Sets

Q1: Multiple Choice Type

i. \( \{x : x \in \mathbb{Z} \text{ and } x^2 – 4x = 0\} \) is equal to:

Step 1: Solve the equation: \( x^2 – 4x = 0 \)
Factor: \( x(x – 4) = 0 \)
Step 2: Equating each factor to zero:
\( x = 0 \) or \( x = 4 \)
Step 3: Both are integers ⇒ \( x \in \mathbb{Z} \)
Answer: b. {0, 4}

ii. \( \{x : x \in \mathbb{Z} \text{ and } x^2 = 9\} \) is equal to:

Step 1: Take square root of both sides:
\( x = \pm\sqrt{9} = \pm3 \)
Step 2: So, \( x = -3 \) or \( x = 3 \)
Answer: d. {-3, 3}

iii. If set A = \( \{x : x \in \mathbb{N},\ x = n – 3,\ n \in \mathbb{N} \text{ and } n < 3 \} \), then set A is:

Step 1: Values of \( n \in \mathbb{N} \) and \( n < 3 \) are: 1, 2
Step 2: Compute x = n – 3:
If n = 1, x = 1 – 3 = -2
If n = 2, x = 2 – 3 = -1
Step 3: But x ∈ ℕ is given — and ℕ contains only positive integers
Answer: b. { } (empty set)

iv. Set \( \{x : x = n^2 – 1,\ x \in \mathbb{Z},\ 2 < n \le 5 \} \) is equal to:

Step 1: Values of \( n \) satisfying \( 2 < n \le 5 \): 3, 4, 5
Step 2: Calculate each:
n = 3 ⇒ \( x = 3^2 – 1 = 9 – 1 = 8 \)
n = 4 ⇒ \( x = 4^2 – 1 = 16 – 1 = 15 \)
n = 5 ⇒ \( x = 5^2 – 1 = 25 – 1 = 24 \)
Answer: a. {8, 15, 24}

v. Set-builder form of set A = \( \left\{\frac{1}{2},\ \frac{2}{3},\ \frac{3}{4} \right\} \) is:

Step 1: Analyze pattern:
\( \frac{1}{2} = \frac{1}{1+1},\ \frac{2}{3} = \frac{2}{2+1},\ \frac{3}{4} = \frac{3}{3+1} \)
Step 2: Each term = \( \frac{n}{n+1} \) for n = 1, 2, 3
Step 3: So the set-builder form is:
\( \{x = \frac{n}{n+1} : n \in \mathbb{N},\ 1 \le n < 4 \} \)
Answer: d. \( \left\{x = \frac{n}{n+1},\ n \in \mathbb{N},\ 1 \le n < 4\right\} \)

Q2: Write the following sets in roster (Tabular) form

i. A1 = \( \{x : 2x + 3 = 11 \} \)

Step 1: Solve the equation \( 2x + 3 = 11 \)
Subtract 3: \( 2x = 8 \)
Divide by 2: \( x = 4 \)
Answer: A1 = {4}

ii. A2 = \( \{x : x^2 – 4x – 5 = 0 \} \)

Step 1: Factor the quadratic:
\( x^2 – 4x – 5 = (x – 5)(x + 1) \)
Step 2: Set each factor to 0:
\( x = 5 \), \( x = -1 \)
Answer: A2 = {-1, 5}

iii. A3 = \( \{x : x \in \mathbb{Z},\ -3 \le x < 4 \} \)

Step 1: List integers from -3 to 3 (4 not included)
Values: -3, -2, -1, 0, 1, 2, 3
Answer: A3 = {-3, -2, -1, 0, 1, 2, 3}

iv. A4 = \( \{x : x \text{ is a two-digit number and sum of digits of x is 7} \} \)

Step 1: Try all two-digit numbers from 10 to 99
List those where digits add up to 7:
16 → 1 + 6 = 7 ✔️
25 → 2 + 5 = 7 ✔️
34 → 3 + 4 = 7 ✔️
43 → 4 + 3 = 7 ✔️
52 → 5 + 2 = 7 ✔️
61 → 6 + 1 = 7 ✔️
70 → 7 + 0 = 7 ✔️
Answer: A4 = {16, 25, 34, 43, 52, 61, 70}

v. A5 = \( \{x : x = 4n,\ n \in \mathbb{W},\ n < 4 \} \)

