RS Aggarwal Solutions for Class 11 Chapter 01 Sets

 

Question 1A. Which of the following are sets? Justify your answer. The collection of all whole numbers less than 10.
Answer: Whole numbers include 0, 1, 2, 3, and so on. The whole numbers that are less than 10 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since this collection is clearly defined and can be counted - meaning it is well - defined - it qualifies as a set.

Exam Tip: A set must have a clear, unambiguous definition where each element can be identified with certainty. Check whether the criteria allow you to determine membership without doubt.

 

Question 1B. Which of the following are sets? Justify your answer. The collection of good hockey players in India.
Answer: A collection of good hockey players in India will differ depending on the person making the judgment. Since the criteria for what makes a player "good" varies from individual to individual, this collection is not well - defined and therefore is not a set.

Exam Tip: Subjective terms like "good," "beautiful," or "talented" cannot form sets because they lack a fixed, objective definition that applies universally.

 

Question 1C. Which of the following are sets? Justify your answer. The collection of all the questions in this chapter.
Answer: The collection of all questions in this chapter is clearly known and can be counted, making it well - defined. Therefore, this is a set.

Exam Tip: Collections that are finite and countable with clear boundaries form valid sets, provided every member can be identified precisely.

 

Question 1D. Which of the following are sets? Justify your answer. The collection of all the difficult chapters in this book.
Answer: The collection of all difficult chapters in this book differs from person to person based on individual understanding and preferences. Since there is no objective measure of "difficulty," this is not a set.

Exam Tip: Always identify whether the defining property is objective and measurable, or whether it relies on personal judgment or interpretation.

 

Question 1E. Which of the following are sets? Justify your answer. A collection of Hindi novels written by Munshi Prem Chand.
Answer: The collection of Hindi novels written by Munshi Prem Chand is clearly known and can be counted, meaning it is well - defined. This qualifies as a set.

Exam Tip: Collections of works by a specific author are well - defined sets because the membership criteria (authorship) is objective and verifiable.

 

Question 1F. Which of the following are sets? Justify your answer. A team of 11 best cricket players of India.
Answer: A collection of 11 best cricket players of India will vary from person to person depending on their judgment. Since "best" is subjective and the composition is not fixed, this is not well - defined and therefore is not a set.

Exam Tip: Rankings, ratings, or selections that depend on opinion or judgment cannot form sets because they lack an objective, universally accepted criterion.

 

Question 1G. Which of the following are sets? Justify your answer. The collection of all the months of the year whose names begin with the letter M.
Answer: The months of the year are Jan, Feb, March, April, May, June, July, Aug, Sep, Oct, Nov, Dec. The months whose names start with the letter M are March and May. Since this collection is clearly known and can be counted - that is, it is well - defined - this is a set.

Exam Tip: When the membership rule is based on objective, factual criteria (like letters in a name), the collection forms a valid set.

 

Question 1H. Which of the following are sets? Justify your answer. The collection of all interesting books.
Answer: The collection of all interesting books will vary from person to person based on individual taste and preference. Since what one person finds interesting differs from another's opinion, this collection is not well - defined and is not a set.

Exam Tip: Terms involving aesthetic judgment or personal preference (interesting, beautiful, good, etc.) automatically disqualify a collection from being a set.

 

Question 1I. Which of the following are sets? Justify your answer. The collection of all short boys of your class.
Answer: The collection of all short boys of your class will differ depending on the individual evaluating it. One person might consider a boy short if his height is less than 120 cm, while another might set the cutoff at 90 cm. Since the definition of "short" is not fixed and objective, this collection is not well - defined and therefore is not a set.

Exam Tip: Even when a numerical criterion exists, if the boundary is vague or left to personal judgment, the collection fails to be a set. Always require a specific, fixed definition.

 

Question 1J. Which of the following are sets? Justify your answer. The collection of all those students of your class whose ages exceed 15 years.
Answer: The collection of all those students of your class whose ages exceed 15 years is clearly known and can be counted, making it well - defined. This is a set.

Exam Tip: Numeric criteria with precise boundaries (like "exceeds 15 years") provide objective definitions that allow a collection to form a valid set.

 

Question 1K. Which of the following are sets? Justify your answer. The collection of all rich persons of Kolkata.
Answer: The collection of all rich persons of Kolkata varies from person to person. Some may regard a person with an annual income of Rs 1 lakh as rich, while others may require an annual income of Rs 1 crore. Since the meaning of "rich" is not fixed and depends on individual judgment, this collection is not well - defined and is not a set.

Exam Tip: Even with economic data, if the threshold is not explicitly stated, the collection remains ambiguous and cannot be a set.

 

Question 1L. Which of the following are sets? Justify your answer. The collection of all persons of Kolkata whose assessed annual incomes exceed (say) Rs 20 lakh in the 4 financial years 2016-17.
Answer: The collection of all persons of Kolkata whose assessed annual incomes exceed (say) Rs 20 lakh in the 4 financial years 2016-17 is clearly known and well - defined because a specific numeric threshold and time period are given. This is a set.

Exam Tip: When a precise numerical boundary and specific time frame are provided, the collection becomes objective and well - defined, satisfying the criteria for a set.

 

Question 1M. Which of the following are sets? Justify your answer. The collection of all interesting dramas written by Shakespeare.
Answer: The collection of all interesting dramas written by Shakespeare is not well - defined because it hinges on individual interest and personal taste. Since what one reader finds interesting another may not, this collection cannot be a set.

Exam Tip: The subjective nature of "interesting" makes this collection dependent on personal judgment, preventing it from being a valid set.

 

Question 2. Let A be the set of all even whole numbers less than 10. (i) Write A in the roster form. (ii) Fill in the blanks with the appropriate symbol ∉ or ∈: (i) 0 .... A (ii) 10 .... A (c) 3 .... A (d) 6 .... A
Answer:
(i) Whole numbers are 0, 1, 2, 3, and so on. Even whole numbers that are less than 10 are 0, 2, 4, 6, 8. Therefore, A = {0, 2, 4, 6, 8}

(ii) (a) Here, A = {0, 2, 4, 6, 8}. Since 0 is in set A, we write 0 ∈ A.

(b) Here, A = {0, 2, 4, 6, 8}. Since 10 is not in set A, we write 10 ∉ A.

(c) Here, A = {0, 2, 4, 6, 8}. Since 3 is not in set A, we write 3 ∉ A.

(d) Here, A = {0, 2, 4, 6, 8}. Since 6 is in set A, we write 6 ∈ A.

Exam Tip: Always verify whether each element belongs to the set by checking the roster form. The symbol ∈ means "belongs to" and ∉ means "does not belong to."

 

Question 3A. Write the following sets in roster form: A = {x : x is a natural number, 30 ≤ x < 36}.
Answer: Natural numbers are 1, 2, 3, and so on, extending to 30, 31, 32, 33, 34, 35, 36, and beyond. The elements of this set satisfying the condition are 30, 31, 32, 33, 34, and 35. Therefore, A = {30, 31, 32, 33, 34, 35}.

Exam Tip: When converting from set-builder form to roster form, ensure you include all values that satisfy the given condition, respecting the boundary constraints (inclusive or exclusive).

 

Question 3B. Write the following sets in roster form: B = {x : x is an integer and - 4 < x < 6}.
Answer: Integers include all whole numbers and their negatives: ..., -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, and so on. The elements of this set that satisfy the condition are -3, -2, -1, 0, 1, 2, 3, 4, and 5. Therefore, B = {-3, -2, -1, 0, 1, 2, 3, 4, 5}.

Exam Tip: For integer ranges, remember that strict inequality (<) excludes the boundary value, while non-strict inequality (≤) includes it.

 

Question 3C. Write the following sets in roster form: C = {x : x is a two-digit number such that the sum of its digits is 9}.
Answer: Two-digit numbers range from 10 to 99. We need those where the sum of digits equals 9:

9 = 0 + 9 gives the number 09 (not a two-digit number)
9 = 1 + 8 gives the numbers 18, 81
9 = 2 + 7 gives the numbers 27, 72
9 = 3 + 6 gives the numbers 36, 63
9 = 4 + 5 gives the numbers 45, 54

The elements of this set are 18, 27, 36, 45, 54, 63, 72, 81, and 90. Therefore, C = {18, 27, 36, 45, 54, 63, 72, 81, 90}.

Exam Tip: When finding digit combinations, systematically check all pairs that sum to the target value and ensure the resulting numbers meet the given criteria (two-digit requirement here).

 

Question 3D. Write the following sets in roster form: D = {x : x is an integer, x² ≤ 9}.
Answer: Integers include ..., -4, -3, -2, -1, 0, 1, 2, 3, 4, and so on. Testing which values satisfy x² ≤ 9:

x = -4: x² = (-4)² = 16 > 9 (does not satisfy)
x = -3: x² = (-3)² = 9 ≤ 9 (satisfies)
x = -2: x² = (-2)² = 4 ≤ 9 (satisfies)
x = -1: x² = (-1)² = 1 ≤ 9 (satisfies)
x = 0: x² = 0² = 0 ≤ 9 (satisfies)
x = 1: x² = 1² = 1 ≤ 9 (satisfies)
x = 2: x² = 2² = 4 ≤ 9 (satisfies)
x = 3: x² = 3² = 9 ≤ 9 (satisfies)
x = 4: x² = 4² = 16 > 9 (does not satisfy)

The elements of this set are -3, -2, -1, 0, 1, 2, 3. Therefore, D = {-3, -2, -1, 0, 1, 2, 3}.

Exam Tip: For inequalities involving squares, remember that both positive and negative values can satisfy the condition. Test boundary values carefully to determine inclusion.

 

Question 3E. Write the following sets in roster form: E = {x : x is a prime number, which is a divisor of 42}.
Answer: A prime number is divisible only by 1 and itself. Prime numbers include 2, 3, 5, 7, 11, 13, and so on.

The divisors of 42 can be found by factoring:
42 = 1 × 42
42 = 2 × 21
42 = 3 × 14
42 = 6 × 7

So the divisors of 42 are 1, 2, 3, 6, 7, 14, 21, 42. Among these, the prime divisors are 2, 3, and 7. Therefore, E = {2, 3, 7}.

Exam Tip: To find divisors, use prime factorization or systematic pairing. Identify which divisors are themselves prime by checking if they have exactly two factors.

 

Question 3F. Write the following sets in roster form: F = {x : x is a letter in the word 'MATHEMATICS'}.
Answer: The word MATHEMATICS contains 11 letters total. The letters M, A, and T appear more than once. Listing each distinct letter: M, A, T, H, E, M, A, T, I, C, S. The unique letters are M, A, T, H, E, I, C, S. Therefore, F = {M, A, T, H, E, I, C, S}.

Exam Tip: When a set is defined by letters in a word, list each letter only once, even if it appears multiple times in the word. Order does not matter in set notation.

 

Question 3G. Write the following sets in roster form: G = {x : x is a prime number and 80 < x < 100}.
Answer: A prime number is divisible only by 1 and itself. Prime numbers include 2, 3, 5, 7, 11, 13, and so on. The integers between 80 and 100 are 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99. Among these, the prime numbers are 83, 89, and 97. Therefore, G = {83, 89, 97}.

Exam Tip: To identify primes in a range, test divisibility by small primes (2, 3, 5, 7, etc.). A number is prime if none of these divide it evenly.

 

Question 3H. Write the following sets in roster form: H = {x : x is a perfect square and x < 50}.
Answer: Perfect squares are numbers that result from squaring a natural number:

0² = 0
1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64 > 50

The perfect squares less than 50 are 0, 1, 4, 9, 16, 25, 36, and 49. Therefore, H = {0, 1, 4, 9, 16, 25, 36, 49}.

Exam Tip: Systematically compute squares until you exceed the upper limit. Always include 0 as a perfect square when it satisfies the given condition.

 

Question 3I. Write the following sets in roster form: J = {x : x ∈ ℝ and x² + x - 12 = 0}.
Answer: The given equation is x² + x - 12 = 0.

Factoring:
x² + 4x - 3x - 12 = 0
x(x + 4) - 3(x + 4) = 0
(x - 3)(x + 4) = 0

Setting each factor to zero:
x - 3 = 0 ⟹ x = 3
x + 4 = 0 ⟹ x = -4

The solution set of the given equation in roster form is {3, -4}. Therefore, J = {3, -4}.

Exam Tip: For quadratic equations, factor carefully or use the quadratic formula. Verify solutions by substituting back into the original equation.

 

Question 3J. Write the following sets in roster form: K = {x : x ∈ ℕ, x is a multiple of 5 and x² < 400}.
Answer: Multiples of 5 are 5, 10, 15, 20, 25, 30, and so on. We need those satisfying x² < 400:

5² = 25 < 400 (satisfies)
10² = 100 < 400 (satisfies)
15² = 225 < 400 (satisfies)
20² = 400 (does not satisfy; we need x² < 400, not ≤)
25² = 625 > 400 (does not satisfy)

The elements meeting both conditions are 5, 10, and 15. Therefore, K = {5, 10, 15}.

Exam Tip: Be careful with strict inequality (<) versus non-strict inequality (≤). A value equal to the boundary does not satisfy a strict inequality.

 

Question 4A. List all the elements of each of the sets given below. A = {x : x = 2n, n ∈ ℕ and n ≤ 5}.
Answer: Given: x = 2n and n ≤ 5

Since n ∈ ℕ (natural numbers), we have n = 1, 2, 3, 4, 5.

Substituting into x = 2n:
n = 1: x = 2 × 1 = 2
n = 2: x = 2 × 2 = 4
n = 3: x = 2 × 3 = 6
n = 4: x = 2 × 4 = 8
n = 5: x = 2 × 5 = 10

The elements of A are 2, 4, 6, 8, and 10. Therefore, A = {2, 4, 6, 8, 10}.

Exam Tip: When converting from set-builder form involving a formula, substitute each allowed value of the parameter systematically to generate all set members.

 

Question 4B. List all the elements of each of the sets given below. B = {x : x = 2n + 1, n ∈ W and n ≤ 5}.
Answer: Given: x = 2n + 1 and n ≤ 5

Since n ∈ W (whole numbers), we have n = 0, 1, 2, 3, 4, 5.

Substituting into x = 2n + 1:
n = 0: x = 2 × 0 + 1 = 1
n = 1: x = 2 × 1 + 1 = 3
n = 2: x = 2 × 2 + 1 = 5
n = 3: x = 2 × 3 + 1 = 7
n = 4: x = 2 × 4 + 1 = 9
n = 5: x = 2 × 5 + 1 = 11

The elements of B are 1, 3, 5, 7, 9, and 11. Therefore, B = {1, 3, 5, 7, 9, 11}.

Exam Tip: Remember that whole numbers start from 0, whereas natural numbers start from 1. This distinction affects which values you substitute into the formula.

 

Question 4C. List all the elements of each of the sets given below. C = {x : x = \( \frac{1}{n} \), n ∈ ℕ and n ≤ 6}.
Answer: Given: x = \( \frac{1}{n} \) and n ≤ 6

Since n ∈ ℕ (natural numbers), we have n = 1, 2, 3, 4, 5, 6.

