Write the following sets in the roster form:
(i) \(A = \{x : x \in \mathbb{R},\; 2x + 11 = 15\}\)
(ii) \(B = \{x \mid x^2 = x,\; x \in \mathbb{R}\}\)
(iii) \(C = \{x \mid x\text{ is a positive factor of a prime number }p\}\)
(i) \(\{2\}\)
(ii) \(\{0,1\}\)
(iii) \(\{1,p\}\)
Write the following sets in the roster form:
(i) \(D = \{t \mid t^3 = t,\; t \in \mathbb{R}\}\)
(ii) \(E = \{w \mid \dfrac{w-2}{w+3} = 3,\; w \in \mathbb{R}\}\)
(iii) \(F = \{x \mid x^4 - 5x^2 + 6 = 0,\; x \in \mathbb{R}\}\)
(i) \(\{0,-1,1\}\)
(ii) \(\{-\tfrac{11}{3}\}\)
(iii) \(\{-\sqrt{3}, -\sqrt{2}, \sqrt{2}, \sqrt{3}\}\)
If \(Y = \{x \mid x\text{ is a positive factor of the number }2^p-1\}\), where \(2^p-1\) is a prime number, write \(Y\) in the roster form.
\(\{1,2,2^2,2^3,\dots,2^{p-1},(2^p-1)\}\)
State which of the following statements are true and which are false. Justify your answer.
(i) \(35 \in \{x \mid x\text{ has exactly four positive factors}\}\).
(ii) \(128 \in \{y \mid \text{the sum of all the positive factors of }y\text{ is }2y\}\).
(iii) \(3 \notin \{x \mid x^4 -5x^3 +2x^2 -112x +6 =0\}\).
(iv) \(496 \notin \{y \mid \text{the sum of all the positive factors of }y\text{ is }2y\}\).
(i) True
(ii) False
(iii) True
(iv) True
Given \(L=\{1,2,3,4\},\; M=\{3,4,5,6\}\) and \(N=\{1,3,5\}\). Verify that \(L-(M\cup N) = (L-M) \cap (L-N)\).
Both sides are equal (verification by computing both sets gives the same result).
If \(A\) and \(B\) are subsets of the universal set \(U\), then show that
(i) \(A \subset A\cup B\).
(ii) \(A \subset B \iff A\cup B = B\).
(iii) \((A\cap B) \subset A\).
Standard subset properties hold; each statement is true (prove by element-wise argument).
Given that \(\mathbb{N}=\{1,2,3,\dots,100\}\). Then write
(i) the subset of \(\mathbb{N}\) whose elements are even numbers.
(ii) the subset of \(\mathbb{N}\) whose elements are perfect square numbers.
(i) \(\{2,4,6,8,\dots,98,100\}\)
(ii) \(\{1,4,9,16,25,36,49,64,81,100\}\)
If \(X=\{1,2,3\}\), if \(n\) represents any member of \(X\), write the following sets containing all numbers represented by
(i) \(4n\)
(ii) \(n+6\)
(iii) \(\tfrac{n}{2}\)
(iv) \(n-1\)
(i) \(\{4,8,12\}\)
(ii) \(\{7,8,9\}\)
(iii) \(\{\tfrac{1}{2},1,\tfrac{3}{2}\}\)
(iv) \(\{0,1,2\}\)
If \(Y=\{1,2,3,\dots,10\}\), and \(a\) represents any element of \(Y\), write the following sets containing all the elements satisfying the given conditions.
(i) \(a\in Y\) but \(a^2 \notin Y\).
(ii) \(a+1=6,\; a\in Y\).
(iii) \(a\) is less than 6 and \(a\in Y\).
(i) \(\{4,5,6,7,8,9,10\}\)
(ii) \(\{5\}\)
(iii) \(\{1,2,3,4,5\}\)
Let \(A,B\) and \(C\) be subsets of the universal set \(U\). If \(A=\{2,4,6,8,12,20\},\; B=\{3,6,9,12,15\},\; C=\{5,10,15,20\}\) and \(U\) is the set of all whole numbers, draw a Venn diagram showing the relation of \(U,A,B\) and \(C\).
Venn diagram showing the three sets \(A,B,C\) inside \(U\) with elements placed according to membership (see provided diagram in workbook).
Let \(U\) be the set of all boys and girls in a school, \(G\) be the set of all girls in the school, \(B\) be the set of all boys in the school, and \(S\) be the set of all students in the school who take swimming. Some, but not all, students in the school take swimming. Draw a Venn diagram showing one of the possible interrelationships among sets \(U,G,B\) and \(S\).