Step 1: \( n \in \mathbb{W} \Rightarrow \) whole numbers: 0, 1, 2, 3
Step 2: Find x = 4n:
n = 0 → x = 0
n = 1 → x = 4
n = 2 → x = 8
n = 3 → x = 12
Answer: A5 = {0, 4, 8, 12}

vi. A6 = \( \{x : x = \frac{n}{n+2},\ n \in \mathbb{N},\ n > 5 \} \)

Step 1: Start with n = 6, 7, 8, 9, 10,…
Calculate first few values:
n = 6 → x = \( \frac{6}{8} = \frac{3}{4} \)
n = 7 → x = \( \frac{7}{9} \)
n = 8 → x = \( \frac{8}{10} = \frac{4}{5} \)
n = 9 → x = \( \frac{9}{11} \)
n = 10 → x = \( \frac{10}{12} = \frac{5}{6} \)
Answer: A6 = {3/4, 7/9, 4/5, 9/11, 5/6, …}

Q3: Write the following sets in set-builder (Rule Method) form

i. B1 = \( \{6, 9, 12, 15, \ldots\} \)

Step 1: Common difference = 3 ⇒ Arithmetic sequence
B1 starts at 6 and increases by 3:
General term: \( x = 3n + 3,\ n \in \mathbb{N} \)
Or simpler: \( x = 3n,\ n \in \mathbb{N},\ n \ge 2 \)
Answer: B1 = \( \{x : x = 3n,\ n \in \mathbb{N},\ n \ge 2 \} \)

ii. B2 = \( \{11, 13, 17, 19\} \)

Step 1: These are all prime numbers between 10 and 20
Answer: B2 = \( \{x : x \text{ is a prime number, } 10 < x < 20\} \)

iii. B3 = \( \left\{\frac{1}{3},\ \frac{3}{5},\ \frac{5}{7},\ \frac{7}{9},\ \frac{9}{11},\ldots \right\} \)

Step 1: Observe pattern:
Numerator: odd numbers → 1, 3, 5, 7, 9 = 2n – 1
Denominator: 2 more than numerator ⇒ 2n + 1
Step 2: General term:
\( x = \frac{2n – 1}{2n + 1},\ n \in \mathbb{N} \)
Answer: B3 = \( \left\{x : x = \frac{2n – 1}{2n + 1},\ n \in \mathbb{N} \right\} \)

iv. B4 = \( \{8, 27, 64, 125, 216\} \)

Step 1: Observe cubes:
\( 2^3 = 8,\ 3^3 = 27,\ 4^3 = 64,\ 5^3 = 125,\ 6^3 = 216 \)
Step 2: So, \( x = n^3,\ n = 2, 3, 4, 5, 6 \)
Answer: B4 = \( \{x : x = n^3,\ n \in \mathbb{N},\ 2 \le n \le 6 \} \)

v. B5 = \( \{-5, -4, -3, -2, -1\} \)

Step 1: These are consecutive negative integers
Step 2: General form: \( x \in \mathbb{Z},\ -5 \le x \le -1 \)
Answer: B5 = \( \{x : x \in \mathbb{Z},\ -5 \le x \le -1 \} \)

vi. B6 = \( \{\ldots, -6, -3, 0, 3, 6, \ldots\} \)

Step 1: Arithmetic sequence with common difference = 3
Starts from 0 and includes both negative & positive multiples of 3
Answer: B6 = \( \{x : x = 3n,\ n \in \mathbb{Z} \} \)

Q4:

i. Is \( \{1, 2, 4, 16, 64\} = \{x : x \text{ is a factor of 32} \} \)? Give reason.

Step 1: Factors of 32 = {1, 2, 4, 8, 16, 32}
Given set = {1, 2, 4, 16, 64}
Step 2: 64 is not a factor of 32 and 8, 32 are missing
Answer: No, the sets are not equal. The given set is missing 8 and 32, and has 64, which is not a factor of 32.

ii. Is \( \{x : x \text{ is a factor of 27} \} \neq \{3, 9, 27, 54\} \)? Give reason.

