Question 1511 Mark
Is x : x is a natural number, x < 5 and, x > 7 null set?
Answer{x : x is a natural number, x< 5 and x > 7} is an empty set because there is no natural number which satisfies simultaneously r < 5 and x > 7.
View full question & answer→Question 1521 Mark
Is set of even prime numbers null set?
AnswerSet of even prime numbers is {2} which is not empty set.
View full question & answer→Question 1531 Mark
Is set of odd natural numbers divisible by 2 null set?
AnswerHere, Set of odd natural numbers divisible by 2.
As we know that a set is a collection of well defined distnict objects.
Let we represent the given set in roaster form:
$⇒$ Set of odd natural numbers divisible by 2 is $\phi$
Because no odd natural number can be divided by 2. Therefore, it is a null set.
View full question & answer→Question 1541 Mark
List the element of the set: E = {x : x is a month of a year not having 31 days}
AnswerE = {x : x is a month of a year not having 31 days}
$\therefore$ E = {February, April, June, September, November}
View full question & answer→Question 1551 Mark
List the element of the set: C = {x : x is an integer, ${x^2} \leq 4$}
AnswerC = {x : x is an integer, ${x^2} \leq 4$}
$\therefore {x^2} \leq 4 \Rightarrow x \leq \pm 2$ $ \Rightarrow - 2 \leq x \leq 2$
$\therefore C = ( - 2, - 1,0,1,2)$
View full question & answer→Question 1561 Mark
List the element of the set: C = {x : x is an integer, $\frac{1}{2}<x<\frac{9}{2}$}
AnswerB = (x : x is an integer, $ - 1/2 < x < 9/2$)
$\therefore$ B = {0, 1, 2, 3, 4}
View full question & answer→Question 1571 Mark
Write the set in the set-builder form: {1, 4, 9, . . . , 100}
AnswerLet E = {1, 4, 9, ....., 100}
All objects ofthe set are perfect squares.
$\therefore D = \{ x:x = {n^2}\,and\,\,1 \leqslant n \leqslant 10\} $
View full question & answer→Question 1581 Mark
Write the set in the set-builder form: {5, 25, 125, 625}
AnswerLet C = {5, 25, 125, 625}
All objects of the set are natural numbers that are powers of 5.
$\therefore C = \{ x:x = {5^n},n \in N\,and\,\,1 \leqslant n \leqslant 4\} $
View full question & answer→Question 1591 Mark
Write the set in the set-builder form: {2, 4, 8, 16, 32}
AnswerLet B = {2, 4,8,16,32}
All objects of the set are natural numbers that are powers of 2.
$\therefore B = \{ x:x = {2^n},n \in N\,and\,\,1 \leqslant n \leqslant 5\} $
View full question & answer→Question 1601 Mark
Write the set in roster form: F = The set of all letters in the word BETTER.
AnswerF = The set of all letters in the word BETTER
$\therefore F$ = {B, E, T, R}
View full question & answer→Question 1611 Mark
Write the set in roster form: E = The set of all letters in the word TRIGONOMETRY
AnswerE = The set of all letters in the word TRIGONOMETRY
$\therefore E$ = {T, R, I, G, O, N, M, E, Y}
View full question & answer→Question 1621 Mark
Write the set in roster form: D = {x : x is a prime number which is divisor of 60}
AnswerD = {x : x is a prime number which is divisor of 60}
$\therefore D$ = {2,3, 5}
View full question & answer→Question 1631 Mark
Write the set in roster form: C = {x : x is a two-digit natural number such that the sum of its digits is 8}
AnswerC = {x : x is a two-digit natural number such that the sum of its digit is 8}
$\therefore$ C = {17, 26, 35, 44, 53, 62, 71, 80}
View full question & answer→Question 1641 Mark
Is the collection of most dangerous animals of the world set? Justify your answer.