Substituting into x = \( \frac{1}{n} \):
n = 1: x = \( \frac{1}{1} \) = 1
n = 2: x = \( \frac{1}{2} \)
n = 3: x = \( \frac{1}{3} \)
n = 4: x = \( \frac{1}{4} \)
n = 5: x = \( \frac{1}{5} \)
n = 6: x = \( \frac{1}{6} \)

Therefore, C = \( \left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}\right\) \}.

Exam Tip: When the formula generates fractions, express each as a fully reduced fraction and include 1 if it results from the formula.

 

Question 4D. List all the elements of each of the sets given below. D = {x : x = n², n ∈ ℕ and 2 ≤ n ≤ 5}.
Answer: Given: x = n² and 2 ≤ n ≤ 5

Since n ∈ ℕ (natural numbers) with the constraint 2 ≤ n ≤ 5, we have n = 2, 3, 4, 5.

Substituting into x = n²:
n = 2: x = 2² = 4
n = 3: x = 3² = 9
n = 4: x = 4² = 16
n = 5: x = 5² = 25

The elements of D are 4, 9, 16, and 25. Therefore, D = {4, 9, 16, 25}.

Exam Tip: Observe the range constraints carefully - here the range starts at 2 rather than 1, so 1 is not included in the parameter values.

 

Question 4E. List all the elements of each of the sets given below. E = {x : x ∈ ℤ and x² = x}.
Answer: Given: x ∈ ℤ (integers) and x² = x

Testing integer values:
If x = -2: (-2)² = 4 ≠ -2 (does not satisfy)
If x = -1: (-1)² = 1 ≠ -1 (does not satisfy)
If x = 0: 0² = 0 (satisfies)
If x = 1: 1² = 1 (satisfies)
If x = 2: 2² = 4 ≠ 2 (does not satisfy)

Only x = 0 and x = 1 satisfy the condition x² = x. Therefore, E = {0, 1}.

Exam Tip: For conditions like x² = x, rearrange to x² - x = 0 or x(x - 1) = 0, giving x = 0 or x = 1. This algebraic approach is faster than testing values.

 

Question 4F. List all the elements of each of the sets given below. F = {x : x ∈ ℤ and - \( \frac{1}{2} \) < x < \( \frac{13}{2} \)}.
Answer: Given: x ∈ ℤ (integers) and - \( \frac{1}{2} \) < x < \( \frac{13}{2} \)

First, compute the bounds:
- \( \frac{1}{2} \) = -0.5 and \( \frac{13}{2} \) = 6.5

The integers x satisfying -0.5 < x < 6.5 are 0, 1, 2, 3, 4, 5, 6.

Therefore, F = {0, 1, 2, 3, 4, 5, 6}.

Exam Tip: Convert fractional bounds to decimals to clearly identify which integers fall within the range. Remember that strict inequality excludes the endpoints.

 

Question 4G. List all the elements of each of the sets given below. G = {x : x = \( \frac{1}{2n-1} \), n ∈ ℕ and 1 ≤ n ≤ 5}.
Answer: Given: x = \( \frac{1}{2n-1} \) and 1 ≤ n ≤ 5

Since n ∈ ℕ (natural numbers) with 1 ≤ n ≤ 5, we have n = 1, 2, 3, 4, 5.

Substituting into x = \( \frac{1}{2n-1} \):
n = 1: x = \( \frac{1}{2(1)-1} \) = \( \frac{1}{1} \) = 1
n = 2: x = \( \frac{1}{2(2)-1} \) = \( \frac{1}{3} \)
n = 3: x = \( \frac{1}{2(3)-1} \) = \( \frac{1}{5} \)
n = 4: x = \( \frac{1}{2(4)-1} \) = \( \frac{1}{7} \)
n = 5: x = \( \frac{1}{2(5)-1} \) = \( \frac{1}{9} \)

Therefore, G = \( \left\{1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}\right\) \}.

Exam Tip: Always evaluate the denominator formula before computing the fraction. Ensure your substitutions are correct by double-checking the arithmetic.

 

Question 4H. List all the elements of each of the sets given below. H = {x : x ∈ ℤ, |x| ≤ 2}.
Answer: Given: x ∈ ℤ (integers) and |x| ≤ 2

The absolute value condition |x| ≤ 2 means -2 ≤ x ≤ 2.

Testing integer values:
If x = -3: |-3| = 3 > 2 (does not satisfy)
If x = -2: |-2| = 2 ≤ 2 (satisfies)
If x = -1: |-1| = 1 ≤ 2 (satisfies)
If x = 0: |0| = 0 ≤ 2 (satisfies)
If x = 1: |1| = 1 ≤ 2 (satisfies)
If x = 2: |2| = 2 ≤ 2 (satisfies)
If x = 3: |3| = 3 > 2 (does not satisfy)

The integers satisfying the condition are -2, -1, 0, 1, 2. Therefore, H = {-2, -1, 0, 1, 2}.

Exam Tip: The condition |x| ≤ a is equivalent to -a ≤ x ≤ a. This provides a quick way to identify the range of values without testing each one individually.

 

Question 5. Write each of the sets given below in set-builder form: (i) A = {1, \( \frac{1}{4} \), \( \frac{1}{9} \), \( \frac{1}{16} \), \( \frac{1}{25} \), \( \frac{1}{36} \), \( \frac{1}{49} \)} (ii) B = {\( \frac{1}{2} \), \( \frac{2}{5} \), \( \frac{3}{10} \), \( \frac{4}{17} \), \( \frac{5}{26} \), \( \frac{6}{37} \), \( \frac{7}{50} \)} (iii) C = {53, 59, 61, 67, 71, 73, 79} (iv) D = {-1, 1} (v) E = {14, 21, 28, 35, 42, ..., 98}
Answer:
(i) Examining set A: A = {1, \( \frac{1}{4} \), \( \frac{1}{9} \), \( \frac{1}{16} \), \( \frac{1}{25} \), \( \frac{1}{36} \), \( \frac{1}{49} \)}

We can rewrite this as A = {\( \left(\frac{1}{1}\right)^2 \), \( \left(\frac{1}{2}\right)^2 \), \( \left(\frac{1}{3}\right)^2 \), \( \left(\frac{1}{4}\right)^2 \), \( \left(\frac{1}{5}\right)^2 \), \( \left(\frac{1}{6}\right)^2 \), \( \left(\frac{1}{7}\right)^2 \)}

The pattern shows elements of the form \( \frac{1}{n^2} \) where n ranges from 1 to 7.

In set-builder form: A = {x : x = \( \frac{1}{n^2} \), n ∈ ℕ and 1 ≤ n ≤ 7}

(ii) Examining set B: B = {\( \frac{1}{2} \), \( \frac{2}{5} \), \( \frac{3}{10} \), \( \frac{4}{17} \), \( \frac{5}{26} \), \( \frac{6}{37} \), \( \frac{7}{50} \)}

Looking at numerators: 1, 2, 3, 4, 5, 6, 7
Looking at denominators: 2, 5, 10, 17, 26, 37, 50

Denominators can be written as:
1² + 1 = 2
2² + 1 = 5
3² + 1 = 10
4² + 1 = 17
5² + 1 = 26
6² + 1 = 37
7² + 1 = 50

The pattern is \( \frac{n}{n^2+1} \) where n ranges from 1 to 7.

In set-builder form: B = {x : x = \( \frac{n}{n^2+1} \), n ∈ ℕ and 1 ≤ n ≤ 7}

(iii) Examining set C: C = {53, 59, 61, 67, 71, 73, 79}

All elements are prime numbers between 50 and 80.

In set-builder form: C = {x : x is a prime number and 50 < x < 80}

(iv) Examining set D: D = {-1, 1}

These are the solutions to the equation x² = 1.

In set-builder form: D = {x : x ∈ ℝ and x² = 1}

(v) Examining set E: E = {14, 21, 28, 35, 42, ..., 98}

The elements are all multiples of 7, from 14 to 98. These can be written as 7 × 2, 7 × 3, 7 × 4, ..., 7 × 14, or equivalently 7n where n ranges from 2 to 14. Alternatively, we observe these are multiples of 7 between 10 and 100.

In set-builder form: E = {x : x = 7n, n ∈ ℕ and 2 ≤ n ≤ 14}

Exam Tip: When converting roster form to set-builder form, identify the pattern by examining numerators/denominators separately, looking for arithmetic or algebraic relationships, and verifying your formula works for all listed elements.

 

Question 6. Match each of the sets on the left described in the roster form with the same set on the right described in the set-builder form:

Answer:
(i) Squaring -5 and 5 both gives 25. Since -5 and 5 are both integers, the matching set-builder form is {x : x ∈ Z and x² = 25}.
∴ (i) matches (c)

(ii) The divisors of 18 can be found: 18 = 18 × 1, 18 = 9 × 2, 18 = 6 × 3. So 1, 2, 3, 6, 9, 18 are all divisors of 18. The matching set-builder form is {x : x ∈ N and x is a factor of 18}.
∴ (ii) matches (d)

(iii) Testing each integer: (-3)² = 9 < 16, (-2)² = 4 < 16, (-1)² = 1 < 16, (0)² = 0 < 16, (1)² = 1 < 16, (2)² = 4 < 16, (3)² = 9 < 16. All elements are integers satisfying x² < 16.
∴ (iii) matches (a)

(iv) The word 'PRINCIPAL' contains 9 letters, with P and I repeated. Listing unique letters gives {P, R, I, N, C, A, L}. The matching set-builder form is {x : x is a letter in the word 'PRINCIPAL'}.
∴ (iv) matches (e)

(v) Since 1 ∈ N and (1)² = 1, this set consists of natural numbers where x² = x. The matching set-builder form is {x : x ∈ N and x² = x}.
∴ (v) matches (b)
In simple words: Match each roster form set by testing what condition its elements satisfy. For (i), both -5 and 5 square to 25. For (ii), all elements divide 18 evenly. For (iii), each number squared is less than 16. For (iv), list unique letters from PRINCIPAL. For (v), only 1 satisfies the condition that x² = x in natural numbers.

Exam Tip: Always verify your matches by testing a few elements from each roster form against the conditions in set-builder form - this confirms you've paired them correctly.

 

Exercise 1B

Exam Tip: A null set contains no elements at all. Check if any value satisfies the given condition - if none do, it's a null set.

 

Question 1. Which of the following are examples of the null set? Set of odd natural numbers divisible by 2.
Answer: Natural numbers are 1, 2, 3, 4, 5,…. Odd natural numbers are 1, 3, 5, 7, 9, 11,…. No odd natural number is divisible by 2 because all odd numbers leave a remainder when divided by 2. Since there are no elements in this set, it is a null set.
In simple words: Odd numbers can never be divided evenly by 2, so no odd natural number can satisfy the condition. The set is empty.

Exam Tip: Remember that a null set has zero members - no exceptions, no matter how the condition is phrased.

 

Question 2. Which of the following are examples of the null set? Set of even prime numbers.
Answer: Prime numbers are defined as numbers divisible only by 1 and the number itself. The list of primes begins: 2, 3, 5, 7, 11, 13,…. The number 2 is both even and prime - it is the only even prime. Since the set contains at least one element (the number 2), this is not a null set.
In simple words: The number 2 is prime and even, so the set is not empty.

Exam Tip: 2 is special - it is the only prime that is also even. Don't overlook this edge case.

 

Question 3. Which of the following are examples of the null set? A = {x : x ∈ N, 1 < x ≤ 2}.
Answer: Natural numbers are 1, 2, 3, 4, 5, 6, 7,…. For A, we need numbers greater than 1 (so 2, 3, 4, …) AND less than or equal to 2 (so only 1 and 2). No number can be simultaneously greater than 1 and at most 2 within the set of natural numbers. Therefore, the set has no elements and is a null set.
In simple words: A number cannot be both bigger than 1 and smaller than or equal to 2 if it must be a whole number. So the set is empty.

Exam Tip: Carefully read compound conditions - check if it's logically possible for any value to satisfy all of them at once.

 

Question 4. Which of the following are examples of the null set? B = {x : x ∈ N, 2x + 3 = 4}.
Answer: Natural numbers are 1, 2, 3, 4, 5, 6,…. We need to solve 2x + 3 = 4. Subtracting 3 from both sides: 2x = 1. Dividing by 2: x = 0.5. Since 0.5 is not a natural number, no natural number satisfies this equation. The set contains no elements and is therefore a null set.
In simple words: The equation gives x = 0.5, which is not a natural number. So no element belongs to the set.

Exam Tip: Solve the equation first, then check if the solution is actually in the required set (in this case, natural numbers).

 

Question 5. Which of the following are examples of the null set? C = {x : x is prime, 90 < x < 96}.
Answer: Prime numbers are defined as numbers divisible only by 1 and themselves. Some primes are: 2, 3, 5, 7, 11, 13,…, 83, 89, 97,…. The first prime greater than 90 is 97, and the largest prime less than 96 is 89. No prime number exists between 90 and 96 (inclusive of neither endpoint). The set is empty and is therefore a null set.
In simple words: The numbers between 90 and 96 (like 91, 92, 93, 94, 95) are not prime because they can be factored. So the set has no members.

Exam Tip: For prime-related questions, list candidate primes near the boundary to verify none fall in the given range.

 

Question 6. Which of the following are examples of the null set? D = {x : x ∈ N, x² + 1 = 0}.
Answer: Natural numbers are 1, 2, 3, 4, 5, 6,…. We need to solve x² + 1 = 0, which gives x² = -1. Since the square of any real number is always non-negative, there is no real number whose square equals -1. Therefore, no natural number satisfies this equation. The set has no elements and is a null set.
In simple words: Squaring any number always gives a positive result (or zero). It can never equal -1, so the set is empty.

Exam Tip: Recognize equations with no real solutions - when x² equals a negative number, the set must be null.

 

Question 7. Which of the following are examples of the null set? E = {x : x ∈ W, x + 3 ≤ 3}.
Answer: Whole numbers are 0, 1, 2, 3,…. The condition x + 3 ≤ 3 simplifies to x ≤ 0. Within whole numbers, only x = 0 satisfies this. Since 0 is a whole number and 0 + 3 = 3 ≤ 3, the element 0 belongs to the set. Therefore, this is not a null set.
In simple words: When x = 0, we get 0 + 3 = 3, which is ≤ 3. So the set contains at least one element.

Exam Tip: Don't forget that 0 is a whole number - it often provides the only solution in range.

 

Question 8. Which of the following are examples of the null set? F = {x : x ∈ Q, 1 < x < 2}.
Answer: Q represents the set of rational numbers. Between any two distinct rational numbers, infinitely many rational numbers exist. For example, the rational number \( \frac{1+2}{2} = \frac{3}{2} = 1.5 \) lies strictly between 1 and 2. Since \( 1 < \frac{3}{2} < 2 \), the set contains at least one element and is not a null set.
In simple words: The fraction 3/2 (or 1.5) is a rational number between 1 and 2. So the set is not empty.

Exam Tip: For rational numbers between two integers, the average of those integers will always be a rational number between them.

 

Question 9. Which of the following are examples of the null set? G = {0}.
Answer: The set G contains exactly one element: the number 0. Since the set has one member, it is not empty. This is a singleton set (a set with exactly one element), not a null set.
In simple words: The set {0} has the number 0 in it, so it is not empty.

Exam Tip: Do not confuse the null set ∅ (or { }, containing nothing) with the singleton set {0} (containing one element: zero).

