Question 13 Marks
If $A = \{6, 7, 8, 9\}, B = \{4, 6, 8,10\}$ and $C = \{x : x \in N : 2 < x \leq 7\}$; Find: $B - (A ∩ C)$.
Answer$A = \{6, 7, 8, 9\}$
$B = \{4, 6, 8, 10\}$
$C =\{x : x \in N : 2 < x \leq 7\}$
$= \{3, 4, 5, 6, 7\}$
$A ∩ C = \{6, 7, 8, 9\} ∩ \{3, 4, 5, 6, 7\}$
$=\{6, 7\}$
$B - (A ∩ C) = \{4, 6, 8, 10\} - \{6,7\}$
$= \{4, 8,10\}$
View full question & answer→Question 23 Marks
If $A = \{6, 7, 8, 9\}, B = \{4, 6, 8,10\}$ and $C = \{x : x ∈ N : 2 < x ≤ 7\} $; Find: $B - (A - C)$.
Answer$A = \{6, 7, 8, 9\}$
$B = \{4, 6, 8, 10\}$
$C = \{x : x \in N : 2 < x \leq 7\}$
$= \{3, 4, 5, 6, 7\}$
$A - C = \{6, 7, 8, 9\} -\{3, 4, 5, 6, 7\}$
$= \{8, 9\}$
$B - (A - C) = \{4, 6, 8, 10\} - \{8, 9\}$
$= \{4, 6, 10\}$
View full question & answer→Question 33 Marks
Given $A = \{0, 1, 2, 4, 5\}, B = \{0, 2, 4, 6, 8\}$ and $C = \{0, 3, 6, 9\}$. Show that $A ∩ (B ∩ C) = (A ∩ B) ∩ C$ i.e. the intersection of sets is associative.
Answer$B ∩ C = \{0, 2, 4, 6, 8\} ∩ \{0, 3, 6, 9\}$
$= \{0,6\}$
Now, $A ∩ (B ∩ C) = \{0, 1, 2, 4, 5\} ∩ \{0,6\}$
$\Rightarrow A ∩ (B ∩ C) = \{0\} ...(I)$
$A ∩ B = \{0, 1, 2, 4, 5\} ∩ \{0, 2, 4, 6, 8\}$
$= \{0,2,4\}$
$\therefore (A ∩ B) ∩ C = \{0,2,4\} ∩ \{0, 3, 6, 9\}$
$\Rightarrow (A ∩ B) ∩ C = \{0\} ...(II)$
From $I$ and $II$ we get
$A ∩ (B ∩ C) = (A ∩ B) ∩ C$
View full question & answer→Question 43 Marks
If $A = \{5, 6, 7, 8, 9\}, B = \{x : 3 < x < 8$ and $x \in W\}$ and $C = \{x : x \leq 5$ and $x \in N\}$. Find: $A ∩ B$ and $(A ∩ B) ∩ C$
Answer$A = (5, 6, 7, 8, 9)$
$B = (4, 5. 6, 7)$
$C = (1, 2, 3, 4, 5)$
$A ∩ B = (5, 6, 7)$
$(A ∩ B) ∩ C = (5)$
View full question & answer→Question 53 Marks
If $A = \{5, 6, 7, 8, 9\}, B = \{x : 3 < x < 8$ and $x \in W\}$ and $C = \{x : x \leq 5$ and $x \in N\}$. Find: $B ∪ C$ and $A ∪ (B ∪ C)$
Answer$A = (5, 6, 7, 8, 9)$
$B = (4, 5. 6, 7)$
$C = (1, 2, 3, 4, 5)$
$B ∪ C = (1, 2, 3, 4, 5, 6, 7)$
$A ∪ (B ∪ C) = (1, 2 , 3, 4, 5, 6, 7, 8, 9)$
View full question & answer→Question 63 Marks
If $A = \{5, 6, 7, 8, 9\}, B = \{x : 3 < x < 8$ and $x \in W\}$ and $C = \{x : x \leq 5$ and $x \in N\}$. Find: $A ∪ B$ and $(A ∪ B) ∪ C$
Answer$A = (5, 6, 7, 8, 9)$
$B = (4, 5. 6, 7)$
$C = (1, 2, 3, 4, 5)$
$A ∪ B = (4, 5, 6, 7, 8, 9)$
$(A ∪ B) ∪ C = (1, 2, 3, 4, 5, 6, 7, 8, 9)$
View full question & answer→Question 73 Marks
Given, universal set $= \{x : x ∈ N, 10 ≤ x ≤ 35\}. A = \{x ∈ N : x ≤ 16\}$ Find: $A'$
AnswerUniversal set $= \{x : x ∈ N, 10 ≤ x ≤ 35\}$
$= \{10, 11, 12, 13, 14, 15,....,34, 35\}$
$A = \{x ∈ N : x ≤ 16\}$
$= \{10, 11, 12, 13, 14, 15, 16\}$
$A' = \{17, 18, 19, 20, 21, 22,.....,33, 34, 35\}$
$= \{x : x ∈ N ; 17 ≤ x ≤ 35\}$
View full question & answer→Question 83 Marks
Find the proper subsets of ${x : x^2 – 9x – 10 = 0}$
Answer$X^2 - 9x - 10 = 0$
$\Rightarrow x^2 - 10x + x - 10 = 0$
$\Rightarrow x(x-10)+1 (x-10) = 0$
$\Rightarrow (x-10)(x+1) = 0$
$\therefore$ Either $x-10 = 0$
$\Rightarrow x = 10$ or $x+1 = 0$
$\Rightarrow x = -1$
Given set $= \{-1, 10\}$
Proper subsets of this set $= φ, \{-1\}, \{10\}$
View full question & answer→Question 93 Marks
State, if the following pair of a set is equal or not: $E = {x : x ^2 + 8x - 9 = 0}$ and $F = {1, - 9}$
Answer$E = \{x : x^2 + 8x - 9 = 0\}$
$x^2 + 8x - 9 = 0$
$\Rightarrow x^2 + 9x - x - 9 = 0$
$\Rightarrow x(x+9) - 1(x+ 9) = 0$
$\Rightarrow (x+1) (x+9) = 0$
$\therefore$ Either $x + 9 = 0$
$\Rightarrow x = -9$ Or $x - 1 = 0$
$\Rightarrow x = 1$
$\therefore E = \{-9, 1\}$
$F = \{1, -9\}$
Now we see that the elements of sets $E$ and $F$ are the same $($identical$)$ Sets $E$ and $F$ are equal.