AnswerA collection of most dangerous animals of the world is not a very clearly defined set as the ranking of the animals keep on altering and their ranking vary from countries to countries.
The collection of distnict objects are not well–defined and don't have universal acceptance as it is.
Therefore,the collection is not set.
View full question & answer→Question 1651 Mark
Is a collection of novels written by the writer Munshi Prem Chand set? Justify your answer
AnswerWe will to explain a collection of novels written by the writer Munshi Prem Chand is a well-defined collection because there are finite numbers of books which Munshi Prem Chand has written. The names of the book could not vary from person to person on the basis of personal choice. The well-defined objects of the collection make it a set.
Therefore, this collection is a set.
View full question & answer→Question 1661 Mark
Is the collection of all the months of a year beginning with the letter J set? Justify your answer
AnswerSet: Collection of well defined and distnict objects.
There are three months of a year which begins with the letter J, rest of the month`s name begin with different letter. Therefore, the given collection has well-defined and distnict objects namely January, June and July.
Therefore, this collection is a set.
{x: x = months of a year beginning with letter J}
Alternatively
{x: x = January, June, July where January, June, July are month of a year}
View full question & answer→Question 1671 Mark
Consider the sets $\phi$, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the symbol $⊂ or ⊄$ between the pair of sets: B . . . C
AnswerB $⊂$ C as each element of B is also an element of C.
View full question & answer→Question 1681 Mark
Consider the sets $\phi$, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the symbol $⊂ or ⊄$ between the pair of set: A . . . C
AnswerSince A $⊂$ C as 1, 3 $∈$ A also belongs to C
View full question & answer→Question 1691 Mark
Consider the sets $\phi$, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the symbol $⊂ or ⊄$ between the pair of set: A . . . B
AnswerA $⊄$ B as 3 $∈$ A and 3 $∉$ B
View full question & answer→Question 1701 Mark
Consider the sets $\phi$, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the symbol $⊂ or ⊄$ between the pair of set: $\phi$ . . . B
AnswerWe have, $\phi ⊂$ B as $\phi$ is a subset of every set
View full question & answer→Question 1711 Mark
Show that the given set that is $A=\left\{n: n \in Z\right.$ and $\left.n^2 \leq 4\right\}$ and $B=\left\{x: x \in R\right.$ and $\left.x^2-3 x+2=0\right\}$ are equal or not? Justify your answer.
Answer$A=\{-2,-1,0,1,2\}, B=\{1,2\}$. Since $0 \in A$ and $0 \notin B$, hence Aand $B$ are not equal sets.
View full question & answer→Question 1721 Mark
Show that the given set that is X, the set of letters in “ALLOY” and B, the set of letters in “LOYAL” are equal? Justify your answer.
AnswerGiven, X = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then X and B are equal sets as repetition of elements in a set do not change a set.
Therefore ,X = {A, L, O, Y} = B
View full question & answer→Question 1731 Mark
Is the set {x : x $\in$ N and x is odd} finite or infinite?
AnswerSince there are infinite number of odd numbers, therefore, the given set is infinite.
View full question & answer→Question 1741 Mark
Is the set {x: x $\in$ N and x is prime} finite or infinite?
AnswerThe given set is the set of all prime numbers and since set of prime numbers is infinite. Therefore, the given set is infinite.
View full question & answer→Question 1751 Mark
Is the set {x : x $\in$ N and 2x –1 = 0} finite or infinite?
AnswerGiven set = $\phi$. Therefore, this is finite.
View full question & answer→Question 1761 Mark
Is the set $\{x : x \in N$ and $x^2 = 4\}$ finite or infinite?
AnswerGiven set $= \{2\}.$ Thus, it is finite.
View full question & answer→Question 1771 Mark
Is the set {x : x $\in$ N and (x – 1) (x – 2) = 0} finite or infinite?
AnswerWe have the set = {1, 2}. Hence, it is finite.