 

Question 10. Which of the following are examples of the singleton set? (i) {x : x ∈ Z, x² = 4}
Answer: Integers are …-3, -2, -1, 0, 1, 2, 3,…. Solving x² = 4 gives x = √4, so x = ±2. Both x = -2 and x = 2 satisfy the equation (since (-2)² = 4 and (2)² = 4). The set contains two elements: {-2, 2}. This is not a singleton set.
In simple words: Both -2 and 2 work, so the set has two elements, not one.

Exam Tip: When squaring, remember that both positive and negative roots satisfy the equation - check both before concluding.

 

Question 11. Which of the following are examples of the singleton set? (ii) {x : x ∈ Z, x + 5 = 0}
Answer: Integers are -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4,…. Solving x + 5 = 0 by subtracting 5 from both sides gives x = -5. Since -5 is an integer, it satisfies the condition. Only one integer solves this equation. The set contains exactly one element: {-5}. This is a singleton set.
In simple words: Only -5 makes the equation true, so the set has just one element.

Exam Tip: For linear equations (no exponents), there is typically only one solution - verify it belongs to the required set.

 

Question 12. Which of the following are examples of the singleton set? (iii) {x : x ∈ Z, |x| = 1}
Answer: Integers are …, -2, -1, 0, 1, 2,…. The absolute value |x| = 1 is satisfied when x = -1 (since |-1| = 1) and when x = 1 (since |1| = 1). The set contains two elements: {-1, 1}. This is not a singleton set.
In simple words: Both -1 and 1 have absolute value 1, so the set has two members.

Exam Tip: Absolute value equations usually have two solutions (one positive, one negative) unless the value is zero.

 

Question 13. Which of the following are examples of the singleton set? (iv) {x : x ∈ N, x² = 16}
Answer: Natural numbers are 1, 2, 3,…. Solving x² = 16 gives x = √16, so x = ±4. Although both x = 4 and x = -4 satisfy the equation mathematically, only x = 4 is a natural number (natural numbers are positive). Therefore, the set contains exactly one element: {4}. This is a singleton set.
In simple words: Even though -4 also squares to 16, it's not a natural number. So only 4 belongs to the set.

Exam Tip: When a condition restricts to a specific set (like N for natural numbers), some mathematical solutions may not qualify - always check.

 

Question 14. Which of the following are examples of the singleton set? (v) {x : x is an even prime number}
Answer: Prime numbers are 2, 3, 5, 7, 11,…. An even prime number is a prime that is also even (divisible by 2). The only even prime is 2 - every other even number has 2 as a factor and is therefore not prime. The set contains exactly one element: {2}. This is a singleton set.
In simple words: The number 2 is the only prime that is even. Every other even number can be divided by 2 and other factors.

Exam Tip: 2 is unique among primes - it's the only even prime number.

 

Question 15. Which of the following are pairs of equal sets? A = set of letters in the word, 'ALLOY.' B = set of letters in the word, 'LOYAL.'
Answer: Two sets are equal when they contain exactly the same elements - the order and repetition do not matter. The word ALLOY contains the letters A, L, L, O, Y. Writing unique letters only: A = {A, L, O, Y}. The word LOYAL contains the letters L, O, Y, A, L. Writing unique letters only: B = {L, O, Y, A}. Both sets contain exactly the same four distinct letters. Therefore, A = B, making them equal sets.
In simple words: Both words contain the same letters (A, L, O, Y), so the sets are identical even though the words spell them differently.

Exam Tip: When forming sets from word letters, list each distinct letter only once and ignore repetition.

 

Question 16. Which of the following are pairs of equal sets? C = set of letters in the word, 'CATARACT.' D = set of letters in the word, 'TRACT.'
Answer: Two sets are equal when they contain exactly the same elements. The word CATARACT contains the letters C, A, T, A, R, A, C, T. Writing unique letters only: C = {C, A, T, R}. The word TRACT contains the letters T, R, A, C, T. Writing unique letters only: D = {T, R, A, C}. Both sets contain exactly the same four distinct letters. Therefore, C = D, making them equal sets.
In simple words: Both words use only the letters C, A, T, and R (though in different arrangements and frequencies). The sets are the same.

Exam Tip: Repetition in a word does not create repeated elements in a set - each letter appears only once in the set.

 

Question 17. Which of the following are pairs of equal sets? E = {x : x ∈ Z, x² ≤ 4} and F = {x : x ∈ Z, x² = 4}
Answer: For E: we need integers where x² ≤ 4. Testing values: (-2)² = 4 ≤ 4 ✓, (-1)² = 1 ≤ 4 ✓, (0)² = 0 ≤ 4 ✓, (1)² = 1 ≤ 4 ✓, (2)² = 4 ≤ 4 ✓. So E = {-2, -1, 0, 1, 2}. For F: we need integers where x² = 4. Testing: (-2)² = 4 ✓, (2)² = 4 ✓. So F = {-2, 2}. Since E contains five elements while F contains only two, E ≠ F. The sets are not equal.
In simple words: E includes all integers whose squares are at most 4, while F includes only those whose squares equal exactly 4. These are different sets.

Exam Tip: Pay close attention to whether a condition uses ≤ (less than or equal) versus = (exactly equal) - they produce different sets.

 

Question 18. Which of the following are pairs of equal sets? G = {-1, 1} and H = {x : x ∈ Z, x² - 1 = 0}
Answer: For H: we need integers satisfying x² - 1 = 0. Solving: x² = 1, so x = ±1. This means x = -1 or x = 1. Therefore, H = {-1, 1}. Comparing the two sets: G = {-1, 1} and H = {-1, 1}. Both sets contain exactly the same two elements. Therefore, G = H. The sets are equal.
In simple words: The equation x² - 1 = 0 has solutions x = -1 and x = 1, which matches set G exactly.

Exam Tip: When an equation factors (like x² - 1 = (x-1)(x+1) = 0), both factors give solutions - don't miss either one.

 

Question 19. Which of the following are pairs of equal sets? J = {2, 3} and K = {x : x ∈ Z, (x² + 5x + 6) = 0}
Answer: For K: we need integers satisfying x² + 5x + 6 = 0. Factoring: x² + 5x + 6 = (x + 2)(x + 3) = 0. This gives x = -2 or x = -3. Therefore, K = {-2, -3}. Comparing: J = {2, 3} and K = {-2, -3}. The sets contain different elements (J has positive integers while K has negative integers). Therefore, J ≠ K. The sets are not equal.
In simple words: The equation produces negative solutions (-2 and -3), while set J contains positive numbers (2 and 3). They don't match.

Exam Tip: Factor quadratic equations carefully - the solutions often surprise you, especially regarding signs.

 

Question 20. Which of the following are pairs of equivalent sets?
(i) A = {-1, -2, 0} and B = {1, 2, 3}
Answer: Two sets are equivalent when they have the same number of elements (same cardinality), even if the elements themselves differ. Set A contains three elements: -1, -2, and 0. Set B contains three elements: 1, 2, and 3. Since both sets have the same cardinality (3 elements each), they are equivalent sets.
In simple words: Both sets have exactly 3 members, so they are equivalent - the actual values don't matter for equivalence.

Exam Tip: Equivalence depends only on the count of elements, not on what those elements are.

 

Question 21. Which of the following are pairs of equivalent sets?
(ii) C = {x : x ∈ N, x < 3} and D = {x : x ∈ W, x < 3}
Answer: For C: Natural numbers less than 3 are 1 and 2. So C = {1, 2}, which has 2 elements. For D: Whole numbers less than 3 are 0, 1, and 2. So D = {0, 1, 2}, which has 3 elements. Since C has 2 elements and D has 3 elements, their cardinalities differ. Therefore, C and D are not equivalent sets.
In simple words: C has 2 members while D has 3 members because whole numbers include 0 but natural numbers do not.

Exam Tip: Remember the key difference: natural numbers start at 1, while whole numbers include 0.

 

Question 22. Which of the following are pairs of equivalent sets?
(iii) E = {a, e, i, o, u} and F = {p, q, r, s, t}
Answer: Two sets are equivalent when they contain the same number of elements. Set E contains five elements: a, e, i, o, u. Set F contains five elements: p, q, r, s, t. Since both sets have the same cardinality (5 elements each), they are equivalent sets.
In simple words: Both sets have exactly 5 members, so they are equivalent even though the letters are completely different.

Exam Tip: Counting elements carefully is crucial for determining equivalence - recount if you're uncertain.

 

Question 23. State whether the given set is finite or infinite: A = Set of all triangles in a plane.
Answer: Within any plane, infinitely many distinct triangles can be drawn with different positions, sizes, and orientations. There is no limit to the number of triangles that can exist in a given plane. The set contains an infinite number of elements, making it an infinite set.
In simple words: You can keep drawing more and more triangles in a plane forever - there's no end to how many you can create.

Exam Tip: If elements can continue indefinitely without exhaustion, the set is infinite.

 

Question 24. State whether the given set is finite or infinite: B = Set of all points on the circumference of a circle.
Answer: The circumference of a circle is a continuous curve composed of infinitely many points. No matter how small the segments you consider, you can always identify more points along the arc. The set contains an infinite number of elements, making it an infinite set.
In simple words: Between any two points on a circle, infinitely many other points exist. The set never ends.

Exam Tip: Continuous geometric objects (curves, lines) always contain infinitely many points.

 

Question 25. State whether the given set is finite or infinite: C = Set of all lines parallel to the y-axis
Answer: Lines parallel to the y-axis are vertical lines of the form x = c, where c can be any real number. Since there are infinitely many real numbers, there are infinitely many such lines. The set contains an infinite number of elements, making it an infinite set.
In simple words: You can draw a vertical line through any point, and there are infinitely many possible locations. The set never ends.

Exam Tip: When a set is defined by a continuously varying parameter (like x = c for all real c), the set is typically infinite.

 

Question 26. State whether the given set is finite or infinite: D = Set of all leaves on a tree
Answer: A tree, though large, has a definite and countable number of leaves. The quantity may be very substantial, but it is not endless - each leaf is a discrete object with a physical limit. By definition, a finite set has a countable number of elements that terminates. Therefore, this set is finite.
In simple words: Even though the number might be very large, a tree will never have an infinite number of leaves. It's a definite, countable amount.

Exam Tip: Physical objects always form finite sets because they have natural limits - don't confuse "very large" with "infinite."

 

Question 27. State whether the given set is finite or infinite: E = Set of all positive integers greater than 500
Answer: Positive integers are 0, 1, 2, 3,…, 500. Those greater than 500 are 501, 502, 503,…. Since integers continue indefinitely without bound, there are infinitely many positive integers greater than 500. The set contains an infinite number of elements, making it an infinite set.
In simple words: You can always add 1 to get the next integer - the sequence never stops, no matter how high you go.

Exam Tip: Unbounded sequences of numbers (like 501, 502, 503,…) are always infinite.

 

Question 28. State whether the given set is finite or infinite: F = {x ∈ R: 0 < x < 1}
Answer: R denotes the set of real numbers, which includes both rational numbers (like 0.5, 0.25) and irrational numbers (like √2/2, π/4). Between any two distinct real numbers, infinitely many other real numbers exist. The interval (0, 1) contains all real numbers strictly between 0 and 1. Therefore, the set contains an infinite number of elements, making it an infinite set.
In simple words: Between 0 and 1, you can find numbers like 0.1, 0.01, 0.001, and infinitely many decimals - they never run out.

Exam Tip: Any interval of real numbers (even a tiny one) contains infinitely many points.

 

Question 5 G. State whether any given set is finite or infinite: G = {x ∈ Z: x < 1}.
Answer: The set of integers is {..., -3, -2, -1, 0, 1, 2, 3, ...}. Integers satisfying the condition x < 1 are {..., -4, -3, -2, -1, 0}. Since there are infinitely many integers less than 1, the given set is infinite.

Exam Tip: Remember that integers extend infinitely in the negative direction, so any constraint like "less than" a fixed number yields an infinite set.

 

Question 5 H. State whether any given set is finite or infinite: H = {x ∈ Z: -15 < x < 15}.
Answer: The set consists of all integers between -15 and 15 (excluding the endpoints). These are {-14, -13, ..., -1, 0, 1, ..., 13, 14}. Since there is a fixed upper and lower bound, the count of integers in this range is limited. Therefore, the given set is finite.

Exam Tip: Whenever both lower and upper bounds exist for an integer set, the set will be finite because integers are discrete.

 

Question 5 I. State whether any given set is finite or infinite: J = {x : x ∈ N and x is prime}.
Answer: This set represents the collection of all prime numbers from the natural numbers. The prime numbers form an unbounded sequence that continues indefinitely. Thus, the given set is infinite.

Exam Tip: Prime numbers never terminate; there are infinitely many primes, a fact proven by Euclid's theorem.

 

Question 5 J. State whether any given set is finite or infinite: K = {x : x ∈ N and x is prime}.
Answer: This set describes all prime numbers within the natural numbers. Since prime numbers are unlimited and extend without end, the set is infinite.

Exam Tip: Be familiar with the distinction between finite sets (with countable elements) and infinite sets (unbounded in size).

 

Question 5 K. State whether any given set is finite or infinite: L = {set of all circles passing through the origin (0, 0)}.
Answer: Infinitely many circles can pass through any given point, including the origin. Each circle is determined by its center and radius, and we can choose the center from infinitely many points in the plane (provided the distance from that center to the origin equals the desired radius). Therefore, the given set is infinite.

Exam Tip: Geometric objects like circles, lines, and planes typically form infinite sets when only one constraint (such as "passing through a point") is imposed.

 

Question 6. Rewrite the following statements using the set notation:
(i) a is an element of set A.
(ii) b is not an element of A.
(iii) A is an empty set and B is a nonempty set.
(iv) A number of elements in A is 6.
(v) 0 is a whole number but not a natural number.
Answer:
(i) Given that a is an element of set A, this translates to: a ∈ A
(ii) Given that b is not an element of A, this translates to: b ∉ A
(iii) Given that A is an empty set and B is a non - empty set, this translates to: A = ∅ and B ≠ ∅
(iv) Given that the total number of elements in set A is 6, this translates to: |A| = 6
(v) Given that 0 is a whole number but not a natural number, this translates to: 0 ∈ W but 0 ∉ N

Exam Tip: Mastering set notation is essential; ∈ denotes membership, ∉ denotes non-membership, ∅ or { } represents the empty set, and |A| denotes the cardinality (number of elements) of set A.

 

Question 1 A. State in each case whether A ⊂ B or A ⊄ B. A = {0, 1, 2, 3}, B = {1, 2, 3, 4, 5}.
Answer: A ⊄ B
Explanation: For A to be a subset of B, every element of A must also be in B. However, 0 ∈ A but 0 ∉ B, so the subset relationship does not hold.

Exam Tip: To disprove a subset claim, find just one element in the first set that does not appear in the second set.

 

Question 1 B. State in each case whether A ⊂ B or A ⊄ B. A = ∅, B = {0}.
Answer: A ⊂ B
Explanation: A is the null (empty) set. The empty set is a subset of every set by definition, therefore A ⊂ B.

Exam Tip: Remember that the empty set is always a subset of any set, including itself.

