, so

f

is a function.

True or False

Correct Answer: True

Solution:

For any integer

a

a - a = 0

is an integer, so

(a, a) \in R

Correct Answer: True

Solution:

The modulus function

f(x) = |x|

is defined for all real numbers

x

because it simply returns the absolute value of

x

, which is always a real number.

Correct Answer: False

Solution:

The modulus function f(x) = |x| is equal to x for non-negative values of x (x \geq 0) and equal to -x for negative values of x (x < 0). Therefore, it is not equal to x for all x \in \mathbb{R}.

Correct Answer: True

Solution:

For each

x

in the set of natural numbers less than 4, there is a unique

y

such that

y = x + 5

. Therefore,

R

is a function.

Correct Answer: False

Solution:

For a relation to be reflexive,

(a, a)

must be in

R

for all

a

. However,

a = b^2

does not hold for all integers

a

, hence

R

is not reflexive.

Correct Answer: True

Solution:

The domain of

f(x) = \sqrt{x-1}

x \geq 1

because the expression under the square root must be non-negative.

Correct Answer: False

Solution:

The greatest integer function

f(x) = [x]

is not continuous because it has jump discontinuities at every integer value of

x

Correct Answer: False

Solution:

The greatest integer function

f(x) = [x]

is not continuous at integer points because it has jump discontinuities at these points.

Correct Answer: True

Solution:

The modulus function is defined as

f(x) = |x|

, which means for

x \geq 0

f(x) = x

, and for

x < 0

f(x) = -x

. This definition matches the description provided.

Correct Answer: False

Solution:

A relation is a function if every element in the domain

A

has a unique image in the codomain

B

. Here, the element

2

A

is associated with both

9

and

11

B

, violating the definition of a function.

Correct Answer: False

Solution:

The function

f(x) = \sqrt{x-1}

is defined only for

x \geq 1

, so its domain is

[1, \infty)

Correct Answer: True

Solution:

The Cartesian product

P \times Q

contains

3 \times 5 = 15

ordered pairs, as there are 3 elements in

P

and 5 elements in

Q

Correct Answer: False

Solution:

The relation

R

is not a function because a single

x

can be paired with multiple

y

values, violating the definition of a function.

Correct Answer: False

Solution:

The modulus function

f(x) = |x|

is not one-to-one because different inputs can produce the same output (e.g.,

f(-1) = f(1) = 1

Correct Answer: True

Solution:

The modulus function

f(x) = |x|

is defined for all real numbers

x

, as it outputs the absolute value of

x

Correct Answer: True

Solution:

The greatest integer function

f(x) = [x]

is defined for all real numbers, as it assigns the greatest integer less than or equal to

x

Correct Answer: True

Solution:

The function

f(x) = x + 1

is a linear function, which means it maps each real number

x

to exactly one real number

x + 1

Correct Answer: True

Solution:

The modulus function

f(x) = |x|

is defined for every real number

x

, as it returns the absolute value of

x

Correct Answer: True

Solution:

The modulus function

f(x) = |x|

is continuous for all real numbers because it does not have any breaks, jumps, or holes in its graph.

Correct Answer: False

Solution:

The Cartesian product

A \times B

is not equal to

B \times A

because the ordered pairs are reversed.

Correct Answer: False

Solution:

The function

f(x) = \sqrt{x-1}

is defined for

x \geq 1

, so it is not defined for all real numbers.

Correct Answer: True

Solution:

A relation is reflexive if every element is related to itself. For

R

, since

a - a = 0

and

0

is an integer,

(a, a) \in R

for all

a \in \mathbb{Z}

. Thus,

R

is reflexive.

Correct Answer: True

Solution:

(a, b) \in R

, then

a - b

is an integer, which implies

b - a

is also an integer. Therefore,

(b, a) \in R

, making the relation symmetric.

Correct Answer: False

Solution:

The cartesian product

P \times Q

consists of ordered pairs where the first element is from

P

and the second is from

Q

Q \times P

reverses this order, so

P \times Q

is not equal to

Q \times P

unless

P = Q

and both are single-element sets.

Correct Answer: True

Solution:

For negative values of

x

, the modulus function

f(x) = |x|

is equal to

-x

Correct Answer: False

Solution:

The relation is not a function because each

x

can correspond to multiple

y

values, violating the definition of a function.

Correct Answer: False

Solution:

The function

f(x) = \sqrt{x-1}

is defined only for

x \geq 1

, as the expression under the square root must be non-negative.

Correct Answer: True

Solution:

The intersection

B \cap C = \{4\}

, so

A \times (B \cap C) = \{(1, 4), (2, 4), (3, 4)\}

Correct Answer: False

Solution:

The relation

R

is defined such that

a = b^2

. For

(a, a)

to be in

R

a

must equal

a^2

, which is only true for

a = 0

a = 1

. Thus,

(a, a)

is not in

R

for all

a \in \mathbb{N}

Correct Answer: False

Solution:

A relation is reflexive if

(a, a) \in R

for all

a \in \mathbb{N}

. In this case,

(a, a) \in R

only if

a = a^2

, which is not true for all natural numbers.