View full question & answer→Question 1781 Mark
Match each of the set on the left described in the roster form with the same set on the right described in the set-builder form :
| (a) $\{P, R, I, N, C, A, L\}$ |
(i) $\{x : x$ is a positive integer and is a divisor of $18\}$ |
| (b) $\{0\}$ |
(ii) $\{x : x$ is an integer and $x^2 – 9 = 0\}$ |
| (c) $\{1, 2, 3, 6, 9, 18\}$ |
(iii) $\{x : x$ is an integer and $x + 1 = 1\}$ |
| (d) $\{3, –3\}$ |
(iv) $\{x : x$ is a letter of the word PRINCIPAL$\}$ |
View full question & answer→Question 1791 Mark
Write the set $\left[\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}\right]$ in the set-builder form.
AnswerHere,we see that each member in the given set has the numerator one less than the denominator. Also, the numerator begin from 1 and do not exceed 6.
Therefore, in the set-builder form the given set is {x : x = $\frac{n}{n+1}$, here n is a natural number and 1 $\leq$ n $\leq$ 6}
View full question & answer→Question 1801 Mark
Write the set $A = \{1, 4, 9, 16, 25, . . . \}$in set-builder form.
AnswerWe have,$A =\{x : x$ is the square of a natural number$\}$
Alternatively, we can write $A = \{x : x = n^2 $, where $n \in N\}$
View full question & answer→Question 1811 Mark
List all the subsets of the set { –1, 0, 1}
AnswerSuppose A = { –1, 0, 1}.
Now, we have to calculate all the subset of A having no element is the empty set $\phi$.
The subsets of A having one element are { –1}, {0}, {1}.
The subsets of
A having two elements are {–1, 0}, {–1, 1} ,{0, 1}.
The subset of A having three elements of A is A itself.
Therefore, all the subsets of A are $\phi$, {–1}, {0}, {1}, {–1, 0}, {–1, 1}, {0, 1} and {–1, 0, 1}
View full question & answer→Question 1821 Mark
There are $200$ individual with a skin disorder, $120$ has been exposed to chemical $C_1, 50$ to chemical $C_2$ and $30$ to both the chemicals $C_1$ and $C_2$. Find the number of individual exposed to chemical $C_1$ or chemical $C_2$
AnswerThe number of individuals exposed to chemical $\mathrm{C}_1$ or chemical $C_2$ is given by $n(A \cup B)$.
Now, we have, $n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$ =120+50-30 $
$ =140$
Therefore, required number of individuals is $140$
View full question & answer→Question 1831 Mark
There are $200$ individual with a skin disorder, $120$ has been exposed to chemical $C_1, 50$ to chemical $C_2$ and $30$ to both the chemicals $C_1$ and $C_2$. Find the number of individual exposed to chemical $C_2$ but not chemical $C_1$
AnswerThe number of individuals exposed to chemical $C_2$ but not chemical $C_1$ is given by $n(\overline A\cap B$).
Now, we have n( $\overline A \cap$ B) $= n (B) - n(A\cap B)$
$= 50 - 30$
$= 20$
Therefore, required number is $20.$
View full question & answer→Question 1841 Mark
There are $200$ individual with a skin disorder, $120$ has been exposed to chemical $C_1, 50$ to chemical $C_2$ and $30$ to both the chemicals $C_1$ and $C_2$. Find the number of individual exposed to chemical $C_1$ but not chemical $C_2$
AnswerThe number of individuals exposed to chemical $C_1$ but not chemical $C_2$ is given by $n (A\cap \overline B).$
Now, we have n(A $\cap \overline B$) $= n(A) - n(A\cap B)$
$= 120 - 30$
$= 90$
Therefore, required number of individuals is $90$
View full question & answer→Question 1851 Mark
Write the set $\{x : x$ is a positive integer and $x^2 < 40\}$ in the roster form.