 

Question 1 C. State in each case whether A ⊂ B or A ⊄ B. A = {1, 2, 3}, B = {1, 2, 4}.
Answer: A ⊄ B
Explanation: Although 1 and 2 are in both sets, 3 ∈ A but 3 ∉ B, so A is not a subset of B.

Exam Tip: Subset membership requires all elements of the first set to appear in the second; a single missing element breaks the relationship.

 

Question 1 D. State in each case whether A ⊂ B or A ⊄ B. A = {x : x ∈ Z, x² = 1}, B = {x : x ∈ N, x² = 1}.
Answer: A ⊄ B
Explanation: Set A consists of integers satisfying x² = 1, giving A = {-1, 1}. Set B consists of natural numbers satisfying x² = 1, giving B = {1}. Since -1 ∈ A but -1 ∉ B, A is not a subset of B.

Exam Tip: When working with set-builder notation, first determine the explicit elements before checking subset relationships.

 

Question 1 E. State in each case whether A ⊂ B or A ⊄ B. A = {x : x is an even natural number}, B = {x : x is an integer}.
Answer: A ⊂ B
Explanation: Set A = {2, 4, 6, 8, ...} comprises all even natural numbers. Set B = {..., -3, -2, -1, 0, 1, 2, 3, ...} comprises all integers. Since every even natural number is also an integer, all elements of A belong to B, so A ⊂ B.

Exam Tip: Understanding the relationships between number systems (naturals, integers, rationals, reals) helps quickly determine subset relationships.

 

Question 1 F. State in each case whether A ⊂ B or A ⊄ B. A = {x : x is an integer}, B = {x : x is a rational number}.
Answer: A ⊂ B
Explanation: Set A = {..., -2, -1, 0, 1, 2, ...} consists of all integers. Set B comprises all rational numbers, which can be expressed as fractions with integer numerators and non-zero denominators. Since every integer can be written as a rational number (e.g., 3 = 3/1), all elements of A are contained in B.

Exam Tip: Integers form a subset of rational numbers because any integer n can be represented as n/1.

 

Question 1 G. State in each case whether A ⊂ B or A ⊄ B. A = {x : x is a real number}, B = {x : x is a complex number}.
Answer: A ⊂ B
Explanation: Set A consists of all real numbers. Set B consists of all complex numbers, which can be expressed in the form a + ib where a and b are real numbers and i is the imaginary unit. Every real number can be written as a complex number with an imaginary part of zero (e.g., 5 = 5 + 0i), so A ⊂ B.

Exam Tip: Real numbers form a subset of complex numbers; every real number is a complex number with zero imaginary component.

 

Question 1 H. State in each case whether A ⊂ B or A ⊄ B. A = {x : x is an isosceles triangle in the plane}, B = {x : x is an equilateral triangle in the same plane}.
Answer: A ⊄ B
Explanation: An isosceles triangle has at least two equal sides, while an equilateral triangle has all three sides equal. Not every isosceles triangle is equilateral (for example, a triangle with sides 5, 5, 8 is isosceles but not equilateral), so the set of isosceles triangles is not contained in the set of equilateral triangles.

Exam Tip: Recall that all equilateral triangles are isosceles, but the reverse is not true, making the subset relationship one-directional.

 

Question 1 I. State in each case whether A ⊂ B or A ⊄ B. A = {x : x is a square in a plane}, B = {x : x is a rectangle in the same plane}.
Answer: A ⊂ B
Explanation: A square is defined as a quadrilateral with four equal sides and four right angles. A rectangle is a quadrilateral with four right angles. Since every square satisfies the definition of a rectangle, all squares are rectangles, making A ⊂ B.

Exam Tip: In geometry, classification hierarchies matter: squares are special cases of rectangles, which are special cases of quadrilaterals.

 

Question 1 J. State in each case whether A ⊂ B or A ⊄ B. A = {x : x is a triangle in a plane}, B = {x : x is a rectangle in the same plane}.
Answer: A ⊄ B
Explanation: A triangle is a polygon with three sides and three angles. A rectangle is a quadrilateral (four sides) with four right angles. No triangle is a rectangle because they have different numbers of sides, so A ⊄ B.

Exam Tip: Geometric shapes with fundamentally different properties (such as different numbers of sides) cannot be subsets of one another.

 

Question 1 K. State in each case whether A ⊂ B or A ⊄ B. A = {x : x is an even natural number less than 8}, B = {x : x is a natural number which divides 32}.
Answer: A ⊄ B
Explanation: Set A = {2, 4, 6} comprises even natural numbers less than 8. Set B = {1, 2, 4, 8, 16, 32} comprises natural numbers that divide 32. Although 2 and 4 are in both sets, 6 ∈ A but 6 ∉ B (since 32 ÷ 6 is not a whole number), so A ⊄ B.

Exam Tip: When comparing sets defined by different conditions, carefully list out or identify all elements that satisfy each condition.

 

Question 2. Examine whether the following statements are true or false:
(i) {a, b} ⊄ {b, c, a}
(ii) {a} ∈ {a, b, c}
(iii) ∅ ⊂ {a, b, c}
(iv) {a, e} ⊂ {x : x is a vowel in the English alphabet}
(v) {x : x ∈ W, x + 5 = 5} = ∅
(vi) a ∈ {{a}, b}
(vii) {a} ⊂ {{a}, b}
(viii) {b, c} ⊂ {a, {b, c}}
(ix) {a, a, b, b} = {a, b}
(x) {a, b, a, b, a, b, ...} is an infinite set.
(xi) If A = set of all circles of unit radius in a plane and B = set of all circles in the same plane then A ⊂ B.
Answer:
(i) False
Explanation: The elements of {a, b} are also elements of {b, c, a}, so {a, b} ⊂ {b, c, a}. The statement claims the opposite, making it false.
(ii) False
Explanation: The set {a} itself is not an element of {a, b, c}. The element a (without braces) is in {a, b, c}, but the set {a} is not.
(iii) True
Explanation: The empty set is a subset of every set by definition.
(iv) True
Explanation: The vowels in the English alphabet are a, e, i, o, u. Both a and e are vowels, so {a, e} ⊂ {a, e, i, o, u}.
(v) False
Explanation: When x = 0 (which is a whole number), we have 0 + 5 = 5. So {0} is not the empty set; instead, the set is {0} ≠ ∅.
(vi) False
Explanation: The elements of {{a}, b} are {a} and b. The individual element a is not an element of this set.
(vii) False
Explanation: The set {{a}, b} has elements {a} and b. Since a is not an element of {{a}, b}, the set {a} is not a subset of it.
(viii) False
Explanation: The set {a, {b, c}} has elements a and {b, c}. The set {b, c} is an element of this set, not a subset. An element cannot also be a subset.
(ix) True
Explanation: In set theory, all elements are treated as distinct. Repetition of elements does not change the set, so {a, a, b, b} = {a, b}.
(x) False
Explanation: Despite the appearance, this set contains only the elements a and b. Repetition does not create new elements, making it the finite set {a, b}.
(xi) True
Explanation: All circles with unit radius form a subset of all circles in the plane, since every unit circle is also a circle in the plane (just with a specific radius constraint).

Exam Tip: Distinguish carefully between ∈ (element of) and ⊂ (subset of); understand that repetition and order are irrelevant in set definitions.

 

Question 3. If A = {1} and B = {{1}, 2} then show that A ⊄ B.
Answer: Set A contains the single element 1. Set B contains two elements: the set {1} and the number 2. For A to be a subset of B, the element 1 must be in B. However, B does not contain 1 as an element; it contains the set {1}, which is different from 1. Therefore, 1 ∈ A but 1 ∉ B, proving that A ⊄ B.

Exam Tip: Pay close attention to the difference between an element x and the set {x}; they are not the same, and this distinction is critical for subset relationships.

 

Question 4. Write down all subsets of each of the following sets:
(i) A = {A}
(ii) B = {a, b}
(iii) C = {-2, 3}
(iv) D = {-1, 0, 1}
(v) E = ∅
(vi) F = {2, {3}}
(vii) G = {3, 4, {5, 6}}
Answer:
(i) The subsets of {A} are ∅ and {A}.
(ii) The subsets of {a, b} are ∅, {a}, {b}, and {a, b}.
(iii) The subsets of {-2, 3} are ∅, {-2}, {3}, and {-2, 3}.
(iv) The subsets of {-1, 0, 1} are ∅, {-1}, {0}, {1}, {-1, 0}, {0, 1}, {-1, 1}, and {-1, 0, 1}.
(v) The empty set ∅ has only one subset: ∅.
(vi) Let x = {3}. Then F = {2, x}. The subsets of {2, x} are ∅, {2}, {x}, and {2, x}, which translates to ∅, {2}, {{3}}, and {2, {3}}.
(vii) Let x = {5, 6}. Then G = {3, 4, x}. The subsets of {3, 4, x} are ∅, {3}, {4}, {x}, {3, 4}, {3, x}, {4, x}, and {3, 4, x}, which translates to ∅, {3}, {4}, {{5, 6}}, {3, 4}, {3, {5, 6}}, {4, {5, 6}}, and {3, 4, {5, 6}}.

Exam Tip: For a set with n elements, the total number of subsets is 2^n. Always include the empty set and the set itself as subsets.

 

Question 5. Express each of the following sets as an interval:
(i) A = {x : x ∈ R, -4 < x < 0}
(ii) B = {x : x ∈ R, 0 ≤ x < 3}
(iii) C = {x : x ∈ R, 2 < x ≤ 6}
(iv) D = {x : x ∈ R, -5 ≤ x ≤ 2}
(v) E = {x : x ∈ R, -3 ≤ x < 2}
(vi) F = {x : x ∈ R, -2 ≤ x < 0}
Answer:
(i) A = (-4, 0)
Explanation: All real numbers strictly between -4 and 0 form an open interval, where neither endpoint is included.
(ii) B = [0, 3)
Explanation: The interval includes 0 (left endpoint) but excludes 3 (right endpoint), making it half - open.
(iii) C = (2, 6]
Explanation: The interval excludes 2 (left endpoint) but includes 6 (right endpoint), making it half - open.
(iv) D = [-5, 2]
Explanation: Both endpoints are included in this closed interval, which contains all real numbers from -5 to 2 inclusive.
(v) E = [-3, 2)
Explanation: The left endpoint -3 is included, while the right endpoint 2 is excluded, creating a half - open interval.
(vi) F = [-2, 0)
Explanation: The left endpoint -2 is included, while the right endpoint 0 is excluded, creating a half - open interval.

Exam Tip: Use parentheses ( ) for open endpoints (not included) and square brackets [ ] for closed endpoints (included) when writing interval notation.

 

Question 6. Write each of the following intervals in the set-builder form:
(i) A = (-2, 3)
(ii) B = [4, 10]
(iii) C = [-1, 8)
(iv) D = (4, 9]
(v) E = [-10, 0)
(vi) F = (0, 5]
Answer:
(i) A = {x : x ∈ R, -2 < x < 3}
(ii) B = {x : x ∈ R, 4 ≤ x ≤ 10}
(iii) C = {x : x ∈ R, -1 ≤ x < 8}
(iv) D = {x : x ∈ R, 4 < x ≤ 9}
(v) E = {x : x ∈ R, -10 ≤ x < 0}
(vi) F = {x : x ∈ R, 0 < x ≤ 5}

Exam Tip: Open parentheses in interval notation correspond to strict inequalities (<, >), while square brackets correspond to non-strict inequalities (≤, ≥).

 

Question 7. If A = {3, {4, 5}, 6} find which of the following statements are true:
(i) {4, 5} ⊄ A
(ii) {4, 5} ∈ A
(iii) {{4, 5}} ⊆ A
(iv) 4 ∈ A
(v) {3} ⊆ A
(vi) {∅} ⊆ A
(vii) ∅ ⊆ A
(viii) {3, 4, 5} ⊆ A
(ix) {3, 6} ⊆ A
Answer:
(i) True
Explanation: Set A has elements 3, {4, 5}, and 6. For {4, 5} to be a subset, its elements 4 and 5 must be in A. However, the individual numbers 4 and 5 are not elements of A; only the set {4, 5} is. Therefore, {4, 5} ⊄ A.
(ii) True
Explanation: The set {4, 5} is one of the three elements of A, so {4, 5} ∈ A.
(iii) True
Explanation: The set {{4, 5}} contains the single element {4, 5}, which is also an element of A. Therefore, {{4, 5}} ⊆ A.
(iv) False
Explanation: The number 4 is not an element of A; only the set {4, 5} is an element.
(v) True
Explanation: The set {3} has 3 as its only element, and 3 is in A, so {3} ⊆ A.
(vi) False
Explanation: The set {∅} contains the empty set as its element. The empty set is not an element of A, so {∅} ⊄ A.
(vii) True
Explanation: The empty set is a subset of every set, including A.
(viii) False
Explanation: For {3, 4, 5} to be a subset of A, all three elements 3, 4, and 5 must be in A. While 3 is in A, the individual numbers 4 and 5 are not; so {3, 4, 5} ⊄ A.
(ix) True
Explanation: Both 3 and 6 are elements of A, so the set {3, 6} ⊆ A.

Exam Tip: When a set contains another set as an element (like {4, 5} in A), do not confuse the set itself with its individual members.

 

Question 8. If A = {a, b, c}, find P(A) and n{P(A)}.
Answer: The power set of A, denoted P(A), is the collection of all subsets of A. The empty set ∅ is a subset of every set. Additionally, {a}, {b}, {c}, {a, b}, {b, c}, and {a, c} are all subsets of {a, b, c}. Every set is also a subset of itself, so {a, b, c} is included. Therefore, the power set consists of eight subsets:

P(A) = {∅, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}}

The cardinality (number of elements) of the power set is given by n{P(A)} = 2^m, where m = n(A) is the number of elements in A. Here, m = 3, so:

n{P(A)} = 2³ = 8

Exam Tip: The power set formula n{P(A)} = 2^n is crucial; memorize it for quick calculations when finding the number of subsets.

 

Question 9. If A = {1, {2, 3}}, find P(A) and n{P(A)}.
Answer: Let x = {2, 3}. Then A = {1, x}. The subsets of A are those that include zero, one, or both elements of A: ∅, {1}, {x}, and {1, x}. Substituting back, the subsets become ∅, {1}, {{2, 3}}, and {1, {2, 3}}.

Thus, P(A) = {∅, {1}, {{2, 3}}, {1, {2, 3}}}.

The cardinality is n{P(A)} = 2^m, where m = n(A) = 2, so:

n{P(A)} = 2² = 4

Exam Tip: Even when a set contains another set as an element, treat that element as a single member when finding the power set.

 

Question 10. If A = ∅ then find n{P(A)}.
Answer: Set A is the empty set, so n(A) = 0 (A has zero elements). The number of subsets of any set is given by the formula n{P(A)} = 2^m, where m = n(A):

n{P(A)} = 2⁰ = 1

The power set P(A) contains exactly one subset: the empty set itself, P(A) = {∅}.

Exam Tip: The empty set has exactly one subset - itself. This is a special case that often appears in exam questions.

 

Question 11. If A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 8} then find the universal set.
Answer: The universal set (for these three sets) comprises all elements found in A, B, and C combined. The union of all elements gives: {1, 3, 5, 2, 4, 6, 0, 8}. When arranged in order, the universal set is:

U = {0, 1, 2, 3, 4, 5, 6, 8}

Exam Tip: The universal set is context-dependent; it includes all elements from all sets under consideration and is determined by the problem statement.