View full question & answer→Question 103 Marks
State, if the following pair of a set is equivalent or not : $P = \{5, 6, 7, 8\}$ and $M = \{x : x \in W$ and $x < 4\}$
Answer$P = \{5,6,7,8\}$
$n (P) = 4$
$M = \{x : x \in W$ and $x \leq 4\}M = \{0, 1, 2, 3, 4\}$
$n (M) = 5$
Now Cardinal number of set $P = 4$ and
Cardinal number of set $M = 5$
These sets are not equivalent.
View full question & answer→Question 113 Marks
State, if the following pair of a set is equivalent or not: $A = \{x : x \in N$ and $11 \geq 2x – 1\}$ and $B = \{y : y \in W$ and $3 \leq y \leq 9\}$
Answer$A=\{x: x \in N$ and $11 \geq 2 x-1\}$
$ 11 \geq 2 x-1$
$ \Rightarrow 11+1 \geq 2 x-1+1$
$ \Rightarrow 12 \geq 2 x$
$ \Rightarrow \frac{12}{2} \geq x$
$ \Rightarrow 6 \geq x$
$ \therefore A=\{1,2,3,4,5,6\}$
$ \therefore n(A)=6$
$ B=\{y: y \in W$ and $3 \leq y \leq 9\}$
$ \because 3 \leq y \leq 9$
$ B=\{3,4,5,6,7,8,9\}$
$ n(B)=7$
Cardinal number of set $A=6$ and cardinal number of set $B=7$
Set $A$ and set $B$ are not equivalent.
View full question & answer→Question 123 Marks
Are the sets $A = \{4, 5, 6\}$ and $B = \{x : x^2 – 5x – 6 = 0\}$ disjoint?
Answer$A = \{4, 5, 6\}$
$B = \{x : x^2 - 5x - 6 = 0\}$
$x^2 - 5x - 6 = 0$
$\Rightarrow x^2 - 6x + x -6 = 0$
$\Rightarrow x(x - 6) +1(x - 6) = 0$
$\Rightarrow (x - 6) (x+1) = 0$
$\therefore $ Either $x - 6 = 0x$
$\Rightarrow x = 6$ Or $x+1 = 0$
$\Rightarrow x = -1$
$\therefore B = \{6, -1\}$
Hence set $A$ and set $B$ are not disjoint because these sets have element $6$ in common.
View full question & answer→Question 133 Marks
Find, if the following sets are empty: $B = \{x : x^2 + 4 = 0, x \in N\}$
Answer$B=\left\{x: x^2+4=0, x \in N\right\}$
$ x^2+4=0$
$ \Rightarrow x^2=-4$
$\Rightarrow x=\sqrt{-4}$ which is not a natural number.
But $x \in N$
$\therefore B=\{\}$
$\therefore$ Given set $B$ is an empty set.
View full question & answer→Question 143 Marks
Find, if the following sets are empty: $A = \{x : x \in N$ and $5 < x < 6\}$
Answer$A = \{x : x \in N$ and $5 < x < 6\}$
As $5 < x \leq 6$
$\therefore x = 6$
$\therefore A = \{6\}$
Hence given set $A$ is not an empty set.
View full question & answer→Question 153 Marks
Find, if the following sets are singleton sets$:B =\{y : 2y + 1 < 3$ and $y \in W\}$
Answer$B=\{y: 2 y+1<3$ and $y \in W\}$
$ 2 y+1<3$
$ \Rightarrow 2 y+1-1<3-1($Subtracting $1$ rom both sides$)$
$ \Rightarrow 2 y<2$
$ \Rightarrow y<\frac{2}{2}($Dividing both sides by $)$
$ \Rightarrow y<1$
$ \therefore B=\{0\}$
Hence it is a singleton set.
