Sets MCQ AP Inter 1st Year Maths Chapter 1

Practice AP Inter 1st Year Maths Study Material Chapter 1 Sets MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Sets MCQ

Question 1.
List of elements of the set {x : x is an integer, x³ ≤ 50}
1) {2, 3, 4}
2) {1, 3, 4}
3) {1, 2, 3}
4) {1, 2, 4}
Solution:
We look integer x such that x³ ≤ 50
Cube of integers : 1³ = 1, 2³ = 8, 3³ = 27, 4³ = 64 (exceeds 50).
So the integers are : x = 1, 2, 3.

Question 2.
Set-builder form of {1, 2, 3, 4} is
1) {x : x is a positive integer, x² < 30}.
2) {x : x is a positive integer, x² + 1 < 20}.
3) {x : x is a positive integer, x² – 1 < 15}.
4) {x : x is a positive integer, 2x < 20}.
Try each :
1) x² < 30 : includes more than {1, 2, 3, 4}.
2) x² + 1 < 20 : x < 4 works, lets check : 1 → 2, 2 → 5, 3 → 10, 4 → 17.
3) x² – 1 < 15: 1 → 0, 2 3, 3 → 8, 4 → 15 but it includes more value like 5.
4) 2x < 20 : includes more values.

Question 3.
Which of the following statement is true ?
1) {a, b} ⊂ {x : x is a vowel in the English alphabet}
2) {1, 2, 3, 4} ⊂ {1, 2, 3}
3) {5} ∈ {4, 5, 6}
4) {2, 3) ∈ {1, {2, 3}, 4, 5}
Solution:
1) X “b” is not a vowel
2) X false
3) X because {5} ∈ {4, 5, 6} but {5} is not an element
4) ✓ true.

Question 4.
The set builder form of interval [-4, 5) is
1) {x : x is an integer -4 < x < 5}
2) {x : x is an integer -4 < x < 5}
3) {x : x is an integer -4 < x < 5}
4) {x : x is an integer -4 < x < 5}
Solution:
Set builder form of interval [- 4, 5)
We need : integers x such that – 4 < x < 5.

Question 5.
A = (1, 2, 3, 4, 5), B = (4, 5, 6, 7, 8), then (A – B) ∪ (B – A)
1) {1, 2, 3, 4, 5, 6, 7, 8}
2) {1, 2, 3, 6, 7, 8}
3) {1, 2, 4, 5, 6}
4) {2, 4, 6, 8}
Solution:
A – B = {1, 2, 3}; B – A = {6, 7, 8}; (A – B) ∪ (B – A) = {1, 2, 3, 6, 7, 8}.

Question 6.
If A = (1, 2, 3, 4}, B = {2, 3, 4}, C = {3, 4, 5, 6} then A ∩ (B ∪ C)
1) {2, 3, 4}
2) {1, 2, 3, 4}
3) {3, 4, 5, 6}
4) {1, 6}
Solution:
B ∪ C = {2, 3, 4, 5, 6}; A ∩ (B ∪ C) = {2, 3, 4}.

Question 7.
Which of the following is false ?
1) (A’)’ = A
2) A ∪ A’ = Φ
3) (A ∪ B)’ = A’ ∩ B’
4) Φ = U (U is universal set)
Solution:
1) (A’)’ = A – True (law of double complement)
2) A ∪ A’ = Φ – False, should be universal set.
3) (A ∪ B)’ = A’ ∩ B’- True (De Morgan’s law)
4) f = U – True

Question 8.
If U = (1, 2, 3, 4, 5) and A = (2, 3), then Φ’ ∩ A =
1) Φ
2) U
3) A
4) {1, 4, 5}
Solution:
Φ = U, so Φ ‘ ∩ A = U ∩ A = A.

Question 9.
U = {x : x is a natural number 1 < x < 15} and A = 1 < x < 15) then A’ = ____________
1) {2, 3, 5, 7, 11, 13}
2) {4, 6, 8, 9, 10, 12, 14}
3) {2, 3, 4, 5, 7, 11, 13}
4) {2, 4, 6, 8, 10, 12, 14}
Solution:
Given : U = {x : x is a natural number and 1 < x < 15}
U = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
A = {x : x is a prime number and 1 < x < 15}
A = {2, 3, 5, 7, 11, 13}
So, A’ = U – A = {4, 6, 8, 9, 10, 12, 14}.

Question 10.
Which of the following is not a subset of (a, b, c, d} ?
1) Φ
2) {a, b}
3) {d, e}
4) {b, d}
Solution:
Let’s analyse :
1) Φ – ✓ Always a subset of any set.
2) {a, b} – ✓ All elements are present in the set.
3) {d, e} – X “e” is not in {a, b, c, d}.
4) {b, d} – ✓ Both elements are present in the set.
∴ {d, e} is not a subset.