 

Question 12. Prove that A ⊆ B, B ⊆ C and C ⊆ A ⇒ A = C.
Answer: Given: A ⊆ B, B ⊆ C, and C ⊆ A.

From the first two conditions, using the transitivity of the subset relation: since A ⊆ B and B ⊆ C, we have A ⊆ C. But we are also given that C ⊆ A. When two sets are subsets of each other (A ⊆ C and C ⊆ A), they must be equal. Therefore, A = C.

Exam Tip: Set equality is proven by showing mutual subset relationships: A = B if and only if A ⊆ B and B ⊆ A.

 

Question 13. For any set A, prove that A ⊆ ∅ ⇔ A = ∅.
Answer: We must prove both directions of the equivalence.

Forward direction (A ⊆ ∅ ⇒ A = ∅): If A is a subset of the empty set, then every element of A is also in ∅. But ∅ has no elements, so A cannot have any elements either. Therefore, A = ∅.

Reverse direction (A = ∅ ⇒ A ⊆ ∅): If A is the empty set, then A is equal to ∅. Since every set is a subset of itself, A ⊆ ∅.

Therefore, for any set A, A ⊆ ∅ if and only if A = ∅.

Exam Tip: To prove an "if and only if" statement, always establish both directions: the forward implication and the reverse implication.

 

Question 14. State whether the given statement is true or false:
(i) If A ⊂ B and x ∉ B then x ∉ A.
(ii) If A ⊆ ∅ then A = ∅.
(iii) If A, B and C are three sets such that A ∈ B and B ⊂ C then A ⊂ C.
(iv) If A, B and C are three sets such that A ⊂ B and B ∈ C then A ∈ C.
(v) If A, B and C are three sets such that A ⊄ B and B ⊄ C then A ⊄ C.
(vi) If A and B are sets such that x ∈ A and A ∈ B then x ∈ B.
Answer:
(i) True
Explanation: Since A ⊂ B, all elements of A are in B. If x ∉ B, then x cannot be in A (because every element of A must be in B). Therefore, x ∉ A.
(ii) True
Explanation: The only set that is a subset of ∅ is ∅ itself, because ∅ contains no elements. If A ⊆ ∅, then A must be the empty set.
(iii) False
Explanation: Consider A = {a}, B = {{a}, b}, and C = {{a}, b, c}. Here, A ∈ B (A is an element of B) and B ⊂ C (B is a subset of C). However, A = {a} is not a subset of C = {{a}, b, c} because {a} is an element of C, not a subset. Elements and subsets are different concepts.
(iv) False
Explanation: Let A = {a}, B = {a, b}, and C = {{a, b}, c}. Then A ⊂ B (A is a subset of B) and B ∈ C (B is an element of C). But A = {a} is not an element of C; {a, b} is. So A ∉ C.
(v) False
Explanation: A counterexample: let A = {a}, B = {b, c}, and C = {a, c}. Since a ∈ A but a ∉ B, we have A ⊄ B. Since b ∈ B but b ∉ C, we have B ⊄ C. However, a ∈ A and a ∈ C, so A ⊂ C, contradicting the claim that A ⊄ C. The absence of one subset relationship does not guarantee the absence of another.
(vi) False
Explanation: Let A = {x}, B = {{x}, y}. Then x ∈ A (the element x is in A) and A ∈ B (the set {x} is an element of B). However, x ∉ B because x is not an element of B; only {x} is. Membership in an element of a set does not guarantee membership in the set itself.

Exam Tip: Always distinguish between ∈ (membership in a set) and ⊂ (subset of); these are fundamentally different relationships.

 

Exercise 1D
Question 1. If A = {a, b, c, d, e, f}, B = {c, e, g, h} and C = {a, e, m, n}, find:

Exam Tip: Exercise 1D continues with set operations on multiple sets; prepare to work with unions, intersections, and complements in the following questions.

 

Question 1. If A = {a, b, c, d, e, f}, B = {c, e, g, h} and C = {a, e, m, n} find:
(i) A ∪ B
(ii) B ∪ C
(iii) B ∪ C
(iv) C ∩ A
(v) A ∩ B
Answer: Given: A = {a, b, c, d, e, f}, B = {c, e, g, h} and C = {a, e, m, n}
(i) A ∪ B = {a, b, c, d, e, f, g, h}
(ii) B ∪ C = {a, c, e, g, h, m, n}
(iii) B ∪ C = {a, c, e, g, h, m, n}
(iv) C ∩ A = {a, e}
(v) A ∩ B = {c, e}

 

Question 2. If A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8} and C = {10, 11, 12, 13, 14} find:
(i) A ∪ B
(ii) B ∪ C
(iii) A ∪ C
(iv) B ∪ D
(v) (A ∪ B) ∪ C
(vi) (A ∪ B) ∩ C
(vii) (A ∩ B) ∪ D
(viii) (A ∩ B) ∪ (B ∩ C)
(ix) (A ∩ C) ∩ (C ∪ D)
Answer: Given: A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8} and C = {10, 11, 12, 13, 14}
(i) A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}
(ii) B ∪ C = {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}
(iii) A ∪ C = {1, 2, 3, 4, 5, 10, 11, 12, 13, 14}
(iv) B ∪ D - (D not defined)
(v) (A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14}
(vi) (A ∪ B) ∩ C = ∅
(vii) (A ∩ B) ∪ D - (D not defined)
(viii) (A ∩ B) ∪ (B ∩ C) = ∅
(ix) (A ∩ C) ∩ (C ∪ D) - (D not defined)

 

Question 3. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13} and C = {11, 13, 15} and D = {15, 17} find:
(i) A ∩ B
(ii) A ∩ C
(iii) B ∩ C
(iv) B ∩ D
(v) B ∩ (C ∪ D)
(vi) A ∩ (B ∪ C)
Answer: Given: A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13} and C = {11, 13, 15} and D = {15, 17}
(i) A ∩ B = {7, 9, 11}
(ii) A ∩ C = {11}
(iii) B ∩ C = {11, 13}
(iv) B ∩ D = ∅
(v) B ∩ (C ∪ D) = {11, 13}
(vi) A ∩ (B ∪ C) = {7, 9, 11}

 

Question 4. If A = {x : x ∈ N}, B = {x : x ∈ N and x is even}, C = {x : x ∈ N and x is odd} and D = {x : x ∈ N and x is prime} then find:
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Answer: Given: A = {x : x ∈ N}, B = {x : x ∈ N and x is even}, C = {x : x ∈ N and x is odd} and D = {x : x ∈ N and x is prime}
(i) A ∩ B = {x : x ∈ N and x is even}
(ii) A ∩ C = {x : x ∈ N and x is odd}
(iii) A ∩ D = {x : x ∈ N and x is prime}
(iv) B ∩ C = ∅
(v) B ∩ D = {2} [∵ 2 is the only even prime number]
(vi) C ∩ D = {x : x ∈ N and x is prime and x ≠ 2}

 

Question 5. If A = {2x : x ∈ N, 1 ≤ x < 4}, B = {x + 2 : x ∈ N and 2 ≤ x < 5} and C = {x : x ∈ N and 4 < x < 8} find:
(i) A ∩ B
(ii) A ∪ B
(iii) (A ∪ B) ∩ C
Answer: Given: A = {2x : x ∈ N, 1 ≤ x < 4}, B = {x + 2 : x ∈ N and 2 ≤ x < 5} and C = {x : x ∈ N and 4 < x < 8}

Based on the given conditions: A = {2, 4, 6}, B = {4, 5, 6} and C = {5, 6, 7}
(i) A ∩ B = {4, 6}
(ii) A ∪ B = {2, 4, 5, 6}
(iii) (A ∪ B) ∩ C = {5, 6}

 

Question 6. If A = {2, 4, 6, 8, 10, 12}, B = {3, 4, 5, 6, 7, 8, 10} find:
(i) (A - B)
(ii) (B - A)
(iii) (A - B) ∪ (B - A)
Answer: Given: A = {2, 4, 6, 8, 10, 12}, B = {3, 4, 5, 6, 7, 8, 10}
(i) (A - B) = {2, 12}
(ii) (B - A) = {3, 5, 7}
(iii) (A - B) ∪ (B - A) = {2, 3, 5, 7, 12}

 

Question 7. If A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g} find:
(i) A ∩ (B - C)
(ii) A - (B ∪ C)
(iii) A - (B ∩ C)
Answer: Given: A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}
(i) A ∩ (B - C) = {a, c}
(ii) A - (B ∪ C) = {d}
(iii) A - (B ∩ C) = {a, b, c, d}

 

Question 8. If A = {1/x : x ∈ N and x < 8} and B = {1/(2x) : x ∈ N and x ≤ 4} find:
(i) A ∪ B
(ii) A ∩ B
(iii) A - B
(iv) B - A
Answer: Given: A = {1/x : x ∈ N and x < 8} and B = {1/(2x) : x ∈ N and x ≤ 4}

Based on the given conditions: A = {1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7} and B = {1/2, 1/4, 1/6, 1/8}
(i) A ∪ B = {1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8}
(ii) A ∩ B = {1/2, 1/4, 1/6}
(iii) A - B = {1, 1/3, 1/5, 1/7}
(iv) B - A = {1/8}

 

Question 9. If R is the set of all real numbers and Q is the set of all rational numbers then what is the set (R - Q)?
Answer: Given: R is the set of all real numbers and Q is the set of all rational numbers.

(R - Q) represents the set of all irrational numbers. These are numbers that cannot be written as a ratio of two integers and include values like π, e, and square roots of non-perfect squares.

Exam Tip: Remember that the complement of rational numbers within the real number system gives you irrational numbers - this is a fundamental classification in number theory.

 

Question 10. If A = {2, 3, 5, 7, 11} and B = ∅ find:
(i) A ∪ B
(ii) A ∩ B
Answer: Given: A = {2, 3, 5, 7, 11} and B = ∅
(i) A ∪ B = {2, 3, 5, 7, 11}
(ii) A ∩ B = ∅

Exam Tip: When one set is empty, the union equals the non-empty set and the intersection is always empty - these are essential properties to memorize for set operations.

 

Question 11. If A and B are two sets such that A ⊆ B then find:
(i) A ∪ B
(ii) A ∩ B
Answer: Given: A and B are two sets such that A ⊆ B.
(i) A ∪ B = B
(ii) A ∩ B = A

Exam Tip: When A is a subset of B, all elements of A are already in B, so their union produces B and their intersection produces A - this is a critical concept for understanding subset relationships.

 

Question 12. Which of the following sets are pairs of disjoint sets? Justify your answer.
(i) A = {3, 4, 5, 6} and B = {2, 5, 7, 9}
(ii) C = {1, 2, 3, 4, 5} and D = {6, 7, 9, 11}
(iii) E = {x : x ∈ N, x is even and x < 8} and F = {x : x = 3n, n ∈ N and x < 4}
(iv) G = {x : x ∈ N, x is even} and H = {x : x ∈ N, x is prime}
(v) J = {x : x ∈ N, x is even} and K = {x : x ∈ N, x is odd}
Answer: Disjoint sets have their intersections as ∅.

(i) A = {3, 4, 5, 6} and B = {2, 5, 7, 9} - These are pairs of disjoint sets. Though they share element 5, they are not disjoint. (Note: This set pair actually shares element 5, so they are NOT disjoint.)

(ii) C = {1, 2, 3, 4, 5} and D = {6, 7, 9, 11} - These are pairs of disjoint sets. Neither set contains any element that appears in the other.

(iii) E = {x : x ∈ N, x is even and x < 8} = {2, 4, 6} and F = {x : x = 3n, n ∈ N and x < 4} = {3} - These are not pairs of disjoint sets. They share no common elements so they are disjoint.

(iv) G = {x : x ∈ N, x is even} and H = {x : x ∈ N, x is prime} - These are not pairs of disjoint sets. Since 2 is an even prime number, their intersection is not ∅.

(v) J = {x : x ∈ N, x is even} and K = {x : x ∈ N, x is odd} - These are pairs of disjoint sets. No number can be both odd and even simultaneously, so their intersection is ∅.

Exam Tip: To verify disjoint sets, check if their intersection equals the empty set - if any element appears in both sets, they are not disjoint.

 

Question 13. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {1, 4, 5, 6} find:
(i) A'
(ii) B'
(iii) C'
(iv) (B')'
(v) (A ∪ B)'
(vi) (A ∩ C)'
(vii) (B - C)'
Answer: Given: U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {1, 4, 5, 6}
(i) A' = {5, 6, 7, 8, 9}
(ii) B' = {1, 3, 5, 7, 9}
(iii) C' = {2, 3, 7, 8, 9}
(iv) (B')' = {2, 4, 6, 8}
(v) (A ∪ B)' = {5, 7, 9}
(vi) (A ∩ C)' = {2, 3, 5, 6, 7, 8, 9}
(vii) (B - C)' = {1, 3, 4, 5, 6, 7, 9}

Exam Tip: The complement of a set includes all universal set elements not in that set - always identify the universal set first before calculating complements.

 

Question 14. If U = {a, b, c} and A = {a, c, d, e} then verify that:
(i) (A ∪ B)' = (A' ∩ B')
(ii) (A ∩ B)' = (A' ∪ B')
Answer: Given: U = {a, b, c} and A = {a, c, d, e} (Note: A contains elements not in U, suggesting set A may be incorrectly defined in the original source.)

(i) This identity represents De Morgan's First Law: the complement of a union equals the intersection of the complements. This means that the elements not in either set A or B are exactly those elements in neither A nor B.

(ii) This identity represents De Morgan's Second Law: the complement of an intersection equals the union of the complements. This means that the elements not in both A and B are those that are in at least one of their complements.

Exam Tip: De Morgan's Laws are fundamental in set theory - practice applying them with concrete examples to strengthen your understanding of how complements interact with union and intersection operations.

 

Question 15. If U is the universal set and A ⊆ U then fill in the blanks.
(i) A ∪ A' = ....
(ii) A ∩ A' = ....
(iii) ∅' ∩ A = ....
(iv) U' ∩ A = ....
Answer: Given: U is the universal set and A ⊆ U
(i) A ∪ A' = U
(ii) A ∩ A' = ∅
(iii) ∅' ∩ A = A
(iv) U' ∩ A = ∅

Exam Tip: These fundamental properties show how complements combine with sets - the union of a set and its complement gives the universal set, while their intersection is always empty.

Exercise 1E

 

Question 1. If A = {a, b, c, d, e}, B = {a, c, e, g}, verify that:
(i) A ∪ B = B ∪ A
(ii) A ∪ C = C ∪ A
(iii) B ∪ C = C ∪ B
(iv) A ∩ B = B ∩ A
(v) B ∩ C = C ∩ B
(vi) A ∩ C = C ∩ A
(vii) (A ∪ B) ∪ C = A ∪ (B ∪ C)
(viii) (A ∩ B) ∩ C = A ∩ (B ∩ C)
Answer:
(i) LHS = A ∪ B = {a, b, c, d, e} ∪ {a, c, e, g} = {a, b, c, d, e, g}
RHS = B ∪ A = {a, c, e, g} ∪ {a, b, c, d, e} = {a, b, c, d, e, g}
Hence proved.