Step 1: Factors of 27 = {1, 3, 9, 27}
Given set = {3, 9, 27, 54}
Step 2: 54 is not a factor of 27 and 1 is missing
Answer: Yes, the sets are not equal. 54 is not a factor of 27 and 1 is missing.

iii. Write the set of even factors of 124.

Step 1: Factors of 124 = {1, 2, 4, 31, 62, 124}
Even factors = those divisible by 2
Answer: {2, 4, 62, 124}

iv. Write the set of odd factors of 72.

Step 1: Factors of 72 = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72}
Odd factors = {1, 3, 9}
Answer: {1, 3, 9}

v. Write the set of prime factors of 3234.

Step 1: Factorize 3234:
3234 ÷ 2 = 1617 (Even)
1617 ÷ 3 = 539
539 ÷ 7 = 77
77 = 7 × 11
So prime factors = 2, 3, 7, 11
Answer: {2, 3, 7, 11}

vi. Is \( \{x : x^2 – 7x + 12 = 0\} = \{3, 4\} \)?

Step 1: Factor: \( x^2 – 7x + 12 = (x – 3)(x – 4) \)
Roots: x = 3, 4
Answer: Yes, both sets are equal since 3 and 4 satisfy the equation.

vii. Is \( \{x : x^2 – 5x – 6 = 0\} = \{2, 3\} \)?

Step 1: Factor: \( x^2 – 5x – 6 = (x – 6)(x + 1) \)
Roots: x = 6, -1
Answer: No, the roots are 6 and -1, so the given set {2, 3} is incorrect.

Q5: Write the following sets in Roster form

i. The set of letters in the word ‘MEERUT’

Step 1: Identify distinct letters from the word MEERUT
Letters = M, E, E, R, U, T → Remove repetition
Answer: {M, E, R, U, T}

ii. The set of letters in the word ‘UNIVERSAL’

Step 1: Extract all letters: U, N, I, V, E, R, S, A, L
All letters are distinct
Answer: {U, N, I, V, E, R, S, A, L}

iii. A = \( \{x : x = y + 3,\ y \in \mathbb{N},\ y > 3 \} \)

Step 1: y ∈ N and y > 3 ⇒ y = 4, 5, 6, 7,…
x = y + 3 = 7, 8, 9, 10,…
Answer: A = {7, 8, 9, 10, 11, …}

iv. B = \( \{p : p \in \mathbb{W},\ p^2 < 20 \} \)

Step 1: Whole numbers (W) = 0, 1, 2, 3, 4…
Check p² < 20:
0² = 0
1² = 1
2² = 4
3² = 9
4² = 16
5² = 25 ❌ → stop
Answer: B = {0, 1, 2, 3, 4}

v. C = \( \{x : x \text{ is a composite number and } 5 \le x \le 21 \} \)

Step 1: Composite numbers between 5 and 21:
Composite = not prime, and > 1
Values: 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21
Answer: C = {6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21}

Q6: List the elements of the following sets

i. \( \{x : x^2 – 2x – 3 = 0\} \)

Step 1: Factor the quadratic:
\( x^2 – 2x – 3 = (x – 3)(x + 1) \)
Set each factor = 0 ⇒ x = 3 or x = -1
Answer: {-1, 3}

ii. \( \{x : x = 2y + 5;\ y \in \mathbb{N} \text{ and} \ 2 \le y < 6 \} \)

Step 1: Plug values of y = 2, 3, 4, 5
y = 2 ⇒ x = 2×2 + 5 = 9
y = 3 ⇒ x = 11
y = 4 ⇒ x = 13
y = 5 ⇒ x = 15
Answer: {9, 11, 13, 15}

iii. \( \{x : x \text{ is a factor of 24} \} \)

Step 1: Find all positive integers dividing 24
Factors = {1, 2, 3, 4, 6, 8, 12, 24}
Answer: {1, 2, 3, 4, 6, 8, 12, 24}

iv. \( \{x : x \in \mathbb{Z},\ x^2 \le 4 \} \)

