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Chapter 1: Sets – Exercise 1.5

Sets – Exercise 1.5 – Q.1

(i) A ∩ B denotes intersection of the two sets A and B, which consists of elements which are common to both A and B.

Since A ⊂ B, every dement of A is already an element of B.

∵ A ∩ B = A

(ii) A ∪ B denotes the union of the sets A and B which consists of elements which are either in A or B or in both A and B.

Since A ⊂ B, every element of A is already an element of B.

∴ A ∩ B = B

Sets – Exercise 1.5 – Q.2(i)

A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

So, A ∪ B = {x : x ϵ A or x ϵ B}

= {1, 2, 3, 4, 5, 6, 7, 8}

Sets – Exercise 1.5 – Q.2(ii)

A ∪ C = {x : x ϵ A x ϵ C}

= {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}

Sets – Exercise 1.5 – Q.2(iii)

B ∪ C = {x : x ϵ B or x ϵ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

Sets – Exercise 1.5 – Q.2(iv)

B ∪ C = {x : x ϵ B or x ϵ D}

= {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

Sets – Exercise 1.5 – Q.2(v)

A ∪ B ∪ C = {x|x ϵ A or x ϵ B or x ϵ C}

= {1, 2, 3, 4, 5, 6, 7, 8,  9, 10,11}

Sets – Exercise 1.5 – Q.2(vi)

A ∪ B ∪ D = {x : x ϵ A or x ϵ B or x ϵ D}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

Sets – Exercise 1.5 – Q.2(vii)

B ∪ C ∪ D = {xlx ϵ B or x ϵ C or x ϵ D}

= {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

Sets – Exercise 1.5 – Q.2(viii)

A ∩(B ∪ C) = all those elements which are common to A and B ∪ C = {xlx ϵ A and x ϵ B ∪ C}

Now, B ∪ C = {4, 5, 6, 7, 8, 9, 10, 11}

∴ A ∩ (B ∪ C) = {1,2,3, 4,5} ∩ {4, 5, 6, 7, 8, 9, 10, 11}

= {4, 5}.

Sets – Exercise 1.5 – Q.2(ix)

(A ∩ B) ∩ (B ∩ C) = {xlx ϵ (A ∩ B) and x ϵ (B ∩ C)}

Now,

A ∩ B = (xlx ϵ A and x ϵ B}

i.e., elements which are common to A & B

∴ A ∩ B = {1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8}

= {4, 5}

Also

B ∩ C = {4, 5, 6, 7, 8} ∩ {7,8, 9, 10,11}

= {7, 8}

Hence, (A ∩ B) ∩ (B ∩ C) = {4, 5} ∩ {7, 8}

= ∮ [∵ there is no element common in {4, 5} and {7, 8}]

Sets – Exercise 1.5 – Q.2(x)

(A ∪ D) ∩ (B ∪ C) = {x|x ϵ (A ∪ D) or x ϵ (B ∪ C)}

Now,

A  ∪ D = {1, 2, 3, 4, 5,10,11,12,13,14}

and B ∪ C = {4, 5, 6, 7, 8, 9, 10, 11}

∴ (A ∪ D) ∩ (B ∪ C) = {4, 5, 10, 11}

Sets – Exercise 1.5 – Q.3(i)

We have,

A = {x : x ϵ N}

= {1, 2, 3,.....}, the set of natural numbers

B = {x : x = 2n, x ϵ N}

= {2, 4, 6, 8,....}, the set of even natural numbers

∴ A ∩ B = {x : x ϵ A and x ϵ B}

= {2, 4, 6,.....}

= B            [∵B ⊂ A]

Sets – Exercise 1.5 – Q.3(ii)

We have,

A = {x : x ϵ N}

= {1, 2, 3,....}, the set of natural numbers

C = {x : x = 2n-1, x ϵ N}

= {1, 3, 5,..} , the set of odd natural numbers

A ∩ C = {x : x ϵ A and x ϵ C}

= C   [∵C ⊂ A]

Sets – Exercise 1.5 – Q.3(iii)

We have,

A = {x : x ϵ N}

= {1, 2, 3,...}, the set of natural numbers

and D = {x : x is a prime natural number}

= {2, 3, 5, 7,...}

A ∩ D = {x : x ϵ A and x ϵ D}

=D         [∵D ⊂ A]

