Problems & Solutions : Sets : Exercise 1(c) : Intermediate
Exercise 1(c) : Problems & Solutions
1. Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:
(i) { 2, 3, 4 } ... { 1, 2, 3, 4, 5 }
Solution :
Given sets { 2, 3, 4 } and { 1, 2, 3, 4, 5 }
Since the set { 1, 2, 3, 4, 5 } containing all the elements 2, 3, 4 , we have
{ 2, 3, 4 } ⊂ { 1, 2, 3, 4, 5 }
(ii) { a, b, c } ... { b, c, d }
Solution :
Given sets are { a, b, c } & { b, c, d }.
Since ' a ' is not in the set { b, c, d }, we have
{ a, b, c } ⊄ { b, c, d }
(iii) {x : x is a student of Class XI of your school} ... {x : x student of your school}
Solutiion :
Given sets are {x : x is a student of Class XI of your school} and {x : x student of your school}
Since the students of Class XI are the students of the their school , we have
{x : x is a student of Class XI of your school} ⊂ {x : x student of your school}
(iv) {x : x is a circle in the plane} ... {x : x is a circle in the same plane with radius 1 unit}
Solution :
Given sets are {x : x is a circle in the plane} &
{x : x is a circle in the same plane with radius 1 unit}
Since a circle in a plane is of any radius , need not be radius 1 , we have
{x : x is a circle in the plane} ⊄ {x : x is a circle in the same plane with radius 1 unit}.
(v) {x : x is a triangle in a plane} ... {x : x is a rectangle in the plane}
Solution :
Given sets are {x : x is a triangle in a plane} & {x : x is a rectangle in the plane}.
Since a rectangle is different from a triangle, we have
{x : x is a triangle in a plane} ⊄ {x : x is a rectangle in the plane}
(vi) {x : x is an equilateral triangle in a plane} ... {x : x is a triangle in the same plane}
Solution :
Given sets are {x : x is an equilateral triangle in a plane} & {x : x is a triangle in the same plane}.
Since an equilateral triangle is a triangle & the set {x : x is a triangle in the same plane} having all types of triangles , we have
{x : x is an equilateral triangle in a plane} ⊂ {x : x is a triangle in the same plane}.
(vii) {x : x is an even natural number} ... {x : x is an integer}
Solution :
Given sets are {x : x is an even natural number} & {x : x is an integer}
Since each even natural number is an integer, we have
{x : x is an even natural number} ⊂ {x : x is an integer}
2. Examine whether the following statements are true or false:
(i) { a, b } ⊄ { b, c, a }
Solution :
Given statement is { a, b } ⊄ { b, c, a }.
Since a and b are in the set { b, c,a,}, the given statement is FALSE
(ii) { a, e } ⊂ {x : x is a vowel in the English alphabet}
Solution :
Given { a, e } ⊂ {x : x is a vowel in the English alphabet}.
Since a and e are vowels , the given statement is TRUE.
(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }
Solution :
Given statement is { 1, 2, 3 } ⊂ { 1, 3, 5 }.
Since ' 2 ' is not in the set { 1 , 3 , 5 }, the given statement is FALSE.
(iv) { a } ⊂ { a, b, c }
Solution :
Given statement is { a } ⊂ { a, b, c }.
Since ' a ' is in the set { a , b , c } , the given statement is TRUE.
(v) { a } ∈ { a, b, c }
Solution :
Given statement is { a } ∈ { a, b, c }.
Since { a } is not an element of { a , b , c } , the given statement is FALSE.
(vi) {x : x is an even natural number less than 6} ⊂ {x : x is a natural number which divides 36}
Solution :
Given statement is
{x : x is an even natural number less than 6} ⊂ {x : x is a natural number which divides 36}.
The even natural numbers less than 6 are 2 , 4.
Since 36 = 2 x 18 and 36 = 4 x 9, we have 2 and 4 are divisors of 36.
∴ The given statement is TRUE.
3. Let (A={1,2,{3,4},5}). Which of the following statements are incorrect and why?
(i) {3,4} ⊂ A.
Solution :
Given {3,4} ⊂ A.
Since { 3 , 4 } is an element of A , not a subset of A.
Hence { 3 , 4 } 𝞊 A.
∴ The given statement is INCORRECT.
(ii) {3,4} 𝞊 A.
Solution :
Given {3,4} 𝞊 A.
Since { 3 , 4 } is an element of A, the statement is CORRECT.
(iii) {{3,4}} ⊂ A
Solution :
Given {{3,4}} ⊂ A.
Since { { 3 , 4 } } containing { 3 , 4 } , and { 3 , 4 } is in A,
the given statement is CORRECT.
(iv) 1 𝞊 A.
Solution :
Given 1 𝞊 A.
Since A is containing ' 1 ' , the given statement is CORRECT.