AnswerThe required numbers are $1, 2, 3, 4, 5, 6.$ Therefore, the given set in the roster form is $\{1, 2, 3, 4, 5, 6\}.$
View full question & answer→Question 1861 Mark
Let V = { a, e, i, o, u } and B = { a, i, k, u}. Find V – B and B – V
AnswerHere ,it is V - B = {e, o}, since the elements e, o belongs to V but not to B and B - V = { k }, since the element k belongs to B but not to V
View full question & answer→Question 1871 Mark
Let A = { 1, 2, 3, 4, 5, 6}, B = { 2, 4, 6, 8 }. Find A – B and B – A.
AnswerHere, A - B = {1, 3, 5}, since the elements 1, 3, 5 belong to A but not to B and also B - A = {8}, since the element 8 belongs to B and not to A.then,
We note that A - B $\ne$ B – A
View full question & answer→Question 1881 Mark
Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are in school hockey team. Let Y = {Geeta, David, Ashok} be the set of students from Class XI who are in the school football team. Find X $\cap$ Y and interpret the set
AnswerWe see that element ‘Geeta’ is the only element common to both. Therefore, X $\cap$ Y = {Geeta}
View full question & answer→Question 1891 Mark
Let A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A $\cap$ B
AnswerWe see that 6, 8 are the only elements which are common to both A and B.
Therefore,A $\cap$ B = { 6, 8 }
View full question & answer→Question 1901 Mark
Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are in school hockey team. Let Y = {Geeta, David, Ashok} be the set of students from Class XI who are in the school football team. Find X $\cup$ Y and interpret the set.
AnswerHere, X $\cup$ Y = {Ram, Geeta, Akbar, David, Ashok}. So,this is the set of students from Class XI who are in the hockey team or the football team or both.
View full question & answer→Question 1911 Mark
Let A = { a, e, i, o, u } and B = { a, i, u }. Show that A $\cup$ B = A
AnswerWe have, A $\cup$ B = { a, e, i, o, u } = A.
This example illustrates that union of sets A and its subset B is the set A We know that if B $\subset$ A, then A $\cup$ B = A.
View full question & answer→Question 1921 Mark
Let A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A $\cup$ B
AnswerIt is given that A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}
$\therefore$ We have A $\cup$ B = { 2, 4, 6, 8, 10, 12}
View full question & answer→Question 1931 Mark
Let A, B and C be three sets. If A $∈$ B and B $⊂$ C, is it true that A $⊂$ C ? If not, give an example.
AnswerWe know that an element of a set can never be a subset of itself.
Suppose A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}.
Here A $∈$ B as A = {1} and B $⊂$ C. But A $⊄$ C as 1 $∈$ A and 1 $∉$ C.
View full question & answer→Question 1941 Mark
Let $A = \{a, e, i, o, u\}$ and $B = \{a, b, c, d\}.$ Is A a subset of $B$? No. $($Why$?)$. Is $B$ a subset of $A?$
Answer$i.$ Is $A \begin{equation} \subset \end{equation} B$
According to the given we can state,
For a set to be a subset of another set, it needs to have all element presents in another set. In the set $A = \{e, i, o, u\}$ elements are present but these are not present in set $B.$
Hence $A \begin{equation} \not \subset \end{equation} B$
$ii.$ Is $b \begin{equation} \subset \end{equation} A$
According to the given we can state,
For this condition to be true, are elements of sets $B$ should be element present in sets $A.$
In the set $B = \{b, c, d\}$ elements are present but these elements are not an element in set $A.$
Hence $B\begin{equation} \not \subset \end{equation} A$
View full question & answer→Question 1951 Mark
Write the solution set of the equation $x^2 + x – 2 = 0$ in roster form.
AnswerHere,the given equation can be written as $-(x – 1) (x + 2) = 0,$ that is $x = 1, – 2$
Therefore, the solution set of the given equation can be written in roster form as $\{1, – 2\}.$
View full question & answer→