(ii) To prove: A ∪ C = C ∪ A. Since the elements of set C are not given, assume x represents any element in C.
LHS = A ∪ C = {a, b, c, d, e} ∪ {x | x ∈ C} = {a, b, c, d, e, x}
RHS = C ∪ A = {x | x ∈ C} ∪ {a, b, c, d, e} = {x, a, b, c, d, e}
Hence proved.

(iii) To prove: B ∪ C = C ∪ B. Since the elements of set C are not given, assume x represents any element in C.
LHS = B ∪ C = {a, c, e, g} ∪ {x | x ∈ C} = {a, c, e, g, x}
RHS = C ∪ B = {x | x ∈ C} ∪ {a, c, e, g} = {x, a, c, e, g}
Hence proved.

(iv) LHS = A ∩ B = {a, b, c, d, e} ∩ {a, c, e, g} = {a, c, e}
RHS = B ∩ A = {a, c, e, g} ∩ {a, b, c, d, e} = {a, c, e}
A ∩ B = B ∩ A Hence proved.

(v) Let x be any element of B ∩ C. Then x ∈ B ∩ C means x ∈ B and x ∈ C. By definition of intersection, this means x ∈ C and x ∈ B, so x ∈ C ∩ B. Therefore, B ∩ C ⊆ C ∩ B.

Similarly, let x be any element of C ∩ B. Then x ∈ C ∩ B means x ∈ C and x ∈ B. By definition of intersection, this means x ∈ B and x ∈ C, so x ∈ B ∩ C. Therefore, C ∩ B ⊆ B ∩ C.

From both directions, B ∩ C = C ∩ B. Hence proved.

(vi) Let x be any element of A ∩ C. Then x ∈ A ∩ C means x ∈ A and x ∈ C. By definition of intersection, this means x ∈ C and x ∈ A, so x ∈ C ∩ A. Therefore, A ∩ C ⊆ C ∩ A.

Similarly, let x be any element of C ∩ A. Then x ∈ C ∩ A means x ∈ C and x ∈ A. By definition of intersection, this means x ∈ A and x ∈ C, so x ∈ A ∩ C. Therefore, C ∩ A ⊆ A ∩ C.

From both directions, A ∩ C = C ∩ A. Hence proved.

(vii) Let x be any element of (A ∪ B) ∪ C. Then x ∈ (A ∪ B) or x ∈ C, which means x ∈ A or x ∈ B or x ∈ C. This is the same as saying x ∈ A or x ∈ (B ∪ C), so x ∈ A ∪ (B ∪ C). Therefore, (A ∪ B) ∪ C ⊆ A ∪ (B ∪ C).

Conversely, let x be any element of A ∪ (B ∪ C). Then x ∈ A or x ∈ (B ∪ C), which means x ∈ A or x ∈ B or x ∈ C. This is the same as saying x ∈ (A ∪ B) or x ∈ C, so x ∈ (A ∪ B) ∪ C. Therefore, A ∪ (B ∪ C) ⊆ (A ∪ B) ∪ C.

From both directions, (A ∪ B) ∪ C = A ∪ (B ∪ C). Hence proved.

(viii) Let x be any element of (A ∩ B) ∩ C. Then x ∈ (A ∩ B) and x ∈ C, which means x ∈ A and x ∈ B and x ∈ C. This is the same as saying x ∈ A and x ∈ (B ∩ C), so x ∈ A ∩ (B ∩ C). Therefore, (A ∩ B) ∩ C ⊆ A ∩ (B ∩ C).

Conversely, let x be any element of A ∩ (B ∩ C). Then x ∈ A and x ∈ (B ∩ C), which means x ∈ A and x ∈ B and x ∈ C. This is the same as saying x ∈ (A ∩ B) and x ∈ C, so x ∈ (A ∩ B) ∩ C. Therefore, A ∩ (B ∩ C) ⊆ (A ∩ B) ∩ C.

From both directions, (A ∩ B) ∩ C = A ∩ (B ∩ C). Hence proved.

Exam Tip: These properties (commutative and associative laws for union and intersection) are fundamental in set theory - demonstrating them using element-based reasoning is a reliable approach that works for all set problems.

 

Question 2. If A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g} verify that:
(i) A ∩ (B - C) = (A ∩ B) - (A ∩ C)
(ii) A - (B ∩ C) = (A - B) ∪ (A - C)
Answer:
(i) B - C represents all elements in B that are not in C. B - C = {a, c}

LHS = A ∩ (B - C) = {a, b, c, d, e} ∩ {a, c} = {a, c}

A ∩ B = {a, c, e}
A ∩ C = {b, e}
RHS = (A ∩ B) - (A ∩ C) = {a, c, e} - {b, e} = {a, c}

LHS = RHS. Hence proved.

(ii) B ∩ C = {e, g}

LHS = A - (B ∩ C) = {a, b, c, d, e} - {e, g} = {a, b, c, d}

(A - B) = {b, d}
(A - C) = {a, c, d}
RHS = (A - B) ∪ (A - C) = {b, d} ∪ {a, c, d} = {a, b, c, d}

LHS = RHS. Hence proved.

Exam Tip: These distributive properties show how intersection and difference operations distribute across union and intersection - practice verifying these with concrete examples to ensure accuracy in your calculations.

 

Question 3. If A = {x : x ∈ N, x ≤ 7}, B = {x : x is prime, x < 8} and C = {x : x ∈ N, x is odd and x < 10} verify that:
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Answer: Natural numbers start from 1. A = {1, 2, 3, 4, 5, 6, 7}, B = {2, 3, 5, 7}, C = {1, 3, 5, 7, 9}

(i) B ∩ C = {3, 5, 7}

LHS = A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6, 7} ∪ {3, 5, 7} = {1, 2, 3, 4, 5, 6, 7}

A ∪ B = {1, 2, 3, 4, 5, 6, 7}
A ∪ C = {1, 2, 3, 4, 5, 6, 7, 9}
RHS = (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6, 7} ∩ {1, 2, 3, 4, 5, 6, 7, 9} = {1, 2, 3, 4, 5, 6, 7}

LHS = RHS. Hence proved.

(ii) B ∪ C = {1, 2, 3, 5, 7, 9}

LHS = A ∩ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7} ∩ {1, 2, 3, 5, 7, 9} = {1, 2, 3, 5, 7}

A ∩ B = {2, 3, 5, 7}
A ∩ C = {1, 3, 5, 7}
RHS = (A ∩ B) ∪ (A ∩ C) = {2, 3, 5, 7} ∪ {1, 3, 5, 7} = {1, 2, 3, 5, 7}

LHS = RHS. Hence proved.

Exam Tip: These distributive laws for union and intersection are critical for simplifying complex set expressions - always verify calculations step-by-step to avoid errors in determining elements.

 

Question 4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7} verify that:
(i) (A ∪ B)' = (A' ∩ B')
(ii) (A ∩ B)' = (A' ∪ B')
Answer:
(i) A ∪ B = {2, 3, 4, 5, 6, 7, 8}
(A ∪ B)' = {1, 9}

A' = {1, 3, 5, 7, 9}
B' = {1, 4, 6, 8, 9}
A' ∩ B' = {1, 9}

(A ∪ B)' = A' ∩ B'. Hence proved.

(ii) A ∩ B = {2}
(A ∩ B)' = {1, 3, 4, 5, 6, 7, 8, 9}

A' ∪ B' = {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}

(A ∩ B)' = A' ∪ B'. Hence proved.

Exam Tip: De Morgan's Laws are essential for simplifying complement operations - (A ∪ B)' converts to an intersection of complements, while (A ∩ B)' converts to a union of complements; always apply the correct form.

 

Question 5. Let A = {a, b, c}, B = {b, c, d, e} and = {c, d, e, f} be subsets of U = {a, b, c, d, e, f}. Then verify that:
(i) (A')' = A
(ii) (A ∪ B)' = (A' ∩ B')
(iii) (A ∩ B)' = (A' ∪ B')
Answer: (i) A' = {d, e, f}
(A')' = {a, b, c} = A Hence proved

(ii) A ∪ B = {a, b, c, d, e}
(A ∪ B)' = {f}
A' = {d, e, f}
B' = {a, f}
A' ∩ B' = {f}
(A ∪ B)' = (A' ∩ B') Hence proved

(iii) A' ∪ B' = {a, d, e, f}
A ∩ B = {b, c}
(A ∩ B)' = {a, d, e, f}
(A ∩ B)' = A' ∪ B' Hence proved
In simple words: To prove these identities, first find the complement of each set (the elements not in that set). Then compute the required operations and check that both sides give the same result.

Exam Tip: Always list out the complement sets first using the universal set U, then carefully apply union and intersection operations to verify both sides match exactly.

 

Question 6. Give an example of three sets A, B, C such that A ∩ C ≠ ϕ, B ∩ C ≠ ϕ, A ∩ B ≠ ϕ, and A ∩ B ∩ C = ϕ
Answer: Let A = {1, 2}
B = {2, 3}
C = {1, 3, 4}

A ∩ B = {2}
A ∩ C = {1}
B ∩ C = {3}
A ∩ B ∩ C = {2} ∩ {1, 3, 4} = ϕ

The three sets satisfy all the given conditions.
In simple words: Choose sets where two pairs share elements, but all three sets have no element in common to their intersection.

Exam Tip: Verify all four conditions (three pairwise intersections and the triple intersection) before finalizing your example.

 

Question 7. For any sets A and B, prove that:
(i) (A - B) ∩ B = ϕ
(ii) A ∪ (B - A) = A ∪ B
(iii) (A - B) ∪ (A ∩ B) = A
(iv) (A ∪ B) - B = A - B
(v) A - (A ∪ B) = A - B
Answer: Two sets are shown with a Venn diagram. Region 1 is shaded yellow (represents A - B). Region 2 is shaded blue (represents B - A). The overlapping region is shaded differently and labeled as region 3 (represents A ∩ B).

(i) A - B represents region 1. B represents regions (2+3). Their intersection yields a null set.
\( \implies \) (A - B) ∩ B = ϕ

(ii) B - A represents region 2. A represents regions (1+3). Their union shows regions (1+2+3), which is the union of A and B.
\( \implies \) A ∪ (B - A) = A ∪ B

(iii) A - B represents region 1. A ∩ B represents region 3. Their union displays regions (1+3), which is set A.
\( \implies \) (A - B) ∪ (A ∩ B) = A

(iv) A ∪ B represents regions (1+2+3). (A ∪ B) - B represents regions (1+2+3) - (2+3) = 1. A - B represents region 1.
\( \implies \) (A ∪ B) - B = A - B

(v) This question appears to be incorrectly stated in the source.
In simple words: Use Venn diagrams to visualize the three regions: A only, B only, and their intersection. Each proof simply shows that the left side shades the same area as the right side.

Exam Tip: Always draw a clear Venn diagram with labeled regions and show which region(s) remain after each operation to verify the identity.

 

Question 8. For any sets A and B, prove that:
(i) A ∩ B' = ϕ \( \implies \) A ⊆ B
(ii) A' ∪ B' = U \( \implies \) A ⊆ B
Answer: (i) The Venn diagram for the given condition is given below. From the diagram, set A is entirely contained within set B, making A a proper subset of B.
\( \implies \) A ⊆ B

(ii) This question is not correctly stated. If A is a proper subset of B, then A' ∪ B' will not equal U.
In simple words: When A has no elements outside of B (the first condition), then every member of A must be in B.

Exam Tip: For subset proofs, always show that every element of the first set belongs to the second set using a Venn diagram or element-by-element reasoning.

 

Exercise 1F

 

Question 1. Let A = {a, b, c, e, f} B = {c, d, e, g} and C = {b, c, f, g} be subsets of the set U = {a, b, c, d, e, f, g, h}. Find:
(i) A ∩ B
(ii) A ∪ (B ∩ C)
(iii) A - B
(iv) B - A
(v) A - (B ∩ C)
(vi) (B - C) ∪ (C - B)
Answer: (i) A ∩ B contains only the elements shared by both A and B.
A ∩ B = {c, e}

(ii) First find B ∩ C = {c, d, g}. Then combine with A.
A ∪ (B ∩ C) = {a, b, c, d, e, f, g}

(iii) A - B includes elements in A that do not appear in B.
A - B = {a, b, f}

(iv) B - A includes elements in B that do not appear in A.
B - A = {d, g}

(v) B ∩ C = {c, d, g}. Elements of A not in this intersection:
A - (B ∩ C) = {a, b, e, f}

(vi) B - C = {d, e} and C - B = {b, f}. Their union:
(B - C) ∪ (C - B) = {b, d, e, f}
In simple words: For intersection, keep only shared elements. For union, combine all elements. For difference, remove elements of the second set from the first.

Exam Tip: Write out each set clearly and use a systematic approach - find intermediate sets first (like B ∩ C) before performing the final operation.

 

Question 2. Let A = {2, 4, 6, 8, 10}, B = {4, 8, 12, 16} and C = {6, 12, 18, 24}. Using Venn diagrams, verify that:
(i) (A ∪ B) ∪ C = A ∪ (B ∪ C)
(ii) (A ∩ B) ∩ C = A ∩ (B ∩ C)
Answer: (i) LHS:
A ∪ B = {2, 4, 6, 8, 10, 12, 16}
(A ∪ B) ∪ C = {2, 4, 6, 8, 10, 12, 16, 18, 24}

RHS:
B ∪ C = {4, 6, 8, 10, 12, 16, 18, 24}
A ∪ (B ∪ C) = {2, 4, 6, 8, 10, 12, 16, 18, 24}

LHS = RHS [Verified]

(ii) LHS:
A ∩ B = {4, 8}
(A ∩ B) ∩ C = { } or ϕ

RHS:
B ∩ C = {12}
A ∩ (B ∩ C) = { }

LHS = RHS [Verified]
In simple words: The union and intersection operations group in the same way regardless of how you arrange the parentheses - this is called the associative property.

Exam Tip: When using Venn diagrams to verify, compute both sides completely before comparing them to show both equal the same final set.

 

Question 3. Let A = {a, e, I, o, u}, B = {a, d, e, o, v} and C = {e, o, t, m}. Using Venn diagrams, verify the following:
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Answer: (i) Given:
A = {a, e, I, o, u}, B = {a, d, e, o, v} and C = {e, o, t, m}
B ∩ C = {e, o}
A ∪ (B ∩ C) = {a, e, I, o, u}

RHS:
A ∪ B = {a, d, e, I, o, u, v}
A ∪ C = {a, e, I, o, u, t, m}
(A ∪ B) ∩ (A ∪ C) = {a, e, I, o, u}

LHS = RHS [Verified]

(ii) Given:
A = {a, e, I, o, u}, B = {a, d, e, o, v} and C = {e, o, t, m}
B ∪ C = {a, d, v, e, o, t, m}
A ∩ (B ∪ C) = {a, e, o}

RHS:
A ∩ B = {a, e, o}
A ∩ C = {e, o}
(A ∩ B) ∪ (A ∩ C) = {a, e, o}

LHS = RHS [Verified]
In simple words: These are the distributive laws for sets - union distributes over intersection and intersection distributes over union, just like multiplication distributes over addition with numbers.

Exam Tip: Compute the intersection or union in parentheses first, then perform the outer operation to avoid errors in ordering.

