Referring to the AP Inter 1st Year Maths Study Material Chapter 1 Sets Exercise 1c Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Sets Solutions Exercise 1c
Question 1.
Make correct statements by filling the symbol a or <z in the blank spaces,
i) {2, 3, 4 }…{1, 2, 3, 4, 5}.
Solution:
⊂
ii) {a, b, c }…{b, c, d}.
Solution:
⊄
iii) {x : x is a student of Class XI of your school}…{x : x is a student of your school}.
Solution:
⊂
iv) {r : x is a circle in the plane} … {x : r is a circle in the same plane with radius 1 unit}.
Solution:
⊄
v) {x : x is a triangle in a plane} … {x : x is a rectangle in the plane}.
Solution:
⊄
vi) (x : x is an equilateral triangle in a plane}…{x : x is a triangle in the same plane}
Solution:
⊂
vii) {x : x is an even natural number} . . . {x : x is an integer}.
Solution:
⊂
Question 2.
Examine whether the following statements are true or false.
i) ( a, b } ⊄ { b, c, a }.
ii) { a, e } ⊂ { x : x is a vowel in the english alphabet}.
iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }.
iv) { a } ⊂ { a, b, c }.
v) { a } ∈ { a, b, c }.
vi) { x : x is an even natural number less than 6} ⊂ { x : x is a natural number which divides 36}.
Solution:
i) False, each element of {a, b} is also an element of {b, c, a}.
ii) {a, e} ⊂ {x : x is a vowel in the english alphabet} is true.
iii) {1, 2, 3} ⊂ {1, 3, 5} is false because 2 ∉ {1, 3, 5}.
iv) True because each element of {a} is also an element of {a, b, c}.
v) False the elements of {a, b, c} are a, b, c therefore {a} ⊂ {a, b, c}.
vi) True {2, 4} ⊂ {1, 2, 3, 4, 6, 9, 12, 18, 36}.
Question 3.
Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?
i) {3, 4} ⊂ A
Solution:
The statement {3, 4} ⊂ A is incorrect because 3e {3, 4} but 3 ∈ A.
ii) {3, 4} ∈ A
Solution:
The statement {3, 4} ∈ A is correct because {3, 4} is an element of A.
iii) {{3, 4}} ⊂ A
Solution:
The statement {{3, 4}} ⊂ A is correct because {3, 4} ∈ {{3, 4}} ∈ A.
iv) 1 ∈ A
Solution:
The statement 1 ∈ A is correct because 1 is an element of A.
v) 1 ⊂ A
Solution:
The statement 1 ⊂ A is incorrect because an element of a set can never be a subset of itself.
vi) {1, 2, 5} ⊂ A
Solution:
The statement {1,2, 5} ⊂ A is correct because each element of {1, 2, 5} is also an element of A.
vii) {1, 2, 5} ∈ A
Solution:
The statement {1, 2, 5} ∈ A is incorrect because {1, 2, 5} is not an element of A.
viii) {1, 2, 3} ⊂ A
Solution:
The statement {1, 2, 3} ⊂ A is incorrect because 3 ∈ {1, 2, 3} but 3 ∉ A.
ix) Φ ∈ A
Solution:
The statement Φ ∈ A is incorrect because Φ is not an element of A.
x) Φ ⊂ A
Solution:
The statement Φ ⊂ A is correct because Φ is a subset of every set.
xi) {Φ} ⊂ A
Solution:
The statement {Φ} ⊂ A is incorrect because Φ ∈ {Φ} but Φ ∉ A.
Question 4.
Write down all the subsets of the following sets.
i) {a}
Solution:
The subsets of { a} are Φ and {a}.
ii) {a, b}
Solution:
The subsets of {a, b} are Φ, { a}, { b } & {a, b}.
iii) {1, 2, 3}
Solution:
The subsets of {1, 2, 3} are (j) , {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3} and {1, 2, 3}.
iv) Φ
Solution:
The subset of Φ is Φ .
Question 5.
Write the following as intervals.
(i) (x : x ∈ R, – 4 < x < 6}.
(ii) {x : x ∈ R, – 12 < x < -10}.
(iii) {x : x ∈ R, 0 < x < 7}.
(iv) {x : x ∈ R, 3 < x < 4}.
Solution:
i) {x : x ∈ R, -4 < x < 6} = (-4, 6]
ii) {x : x ∈ R, -12 < x < – 10} = (-12, -10)
iii)(x : x ∈ R, 0 < x < 7} = [0, 7).
iv) {x : x ∈ R, 3 < x < 4} = [3, 4],
Question 6.
Write the following intervals in set-builder form.
(i) (- 3, 0)
(ii) [6,12]
(iii) (6, 12]
(iv) [-23, 5)
Solution:
i) (-3, 0) = {x : x ∈ R and -3 < x < 0 }.
ii) [6, 12] = {x : x ∈ R and 6 < x < 12 }.
iii) (6, 12] = {x : x ∈ R and 6 < x < 12}.
iv) [-23, 5) = {x : x ∈ R and -23 < x < 5}.
Question 7.
What universal set(s) would you propose for each of the following?
i) The set of right triangles.
ii) The set of isosceles triangles.
Solution:
i) The universal set for the set of right triangles is the set of all triangles.
ii) The universal set for the set of isosceles triangles is the set of all polygons.
Question 8.
Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set (s) for all the three sets A, B and C.
i) (0, 1, 2, 3, 4, 5, 6).
ii) Φ
iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
iv) {1, 2, 3, 4, 5, 6, 7, 8}.
Solution:
i) A ⊂ {0, 1, 2, 3, 4, 5, 6}, B ⊂ {0, 1, 2, 3, 4, 5, 6}, C ⊄ {0, 1, 2, 3, 4, 5, 6}
∴ {0, 1, 2, 3, 4, 5, 6} cannot be the universal set of the sets A, B, & C.
ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ
∴ Φ cannot be the universal set of the sets A, B & C.
iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
∴ The set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets A, B & C.
iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8};
B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}
The set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B & C.