Question 1

Let A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8}. A relation R is defined from A to B such that R = {(a, b) : a ∈ A, b ∈ B, and a divides b}. Which of the following statements is correct?

Accepted Answer

We need to find the relation R where (a, b) ∈ R if and only if a divides b (written as a | b).

Step 1: Check divisibility for each element of A with elements of B.

For a = 1: 1 divides all elements of B → (1,2), (1,4), (1,6), (1,8)
For a = 2: 2 divides 2, 4, 6, 8 → (2,2), (2,4), (2,6), (2,8)
For a = 3: 3 divides 6 only → (3,6)
For a = 4: 4 divides 4, 8 → (4,4), (4,8)
For a = 5: 5 divides none of {2,4,6,8} → no pairs

Step 2: Write the complete relation.
R = {(1,2), (1,4), (1,6), (1,8), (2,2), (2,4), (2,6), (2,8), (3,6), (4,4), (4,8)}

Step 3: Count the number of ordered pairs.
Total ordered pairs in R = 11

Step 4: Verify domain and range.
Domain of R = {1, 2, 3, 4} (all elements of A that have at least one relation to B)
Range of R = {2, 4, 6, 8} (all elements of B that are related to some element of A)

Note: The domain excludes 5 because 5 does not divide any element in B.
The range includes all elements of B because: 1|2, 2|4, 3|6, 4|8.

Question 2

Let f: ℝ → ℝ be defined by f(x) = x² + 2x + 3. Which of the following statements is correct?

Accepted Answer

Setup: We analyze the function f(x) = x² + 2x + 3 to determine its range and injectivity/surjectivity properties.

Solution:
Complete the square:
f(x) = x² + 2x + 3 = (x² + 2x + 1) + 2 = (x + 1)² + 2.

Since (x + 1)² ≥ 0 for all x ∈ ℝ, we have:
f(x) = (x + 1)² + 2 ≥ 0 + 2 = 2.

The minimum value is f(-1) = 0 + 2 = 2.

Therefore:
- Range of f = [2, ∞), not all of ℝ.
- f is NOT surjective (onto) because no x maps to values less than 2.
- f is NOT injective (one-to-one) because f(0) = 3 and f(-2) = 4 - 4 + 3 = 3, so two different inputs give the same output.
- f is NOT bijective.

The correct statement is: f is neither injective nor surjective.

Question 3

Let A and B be two finite sets such that n(A) = 5, n(B) = 3, and n(A ∩ B) = 2. Then n(A ∪ B) is equal to:

Accepted Answer

Setup: We use the fundamental principle of set union cardinality.

Solution:
By the inclusion-exclusion principle:
n(A ∪ B) = n(A) + n(B) - n(A ∩ B).

Substituting the given values:
n(A ∪ B) = 5 + 3 - 2 = 6.

Verification: 
- Set A has 5 elements.
- Set B has 3 elements.
- They share 2 common elements.
- When we add n(A) + n(B) = 8, we count the 2 common elements twice.
- So we subtract n(A ∩ B) = 2 once to correct for the double-counting.
- Result: 8 - 2 = 6 elements in A ∪ B.

Question 4

Let A = {x ∈ ℤ : −3 ≤ x ≤ 3} and B = {x ∈ ℤ : x² < 9}. Then A ∩ B is equal to:

Accepted Answer

Setup: Finding the intersection of two sets defined by conditions on integers — requires understanding set-builder notation and solving inequalities.

Solution:
Step 1: Determine set A
A = {x ∈ ℤ : −3 ≤ x ≤ 3}
A = {−3, −2, −1, 0, 1, 2, 3}

Step 2: Determine set B
B = {x ∈ ℤ : x² < 9}
We need x² < 9, which means −3 < x < 3 (taking square roots and considering both positive and negative roots)
For integers: x ∈ {−2, −1, 0, 1, 2}
B = {−2, −1, 0, 1, 2}

Step 3: Find A ∩ B
A ∩ B consists of elements in both A and B
A ∩ B = {−2, −1, 0, 1, 2}

Question 5

Let f: {1, 2, 3, 4} → {a, b, c, d} be a function defined by f(1) = a, f(2) = b, f(3) = c, f(4) = d. Which of the following statements is correct?