(v) 1 ⊂ A
Solution :
Given 1 ⊂ A
Since 1 is not a set, the given statement is INCORRECT.
(vi) {1,2,5} ⊂ A
Solution :
Given {1,2,5} ⊂ A.
Since { 1 , 2 , 5 } having the numbers 1 , 2 , 5 and are lies in A,
the given statement is CORRECT.
(vii) {1,2,5} 𝞊 A
Solution :
Given {1,2,5} 𝞊 A
Since {1,2,5} is a set, not an element, the given statement is INCORRECT.
(viii) {1,2,3} ⊂ A
Solution :
Given {1,2,3} ⊂ A.
Since 3 is not separately in A but in { 1 , 2 , 3 } the given statement is INCORRECT.
(ix) ⲫ 𝞊 A
Solution :
Given ⲫ 𝞊 A.
Since the empty set ⲫ is not an element, the given statement is INCORRECT.
(x) ⲫ ⊂ A
Solution :
Given ⲫ ⊂ A.
Since the empty set ⲫ is subset to every set, the given statement is CORRECT.
(xi) { ⲫ } ⊂ A
Solution :
Given { ⲫ } ⊂ A.
Given statement is true when ⲫ 𝞊 A.
Since A does not containing ⲫ, the given statement is INCORRECT.
4. Write down all the subsets of the following sets:
(i) {a}
Solution :
Given set is {a}.
Since ⲫ is a subset to each set , the subsets of the given set are ⲫ and { a } .
(ii) {a,b}
Solution :
Given set is { a, b }.
All the subsets of the given set are ⲫ, {a} , { b } , { a , b }
(iii) {1,2,3}
Solution :
Given set is { 1 , 2 , 3 }
All the subsets of the given set are ⲫ, { 1 } , { 2 } , { 3 } , { 1 , 2 } , { 1 , 3 } , { 2 , 3 } and
{ 1 , 2 , 3 }.
(iv) ⲫ
Solution :
Given set is the empty set ⲫ.
Since the empty set does not contains any elements, the only subset of ⲫ is ⲫ itself.
5. Write the following as intervals:
(i) {x: x 𝞊 R ,-4 < x ≤ 6}
Solution :
Given {x: x 𝞊 R ,-4 < x ≤ 6}
= ( -4 , 6 ]
(ii) {x: x 𝞊 R, -12 < x < -10}
Solution :
Given {x: x 𝞊 R, -12 < x < -10}
= ( -12 , -10 )
(iii) {x: x 𝞊R,0 ≤ x <7 }
Solution :
Given {x: x 𝞊R,0 ≤ x <7 }
= [ 0 , 7 )
(iv) {x: x 𝞊 R, 3 ≤ x ≤ 4 }
Solution :
Given {x: x 𝞊 R, 3 ≤ x ≤ 4 }
= [ 3 , 4 ]
6. Write the following intervals in set-builder form:
(i) (-3,0)
Solution :
Given ( -3 , 0 )
= {x: x 𝞊 R, -3 < x < 0 } | Since -3 and 0 are not in the set |
(ii) [6,12]
Solution :
Given [ 6 , 12 ]
={x: x 𝞊 R, 6 ≤ x ≤ 12 } | Since 6 and12 are in the set |
(iii) (6,12]
Solution :
Given ( 6 , 12 ]
= {x: x 𝞊 R, 6 < x ≤ 12 } | Since 6 is not in the set and 12 is in the set |
(iv) [-23,5)
Solution :
Given [ -23 , 5 )
= {x: x 𝞊 R, -23 ≤ x < 5 } | Since -23 is in the set and 5 is not in the set |
7. What universal set(s) would you propose for each of the following:
(i) The set of right triangles.
Solution :
Given set is the set of right angled triangles
Since right angled triangle is one type of triangle , the set of all triangles is the
universal set.
(ii) The set of isosceles triangles.
Solution :
Given set is the set of all isosceles triangles.
Since isosceles triangle is one type of triangle, the set of all triangles is the
universal set.
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}
Solution ;
Given set is { 0,1,2,3,4,5,6 }
Universal set contains all the elements.
Since for C, the number 8 is not in the given set, given set is not universal set.
(ii) ⲫ.
Solution :
Given set is the empty set ⲫ.
Since the empty set ⲫ is a subset to every set, ⲫ is not universal set.
(iii) {0,1,2,3,4,5,6,7,8,9,10}
Solution :
Given set is {0,1,2,3,4,5,6,7,8,9,10}.
Since the given set contains all the elements of A, B and C, given set is universal set.
(iv) {1,2,3,4,5,6,7,8}
Solution :
Given set is {1,2,3,4,5,6,7,8}
Universal set contains all the elements.
Since for the set C, the number 0 is not in the given set, the given set is not universal set.
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