Sets – Exercise 1.5 – Q.3(iv)

We have,

B = {x : x = 2n,x ϵ N}

= {2, 4, 6, 8,....}, the set of even natural numbers

and

C = {x: x = 2n-1, x ϵ N}

= {1, 3, 5,...}, the set of odd natural numbers

B ∩ C = {x : x ϵ B and x ϵ C)

= ∮   [∵ B and C are disjoint sets, i.e., have no elements in common]

Sets – Exercise 1.5 – Q.3(v)

Here,

B = {x : x = 2n, x ϵ N)

= (2, 4, 6, 8....}, the set of even natural numbers

and D = {x : x is a prime natural number}

= {2, 3, 5, 7,..}

B ∩ D = {x : x ϵ B and x ϵ D}

= {2}

Sets – Exercise 1.5 – Q.3(vi)

Here,

C = {x : x = 2n-1,x ϵ N}

= {1,3,5,...}, the set of odd natural numbers

and D = {x : x is a prime natural number)

= {2, 3, 5, 7}

C ∩ D = {x : x ϵ C and x ϵ D}

we observe that except, the element 2, every other element in 0 is an odd natural number.

Hence, C ∩ D = D - {2}

= {x ϵ D : x ≠ 2}

Sets – Exercise 1.5 – Q.4

We have,

A = {3, 6, 12, 15, 18, 21}

B = {4, 8, 12, 16, 20}

C = {2, 4, 6, 8, 10, 12, 14, 16}

D = {5, 10, 15, 20}

If A and B are two sets, then the set A - B is defined as

A - B = {x ϵ A : x ∉ B}

(i) A - B = {x ϵ A : x ∉ B) = {3, 6, 15, 18, 21}

(ii) A - C = {x ϵ A ϵ C} = {3, 15,18, 21}

(iii) A - D = {x ϵ A : x ∉ D} = {3, 6,12,18, 21}

(iv) B - A = {x ϵ A : x ∉ A} = {4, 8, 16, 20}

(v) C- A = {x ϵ C : x ∉ A} = {2, 4, 8, 10, 14,16}

(vi) D - A = {x ϵ D : x ∉ A} = {5,10 ,20}

(vii) B - C = {x ϵ B : x ∉ C} = {20}

(viii) B - D = (x ϵ B : x ∉ D} = {4, 8,12,16}

Sets – Exercise 1.5 – Q.5

(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8}, C = {3, 4, 5, 6}

By the complement of a set A, which respect to the universal set U, denoted by A' or Ac or U - A, we mean {x ϵ U : x ∉ A}.

Hence, A' = (x ϵ U : x ∉ A} = {5, 6, 7, 8, 9}

(ii) B' = {x ϵ U : x ∉ B} = {1, 3, 5, 7, 9}

(iii) (A ∩ C)' = {x ϵ U : x ∉ A ∩ C}

Now,

A ∩ C = {x : x ϵ A and x ϵ C} = {3,4}

∴ (A ∩ C)' = {1, 2, 5, 6, 7, 8, 9}

Sets – Exercise 1.5 – Q.6

(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

B = {2, 3, 5, 8}

B = {2, 3, 5, 7}

We have,

A ∪ B = {x : x ϵ A or x ϵ B}

= {2, 3, 4, 5, 6, 7, 8}

∴ (A ∪ B)' = {x ϵ U : x ∉ A ∪ B}

= {1, 9}

A' = {x ϵ U : x ∉, A}

= {1, 3, 5, 7, 9}

B' = {x ϵ U : x ϵ B}

= {1, 4, 6, 8, 9}

Hence, A' ∩ B' = {1, 9}

Hence, (A ∪ B)' = A' ∩ B' = {1, 9}

(ii) A ∩ B = {x : x ϵ A and x ϵ B}

= {2}

∴ (A ∩ B)' = {x ϵ U : x ∉ A ∩ B}

= {1, 3, 4, 5, 6, 7, 8, 9}

Also ,

A' ∪ B' = {x :x ϵ A' or x ϵ B'}

= {1, 3, 4, 5, 6, 7, 8, 9}

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

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