Step 1: Find integers x such that x² ≤ 4
x² = 0 ⇒ x = 0
x² = 1 ⇒ x = ±1
x² = 4 ⇒ x = ±2
Only values: x ∈ {–2, –1, 0, 1, 2}
Answer: {-2, -1, 0, 1, 2}

v. \( \{x : 3x – 2 \le 10,\ x \in \mathbb{N} \} \)

Step 1: Solve the inequality:
3x – 2 ≤ 10 ⇒ 3x ≤ 12 ⇒ x ≤ 4
Since x ∈ N, x = 1, 2, 3, 4
Answer: {1, 2, 3, 4}

vi. \( \{x : 4 – 2x > -6,\ x \in \mathbb{Z} \} \)

Step 1: Solve the inequality:
4 – 2x > -6
Subtract 4: -2x > -10
Divide: x < 5
So x ∈ Z and x < 5
Answer: {…, -3, -2, -1, 0, 1, 2, 3, 4}

Q7: State which of the following sets are finite and which are infinite

i. A = \( \{x : x \in \mathbb{Z} \text{ and} \ x < 10 \} \)

Step 1: x belongs to integers and x < 10
This means x can be …, -3, -2, -1, 0, 1, …, 9 (infinite on the left side)
Answer: Infinite set

ii. B = \( \{x : x \in \mathbb{W} \text{ and} \ 5x – 3 \le 20 \} \)

Step 1: Solve the inequality:
5x – 3 ≤ 20 ⇒ 5x ≤ 23 ⇒ x ≤ 4.6
x ∈ W ⇒ x = 0, 1, 2, 3, 4
Answer: Finite set

iii. P = \( \{y : y = 3x – 2,\ x \in \mathbb{N} \text{ and} \ x > 5 \} \)

Step 1: x ∈ N and x > 5 ⇒ x = 6, 7, 8, 9, … (infinite values)
Each x gives a new y: y = 3x – 2 ⇒ Infinite y values
Answer: Infinite set

iv. M = \( \{x : x = \frac{3}{n},\ n \in \mathbb{W} \text{ and} \ 6 < n \le 15 \} \)

Step 1: n ∈ W and 6 < n ≤ 15 ⇒ n = 7 to 15
n = 7, 8, 9, …, 15 ⇒ 9 values
So x = 3/7, 3/8, 3/9, …, 3/15
Answer: Finite set

Q8: Find which of the following sets are singleton sets

i. The set of points of intersection of two non-parallel straight lines on the same plane.

Step 1: Two non-parallel straight lines in the same plane intersect at only one point.
This means the set contains exactly one element.
Answer: Singleton set

ii. A = \( \{x : 7x – 3 = 11 \} \)

Step 1: Solve the equation:
7x – 3 = 11 ⇒ 7x = 14 ⇒ x = 2
Only one solution exists ⇒ Set = {2}
Answer: Singleton set

iii. B = \( \{y : 2y + 1 < 3 \text{ and} \ y \in \mathbb{W} \} \)

Step 1: Solve the inequality:
2y + 1 < 3 ⇒ 2y < 2 ⇒ y < 1
y ∈ W ⇒ Whole numbers = {0, 1, 2, …}
Only y = 0 satisfies the condition
Answer: Singleton set

Q9: Find which of the following sets are empty

i. The set of points of intersection of two parallel lines

Step 1: Two parallel lines never meet.
So, there is no point of intersection.
Answer: Empty set

ii. A = \( \{x : x \in \mathbb{N} \text{ and} \ 5 < x \le 6 \} \)

Step 1: x must be a natural number > 5 and ≤ 6
Only possible x = 6 ⇒ x = 6 ∈ N ✅
Answer: Not an empty set

iii. B = \( \{x : x^2 + 4 = 0 \text{ and} \ x \in \mathbb{N} \} \)