One possible Venn diagram: \(U\) as the universal rectangle, two disjoint circles for \(B\) and \(G\) (or overlapping if some students counted in both), and \(S\) as a circle overlapping both showing swimmers among boys and girls.
For all sets \(A,B\) and \(C\), show that \((A-B) \cap (C-B) = A - (B\cup C)\).
Determine whether each of the statements in Exercises 13–17 is true or false. Justify your answer.
The equality holds (prove by element-wise argument).
For all sets \(A\) and \(B\), show whether \((A-B) \cup (A\cap B) = A\) is true or false.
True
For all sets \(A,B\) and \(C\), determine whether \(A-(B-C) = (A-B)-C\) is true or false.
False
For all sets \(A,B\) and \(C\), if \(A \subset B\), then is \(A\cap C \subset B\cap C\)?
True
For all sets \(A,B\) and \(C\), if \(A \subset B\), then is \(A\cup C \subset B\cup C\)?
True
For all sets \(A,B\) and \(C\), if \(A \subset C\) and \(B \subset C\), then is \(A\cup B \subset C\)?
True
Using properties of sets, prove the statement: For all sets \(A\) and \(B\), \(A\cup (B-A) = A\cup B\).
True; prove by showing each side contains the same elements (element-wise argument).
Using properties of sets, prove: For all sets \(A\) and \(B\), \(A - (A - B) = A\cap B\).
True; standard set identity (verify element-wise).
Using properties of sets, prove: For all sets \(A\) and \(B\), \(A - (A\cap B) = A - B\).
True; follows from definitions of set difference and intersection.
Using properties of sets, prove: For all sets \(A\) and \(B\), \((A\cup B) - B = A - B\).
True; verify by element-wise consideration.
Let \(T = \{x \mid \dfrac{x+5}{x-7} - 5 = \dfrac{4x-40}{13-x}\}\). Is \(T\) an empty set? Justify your answer.
\(T = \{10\}\).
Let \(A\), \(B\) and \(C\) be sets. Then show that
\[ A \cap (B \cup C) = (A \cap B) \cup (A \cap C). \]
To prove: \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\).
Proof:
(1) Show that \(A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)\)
Let \(x \in A \cap (B \cup C)\). Then:
Since \(x \in B \cup C\), \(x\) is in either \(B\) or \(C\).
Case 1: If \(x \in B\), then \(x \in A\) and \(x \in B\), so \(x \in A \cap B\).
Case 2: If \(x \in C\), then \(x \in A\) and \(x \in C\), so \(x \in A \cap C\).
Thus in either case, \(x \in (A \cap B) \cup (A \cap C)\).
(2) Show that \((A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)\)
Let \(x \in (A \cap B) \cup (A \cap C)\). Then:
In both cases, \(x \in A\). Also:
If \(x \in B\), then \(x \in B \cup C\).
If \(x \in C\), then \(x \in B \cup C\).
Thus \(x \in A\) and \(x \in B \cup C\), which means:
\[ x \in A \cap (B \cup C). \]
Hence, both sides are equal.
\[ A \cap (B \cup C) = (A \cap B) \cup (A \cap C). \]
Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 7 in Mathematics and Science, 4 in English and Science, and 4 in all the three. Find how many passed:
Given:
1. English and Mathematics but not Science
\[ (E \cap M) - S = 6 - 4 = 2 \]
2. Mathematics and Science but not English
\[ (M \cap S) - E = 7 - 4 = 3 \]
3. Mathematics only
\[ M - (E \cap M) - (M \cap S) + (E \cap M \cap S) \]
\[ = 12 - 6 - 7 + 4 = 3 \]
4. More than one subject only
Sum of students in exactly two-subject combinations:
\[ (E \cap M - E \cap M \cap S) + (M \cap S - E \cap M \cap S) + (E \cap S - E \cap M \cap S) \]
\[ = (6-4) + (7-4) + (4-4) = 2 + 3 + 0 = 5 \]
Thus the number of students who passed more than one subject only = \(9\).
In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Find the number of students who play neither.
Given:
Using the formula:
\[ |C \cup T| = |C| + |T| - |C \cap T| \]
\[ = 25 + 20 - 10 = 35 \]
Number who play neither:
\[ 60 - 35 = 25 \]
Thus, 25 students play neither game.
In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects.
Given:
Students studying at least one subject:
\[ 200 - 20 = 180 \]
Using the three-set union formula:
\[ |M \cup P \cup C| = M + P + C - (M \cap P) - (P \cap C) - (C \cap M) + (M \cap P \cap C) \]
Substituting:
\[ 180 = 120 + 90 + 70 - 40 - 30 - 50 + x \]
\[ 180 = 160 + x \]
\[ x = 20 \]
Thus, 20 students study all three subjects.
