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 $f(2) = 2 + 1 = 3$.

If $f(x) = 2x - 1$, what is $f(2)$?

Correct Option: C. Solution: Substitute $x = 2$ into the function: $f(2) = 2(2) - 1 = 4 - 1 = 3$.

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.

Let $A = \{1, 2, 3, 4, 6\}$ and $R$ be a relation on $A$ defined by $R = \{(a, b) : a, b \in A, b$ is exactly divisible by $a\}$. What is the range of $R$?

Correct Option: A. Solution: The range of $R$ includes all elements $b$ in $A$ that are exactly divisible by some $a$ in $A$. Since every element in $A$ is divisible by itself, the range is $\{1, 2, 3, 4, 6\}$.

If a function $$g(x)$$ is defined such that $$g(x) = 3x^2 - 2x + 1$$, what is the value of $$g'(x)$$, the derivative of $$g(x)$$?

Correct Option: A. Solution: The derivative of $$g(x) = 3x^2 - 2x + 1$$ is $$g'(x) = \frac{d}{dx}(3x^2) - \frac{d}{dx}(2x) + \frac{d}{dx}(1) = 6x - 2$$.

Relations and Functions

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Summary

Chapter Summary: Relations and Functions

Key Concepts

Pattern Recognition: Mathematics involves finding patterns and relationships between quantities.
Relations: A relation involves pairs of objects in a certain order, such as (m, n) where m is related to n.
Functions: A special type of relation where each element in the domain corresponds to exactly one element in the codomain.

Important Definitions

Ordered Pair: A pair of elements grouped in a specific order.
Cartesian Product: For sets A and B, the Cartesian product A x B is defined as A x B = {(a, b): a ∈ A, b ∈ B}.
Domain: The set of all first elements of ordered pairs in a relation.
Range: The set of all second elements of ordered pairs in a relation.
Function: A relation f from set A to set B where every element x in A has one and only one image y in B, denoted as f: A → B.

Examples

Example of Cartesian Product: If A = {red, blue} and B = {b, c, s}, then A x B = {(red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s)}.
Example of Function: Let f: A → B be defined by f(n) = the highest prime factor of n.

Common Pitfalls

Not Recognizing Functions: Ensure that each element in the domain maps to only one element in the codomain to qualify as a function.
Confusing Relations with Functions: Not all relations are functions; check for unique mappings.

Tips for Exam Preparation

Review definitions and properties of relations and functions.
Practice identifying functions from given relations.
Familiarize yourself with Cartesian products and how to compute them.

Learning Objectives

Understand the concept of relations and functions in mathematics.
Identify ordered pairs and Cartesian products of sets.
Define relations as subsets of Cartesian products.
Distinguish between general relations and functions.
Determine the domain and range of a relation or function.
Recognize the importance of functions in establishing a unique correspondence between sets.
Apply the definitions of relations and functions to solve mathematical problems.

Detailed Notes

Chapter 2: Relations and Functions

2.1 Introduction

Mathematics involves finding patterns and relationships between quantities.
Examples of relations in daily life: brother-sister, father-son, teacher-student.
In mathematics: relations like m < n, line l || line m, set A ⊆ set B.
This chapter covers linking pairs of objects from two sets and special relations qualifying as functions.

2.2 Cartesian Products of Sets

Definition: The Cartesian product A × B of two sets A and B is given by:
- A × B = {(a, b): a ∈ A, b ∈ B}
Example: If A = {red, blue} and B = {b, c, s}, then:
- Distinct pairs: (red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s) → 6 pairs.
Properties:
- n(A) = p, n(B) = q → n(A × B) = p × q.
- A × ∅ = ∅.
- A × B ≠ B × A generally.

2.3 Relations

Definition: A relation R from set A to set B is a subset of the Cartesian product A × B.
Domain: Set of all first elements of ordered pairs in R.
Range: Set of all second elements of ordered pairs in R.
Function: A relation where every element x in set A has one and only one image y in set B.

2.4 Functions

Definition: A function f from set A to set B is a relation where:
- Each element of A has one and only one image in B.
- Denoted as f: A → B.
Examples:
- Modulus function: f(x) = |x|, where:
  - f(x) = x for x ≥ 0
  - f(x) = -x for x < 0
- Signum function: f(x) = 1 if x > 0, f(x) = 0 if x = 0, f(x) = -1 if x < 0.