Accepted Answer

Setup: Understanding properties of functions — injective (one-to-one), surjective (onto), and bijective (both).

Solution:
Given: f: {1, 2, 3, 4} → {a, b, c, d} with f(1) = a, f(2) = b, f(3) = c, f(4) = d

Step 1: Check if f is injective (one-to-one)
For f to be injective, different inputs must map to different outputs.
f(1) = a, f(2) = b, f(3) = c, f(4) = d
All four outputs are distinct, so f IS injective. ✓

Step 2: Check if f is surjective (onto)
For f to be surjective, every element in the codomain {a, b, c, d} must be the image of at least one element in the domain.
Every element a, b, c, d appears exactly once as an output, so f IS surjective. ✓

Step 3: Conclusion
Since f is both injective and surjective, f is bijective.
Also, the domain and codomain have the same cardinality (both have 4 elements).

Question 6

Let R be a relation on the set A = {1, 2, 3, 4} defined by R = {(a, b) : a − b is even}. Which of the following is true about R?

Accepted Answer

Setup: Analyzing properties of relations — reflexivity, symmetry, and transitivity.

Solution:
R = {(a, b) : a − b is even}, where a, b ∈ A = {1, 2, 3, 4}

Step 1: List all pairs in R
For a − b to be even, both a and b must have the same parity (both odd or both even).
• Odd elements in A: {1, 3}
• Even elements in A: {2, 4}

Pairs where both are odd: (1,1), (1,3), (3,1), (3,3)
Pairs where both are even: (2,2), (2,4), (4,2), (4,4)

R = {(1,1), (1,3), (3,1), (3,3), (2,2), (2,4), (4,2), (4,4)}

Step 2: Check reflexivity
For reflexivity, (a, a) must be in R for all a ∈ A.
(1,1), (2,2), (3,3), (4,4) are all in R. ✓ R is reflexive.

Step 3: Check symmetry
For symmetry, if (a, b) ∈ R, then (b, a) ∈ R.
If a − b is even, then b − a = −(a − b) is also even.
✓ R is symmetric.

Step 4: Check transitivity
For transitivity, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
If a − b is even and b − c is even, then a − c = (a − b) + (b − c) is even.
✓ R is transitive.

Conclusion: R is reflexive, symmetric, and transitive, so R is an equivalence relation.

Question 7

Let f: ℝ → ℝ be defined by f(x) = 2x + 3. If g: ℝ → ℝ is defined by g(x) = x² − 1, then (f ∘ g)(2) is equal to:

Accepted Answer

Setup: Composition of functions — (f ∘ g)(x) means apply g first, then apply f to the result.

Solution:
By definition, (f ∘ g)(2) = f(g(2)).

Step 1: Find g(2)
g(2) = (2)² − 1 = 4 − 1 = 3

Step 2: Apply f to the result
f(g(2)) = f(3) = 2(3) + 3 = 6 + 3 = 9

Therefore, (f ∘ g)(2) = 9

Question 8

Let A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8}. If R is a relation from A to B defined by R = {(a, b) : a ∈ A, b ∈ B, and b = 2a}, then the number of elements in R is:

Accepted Answer

Setup: We need to find all ordered pairs (a, b) where a ∈ A, b ∈ B, and b = 2a.

Solution:
For each element a in A, check if 2a exists in B:
- a = 1: 2a = 2. Is 2 ∈ B? Yes. So (1, 2) ∈ R.
- a = 2: 2a = 4. Is 4 ∈ B? Yes. So (2, 4) ∈ R.
- a = 3: 2a = 6. Is 6 ∈ B? Yes. So (3, 6) ∈ R.
- a = 4: 2a = 8. Is 8 ∈ B? Yes. So (4, 8) ∈ R.
- a = 5: 2a = 10. Is 10 ∈ B? No. So (5, 10) ∉ R.