Step 1: The expression is incomplete. Assuming the intention was a condition like “x² + 4 = 0” or “x² + 4 < 0” — but these have no natural number solutions.
Let’s assume the correct condition was:
→ If \( x^2 + 4 = 0 \) ⇒ \( x^2 = -4 \) ❌ Not possible in real numbers
Answer: Empty set

iv. C = {even numbers between 6 and 10}

Step 1: Even numbers between 6 and 10 = 8 only
Answer: Not an empty set

v. D = {prime numbers between 7 and 11}

Step 1: Primes between 7 and 11 (excluding 7 and 11) =
Check numbers: 8 ❌, 9 ❌, 10 ❌ → No primes
Answer: Empty set

Q10:

i. Are the sets A = {4, 5, 6} and B = \( \{x : x^2 – 5x – 6 = 0\} \) disjoint?

Step 1: Solve the quadratic equation:
\( x^2 – 5x – 6 = 0 \)
Factor: \( (x – 6)(x + 1) = 0 \)
⇒ x = 6, -1
So, B = {6, -1}
Step 2: Set A = {4, 5, 6}
Set B = {6, -1}
Common element = 6
Answer: No, the sets are not disjoint. They are joint sets because they have common element 6.

ii. Are the sets A = {b, c, d, e} and B = {x : x is a letter in the word ‘MASTER’} joint?

Step 1: Set A = {b, c, d, e}
Letters in ‘MASTER’ = M, A, S, T, E, R → Set B = {m, a, s, t, e, r}
Step 2: Common elements?
Check A ∩ B = {e} ✅
Answer: Yes, the sets are joint because they share common element ‘e’.

Q11: State whether the following pairs of sets are equivalent or not

i. A = \( \{x : x \in \mathbb{N} \text{ and} \ 11 \ge 2x – 1 \} \) and B = \( \{y : y \in \mathbb{W} \text{ and} \ 3 \le y \le 9 \} \)

Step 1: Solve for A:
\( 2x – 1 \le 11 \Rightarrow 2x \le 12 \Rightarrow x \le 6 \)
Since x ∈ N ⇒ x = 1 to 6
⇒ A = {1, 2, 3, 4, 5, 6} → 6 elements
Step 2: Set B = {3, 4, 5, 6, 7, 8, 9} → 7 elements
Answer: Not equivalent (A has 6 elements, B has 7)

ii. Set of integers and set of natural numbers

Step 1: Integers (Z) = {…, -3, -2, -1, 0, 1, 2, …} → Infinite
Natural numbers (N) = {1, 2, 3, 4, …} → Infinite
But both have infinite elements, still not equivalent because their elements differ in nature.
Answer: Not equivalent

iii. Set of whole numbers and set of multiples of 3

Step 1: Whole numbers (W) = {0, 1, 2, 3, …} → Infinite
Multiples of 3 = {0, 3, 6, 9, …} → Infinite
Both infinite, but not same kind of elements
Answer: Not equivalent

iv. P = {5, 6, 7, 8} and M = \( \{x : x \in \mathbb{W} \text{ and} \ x \le 4 \} \)

Step 1: M = {0, 1, 2, 3, 4} → 5 elements
P = {5, 6, 7, 8} → 4 elements
Answer: Not equivalent

Q12: State whether the following pairs of sets are equal or not

i. A = {2, 4, 6, 8} and B = \( \{2n : n \in \mathbb{N} \text{ and} \ n < 5\} \)

Step 1: Values of n ∈ N and n < 5 ⇒ n = 1, 2, 3, 4
Then: 2n = 2, 4, 6, 8 ⇒ B = {2, 4, 6, 8}
Set A = {2, 4, 6, 8}
Answer: Equal sets

ii. M = \( \{x : x \in \mathbb{W} \text{ and} \ x + 3 < 8 \} \) and N = \( \{y : y = 2n – 1,\ n \in \mathbb{N} \text{ and} \ n < 5 \} \)

