Y={10}

A - (B - C) = A - (B \(\cap\) C') [\(\because\) A - B = A \(\cap\) B']
= A \(\cap\) (B \(\cap\) C')'
= A \(\cap\) (B' \(\cup\) C) [\(\because\) (A \(\cap\) B)'=A' \(\cup\) B']
= (A \(\cap\) B') \(\cup\) (A \(\cap\) C) = (A - B) \(\cup\) (A \(\cap\) C)

Here
A - (B \(\cup\) C) = A \(\cap\) (B \(\cup\) C)' [\(\because\) A-B=AnB']
\(\Rightarrow\) A \(\cap\) (B' \(\cap\) C') [\(\because\) (A\(\cup\)B)'=A' \(\cap\) B']
(A \(\cap\) B') \(\cap\) (A \(\cap\) C')
\(\Rightarrow\) (A - B) \(\cap\) (A - C)
Thus A - (B \(\cup\) C) = (A - B) \(\cap\) (A - C).

Here A = {x : x ∈ R and x satisfies x² - 8x + 12 = 0}
= {x:x ∈ Rand (x-6)(x-2)=0}
= {2, 6}
B = {2, 4, 6}
C = {2, 4, 6, 8, } and D = {6}
Now A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B and D ⊂ C.

Max.12 and Min. 4

200

Here
n( S) = 21, n(T) = 32
and n(S \(\cap\) T) = 11
We know that
n(S \(\cup\) T) = n(S) + n (T) - n(S \(\cap\) T)
\(\therefore\) n(S \(\cup\) T) = 21 + 32 -11 = 42.

Given sets are A = {1,3,5,6,7} and B={3,7,8,9}.
Now, A-B={1,3,5,6,7} - {3,7,8,9} ={1,5,6}
[set of those elements of A, which are not present in B]
and, B-A={3,7,8,9} - {1,3,5,6,7} ={8,9}
[set of those elements of B, which are not present in A]
\(\because \) Required symmetric difference,
\(A\triangle B=(A-B)\cup (B-A)\quad =\{ 1,5,6\} U\{ 8,9\} =\{ 1,5,6,8,9\} \)

Here, x is a positive and a divisor of 9.
So, x can take values 1, 3, 9.
{x: X is a positive integer and a divisor of 9} = {1, 3, 9}

Let C and T respectively donote the set of students who play Cricket and Tennis, and set U donotes the set of all students in a class.
Then n(C) = 25, n(T) = 20, n(C\(​​\)\(\cap \)T) = 10 and n (U)= 60
We know that, n(C\(​​\)\(\cup \)T) = n(C) + n(T) - n(C\(​​\)\(\cap \)T)
\(         \Rightarrow \) n(C\(​​\)\(\cup \) T) = 25 + 20 - 10 = 45 - 10 = 35
Now the Number of students who play neither game
n(C'\(\cap \)T' )   = n(C'\(\cup \)T' ) 
= n (U) -  n(C\(\cup \)T) = 60 - 35= 25
Hence, 25 students play neither games.

Given, n (A) = 4, n(B) =5, n(U) =7 
and n(A\(\cup \)B) = 2
ஃ  n(A\(\cup \)B) = n(A) + n(B) - n(A\(\cap \)B)
ஃ  n(A\(\cup \)B) = 4+5-2 = 7
Now,n(A\(\cup \)B)' = n(U)-n(A\(\cup \)B) = 7-7 = 0

Wehave, A={1,2,{3,4},5}
Since,{3,4} is a member of set A.
\(\therefore\)      {3,4}\(\in\) A
Hence,{3,4} \(\subset\) A is incorrect.