Therefore, R = {(1, 2), (2, 4), (3, 6), (4, 8)}.

The number of elements in R is 4.

Question 9

Let f: ℝ → ℝ be defined by f(x) = x² − 4x + 5. Which of the following statements is correct?

Accepted Answer

Setup: We need to analyze the function f(x) = x² − 4x + 5 to determine its properties (domain, range, injectivity, surjectivity).

Solution:
Complete the square:
f(x) = x² − 4x + 5
f(x) = (x² − 4x + 4) + 1
f(x) = (x − 2)² + 1

From this form:
• The vertex is at (2, 1)
• Since the coefficient of (x−2)² is positive, the parabola opens upward
• The minimum value of f(x) is 1, achieved at x = 2
• Therefore, the range of f is [1, ∞)

For injectivity: f(0) = 5 and f(4) = 16 − 16 + 5 = 5. Since f(0) = f(4) with 0 ≠ 4, the function is NOT injective.

For surjectivity: Since the range is [1, ∞) but the codomain is ℝ, the function is NOT surjective (no x maps to 0, for example).

Correct statement: The range of f is [1, ∞).

Question 10

Let A = {1, 2, 3} and B = {a, b}. The number of relations from A to B is:

Accepted Answer

Setup: A relation from set A to set B is any subset of the Cartesian product A × B. We need to count all possible subsets of A × B.

Solution:
First, find A × B:
A × B = {(1,a), (1,b), (2,a), (2,b), (3,a), (3,b)}

The number of elements in A × B is |A| × |B| = 3 × 2 = 6.

A relation from A to B is any subset of A × B. The number of subsets of a set with n elements is 2ⁿ.

Therefore, the number of relations from A to B = 2⁶ = 64.

Note: This includes the empty relation (no pairs) and the universal relation (all 6 pairs).

Question 11

Let f: {1, 2, 3, 4} → {a, b, c} be a function. If f(1) = a, f(2) = b, f(3) = c, and f(4) = a, then which of the following is true?

Accepted Answer

Setup: We are given an explicit function f from a 4-element set to a 3-element set. We need to determine properties of this function (injectivity, surjectivity, bijectivity).

Solution:
Given: f(1) = a, f(2) = b, f(3) = c, f(4) = a

Check injectivity (one-to-one):
A function is injective if f(x₁) = f(x₂) implies x₁ = x₂.
Here, f(1) = a and f(4) = a, but 1 ≠ 4.
Therefore, f is NOT injective.

Check surjectivity (onto):
A function is surjective if every element in the codomain is the image of at least one element in the domain.
Range of f = {f(1), f(2), f(3), f(4)} = {a, b, c}
Codomain = {a, b, c}
Since range = codomain, f IS surjective.

Conclusion: f is surjective but NOT injective. Therefore, f is NOT bijective.

Correct statement: f is surjective (onto).

Question 12

Let R be a relation on the set A = {1, 2, 3, 4} defined by R = {(a, b) : a − b is even}. Which of the following ordered pairs is NOT in R?

Accepted Answer

Setup: A relation R is defined by the condition that (a, b) ∈ R if and only if a − b is even. We need to determine which ordered pair does NOT satisfy this condition.

Solution:
Recall: a − b is even if and only if a and b have the same parity (both odd or both even).