In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, find:
Convert percentages to numbers:
Total families = 10,000
1. Families buying A only
\[ |A| - (|A \cap B| + |A \cap C|) + |A \cap B \cap C| \]
\[ = 4000 - (500 + 400) + 200 = 3300 \]
2. Families buying none
Use union formula:
\[ |A \cup B \cup C| = A + B + C - (AB + BC + CA) + ABC \]
\[ = 4000 + 2000 + 1000 - (500 + 300 + 400) + 200 \]
\[ = 6000 - 1200 + 200 = 5000 \]
Thus families buying none:
\[ 10000 - 5000 = 4000 \]
In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows:
French = 17, English = 13, Sanskrit = 15
French and English = 9, English and Sanskrit = 4, French and Sanskrit = 5
English, French and Sanskrit = 3
Find the number of students who study:
Given:
Compute the exact two-language intersections:
\(F \cap E \text{ only} = 9 - 3 = 6\)
\(E \cap S \text{ only} = 4 - 3 = 1\)
\(F \cap S \text{ only} = 5 - 3 = 2\)
1. French only
\[ 17 - (6 + 2 + 3) = 6 \]
2. English only
\[ 13 - (6 + 1 + 3) = 3 \]
3. Sanskrit only
\[ 15 - (2 + 1 + 3) = 9 \]
4. English and Sanskrit but not French
\[ 1 \]
5. French and Sanskrit but not English
\[ 2 \]
6. French and English but not Sanskrit
\[ 6 \]
7. At least one language
Sum all regions:
\[ 6 + 3 + 9 + 6 + 1 + 2 + 3 = 30 \]
8. None
\[ 50 - 30 = 20 \]
Suppose \(A_1, A_2, \dots, A_{30}\) are thirty sets each having 5 elements and \(B_1, B_2, \dots, B_n\) are \(n\) sets each with 3 elements. Let \(\bigcup_{i=1}^{30} A_i = \bigcup_{j=1}^n B_j = S\) and each element of \(S\) belongs to exactly 10 of the \(A_i\)'s and exactly 9 of the \(B_j\)'s. Then \(n\) is equal to
15
3
45
35
Two finite sets have \(m\) and \(n\) elements. The number of subsets of the first set is 112 more than that of the second set. The values of \(m\) and \(n\) are, respectively,
4, 7
7, 4
4, 4
7, 7
The set \((A \cap B') \cup (B \cap C)\) is equal to
\(A' \cup B \cup C\)
\(A' \cup B\)
\(A' \cup C'\)
\(A' \cap B\)
Let \(F_1\) be the set of parallelograms, \(F_2\) the set of rectangles, \(F_3\) the set of rhombuses, \(F_4\) the set of squares and \(F_5\) the set of trapeziums in a plane. Then \(F_1\) may be equal to
\(F_2 \cap F_3\)
\(F_3 \cap F_4\)
\(F_2 \cup F_5\)
\(F_2 \cup F_3 \cup F_4 \cup F_1\)
Let \(S\) = set of points inside the square, \(T\) = set of points inside the triangle and \(C\) = set of points inside the circle. If the triangle and circle intersect each other and are contained in a square, then
\(S \cap T \cap C = \varnothing\)
\(S \cup T \cup C = C\)
\(S \cup T \cup C = S\)
\(S \cup T = S \cap C\)
Let \(R\) be set of points inside a rectangle of sides \(a\) and \(b\) (\(a,b>1\)) with two sides along the positive direction of \(x\)-axis and \(y\)-axis. Then
\(R=\{(x,y):0\le x\le a,\;0\le y\le b\}\)
\(R=\{(x,y):0\le x
\(R=\{(x,y):0\le x\le a,\;0< y< b\}\)
\(R=\{(x,y):0< x< a,\;0< y< b\}\)
In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Then, the number of students who play neither is
0
25
35
45
In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Then the number of persons who read neither is
210
290
180
260
If \(X=\{8^n-7n-1\mid n\in \mathbb{N}\}\) and \(Y=\{49n-49\mid n\in \mathbb{N}\}\). Then
\(X \subset Y\)
\(Y \subset X\)
\(X = Y\)
\(X \cap Y = \varnothing\)
A survey shows that 63% of the people watch a News Channel whereas 76% watch another channel. If \(x\%\) of the people watch both channels, then
\(x = 35\)
\(x = 63\)
\(39 \le x \le 63\)
\(x = 39\)
If sets \(A\) and \(B\) are defined as \(A=\{(x,y)\mid y=\tfrac{1}{x},\;0\ne x\in\mathbb{R}\}\) and \(B=\{(x,y)\mid y=-x,\;x\in\mathbb{R}\}\), then
\(A \cap B = A\)
\(A \cap B = B\)
\(A \cap B = \varnothing\)
\(A \cup B = A\)
If \(A\) and \(B\) are two sets, then \(A \cap (A \cup B)\) equals
\(A\)
\(B\)
\(\varnothing\)
\(A \cap B\)
If \(A=\{1,3,5,7,9,11,13,15,17\},\; B=\{2,4,\dots,18\}\) and \(\mathbb{N}\) the set of natural numbers is the universal set, then \(A' \cup (A \cup B) \cap B'\) is
\(\varnothing\)
\(\mathbb{N}\)
\(A\)
\(B\)
Let \(S=\{x\mid x\text{ is a positive multiple of }3\text{ less than }100\}\). \(P=\{x\mid x\text{ is a prime number less than }20\}\). Then \(n(S)+n(P)\) is
34
31
33
30
If \(X\) and \(Y\) are two sets and \(X'\) denotes the complement of \(X\), then \(X \cap (X \cup Y)'\) is equal to
\(X\)
\(Y\)
\(\varnothing\)
\(X \cap Y\)
The set \( \{ x \in \mathbb{R} : 1 \le x < 2 \} \) can be written as ____.