Correct Answer: True

Solution:

The modulus function

f(x) = |x|

returns the absolute value of

x

, which is the definition of the modulus function.

Correct Answer: True

Solution:

A function is a relation where each element in the domain is associated with exactly one element in the range. Here, for each

x

in the domain (1, 2, 3), there is a unique

y

in the range, satisfying the condition

y = x + 5

. Therefore,

R

is a function.

Correct Answer: True

Solution:

The modulus function

f(x) = |x|

takes any real number

x

and returns its absolute value, which is always defined for all real numbers.

Correct Answer: True

Solution:

The Cartesian product

A \times B

has

3 \times 2 = 6

ordered pairs, as each element of

A

pairs with each element of

B

Correct Answer: True

Solution:

The greatest integer function, also known as the floor function, is represented graphically by horizontal line segments at integer values on the y-axis.

Correct Answer: False

Solution:

The domain of the function

f(x) = x^2 + 5x + 4

is all real numbers, as it is a polynomial function and defined for all

x \in \mathbb{R}

Correct Answer: True

Solution:

A relation is reflexive if for every element a in the set, (a,a) is in the relation. Since a - a = 0, which is an integer, (a,a) is in R for all a in \mathbb{Z}. Therefore, R is reflexive.

Correct Answer: False

Solution:

The Cartesian product

A \times B

consists of ordered pairs

(a, b)

where

a \in A

and

b \in B

B \times A

consists of ordered pairs

(b, a)

where

b \in B

and

a \in A

. These are not the same unless

A = B

and the order does not matter.

Correct Answer: True

Solution:

This relation is a function because each prime number less than 10 has a unique cube, hence there are no repeated first elements in the ordered pairs.

Correct Answer: False

Solution:

The cartesian product

A \times B

consists of ordered pairs where the first element is from

A

and the second is from

B

B \times A

reverses this order. Unless

A = B = \emptyset

A \times B \neq B \times A

Correct Answer: False

Solution:

The relation

R = \{(a, b) : a, b \in \mathbb{Z}, a = b^2\}

is not a function because a single value of

b

can correspond to multiple values of

a

(e.g.,

b = 2

gives

a = 4

and

b = -2

also gives

a = 4

Correct Answer: False

Solution:

The Cartesian product

P \times Q

is not equal to

Q \times P

because

(a, r) \neq (r, a)

. They are different ordered pairs.

Correct Answer: True

Solution:

The function

f(x) = x^2 + 5x + 4

is undefined at

x = 4

and

x = 1

due to division by zero.

Correct Answer: False

Solution:

A relation is symmetric if for every

(a, b) \in R

(b, a) \in R

. Here, if

a = b^2

, it does not imply

b = a^2

. Therefore,

R

is not symmetric.

Correct Answer: False

Solution:

The cartesian product

P \times Q

is not equal to

Q \times P

because the order of elements in the ordered pairs matters.

Correct Answer: True

Solution:

The modulus function

f(x) = |x|

returns

x

for non-negative values of

x

and

-x

for negative values of

x

. Therefore, for

x \geq 0

f(x) = x

Correct Answer: True

Solution:

The relation is symmetric because if

(a, b) \in R

, then

a - b

is an integer, which implies

b - a

is also an integer, hence

(b, a) \in R

Correct Answer: True

Solution:

The greatest integer function

f(x) = [x]

is discontinuous at integer values because it jumps from one integer value to the next without covering any intermediate values.

Correct Answer: False

Solution:

The domain of the function

f(x) = \sqrt{x-1}

x \geq 1

, as the expression under the square root must be non-negative.

Correct Answer: False

Solution:

The greatest integer function

f(x) = [x]

is not continuous at integer points because it has jump discontinuities at these points.

Correct Answer: True

Solution:

The function

f(x) = \lfloor x \rfloor

returns the greatest integer less than or equal to

x

, hence it is called the greatest integer function.

Correct Answer: True

Solution:

A linear function is of the form

f(x) = mx + c

. The function

f(x) = x + 1

can be written as

f(x) = 1 \cdot x + 1

, which is linear with slope

m = 1

and intercept

c = 1

Correct Answer: True

Solution:

For any

a \in \mathbb{Q}

a - a = 0 \in \mathbb{Z}

, so

(a, a) \in R

. Therefore,

R

is reflexive.

Correct Answer: True

Solution:

For non-negative values of

x

f(x) = |x|

is equal to

x

Relations and Functions

Learning Objectives

Learning Objectives

Revision Notes & Summary

Chapter Notes

Ordered Pair

Cartesian Product

Relation

Function

Domain and Range

Real Valued Function

Algebra of Functions

Arrow Diagram

Set-Builder and Roster Method

Modulus and Signum Functions

Exam Tips & Common Mistakes

Common Mistakes & Exam Tips

Ordered Pair

Cartesian Product

Relation

Function

Domain and Range

Real Valued Function

Algebra of Functions

Arrow Diagram

Set-Builder and Roster Method

Modulus and Signum Functions

AI Learning Assistant

Practice Test – MCQs, True/False

Multiple Choice Questions

Q1.Let f(x)=x+1f(x) = x + 1f(x)=x+1. What is the value of f(3)f(3)f(3)?