In A = {1, 2, 3, 4}:
• Odd elements: {1, 3}
• Even elements: {2, 4}

For (a, b) ∈ R, we need a − b to be even:
• If a is odd and b is odd: a − b is even ✓
• If a is even and b is even: a − b is even ✓
• If a is odd and b is even: a − b is odd ✗
• If a is even and b is odd: a − b is odd ✗

Therefore, R consists of all pairs where both elements have the same parity:
R = {(1,1), (1,3), (3,1), (3,3), (2,2), (2,4), (4,2), (4,4)}

Now check the options:
• (1, 2): 1 − 2 = −1 (odd) → NOT in R ✓
• (2, 3): 2 − 3 = −1 (odd) → NOT in R
• (3, 4): 3 − 4 = −1 (odd) → NOT in R
• (1, 1): 1 − 1 = 0 (even) → IN R

The answer depends on which option is presented. If (1, 2) is an option, it is NOT in R.

Question 13

Let R be a relation on the set ℕ (natural numbers) defined by R = {(a, b) : a, b ∈ ℕ and a divides b}. Which of the following properties does R satisfy?

Accepted Answer

Setup: We check whether the divisibility relation satisfies reflexivity, symmetry, and transitivity.

Solution:

1. Reflexivity: Is (a, a) ∈ R for all a ∈ ℕ?
   Does a divide a? Yes, because a = a × 1.
   So R is reflexive. ✓

2. Symmetry: If (a, b) ∈ R, is (b, a) ∈ R?
   If a divides b, does b divide a?
   Counterexample: (2, 4) ∈ R because 2 divides 4.
   But (4, 2) ∉ R because 4 does not divide 2.
   So R is NOT symmetric. ✗

3. Transitivity: If (a, b) ∈ R and (b, c) ∈ R, is (a, c) ∈ R?
   If a divides b and b divides c, does a divide c?
   If a | b, then b = ka for some k ∈ ℕ.
   If b | c, then c = mb for some m ∈ ℕ.
   Then c = m(ka) = (mk)a, so a | c.
   So R is transitive. ✓

Conclusion: R is reflexive and transitive, but not symmetric. Therefore, R is a quasi-order (or preorder).

Question 14

Let f: A → B and g: B → C be two functions. If g ∘ f is injective, which of the following must be true?

Accepted Answer

Setup: We analyze the properties of composite functions and injectivity. If g ∘ f is injective, what can we conclude about f and g?

Solution:

Given: (g ∘ f) is injective, meaning (g ∘ f)(x₁) = (g ∘ f)(x₂) ⟹ x₁ = x₂ for all x₁, x₂ ∈ A.

This means: g(f(x₁)) = g(f(x₂)) ⟹ x₁ = x₂.

Claim: f must be injective.

Proof: Suppose f(x₁) = f(x₂) for some x₁, x₂ ∈ A.
Then g(f(x₁)) = g(f(x₂)) (applying g to both sides).
Since g ∘ f is injective, this implies x₁ = x₂.
Therefore, f is injective. ✓

Claim: g must be injective.

Counterexample: Let A = {1}, B = {2, 3}, C = {4}.
Define f(1) = 2 and g(2) = g(3) = 4.
Then (g ∘ f)(1) = g(f(1)) = g(2) = 4.
The composite g ∘ f is injective (only one element in domain).
But g is not injective (g(2) = g(3) = 4).
So g need not be injective. ✗

Conclusion: If g ∘ f is injective, then f must be injective, but g need not be.

Question 15

Let A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8}. If R is a relation from A to B defined by R = {(a, b) : a ∈ A, b ∈ B, and a divides b}, then the number of elements in R is:

Accepted Answer

Setup: A relation from set A to set B is a subset of A × B. We must find all ordered pairs (a, b) where a divides b evenly.

Solution:
For each element a ∈ A, check which elements b ∈ B satisfy 'a divides b':

• a = 1: 1 divides 2, 4, 6, 8 → pairs: (1,2), (1,4), (1,6), (1,8) → 4 pairs
• a = 2: 2 divides 2, 4, 6, 8 → pairs: (2,2), (2,4), (2,6), (2,8) → 4 pairs
• a = 3: 3 divides 6 only → pairs: (3,6) → 1 pair
• a = 4: 4 divides 4, 8 → pairs: (4,4), (4,8) → 2 pairs
• a = 5: 5 divides none in B → 0 pairs

Total elements in R = 4 + 4 + 1 + 2 + 0 = 11

Question 16

Let R be a relation on the set A = {1, 2, 3, 4} defined by R = {(a, b) : a² - b² = a - b}. Which of the following is true about R?