 

Question 4. Let A ⊂ B ⊂ U. Exhibit it in a Venn diagram.
Answer: Given: A ⊂ B ⊂ U, which means set A is fully contained inside set B, and set B is fully contained inside the universal set U.

The Venn diagram displays three concentric regions: the smallest circle (innermost) represents A, the medium circle surrounding it represents B, and the largest rectangle (outermost) represents U. This arrangement shows the nested relationship where every element of A belongs to B, and every element of B belongs to U.
In simple words: Draw A as a small circle inside B, and draw B inside a large rectangle representing U. This shows that everything in A is also in B, and everything in B is also in U.

Exam Tip: For subset relationships, always use concentric (nested) circles or regions - the smaller set should be drawn completely inside the larger one.

 

Question 5. Let A = {2, 3, 5, 7, 11, 13}, B = {5, 7, 9, 11, 15} be subsets of U = {2, 3, 5, 7, 9, 11, 13, 15}. Using Venn diagrams, verify that:
(i) (A ∪ B') = (A' ∩ B')
(ii) (A ∩ B)' = (A' ∪ B')
Answer: (i) A yellow region denotes the portion of A not in B (set A - B). A green region denotes the portion of B not in A (set B - A). The overlapping region represents A ∩ B, and the outer region represents the universal set U.

From the diagram, (A ∪ B') = {2, 3, 5, 7, 11, 13} (B' is the set excluding those elements in B, namely the A - B region).
A' = {9, 15}
B' = {2, 3, 13}
A' ∩ B' = { }

Therefore, (A ∪ B') ≠ (A' ∩ B') [Not verified as stated]

(ii) From the diagram, (A ∩ B)' = {2, 3, 9, 13, 15} (elements except those in A ∩ B)
A' ∪ B' = {2, 3, 9, 13, 15}

Therefore, (A ∩ B)' = (A' ∪ B') [Verified]
In simple words: For the complement of an intersection, take all elements NOT in both sets - this equals the union of all elements not in A plus all elements not in B (De Morgan's law).

Exam Tip: When using De Morgan's laws with complements, remember that the complement of an intersection becomes a union, and vice versa.

 

Question 6. Using Venn diagrams, show that (A - B), (A ∩ B) and (B - A) are disjoint sets, taking A = {2, 4, 6, 8, 10, 12} and B = {3, 6, 9, 12, 15}.
Answer: (A - B) is shown by the yellow region only (elements in A but not in B). (B - A) is shown by the blue region only (elements in B but not in A). (A ∩ B) is shown by the overlapping region (elements in both). Since these three regions do not overlap with each other, the three sets have no common elements.

A - B = {2, 4, 8, 10}
A ∩ B = {6, 12}
B - A = {3, 9, 15}

These three regions have no shared elements, so they are disjoint sets.
In simple words: Disjoint sets have no elements in common. The yellow, green, and overlapping regions of the Venn diagram do not share any area, showing the three sets are completely separate.

Exam Tip: To verify disjoint sets, check that every pair has an empty intersection - (A - B) ∩ (A ∩ B) = ϕ, (A - B) ∩ (B - A) = ϕ, and (A ∩ B) ∩ (B - A) = ϕ.

 

Exercise 1G

 

Question 1. If A and B are two sets such that n(A) = 37, n(B) = 26 and n(A ∪ B) = 51, find n(A ∩ B).
Answer: Given:
n(A) = 37
n(B) = 26
n(A ∪ B) = 51

To Find: n(A ∩ B)

We know that the formula relating these quantities is:
\( |A \cup B| = |A| + |B| - |A \cap B| \) (where A and B are two finite sets)

Therefore,
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
51 = 37 + 26 - n(A ∩ B)
n(A ∩ B) = 63 - 51 = 12

Therefore, n(A ∩ B) = 12
In simple words: The inclusion-exclusion principle tells us that the total size of two sets equals the sum of their individual sizes minus their overlap, since we counted the intersection twice.

Exam Tip: Always use the inclusion-exclusion formula directly and substitute values carefully - watch for arithmetic errors when computing the final answer.

 

Question 2. If A and B are two sets such that n(A) = 24, n(B) = 22 and n(A ∩ B) = 8, find:
(i) n(A ∪ B)
(ii) n(A - B)
(iii) n(B - A)
Answer: Given:
n(A) = 24
n(B) = 22
n(A ∩ B) = 8

To Find:
(i) n(A ∪ B)
Using the inclusion-exclusion principle:
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
n(A ∪ B) = 24 + 22 - 8 = 38

Therefore, n(A ∪ B) = 38

(ii) n(A - B)
We use the formula:
n(A - B) = n(A) - n(A ∩ B)
n(A - B) = 24 - 8 = 16

Therefore, n(A - B) = 16

(iii) n(B - A)
We use the formula:
n(B - A) = n(B) - n(A ∩ B)
n(B - A) = 22 - 8 = 14

Therefore, n(B - A) = 14
In simple words: The union includes all unique elements from both sets. Elements in A only equals A's size minus the overlap. Elements in B only equals B's size minus the overlap.

Exam Tip: Remember the three key formulas: union uses subtraction, while A - B and B - A each require subtracting the intersection from the respective set size.

 

Question 3. If A and B are two sets such that n(A - B) = 24, n(B - A) = 19 and n(A ∩ B) = 11, find:
(i) n(A)
(ii) n(B)
(iii) n(A ∪ B)
Answer: Given:
n(A - B) = 24
n(B - A) = 19
n(A ∩ B) = 11

To Find:
(i) n(A)
We know that:
n(A) = n(A - B) + n(A ∩ B)
n(A) = 24 + 11 = 35

Therefore, n(A) = 35 ...(1)

(ii) n(B)
We know that:
n(B) = n(B - A) + n(A ∩ B)
n(B) = 19 + 11 = 30

Therefore, n(B) = 30 ...(2)

(iii) n(A ∪ B)
We know that:
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
n(A ∪ B) = 35 + 30 - 11 = 54

Therefore, n(A ∪ B) = 54
In simple words: To rebuild the original set sizes, add back the intersection to each difference. The union then follows from adding the two set sizes and subtracting the overlap once.

Exam Tip: When given differences and intersections, build the original sets first, then compute the union using the standard formula.

 

Question 4. In a committee, 50 people speak Hindi, 20 speak English and 10 speak both Hindi and English. How many speak at least one of these two languages?
Answer: Given:
People who speak Hindi = 50
People who speak English = 20
People who speak both English and Hindi = 10

To Find: People who speak at least one of these two languages

Let us consider:
People who speak Hindi = n(H) = 50
People who speak English = n(E) = 20
People who speak both Hindi and English = n(H ∩ E) = 10
People who speak at least one of the two languages = n(H ∪ E)

Using the inclusion-exclusion principle:
n(H ∪ E) = n(H) + n(E) - n(H ∩ E)
n(H ∪ E) = 50 + 20 - 10 = 60

Therefore, people who speak at least one of the two languages total 60.
In simple words: Add speakers of Hindi and speakers of English, then subtract those counted twice (both languages) to get the total speaking at least one language.

Exam Tip: "At least one" always means union. Subtract the overlap to avoid double-counting people bilingual in Hindi and English.

 

Question 5. In a group of 50 persons, 30 like tea, 25 like coffee and 16 like both. How many like
(i) either tea or coffee?
(ii) neither tea nor coffee?
Answer: Given:
In a group of 50 persons:
- 30 like tea
- 25 like coffee
- 16 like both tea and coffee

To Find:
(i) People who like either tea or coffee.

Let us consider:
Total number of people = n(X) = 50
People who like tea = n(T) = 30
People who like coffee = n(C) = 25
People who like both tea and coffee = n(T ∩ C) = 16

Using the inclusion-exclusion principle:
n(T ∪ C) = n(T) + n(C) - n(T ∩ C)
n(T ∪ C) = 30 + 25 - 16 = 39

Therefore, 39 people like either tea or coffee.

(ii) People who like neither tea nor coffee.

People who like neither = Total people - People who like at least one
People who like neither = n(X) - n(T ∪ C)
People who like neither = 50 - 39 = 11

Therefore, 11 people like neither tea nor coffee.
In simple words: For "either/or," use union and apply inclusion-exclusion. For "neither," subtract those who like at least one from the total group.

Exam Tip: "Either...or" (union) is found using inclusion-exclusion. "Neither" is the complement - subtract the union from the total to find those outside both sets.

 

Question 1. In a survey, 30 people liked tea and 25 people liked coffee. If 16 people liked both tea and coffee, and the total number of people surveyed was 50, how many people liked either tea or coffee, and how many liked neither?
Answer: Given that 30 people enjoyed tea and 25 enjoyed coffee, with 16 enjoying both beverages. Using the set union formula, \( n(T \cup C) = n(T) + n(C) - n(T \cap C) = 30 + 25 - 16 = 39 \). Therefore, 39 people enjoyed either tea or coffee. To determine those who enjoyed neither, we subtract: \( 50 - 39 = 11 \). So, 11 people liked neither tea nor coffee.
In simple words: Add the tea lovers and coffee lovers, then subtract those counted twice (both). This gives people who liked at least one drink. Subtract from the total to find those who liked neither.

Exam Tip: Always use the inclusion-exclusion principle when finding unions: add the individual sets and subtract the intersection to avoid double counting.

 

Question 2. There are 200 individuals with a skin disorder. 120 had been exposed to chemical C1, 50 to chemical C2, and 30 to both the chemicals C1 and C2. Find the number of individuals exposed to (i) Chemical C1 but not chemical C2 (ii) Chemical C2 but not chemical C1 (iii) Chemical C1 or chemical C2
Answer:
(i) Using the set difference formula, the count of those exposed only to C1 is \( n(C1 - C2) = n(C1) - n(C1 \cap C2) = 120 - 30 = 90 \). Therefore, 90 individuals were exposed to chemical C1 only.
(ii) For those exposed only to C2, we calculate \( n(C2 - C1) = n(C2) - n(C1 \cap C2) = 50 - 30 = 20 \). So, 20 individuals were exposed to chemical C2 only.
(iii) For those exposed to either C1 or C2, \( n(C1 \cup C2) = n(C1) + n(C2) - n(C1 \cap C2) = 120 + 50 - 30 = 140 \). Therefore, 140 individuals were exposed to chemical C1 or C2.
In simple words: To find "only one chemical," subtract the overlap from that chemical's total. To find "either chemical," add both totals and subtract the overlap.

Exam Tip: Draw a Venn diagram with two circles - it clarifies which regions represent "only one," "both," and "at least one" instantly.

 

Question 3. In a class of a certain school, 50 students offered mathematics, 42 offered biology and 24 offered both the subjects. Find the number of students offering (i) mathematics only, (ii) biology only, (iii) any of the two subjects.
Answer:
(i) To find those studying mathematics alone, we use \( n(M - B) = n(M) - n(M \cap B) = 50 - 24 = 26 \). Thus, 26 students studied mathematics only.
(ii) For those studying biology alone, \( n(B - M) = n(B) - n(M \cap B) = 42 - 24 = 18 \). So, 18 students studied biology only.
(iii) For those taking either subject, \( n(M \cup B) = n(M) + n(B) - n(M \cap B) = 50 + 42 - 24 = 68 \). Therefore, 68 students took at least one subject.
In simple words: Subtract the shared students from each subject to get "only" counts. Add both subjects and subtract the overlap to get "at least one."

Exam Tip: Always label your Venn diagram regions clearly - this prevents mixing up "only one" with "at least one" in your calculations.

 

Question 4. In an examination, 56% of the candidates failed in English and 48% failed in science. If 18% failed in both English and science, find the percentage of those who passed in both the subjects.
Answer: We know that 56% failed English, 48% failed science, and 18% failed both. Using the inclusion-exclusion principle, the total failing at least one subject is \( n(E - S) + n(S - E) + n(E \cap S) = (56 - 18) + (48 - 18) + 18 = 38 + 30 + 18 = 86\% \). Since the total is 100%, the percentage who passed both subjects is \( 100 - 86 = 14\% \).
In simple words: Find those who failed at least one subject, then subtract from 100% to get those who passed both.

Exam Tip: "Failed both" and "passed both" are complementary outcomes - use subtraction from 100% to verify your answer quickly.

 

Question 5. In a group of 65 people, 40 like cricket and 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Answer: Given that 65 people form the union, 40 like cricket, and 10 like both sports. Using the formula \( n(C \cup T) = n(C) + n(T) - n(C \cap T) \), we substitute: \( 65 = 40 + n(T) - 10 \), which gives \( n(T) = 35 \). Therefore, 35 people like tennis. To find those liking tennis only: \( n(T - C) = n(T) - n(C \cap T) = 35 - 10 = 25 \). Thus, 25 people like tennis only.
In simple words: First use the union formula to find the total tennis lovers. Then subtract those who like both to find "tennis only."

Exam Tip: When one unknown appears in the union formula, rearrange it algebraically to solve before finding the "only" region.

 

Question 6. A school awarded 42 medals in hockey, 18 in basketball and 23 in cricket. If these medals were bagged by a total of 65 students and only 4 students got medals in all the three sports, how many students received medals in exactly two of the three sports?
Answer: The total number of medals distributed is \( 42 + 18 + 23 = 83 \). Since 4 students earned medals in all three sports, those 4 collectively received \( 4 \times 3 = 12 \) medals. This leaves \( 83 - 12 = 71 \) medals for the remaining 61 students. Since each of the 61 must have earned at least 1 medal, the surplus is \( 71 - 61 = 10 \). These 10 extra medals represent students who earned exactly 2 medals (more than 1 but less than 3). Therefore, 10 students got medals in exactly two sports.
In simple words: Count total medals, subtract what the "all three" students got, then find how many extras remain - those extras become the "exactly two" count.

Exam Tip: Break three-set problems into groups: those with all three, exactly two, exactly one. The "extra medals" technique cleanly isolates the "exactly two" group.

 

Question 7. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read the newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, and 3 read all the three newspapers. Find (i) The number of people who read at least one of the newspapers, (ii) The number of people who read exactly one newspaper.
Answer:
(i) Using the three-set inclusion-exclusion formula: \( n(H \cup T \cup I) = n(H) + n(T) + n(I) - n(H \cap I) - n(H \cap T) - n(T \cap I) + n(H \cap T \cap I) = 25 + 26 + 26 - 9 - 11 - 8 + 3 = 52 \). Therefore, 52 people read at least one newspaper.
(ii) Let p, q, r, s denote those reading H and T only, H and I only, T and I only, and all three, respectively. We have: \( p + s = 11 \), \( q + s = 9 \), \( r + s = 8 \). Adding these: \( p + q + r + 3s = 28 \), so \( p + q + r + s = 28 - 6 = 22 \). Those reading exactly one newspaper = \( 52 - 22 = 30 \). Thus, 30 people read exactly one newspaper.
In simple words: The inclusion-exclusion formula gives those reading at least one. Subtract the number reading two or more from this total to get those reading exactly one.

Exam Tip: For three sets, always expand the "at least two" regions using pairwise intersections - this avoids mixing up overlaps.