2.5 Important Properties of Functions

The domain of a function is all real numbers except where it is undefined.
The range of a function is the set of images produced by the function.
Algebra of Functions:
- (f + g)(x) = f(x) + g(x)
- (f - g)(x) = f(x) - g(x)
- (f.g)(x) = f(x) * g(x)
- (kf)(x) = k * f(x), where k is a real number.

2.6 Examples and Exercises

Example of a function: f = {(1,1), (2,3), (0,-1), (-1,-3)}.
Determine if a relation is a function by checking if each input has a unique output.
Exercises include finding domains, ranges, and determining if given relations are functions.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Functions: Students often confuse relations with functions. Remember, a function must have exactly one output for each input.
Domain and Range Errors: Failing to correctly identify the domain and range of a function can lead to incorrect answers. Always check the conditions of the function.
Cartesian Product Confusion: When forming Cartesian products, students may forget that the order matters. A x B is not the same as B x A.
Ignoring Restrictions: When dealing with functions like square roots or logarithms, students often ignore restrictions on the domain.

Exam Tips

Clarify Definitions: Be clear on definitions of terms like relation, function, domain, and range. Write them down if necessary.
Use Roster Form: When asked to define relations, use roster form to clearly show the elements involved.
Check for Uniqueness: For a relation to be a function, ensure each input maps to one and only one output.
Practice with Examples: Work through examples of functions and non-functions to solidify understanding. Use provided examples to guide your reasoning.
Draw Graphs: If applicable, sketch graphs of functions to visualize their behavior, especially for piecewise functions.

Practice & Assessment

Multiple Choice Questions

Correct Answer: C

Solution:

Substituting

x = 2

into

f(x) = x + 1

, we get

f(2) = 2 + 1 = 3

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

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

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.

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

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.

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

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

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

\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: 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.

Correct Answer: C

Solution:

Substitute

x = 2

into the function:

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

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

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.

\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)

2.7

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: 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

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}

Correct Answer: B

Solution:

Substitute

x = 5

into the function:

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

-5

Correct Answer: A

Solution:

The modulus function

f(x) = |x|

returns the non-negative value of

x

. Therefore,

f(-5) = 5

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.

Correct Answer: B

Solution:

The Cartesian product

A \times B

has

3 \times 2 = 6

elements.

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 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.

Correct Answer: B

Solution:

Substitute

x = 3

into the function:

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

. Therefore,

x = 3

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

(-\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)

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}

-3

Correct Answer: B

Solution:

The modulus function

f(x) = |x|

returns the absolute value of

x

. Therefore,

f(-3) = 3

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.

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.

\{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\}

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.

[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)

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

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

\{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\}

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

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

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.

(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.

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

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.

-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

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

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

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.

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

(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.

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

{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}.

-3

Correct Answer: A

Solution:

Substituting

x = 0

into

f(x) = 2x - 3

gives

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

(-\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)

\{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\}

(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.

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.

-1

Correct Answer: A

Solution:

Substitute

x = 0

into the function:

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

Correct Answer: A

Solution:

Substitute

x = 3

into the function:

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

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.

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

{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\}

(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.

(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: C

Solution:

The function

f(x) = x + 1

implies

f(4) = 4 + 1 = 5

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

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

(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

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

\{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\}

Correct Answer: B

Solution:

Substituting

x = 4

into

f(x) = 2x - 3

, we get

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

True or False

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 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 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: 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 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 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:

For any integer

a

a - a = 0

is an integer, so

(a, a) \in R

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 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: 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: 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: 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: 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 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 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 = \{(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 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 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: 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: 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: True

Solution:

For non-negative values of

x

f(x) = |x|

is equal to

x

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:

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 intersection

B \cap C = \{4\}

, so

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

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: 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 domain of

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

x \geq 1

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

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 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 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: 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:

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: 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: 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 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: 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:

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 negative values of

x

, the modulus function

f(x) = |x|

is equal to

-x

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: 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:

The function

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

is undefined at

x = 4

and

x = 1

due to division by zero.