Accepted Answer

Setup: We need to find all pairs (a,b) satisfying a² - b² = a - b, then determine properties of R (reflexive, symmetric, transitive, etc.).

Solution:
Start with the equation: a² - b² = a - b
Factor the left side: (a-b)(a+b) = a - b
Rearrange: (a-b)(a+b) - (a-b) = 0
Factor: (a-b)(a+b-1) = 0

This gives us two cases:
Case 1: a - b = 0 → a = b
Case 2: a + b - 1 = 0 → a + b = 1

For A = {1, 2, 3, 4}:
Case 1 (a = b): (1,1), (2,2), (3,3), (4,4)
Case 2 (a + b = 1): No solutions since a,b ≥ 1 implies a+b ≥ 2

Therefore: R = {(1,1), (2,2), (3,3), (4,4)}

Properties:
- Reflexive: ✓ (all (a,a) are in R)
- Symmetric: ✓ (if (a,b) ∈ R, then (b,a) ∈ R; here all pairs are (a,a))
- Transitive: ✓ (if (a,b) and (b,c) in R, then (a,c) in R; verified: (a,a) and (a,a) gives (a,a))

R is an equivalence relation (specifically, the identity relation).

Question 17

Let f: ℝ → ℝ be defined by f(x) = (x² - 4)/(x - 2) for x ≠ 2. If we extend f to a function g: ℝ → ℝ by defining g(2) = L, then g is continuous at x = 2. The value of L is:

Accepted Answer

Setup: We are testing removable discontinuity and the concept of continuous extension via limits. The function f has a removable discontinuity at x = 2.

Solution:
For x ≠ 2, we simplify f(x):
f(x) = (x² - 4)/(x - 2) = [(x - 2)(x + 2)]/(x - 2) = x + 2 (for x ≠ 2)

To make g continuous at x = 2, we need:
g(2) = lim(x→2) f(x) = lim(x→2) (x + 2) = 2 + 2 = 4

Verification: With g(2) = 4, the function g(x) = x + 2 for all x ∈ ℝ, which is continuous everywhere.

Trap explanation: Students may incorrectly substitute x = 2 directly into the original form (x² - 4)/(x - 2), getting 0/0 and concluding no limit exists. Others may forget to simplify before taking the limit, or may confuse the limit value with the original function's behavior.

Question 18

Let R be a relation on the set A = {1, 2, 3, 4} defined by R = {(a, b) : |a - b| is even}. Which of the following statements is correct?

Accepted Answer

Setup: We test properties of relations (reflexive, symmetric, transitive) by analyzing when |a - b| is even.

Solution:
First, determine when |a - b| is even:
|a - b| is even ⟺ a - b is even ⟺ a and b have the same parity (both odd or both even)

For A = {1, 2, 3, 4}:
- Odd elements: {1, 3}
- Even elements: {2, 4}

R = {(1,1), (1,3), (3,1), (3,3), (2,2), (2,4), (4,2), (4,4)}

Check properties:
1. Reflexive: For all a ∈ A, (a,a) ∈ R? |a-a| = 0 (even) ✓ YES
2. Symmetric: If (a,b) ∈ R, is (b,a) ∈ R? |a-b| = |b-a| ✓ YES
3. Transitive: If (a,b) ∈ R and (b,c) ∈ R, is (a,c) ∈ R? 
   If a,b same parity AND b,c same parity, then a,c same parity ✓ YES

R is reflexive, symmetric, and transitive (equivalence relation).