Correct Answer: C

Q2.Given the function f(x)=x2+5x+4f(x) = x^2 + 5x + 4f(x)=x2+5x+4, for which values of xxx is f(x)f(x)f(x) defined?

Correct Answer: B

Q3.Let f(x)=2x−1f(x) = 2x - 1f(x)=2x−1. What is the value of f(3)f(3)f(3)?

Correct Answer: A

Q4.Let f(x)=x2−5x+4f(x) = x^2 - 5x + 4f(x)=x2−5x+4. What are the values of xxx for which f(x)=0f(x) = 0f(x)=0?

Correct Answer: A

Q5.Consider the function f(x)=[x]f(x) = [x]f(x)=[x], where [x][x][x] is the greatest integer less than or equal to xxx. What is f(2.7)f(2.7)f(2.7)?

Correct Answer: A

Q6.Let RRR be a relation from NNN to NNN defined by R={(a,b):a=b2}R = \{(a, b) : a = b^2\}R={(a,b):a=b2}. Which of the following is true?

Correct Answer: D

Q7.Let f(x)=x2+5x+6f(x) = x^2 + 5x + 6f(x)=x2+5x+6. For which values of xxx is f(x)f(x)f(x) not defined?

Correct Answer: D

Q8.Let f(x)=x2+5x+6f(x) = x^2 + 5x + 6f(x)=x2+5x+6. For which values of xxx does f(x)f(x)f(x) attain its minimum value?

Correct Answer: A

Q9.Let RRR be a relation from the set of natural numbers NNN to NNN defined by R={(a,b):a=b2}R = \{(a, b) : a = b^2\}R={(a,b):a=b2}. Which of the following statements is true?

Correct Answer: D

Q10.Consider the function f(x)=∣x∣f(x) = |x|f(x)=∣x∣. What is the range of this function?

Correct Answer: B

Q11.Consider the greatest integer function f(x)=[x]f(x) = [x]f(x)=[x]. What is f(2.5)f(2.5)f(2.5)?

Correct Answer: A

Q12.Let f(x)=x−1f(x) = \sqrt{x - 1}f(x)=x−1​. What is the domain of f(x)f(x)f(x)?

Correct Answer: A

Q13.Let f(x)=∣x∣f(x) = |x|f(x)=∣x∣ be the modulus function. Which of the following correctly describes the range of f(x)f(x)f(x)?

Correct Answer: B

Q14.Consider the function defined by f(x)=x2−4x+3f(x) = x^2 - 4x + 3f(x)=x2−4x+3 for x<2x < 2x<2 and f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 for x≥2x \geq 2x≥2. What is the value of f(3)f(3)f(3)?

Correct Answer: A

Q15.Let the function f(x)=∣x∣f(x) = |x|f(x)=∣x∣ be defined for all real numbers. Which of the following statements is true about its graph?

Correct Answer: B

Q16.Given the function f(x)=x2+5x+6f(x) = x^2 + 5x + 6f(x)=x2+5x+6, what is the range of f(x)f(x)f(x) for x∈Rx \in \mathbb{R}x∈R?

Correct Answer: B

Q17.Let A={1,2,3}A = \{1, 2, 3\}A={1,2,3} and B={4,5}B = \{4, 5\}B={4,5}. How many relations can be formed from set AAA to set BBB?

Correct Answer: D

Q18.If f(x)=⌊x⌋f(x) = \lfloor x \rfloorf(x)=⌊x⌋, what is the value of f(2.7)f(2.7)f(2.7)?

Correct Answer: A

Q19.Let A={1,2,3}A = \{1, 2, 3\}A={1,2,3} and B={4,5}B = \{4, 5\}B={4,5}. How many relations can be formed from AAA to BBB?

Correct Answer: D

Q20.Let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R be defined by f(x)=∣x∣f(x) = |x|f(x)=∣x∣. Which of the following is the range of f(x)f(x)f(x)?

Correct Answer: B

Q21.Let f(x)=x+1f(x) = x + 1f(x)=x+1. What is the value of f(4)f(4)f(4)?

Correct Answer: C

Q22.The greatest integer function f(x)=[x]f(x) = [x]f(x)=[x] is defined as the greatest integer less than or equal to xxx. What is the value of f(−2.5)f(-2.5)f(−2.5)?

Correct Answer: A

Q23.Let f(x)=x−1f(x) = \sqrt{x - 1}f(x)=x−1​. What is the range of this function?

Correct Answer: A

Q24.Consider the relation R={(a,b):a,b∈N and a=b2}R = \{(a, b) : a, b \in \mathbb{N} \text{ and } a = b^2\}R={(a,b):a,b∈N and a=b2}. Which of the following statements is true?

Correct Answer: D

Q25.If the function g(x)=2x−3g(x) = 2x - 3g(x)=2x−3 is defined on the set of real numbers, what is the range of g(x)g(x)g(x)?

Correct Answer: A