If A and B are two sets such that A ⊂ B, then A ∪ B = B.

Given n(A) = 3
and n (B) = 6
We know that,
n ( A \(\cup \) B) = n(A) + n(B) = n ( A \(\cap\) B)
Clearly, n ( A \(\cup \) B) will be maximum when n ( A \(\cap\) B) is minimum and it will be minimum n ( A \(\cap\) B) is minimum. Therefore, two cases arise
Case I when n ( A \(\cap\) B) is minimum
The minimum value of n ( A \(\cap\) B) = 0
\(\therefore\)  n ( A  \(\cup \) B) = n(A) + n(B) = 3 + 6 =9 
Then, maximum value of    n ( A  \(\cup \) B) is 9
Case II When n ( A \(\cap\) B) is maximum
n ( A \(\cap\) B) will be maximum, if A \(\subseteq \) B
In this case, n ( A \(\cap\) B) = 3
ஃ   n ( A \(\cap\) B) = 3 + 6 -3 = 6
Thus, the minimum number of elements in A \(\cup \) B is 6.

All rectangles, Rhombus and square are parallelograms because its opposite sides are equal and parallel.
Therefore \({ F }_{ 2 }\subset { F }_{ 1 },{ F }_{ 3 }\subset { F }_{ 1 }\) and \({ F }_{ 4 }\subset { F }_{ 1 }\)
\({ F }_{ 1 }={ F }_{ 2 }\cup { F }_{ 3 }\subset { F }_{ 4 }\)

Given, n(X)=40, n(X\(\cup \)Y)=60
and  n(X\(\cap \)Y) = 10
clearly, n(X\(\cup \)Y)= n(X) + n(Y) - n(X\(\cap \)Y)
\(\Rightarrow \) 60= 40 + n(Y)-10
\(\Rightarrow \) 60= 30 + n(Y)
\(\Rightarrow \) n(Y)= 60-30
\(\therefore\)  n(Y)=30
Hence, Y have 30 elements

Given, A = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Since, 12 is not an element of A, therefore 12 \(\notin \) A

We are sure that members of this collection are January, June, and July. So, this collection is well-defined. Hence, it is a set.

(i) (A ∪ B)'

(ii).  A′ ∩ B′

(iii).  (A ∩ B)′

(iv).  A′ ∪ B′

(i) { 2, 3, 4 } . . . { 1, 2, 3, 4,5 }
{2, 3, 4} ⊂ {1, 2, 3, 4, 5}
(ii) { a, b, c } . . . { b, c, d }
{a, b, c} ⊄ {b, c, d}
(iii) {x : x is a student of Class XI of your school}. . .{x : x student of your school}
{x: x is a student of class XI of your school}⊂ {x: x is 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}
{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}
{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}
{x: x is an equilateral triangle in a plane}⊂ {x: x in a triangle in the same plane}
(vii) {x : x is an even natural number} . . . {x : x is an integer}
{x: x is an even natural number} ⊂ {x: x is an integer}

(1) All the elements of this set are natural numbers as well as the divisors of 6. Therefore, (1) matches with (c).
(2) It can be seen that 2 and 3 are prime numbers. They are also the divisors of 6.
Therefore, (2) matches with (a).
(3) All the elements of this set are letters of the word MATHEMATICS. Therefore, (3) matches with (d).
(4) All the elements of this set are odd natural numbers less than 10. Therefore, (4) matches with (b)

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

Let F, B and C denote the set of men who received medals in football, basketball and cricket, respectively.
Then n ( F ) = 38, n ( B ) = 15, n ( C ) = 20
n (F \(\cup\) B \(\cup\) C ) = 58 and n (F \(\cap\) B \(\cap\) C ) = 3
Therefore, n (F \(\cup\) B \(\cup\) C ) = n ( F ) + n ( B )
+ n ( C ) – n (F \(\cap\) B ) – n (F \(\cap\) C ) – n (B \(\cap\) C ) + n ( F \(\cap\) B \(\cap\) C ),
gives n ( F \(\cap\) B ) + n ( F \(\cap\) C ) + n ( B \(\cap\) C ) = 18
Consider the Venn diagram.
Here, a denotes the number of men who got medals in football and basketball only, b denotes the number of men who got medals in football and cricket only, c denotes the number of men who got medals in basket ball and cricket only and d denotes the number of men who got medal in all the three. Thus, d = n ( F \(\cap\) B \(\cap\) C ) = 3 and
a + d + b + d + c + d = 18
Therefore a + b + c = 9,
which is the number of people who got medals in exactly two of the three sports.

(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) {1,7,9}
(ii) {1,3,5,7,9}
(iii) ф
(iv) {2,4,6,8,10}

Step I Let P(n) be the given statement
i.e.P(n) : n(n + 1) (n + 5) is a multiple of 3.
Step II Forn = 1,we have P(1) : 1(1 + 1)(1 + 5) = 1 × 2 × 6 =12=3×4
Which is a multiple of 3.
So, P(1) is true.
Step III Let P(k) be true form = k.
i.e.P(k) : k(k + 1)(k + 5) is a multiple of 3.
Then,k(k + 1)(k + 5) = 3λ
⇒k(k² + 5k + k5) = 3λ
⇒k³ + 6k² + 5k = 3λ...(i)
Step IV Now,we shall prove the statement form = k + 1.
For this,we have to show that (k + 1)(k + 1 + 1)(k + 1+5)is a multiple of 3.
Then,(k+1)(k+1+1)(k+1+5)=(k+1)(k+2)(k+6)
=(k²+2k+k+2)(k+6)=(k²+3k+2)(k+6)
=k³+6k²+3k²+18k+2k+12
=k³+9k²+20k+12
=(3λ−6k²−5k)+9k²+20k+12 [fromEq.(i),k³=3λ−6k²−5k]
=3λ+3k²+15k+12=3(λ+k²+5k+4)
Which is a multiple of 3 .
Thus,P(k+1) is true, whenever P(k) is true.
Hence,by principle of mathematical induction,the statement is true for all natural numbers n.