Given, A ={x :x is a natural number} ={ 1, 2,3, 4.....} and B ={x : x is an even naturai number} = 2, 4, 6,.......} We observe that 2, 4, 6...... are the elements which are common to both the sets A and B.
A \(\cap\) B={2, 4, 6, ..........}=B

\(\phi \)

Here, A= {2,3} and B= {x : x is a solution of x² + 5x + 6 = 0}
Firstly, we find the solution of  x² + 5x + 6 = 0.
By splitting the middle term, we get
x² + 3x + 2x +  6 = 0
\(\Rightarrow \)  x(x + 3) + 2(x + 3) = 0
\(\Rightarrow \)  (x + 2)(x + 3) = 0 \(\Rightarrow \) x= -2, -3
\(\therefore \)  B= {-2, -3}
Since, the elements of sets A and B are not same, therefore A\(\neq \)B.

Given, A = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Since, 9 is an element of A, therefore 9 \(\in \) A.

Given, A= {1, 2, 3} \(\Rightarrow \) n(A) = 3
B= {t, p, q, r, s} \(\Rightarrow \) n(B) = 5
C= {\(\alpha ,\beta ,\gamma \)} \(\Rightarrow \) n(C) = 3
D= {a, e, i, o, u} \(\Rightarrow \) n(D) = 5
Here n(A) = n(C) = 3 and n(B) = n(D) = 5
\(\therefore \) The sets A, C and B, D are equivalent sets.

{0,1,2}

n(A\(\cup \)B)\(\ge \)n(B)=6

n(E)=15,n(M)=12,n(s)=8,\(n(E\cap M)=6\)
\(n(M\cap S)=7,n(E\cap S)=4,n(E\cap M\cap S)=4\)
(i)\(n(E\cap M\cap \overline { S } )=n(E\cap M)-n(E\cap M\cap S)\)
=6-4=2

(A \(\cap\) B) \(\cup\) (A - B)
= (A \(\cap\) B) \(\cup\) (A \(\cap\) B') = A \(\cap\) (B \(\cup\) B') (By distributive law)
=A\(\cap\)U=A
Hence A = (A\(\cap\) B) \(\cup\) (A - B)
Also A \(\cup\) (B - A)
=A\(\cup\)(B\(\cap\)A')
= (A \(\cup\) B) \(\cap\) (A \(\cup\) A') (By distributive law)
= (A\(\cup\)B) \(\cap\) U
=A\(\cup\)B
Hence A \(\cup\) (B - A) = A \(\cup\) B.

(i) {3, 4} is a member of set A.
\(\therefore\) {3, 4} ∈ A
Hence {3, 4}⊂ A is incorrect.
(ii) {3, 4} is a member of set A.
\(\therefore\){3, 4} ∈ A is correct.
(iii) Here {3, 4} is a member of set A.
\(\therefore\) {{3,4}} is a set
\(\therefore\) {{3,4}} ⊂ A is correct.
(iv) 1 is a member of set A.
\(\therefore\) 1 ∈ A is correct.
(v) 1 is not a set, it is a member of set A.
\(\therefore\) 1⊂ A is incorrect.
(vi) 1, 2, 5 are members of set A.
\(\therefore\) {1, 2, 5} is a subset of set A.
\(\therefore\) {1, 2, 5} ⊂ A is correct.
(vii) 1, 2, 5 are members of set A.
\(\therefore\) {1, 2, 5} is a subset of set A.
\(\therefore\) {1, 2, 5} ∈ A is incorrect.
(viii) 3 is not a member of set A.
\(\therefore\) {1, 2, 3} is not a subset of set A.
\(\therefore\) {1, 2, 3} ⊂ A is incorrect.
(ix) ф is not a member of set A.
\(\therefore\) ф ∈ A is incorrect.
(x) Since ф is subset of every set,
\(\therefore\)ф ⊂ A is correct.
(xi)ф​​​​​​​ is not a member of set A.
\(\therefore\) {ф} ⊂ A is incorrect.