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

Relations and Functions

AI Learning Assistant

Summary

Chapter Summary: Relations and Functions

Key Concepts

Important Definitions

Examples

Common Pitfalls

Tips for Exam Preparation

Learning Objectives

Detailed Notes

Chapter 2: Relations and Functions

2.1 Introduction

2.2 Cartesian Products of Sets

2.3 Relations

2.4 Functions

2.5 Important Properties of Functions

2.6 Examples and Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Exam Tips

Practice & Assessment

Multiple Choice Questions

Q1.Given the function f(x)=x+1f(x) = x + 1f(x)=x+1, what is f(2)f(2)f(2)?

Correct Answer: C

Q2.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

Q3.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

Q4.Let A={1,2,3,4}A = \{1, 2, 3, 4\}A={1,2,3,4} and B={1,5,9,11,15,16}B = \{1, 5, 9, 11, 15, 16\}B={1,5,9,11,15,16}. A relation fff from AAA to BBB is given by f={(1,5),(2,9),(3,1),(4,5)}f = \{(1, 5), (2, 9), (3, 1), (4, 5)\}f={(1,5),(2,9),(3,1),(4,5)}. Is fff a function?

Correct Answer: A

Q5.Consider the function f(x)=x2−5x+4x−1f(x) = \frac{x^2 - 5x + 4}{x - 1}f(x)=x−1x2−5x+4​. For which value of xxx is the function not defined?

Correct Answer: A

Q6.Let f(x)=∣x∣f(x) = |x|f(x)=∣x∣ be the modulus function. What is the value of f(−5)f(-5)f(−5)?

Correct Answer: B

Q7.Consider the function f(x)={1−x,if x<0x+1,if x>0f(x) = \begin{cases} 1 - x, & \text{if } x < 0 \\ x + 1, & \text{if } x > 0 \end{cases}f(x)={1−x,x+1,​if x<0if x>0​. What is the value of f(−2)+f(3)f(-2) + f(3)f(−2)+f(3)?

Correct Answer: A

Q8.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

Q9.If f(x)=∣x∣f(x) = |x|f(x)=∣x∣ is the modulus function, which of the following statements is true about its graph?

Correct Answer: C

Q10.Consider the function f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1​ for x≠1x \neq 1x=1. What is the limit of f(x)f(x)f(x) as xxx approaches 1?

Correct Answer: C

Q11.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

Q12.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

Q13.If f(x)=2x−1f(x) = 2x - 1f(x)=2x−1, what is f(2)f(2)f(2)?

Correct Answer: C

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.Consider the function f(x)=x2+5x+4f(x) = x^2 + 5x + 4f(x)=x2+5x+4. What is the domain of this function?

Correct Answer: B

Q16.Consider the function f(x)=∣x∣f(x) = |x|f(x)=∣x∣. What is the range of this function?

Correct Answer: B

Q17.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

Q18.Let f(x)=x+1f(x) = x + 1f(x)=x+1 and g(x)=2x−3g(x) = 2x - 3g(x)=2x−3. What is the value of g(f(2))g(f(2))g(f(2))?

Correct Answer: C

Q19.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

Q20.If f(x)=2x−3f(x) = 2x - 3f(x)=2x−3, what is the value of f(5)f(5)f(5)?

Correct Answer: B

Q21.Consider the function f(x)=∣x∣f(x) = |x|f(x)=∣x∣. What is the value of f(−5)f(-5)f(−5)?

Correct Answer: A

Q22.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

Q23.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 elements are there in the Cartesian product A×BA \times BA×B?

Correct Answer: B

Q24.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

Q25.What is the domain of the function f(x)=1x−2f(x) = \frac{1}{x-2}f(x)=x−21​?

Correct Answer: A

Q26.If f(x)=2x−1f(x) = 2x - 1f(x)=2x−1 and f(3)=5f(3) = 5f(3)=5, what is the value of xxx?

Correct Answer: B

Q27.Let the function f(x)=x−1f(x) = \sqrt{x-1}f(x)=x−1​ be defined for real numbers. What is the domain of this function?

Correct Answer: A

Q28.Consider the function f(x)=∣x∣f(x) = |x|f(x)=∣x∣. What is the range of f(x)f(x)f(x)?

Correct Answer: B