 

Question 8. In a survey of 100 students, the number of students studying the various languages is found as English only 18; English but not Hindi 23; English and Sanskrit 8; Sanskrit and Hindi 8; English 26; Sanskrit 48 and no language 24. Find (i) how many students are studying Hindi? (ii) how many students are studying English and Hindi both?
Answer:
(i) Define variables: a = English only = 18, b = Sanskrit only, c = Hindi only, d = Sanskrit and Hindi only, e = English and Sanskrit only, f = English and Hindi only, g = all three. From the constraints: \( a + e = 18 + e = \) English only given, and \( a + e + f + g = 26 \) (total English). This yields \( 18 + e + f + g = 26 \). Also, \( e + g = 8 \) (English and Sanskrit), and \( a + e = 23 \) (English but not Hindi). From \( 18 + e = 23 \), we get \( e = 5 \). From \( 5 + g = 8 \), we get \( g = 3 \). From \( 18 + 8 + f = 26 \), we get \( f = 0 \). From \( d + g = 8 \), we get \( d = 5 \). For Sanskrit total: \( b + e + d + g = 48 \), so \( b + 5 + 5 + 3 = 48 \), giving \( b = 35 \). For Hindi only: \( c = 100 - (18 + 5 + 35 + 5 + 0 + 3) - 24 = 100 - 66 - 24 = 10 \). Total studying Hindi: \( c + f + g + d = 10 + 0 + 3 + 5 = 18 \). Therefore, 18 students studied Hindi.
(ii) Those studying both English and Hindi: \( f + g = 0 + 3 = 3 \). Thus, 3 students studied both English and Hindi.
In simple words: Build up a system of equations from each constraint. Solve for one variable at a time, then use substitution to find all regions. Sum appropriately to answer each part.

Exam Tip: Sketch the Venn diagram and label each region with a variable - solving becomes systematic rather than guesswork.

 

Question 9. In a town of 10,000 families, it was found that 40% of the families buy newspaper A, 20% buy newspaper B, 10% buy newspaper C, 5% buy A and B; 3% buy B and C, and 4% buy A and C. If 2% buy all the three newspapers, find the number of families which buy (i) A only, (ii) B only, (iii) none of A, B, and C.
Answer: Convert percentages to counts: \( n(A) = 4000 \), \( n(B) = 2000 \), \( n(C) = 1000 \), \( n(A \cap B) = 500 \), \( n(B \cap C) = 300 \), \( n(A \cap C) = 400 \), \( n(A \cap B \cap C) = 200 \). Using region labels p, q, r, s, t, u, v for A only, B only, C only, A and B only, A and C only, B and C only, and all three respectively:
(i) For A only: \( v + t = 500 \) and \( v = 200 \), so \( t = 300 \). Also, \( u + v = 400 \), giving \( u = 200 \). From \( p + t + u + v = 4000 \), we get \( p = 4000 - 300 - 200 - 200 = 3300 \). Therefore, 3300 families buy newspaper A only.
(ii) For B only: \( t + v = 500 \) and \( s + v = 300 \), giving \( s = 100 \). From \( q + t + s + v = 2000 \), we get \( q = 2000 - 300 - 100 - 200 = 1400 \). Thus, 1400 families buy newspaper B only.
(iii) For none: Using inclusion-exclusion, those buying at least one = \( 4000 + 2000 + 1000 - 500 - 400 - 300 + 200 = 6000 \). Therefore, none buy any newspaper = \( 10000 - 6000 = 4000 \).
In simple words: Convert percentages to actual counts. Set up a system of equations from pairwise and triple intersections. Solve for each region step by step.

Exam Tip: In percentage-based problems, always convert to absolute numbers first - it reduces rounding errors and makes arithmetic clearer.

 

Question 1. If a set A has n elements then find the number of elements in its power set P(A).
Answer: The power set of set A refers to a collection of all subsets of A. For instance, if set A is {1, 2}, then all possible subsets of A would be {} (the empty set), {1}, {2}, {1, 2}. Therefore, the power set P(A) would be {ϕ, {1}, {2}, {1,2}}. When the number of elements in set A is n, the number of elements in the power set P(A) becomes \( 2^n \).
In simple words: The power set contains every possible combination of elements you can make from set A, including the empty set and A itself. The total count is always \( 2^n \).

Exam Tip: Remember that every set always contains two subsets - the empty set and the set itself. For a set with n elements, you always get \( 2^n \) total subsets.

 

Question 2. If A = ϕ then write P(A).
Answer: The power set of set A represents a collection of all subsets of A. Here, A = {ϕ} (the empty set). The only subset of A is the null set ϕ itself. Therefore, P(A) = {ϕ}.
In simple words: When A is empty, the only subset is emptiness itself, so the power set has just one element - the empty set.

Exam Tip: Do not confuse A = ϕ with P(A) = ϕ. When A is empty, P(A) is not empty - it contains one element, which is the empty set.

 

Question 3. If n(A) = 3 and n(B) = 5, find:
(i) The maximum number of elements in A ∪ B,
(ii) The minimum number of elements in A ∪ B.
Answer: The number of elements in set A is n(A) = 3 and the number of elements in set B is n(B) = 5. The number of elements in A ∪ B is represented by n(A ∪ B).

(i) For the count of elements in A ∪ B to reach its maximum, the two sets should have no overlap - that is, A and B must be disjoint sets. In this case, the number of elements in A ∪ B is n(A ∪ B) = n(A) + n(B). Substituting the values: n(A ∪ B) = 3 + 5, which gives n(A ∪ B) = 8. Therefore, the maximum number of elements in A ∪ B is 8.

(ii) For the count of elements in A ∪ B to be at its minimum, there must be an overlap between sets A and B so that some elements are shared. The count reaches its minimum when all elements from set A are also in set B; however, the reverse is not possible since n(A) - n(B). If all 3 elements of A exist in the intersection of A and B, then the number of elements only in B would be 2 (since n(B) = 5). Therefore, the minimum number of elements in A ∪ B = 5.
In simple words: Maximum union happens when sets do not overlap at all, so you add them: 3 + 5 = 8. Minimum union happens when the smaller set fits inside the larger one, giving you just 5 elements total.

Exam Tip: Always check: maximum union = sum of cardinalities (if disjoint); minimum union = the larger cardinality (if one is a subset of the other).

 

Question 4. If A and B are two sets such that n(A) = 8, n(B) = 11 and n(A ∪ B) = 14 then find n(A ∩ B).
Answer: Given: n(A) = 8, n(B) = 11, n(A ∪ B) = 14. Using the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B), we substitute: 14 = 8 + 11 - n(A ∩ B). Simplifying: 14 = 19 - n(A ∩ B). Rearranging: n(A ∩ B) = 19 - 14. Therefore, n(A ∩ B) = 5.
In simple words: Use the union formula rearranged to find the intersection. The overlap between the two sets has 5 elements.

Exam Tip: Always use the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B) to find any missing piece - rearrange as needed.

 

Question 5. If A and B are two sets such that n(A) = 23, n(B) = 37 and n(A - B) = 8 then find n(A ∪ B).
Answer: Given: n(A) = 23, n(B) = 37, n(A - B) = 8. Using the relationship n(A) = n(A - B) + n(A ∩ B), we get: 23 = 8 + n(A ∩ B). Solving: n(A ∩ B) = 23 - 8 = 15. Now applying the union formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B): n(A ∪ B) = 23 + 37 - 15 = 45. Therefore, n(A ∪ B) = 45.
In simple words: First find how many elements are in both sets (the overlap) using the difference. Then use that to find the total union size.

Exam Tip: When given set difference, always use it to first determine the intersection, then find the union.

 

Question 6. If A and B are two sets such that n(A) = 54, n(B) = 39 and n(B - A) = 13 then find n(A ∪ B).
Answer: Given: n(A) = 54, n(B) = 39, n(B - A) = 13. Using the relationship n(B) = n(B - A) + n(A ∩ B), we get: 39 = 13 + n(A ∩ B). Solving: n(A ∩ B) = 39 - 13 = 26. Now applying the union formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B): n(A ∪ B) = 54 + 39 - 26 = 67. Therefore, n(A ∪ B) = 67.
In simple words: Find the shared elements from the difference given for B. Then add both set sizes and subtract the overlap to get the union count.

Exam Tip: The structure of this problem mirrors Question 5 - identify the intersection first, then compute the union using the standard formula.

 

Question 7. If A ⊂ B, prove that B' ⊂ A'.
Answer: Since A ⊂ B, set A is completely inside set B. This means A ∪ B = B. Taking the complement of both sides: (A ∪ B)' = B'. Using De Morgan's law, (A ∪ B)' = A' ∩ B', so we have A' ∩ B' = B'. The equation A' ∩ B' = B' indicates that set B' is contained within set A'. Therefore, B' ⊂ A', which completes the proof.
In simple words: When one set is inside another, their complements reverse the order - the complement of the larger set sits inside the complement of the smaller set.

Exam Tip: Remember that taking complements reverses subset relationships - this is a key property of set complements and complements of unions.

 

Question 8. If A ⊂ B, show that (B' - A') = ϕ.
Answer: Since A ⊂ B, set A is completely inside set B. Therefore, A ∪ B = B. Taking the complement of both sides: (A ∪ B)' = B'. Using De Morgan's law, (A ∪ B)' = A' ∩ B', which gives us A' ∩ B' = B'. We know that B' can be decomposed as B' = (B' - A') + (A' ∩ B'). Substituting the earlier result: B' = (B' - A') + B'. Solving for (B' - A'): (B' - A') = B' - B' = ϕ. Thus (B' - A') = {ϕ} or the empty set, completing the proof.
In simple words: When A is inside B, the complement of B has no elements outside the complement of A, making their difference empty.

Exam Tip: Set difference A - B represents elements in A but not in B. When showing a difference equals the empty set, decompose the larger set to isolate the difference term.

 

Question 9. Let A = {x : x = 6n, n ∈ N} and B = {x : x = 9n, n ∈ N}, find A ∩ B.
Answer: For set A where x = 6n: when n = 1, 2, 3, 4, 5, 6..., we get x = 6, 12, 18, 24, 30, 36... Therefore, A = {6, 12, 18, 24, 30, 36...}. For set B where x = 9n: when n = 1, 2, 3, 4..., we get x = 9, 18, 27, 36... Therefore, B = {9, 18, 27, 36...}. The intersection A ∩ B contains elements that appear in both sets. The common elements are 18, 36, 54, ... These elements are all multiples of 18. Therefore, A ∩ B = {x: x = 18n, n ∈ N}.
In simple words: List out elements of each set by substituting values. Find which numbers appear in both lists. Notice they follow a pattern - they are all multiples of 18.

Exam Tip: When finding intersections of sets defined by formulas, write out the first few elements to spot the pattern, then express the intersection using the correct generating formula.

 

Question 10. If A = {5, 6, 7}, find P(A).
Answer: The power set P(A) represents a collection of all possible subsets of A. The possible subsets of A are: {ϕ} (the empty set), {5}, {6}, {7}, {5,6}, {5,7}, {6,7}, and {5, 6, 7}. Therefore, the power set P(A) = {{ϕ}, {5}, {6}, {7}, {5,6}, {5,7}, {6,7}, {5,6,7}}. This power set contains \( 2^3 = 8 \) elements in total.
In simple words: List every possible combination - single elements, pairs, the full set, and the empty set. Each combination becomes an element of the power set.

Exam Tip: For a set with n elements, the power set always has exactly \( 2^n \) subsets. Count them systematically: subsets of size 0 (just the empty set), size 1, size 2, and so on up to size n.

 

Question 11. If A = {3, {2}}, find P(A).
Answer: The power set P(A) represents a collection of all possible subsets of A. The elements of A are 3 and {2} (note that {2} is itself a set, treated as a single element). The possible subsets of A are: {ϕ} (the empty set), {3}, {{2}}, and {3, {2}}. Therefore, the power set P(A) = {{ϕ}, {3}, {{2}}, {3, {2}}}. This power set contains \( 2^2 = 4 \) elements in total.
In simple words: Set A has two elements: the number 3 and the set {2}. Form all combinations using these two elements to build the power set.

Exam Tip: Be careful when a set contains another set as an element. Count elements correctly: {2} is one element, not two. The power set counts combinations of these elements, whatever their nature.

 

Question 12. Prove that A ∩ (A ∪ B)' = ϕ
Answer: Let LHS = A ∩ (A ∪ B)'. Using De Morgan's law, (A ∪ B)' = A' ∩ B'. So: LHS = A ∩ (A' ∩ B') = (A ∩ A') ∩ (A ∩ B'). Since the intersection of any set with its complement equals the empty set, A ∩ A' = ϕ. Therefore: LHS = ϕ ∩ (A ∩ B'). The intersection of the empty set with any set yields the empty set, so: LHS = ϕ. Since LHS = ϕ and RHS = ϕ, the statement is proved: LHS = RHS.
In simple words: Start with A and the complement of (A or B). The complement removes everything in the union, so nothing from A can remain. The result is always empty.

Exam Tip: Apply De Morgan's law early, then use the fact that X ∩ X' = ϕ for any set X. These two properties together simplify the expression to the empty set.

 

Question 13. Find the symmetric difference A Δ B, when A = {1, 2, 3} and B = {3, 4, 5}.
Answer: The symmetric difference A Δ B is defined as (A - B) ∪ (B - A). First, find A - B (elements in A but not in B): A - B = {1, 2}. Next, find B - A (elements in B but not in A): B - A = {4, 5}. The symmetric difference is the union of these two sets: A Δ B = {1, 2} ∪ {4, 5} = {1, 2, 4, 5}. The symmetric difference contains elements that belong to exactly one of the two sets, not to both.
In simple words: Take elements that appear in A or B, but not in both. Remove the common element (3 in this case) and combine the remaining parts.

Exam Tip: Symmetric difference excludes the intersection. Always compute A - B and B - A separately, then combine them with union.

 

Question 14. Prove that A - B = A ∩ B'.
Answer: Let x be an arbitrary element in set A - B, so x ∈ (A - B). By definition of set difference, this means x ∈ A and x ∉ B. Since x ∉ B, it follows that x ∈ B'. Combined with x ∈ A, we can write x ∈ A and x ∈ B', which means x ∈ (A ∩ B'). Since x was arbitrary and any element in (A - B) belongs to (A ∩ B'), we have shown A - B = A ∩ B'.
In simple words: Elements in A - B are those in A but not in B. These are exactly the elements in A that are also in the complement of B.

Exam Tip: To prove set equality, show that any element in one set also belongs to the other. Element-by-element logical reasoning works best for this type of proof.

 

Question 15. If A = {x : x ∈ R, x - 5} and B = {x : x ∈ R, x > 4}, find A ∩ B.
Answer: For set A, where x takes all real values less than 5, the set contains all numbers from negative infinity to 5 (not including 5): A = (-∞, 5). For set B, where x takes all real values greater than 4, the set contains values from 4 (not including 4) extending to positive infinity: B = (4, -∞). The intersection A ∩ B contains values that satisfy both conditions: they must be less than 5 AND greater than 4. Therefore, the intersection is the interval from 4 to 5 (not including either endpoint): A ∩ B = (4, 5).
In simple words: Set A goes up to 5. Set B starts after 4. Their overlap is the stretch between 4 and 5, excluding the endpoints.

Exam Tip: When finding intersections of intervals, identify the region where both conditions hold simultaneously. Use interval notation carefully - parentheses mean the endpoint is excluded, brackets mean it is included.