Trap explanation: Students may confuse the condition |a-b| is even with |a-b| is odd, or they may check only one property and miss that R is an equivalence relation.

Question 19

Let A = {x ∈ ℝ : x² - 5x + 6 ≤ 0} and B = {x ∈ ℝ : |x - 2| < 2}. Then A ∩ B is equal to:

Accepted Answer

Setup: We test set operations and interval notation by finding the intersection of two sets defined by inequalities. Solution: Find set A: x² - 5x + 6 ≤ 0 Factor: (x - 2)(x - 3) ≤ 0 The roots are x = 2 and x = 3. For a product to be ≤ 0, factors must have opposite signs or one is zero. Test intervals: - x < 2: (−)(−) = (+) > 0 ✗ - 2 ≤ x ≤ 3: (0 or +)(0 or −) ≤ 0 ✓ - x > 3: (+)(+) = (+) > 0 ✗ A = [2, 3] Find set B: |x - 2| < 2 -2 < x - 2 < 2 0 < x < 4 B = (0, 4) Find A ∩ B: A ∩ B = [2, 3] ∩ (0, 4) = [2, 3] (The interval [2, 3] is entirely contained in (0, 4), so the intersection is [2, 3].) Trap explanation: Students may incorrectly factor the quadratic, misinterpret the inequality direction, or confuse closed vs. open intervals. Some may forget that the intersection of [2,3] with (0,4) is [2,3], not (2,3).

Question 20

Let f: ℝ → ℝ be defined by f(x) = (2x + 3)/(x - 1). The range of f is:

Accepted Answer

Setup: To find the range of a rational function, we solve y = f(x) for x in terms of y and determine which y-values are achievable.

Solution:
Let y = (2x + 3)/(x - 1)
Multiply both sides by (x - 1):
y(x - 1) = 2x + 3
yx - y = 2x + 3
yx - 2x = y + 3
x(y - 2) = y + 3
x = (y + 3)/(y - 2)

For x to be defined in ℝ, the denominator (y - 2) ≠ 0, so y ≠ 2.

Also, the original function f(x) is undefined at x = 1 (vertical asymptote), and as x → ∞, f(x) → 2 (horizontal asymptote).

Therefore, the range is ℝ \ {2} or (-∞, 2) ∪ (2, ∞).

Trap: Students often confuse the domain restriction (x ≠ 1) with the range restriction. The domain excludes x = 1, but the range excludes y = 2 (the horizontal asymptote).

Question 21

Let A = {x ∈ ℝ : x² - 5x + 6 ≤ 0} and B = {x ∈ ℝ : |x - 2| < 2}. Then A ∩ B is:

Accepted Answer

Setup: Find sets A and B by solving inequalities, then compute their intersection. Solution: For set A: x² - 5x + 6 ≤ 0 Factor: (x - 2)(x - 3) ≤ 0 The roots are x = 2 and x = 3. For a product to be ≤ 0, the factors must have opposite signs or one is zero. Test intervals: x < 2: (−)(−) = (+) > 0 ✗ 2 ≤ x ≤ 3: (0 or +)(0 or −) ≤ 0 ✓ x > 3: (+)(+) = (+) > 0 ✗ A = [2, 3] For set B: |x - 2| < 2 This means -2 < x - 2 < 2 Add 2 to all parts: 0 < x < 4 B = (0, 4) Intersection A ∩ B: A = [2, 3] (closed interval) B = (0, 4) (open interval) A ∩ B = [2, 3] (the overlap, inheriting the closed endpoints from A) Trap: Students often forget whether endpoints are included. Set A includes 2 and 3 (≤ inequality), while B excludes 0 and 4 (< inequality). The intersection includes 2 and 3.

CDS Sets, Relations & Functions

CDS Sets, Relations & Functions — Past Exam Questions

Sets, Relations & Functions— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Sets, Relations & Functions — Practice Questions

60-Second Revision — Sets, Relations & Functions