[1,2)
When \(A = \varnothing\), then number of elements in \(P(A)\) is ____.
1
If \(A\) and \(B\) are finite sets such that \(A \subset B\), then \(n(A \cup B) = ____\).
n(B)
If \(A\) and \(B\) are any two sets, then \(A - B\) is equal to ____.
A \cap B'
Power set of the set \(A = \{1,2\}\) is ____.
{∅, {1}, {2}, {1,2}}
Given the sets \(A = \{1,3,5\}\), \(B = \{2,4,6\}\) and \(C = \{0,2,4,6,8\}\), the universal set of all the three sets \(A,B,C\) can be ____.
{0,1,2,3,4,5,6,8}
For all sets \(A\) and \(B\), \(A - (A \cap B)\) is equal to ____.
A ∪ B'
If \(U = \{1,2,3,4,5,6,7,8,9,10\}\), \(A = \{1,2,3,5\}\), \(B = \{2,4,6,7\}\) and \(C = \{2,3,4,8\}\), then \((B \cup C)'\) is ____.
{1,5,9,10}
If \(U = \{1,2,3,4,5,6,7,8,9,10\}\), \(A = \{1,2,3,5\}\), \(B = \{2,4,6,7\}\) and \(C = \{2,3,4,8\}\), then \((C - A)'\) is ____.
{1,2,3,5,6,7,9,10}
Match the items in Column A with Column B using the table below. Do NOT start the question text with a number.
| Column A | Column B |
|---|---|
(i) \(((A' \cup B') - A)'\) | (a) \(A - B\) |
(ii) \([B' \cup (B' - A)]'\) | (b) \(A\) |
(iii) \((A - B) - (B - C)\) | (c) \(B\) |
(iv) \((A - B) \cap (C - B)\) | (d) \((A \times B) \cap (A \times C)\) |
(v) \(A \times (B \cap C)\) | (e) \((A \times B) \cup (A \times C)\) |
(vi) \(A \times (B \cup C)\) | (f) \((A \cap C) - B\) |
| Column A | Matched Item from Column B |
|---|---|
(i) \(((A' \cup B') - A)'\) | (b) \(A\) |
(ii) \([B' \cup (B' - A)]'\) | (c) \(B\) |
(iii) \((A - B) - (B - C)\) | (a) \(A - B\) |
(iv) \((A - B) \cap (C - B)\) | (f) \((A \cap C) - B\) |
(v) \(A \times (B \cap C)\) | (d) \((A \times B) \cap (A \times C)\) |
(vi) \(A \times (B \cup C)\) | (e) \((A \times B) \cup (A \times C)\) |
If A is any set, then \(A \subset A\).
True
Given that \(M = \{1,2,3,4,5,6,7,8,9\}\) and if \(B = \{1,2,3,4,5,6,7,8,9\}\), then \(B \not\subset M\).
False
The sets \(\{1,2,3,4\}\) and \(\{3,4,5,6\}\) are equal.
False
\(Q \cup Z = Q\), where \(Q\) is the set of rational numbers and \(Z\) is the set of integers.
True
Let sets R and T be defined as \(R = \{x \in \mathbb{Z} \mid x \text{ is divisible by 2}\}\) and \(T = \{x \in \mathbb{Z} \mid x \text{ is divisible by 6}\}\). Then \(T \subset R\).
True
Given \(A = \{0,1,2\}\), \(B = \{x \in \mathbb{R} \mid 0 \le x \le 2\}\). Then \(A = B\).
False