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Three sets \(A,B\) and \(C\) are such that \(A = B \cap C\) and \(B = C \cap A,\) then
(Association law)
For any two sets \(A\) and \(B,\) if \(A \cap X = B \cap X = \phi \) and \(A \cup X = B \cup X\) for some sets \(X,\) then
From the above condition, I found that and are disjoint sets.
Given
If \(S\) is a set with \(10\) elements and \(A = \{ \left. {(x,y)} \rightx, y\in s\), \(x \ne y\} ,\) then the number of elements in \(A\) is
,
Number of elements in
As, yes and then
Number of elements in .
If \(n(A) = 4,n(B) = 5\) and \(n(A \cap B) = 3,\) then \(n[(A \times B) \cap (B \times A)]\) is
If \(A = \{ 1,2,3,4\} ,B = \{ 2,4,6\} ,\) then the number of possible sets that is, \(C\), such that \(A \cap B \subseteq C \subseteq A \cup B\).
.
If \(A\) and \(B\) are two sets, then \({(A \cup B)^ \subset } \cap ({A^ \subset } \cap B)\) is equal to
(i) For disjoint sets : –
(Associative law)
(ii) For connected sets: –
The answer is .
If \(A = \{ 1,2\} ,B = \{ \{ 1\} ,\{ 2\} \} ,C = \{ \{ 1,2\} \} \), then which of the following relation is correct?
Here, contains .
Hence, is an element of .
For any two sets \(A\) and \(B\), \(A – (A – B)\) is equal to,
Out of \(64\) students, the number of students taking mathematics is \(45\) and number of students taking both mathematics and Biology is \(10\). Then, the number of students taking only Biology is
Let and denote the number of students taking only Mathematics and only Biology respectively,
So,
Number of students taking mathematics โmay be some taking Biologyโ
If \(A = \{ 1,2,3\} ,B = \{ 3,4\} \) and \(C = \{ 4,5,6,\} \), then \(A \cup (B \cup C)\) is equal to
A survey shows that \(63\% \) of the Americans like cheese where as \(76\% \) like apples. If \(x\% \) of the Americans like both cheese and apples, then
Let be the Americanโs who like cheese and be the Americanโs who like apples.
.
In a class of \(60\) students, if \(25\) students play cricket, \(20\) students play tennis and \(10\) students play both the games, then the number of students who play neither is
Let number of students play cricket and number of students play tennis.
Let be total number of students.
So, number of students who play neither game
The set \(A = \{ \left. x \rightx \in R,{x^2} = 16\) and \(2x = 6\} \) is equal to
and
And
So, has no value which is satisfied.
The shaded region in the figure represents
.
If \(A \subseteq B,\) then \(B \cup A\) is equal to
.
Two finite sets \(A\) and \(B\) have \(m\) and \(n\) elements respectively. If total number of subset of \(A\) is \(112\) more than the total number of subsets of \(B\), then the value of \(m\) is
and
and
.
If \(A = \{ x,y\} ,\) then the power set of \(A\) is
set of all possible subset is called power set
.
If \(n(A) = 4,n(B) = 3\) and \(n(A \times B \times C) = 24,\) then \(n(C)\) is equal to
\(\{ n(n + 1)(2x + 1):n \in Z\} \) is a subset of
For
If \(A = \{ x:x\) is a multiple of \(3\} \) and \(B = \{ x:x\) is a multiple of \(5\} \), then \(A \cap B\) is given by
is a multiple of
is a multiple of
is a multiple of
.
The total number of subsets of a finite set \(A\) has \(56\) more elements than the total number of subsets of another finite set \(B\). What is the number of elements in the set \(A\)?
If and have and elements respectively.
and
and
.
If \(X\) and \(Y\) are the sets of all positive division of \(400\) and \(100\) respectively (including \(1\) and the number). Then, \(n(X \cap Y)\) is equal to
.
The number of elements in the set \(\{ (a,b):2{a^2} + 3{b^2} = 35,a,b \in Z\} ,\) where \(Z\) is the set of all integers, is
Or
, it is only possible for
or
So, solution set is given by
or .
If \(X = \{ {4^n} – 3n – 1:n \in N\} \) and \(Y = \{ 9(n – 1):n \in N\} \), where \(N\) is the set of natural numbers, then \(X \cup Y\) is equal to
.
The set of \(A = \{ x:12x + 3K7\} \) is equal to the set of
.
If \(A = \{ (x,y):y = {e^{ – x}}\} \) and \(B = \{ (x,y):y = – x\} ,\) then
&
If possible, let for any value
That is,
Which is not possible, since for any value of .
Hence, .
\(A = \{ x \in R:\frac{{2x – 1}}{{{x^3} + 4{x^2} + 3x}}\} \) is equal to
.
If \(n(A) = 2,n(B) = 3,\) then the number of relations from \(A\) and \(B\) is
The number of relations from to .
If \(A = \{ a,b,c\} ,B = \{ b,c,d\} \) and \(C = \{ a,d,c\} ,\) then \((A – B) \times (B \cap C)\) is equal to
If \(A\) and \(B\) are two disjoint sets, \(n(A) = 7,n(B) = x,n(A \cup B) = 16,n(A \cap B) = y\). Then, find \(x\) and \(y\).
Since, and are disjoint sets
.