Step 1: M ⇒ x + 3 < 8 ⇒ x < 5 ⇒ x ∈ W ⇒ x = 0, 1, 2, 3, 4
So, M = {0, 1, 2, 3, 4}
Step 2: N ⇒ n ∈ N and n < 5 ⇒ n = 1, 2, 3, 4
Then y = 2n – 1 = 1, 3, 5, 7
So, N = {1, 3, 5, 7}
Answer: Not equal

iii. E = \( \{x : x^2 + 8x – 9 = 0 \} \) and F = {1, -9}

Step 1: Solve the quadratic:
x² + 8x – 9 = 0 Use quadratic formula:
\( x = \frac{-8 \pm \sqrt{(8)^2 + 4(1)(9)}}{2(1)} = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2} \)
⇒ x = 1, -9
So, E = {1, -9} = F
Answer: Equal sets

iv. A = \( \{x : x \in \mathbb{N},\ x < 3 \} \) and B = \( \{y : y^2 – 3y + 2 = 0 \} \)

Step 1: A ⇒ x ∈ N and x < 3 ⇒ A = {1, 2}
Step 2: Solve: \( y^2 – 3y + 2 = 0 \) ⇒ (y – 1)(y – 2) = 0 ⇒ y = 1, 2 ⇒ B = {1, 2}
Answer: Equal sets

Q13: State whether each of the following sets is a finite set or an infinite set

i. The set of multiples of 8

Step 1: Multiples of 8 = {8, 16, 24, 32, …} continue forever.
Answer: Infinite set

ii. The set of integers less than 10

Step 1: Integers less than 10 = {…, -3, -2, -1, 0, 1, …, 9}
There is no lower bound ⇒ Infinite
Answer: Infinite set

iii. The set of whole numbers less than 12

Step 1: Whole numbers = {0, 1, 2, …, 11} ⇒ Countable
12 not included
Answer: Finite set

iv. \( \{x : x = 3n – 2,\ n \in \mathbb{W},\ n \le 8 \} \)

Step 1: W = {0, 1, 2, …, 8}
Apply formula for each n:
n = 0 ⇒ x = -2
n = 1 ⇒ x = 1
n = 2 ⇒ x = 4
…
n = 8 ⇒ x = 22
Only 9 values possible
Answer: Finite set

v. \( \{x : x = 3n – 2,\ n \in \mathbb{Z},\ n \le 8 \} \)

Step 1: n ∈ Z and n ≤ 8 ⇒ {…, -3, -2, …, 7, 8}
Infinite values of n possible
Answer: Infinite set

vi. \( \{x : x = \frac{n – 2}{n + 1},\ n \in \mathbb{W} \} \)

Step 1: n ∈ W = {0, 1, 2, 3, …} infinite
Each value gives different fraction:
n = 0 ⇒ x = -2
n = 1 ⇒ x = -0.33
n = 2 ⇒ x = 0
n = 3 ⇒ x = 0.25
… continues endlessly
Answer: Infinite set

Q14: Answer whether the following statements are true or false. Give reasons.

i. The set of even natural numbers less than 21 and the set of odd natural numbers less than 21 are equivalent sets.

Step 1: Even natural numbers less than 21 = {2, 4, 6, …, 20} ⇒ 10 elements
Odd natural numbers less than 21 = {1, 3, 5, …, 19} ⇒ 10 elements
Answer: True

ii. If E = {factors of 16} and F = {factors of 20}, then E = F.

Step 1: Factors of 16 = {1, 2, 4, 8, 16}
Factors of 20 = {1, 2, 4, 5, 10, 20}
Clearly different elements
Answer: False

iii. If set A = {Integers less than 20} is a finite set.

Step 1: Integers less than 20 = {…, -3, -2, -1, 0, 1, 2, …, 19} → Infinite below
Answer: False

iv. If A = {x : x is an even prime number}, then set A is empty.

Step 1: 2 is an even prime number ⇒ A = {2}
Answer: False

v. The set of odd prime numbers is the empty set.

Step 1: Odd primes = {3, 5, 7, 11, …} infinite
Answer: False

vi. The set of squares of integers and the set of whole numbers are equal sets.

Step 1: Squares of integers = {0²=0, 1²=1, 2²=4, 3²=9, …}
Whole numbers = {0, 1, 2, 3, 4, 5, 6, …}
Clearly, 2 ∈ whole numbers but not in squares of integers
Answer: False