View full question & answer→Question 163 Marks
Find, if the following sets are singleton sets: $A = \{x : 7x – 3 = 11\}$
Answer$A=\{x: 7 x-3=11\}$
$ 7 x-3=11$
$ \Rightarrow 7 x=11+3$
$ \Rightarrow 7 x=14$
$ \Rightarrow x=\frac{14}{7}=2$
$ \therefore A=\{2\}$
Hence given set $\mathrm{A}$ is a singleton set.
View full question & answer→Question 173 Marks
State the following sets are finite or infinite: $B =\{x : x ∈ W$ and $5x -3 ≤ 20\}$
Answer$B=\{x: x \in W$ and $5 x-3 \leq 20\}$
$5 x-3 \leq 20$
$\Rightarrow 5 x-3+3 \leq 20+3 ($Adding $3$ to both sides$)$
$\Rightarrow 5 x \leq 20+3$
$ \Rightarrow 5 x \leq 23$
$\Rightarrow \mathrm{x} \leq \frac{23}{5}$
$($Dividing both sides by $5)$
$\Rightarrow x \leq 4.6$
$\therefore B=\{0,1,2,3,4\}$
$\therefore$ It is a finite set.$c$
View full question & answer→Question 183 Marks
Find the cardinal number of the following sets: $A_3 =\{p : p \in W$ and $2P - 3 < 8\}$
Answer$A_3=\{P: P \in W$ and $2 P-3<8\}$
$2 P-2<8$
$\Rightarrow 2 \mathrm{P}-3+3<8+3$
$($Adding $3$ to both sides$)$
$\Rightarrow 2 \mathrm{P}<11$
$\Rightarrow \mathrm{P}<\frac{11}{2}$
$($Dividing both sides by $2)$
$\Rightarrow \mathrm{P}<5.5$
$\therefore A_3=\{0,1,2,3,4,5\}$
$\therefore$ Cardinal number of set $\mathrm{A}_3=6$
View full question & answer→Question 193 Marks
List the elements of the following sets: $\{x : 3x\ – 2 \leq 10, x \in N\}$
Answer$\{x: 3 x-2 \leq 10, x \in N\}$
$ 3 x-2 \leq 10$
$ \Rightarrow 3 x \leq 10+2$
$ \Rightarrow 3 x \leq 12$
$ \Rightarrow x \leq \frac{12}{3}$
$ \Rightarrow x \leq 4$
$\therefore$ Elements of the given set $\{x: 3 x-2 \leq 10, x \in N\}$ are $1,2,3$ and $4$
View full question & answer→Question 203 Marks
List the elements of the following sets: $\{x : x \in Z$ and $x^2 \leq 4\}$
Answer${x: x \in Z}$ and ${x^2 \leq }4$
When $x^2=4$
$x= \pm \sqrt{4}= \pm 2$
When $x^2=1$
$x= \pm \sqrt{1}= \pm 1$
When $x^2=0$
$x=\sqrt{0}=0$
$\therefore$ Elements of the given set $\left\{x: x \in Z\right.$ and $\left.x^2 \leq 4\right\}$ are $+2,-2,+1,-1,0$ or are $-2,-1,0,1,2$
View full question & answer→Question 213 Marks
List the elements of the following sets :$\{x : x$ is a factor of $24\}$
Answer$\{x : x$ is a factor of $24\}$
$24 = 1 \times 24$
$24 = 2 \times 12$
$24 = 3 \times 8$
$24 = 4 \times 6$
$\therefore$ Elements of the given set $\{x : x$ is a factor of $24\}$ are $1, 2. 3, 4, 6, 8, 12, 24$
View full question & answer→Question 223 Marks
List the elements of the following sets: ${x : x^2 – 2x – 3 = 0}$
Answer${x : x^2 – 2x – 3 = 0}$
$x^2 - 2x - 3 = 0$
$\Rightarrow x^2 - 3x + x - 3 = 0$
$\Rightarrow x (x-3) + 1 (x-3) = 0$
$\Rightarrow (x+1) (x - 3) = 0$
$\therefore $ Either $x - 3 = 0 x$
$x = 3$ Or $x + 1 = 0$
$x = -1$
$\therefore $ Elements of the set ${x : x^2 - 2x - 3 = 0}$ are $3$ and $-1$
View full question & answer→Question 233 Marks
Write the following sets in Roster form: $B = \{p : p \in $ W and $p^2 < 20\}$
Answer$B=\left\{P: P \in W \text { and } p^2<20\right\}$
When $p^2=0$
$P=\sqrt{0}=0$
When $p^2=1$
$P=\sqrt{1}=1$
When $\mathrm{p}^2=4$
$P=\sqrt{4}=2$
When $\mathrm{p}^2=9$
$P=\sqrt{9}=3$
When $p^2=16$
$P=\sqrt{16}=4$
$\therefore$ Roster form of the given set $B=\{0,1,2,3,4\}$
View full question & answer→Question 243 Marks
Write the following sets in Roster form: $A = \{x : x = y + 3, y \in N$ and $y > 3\}$
Answer$A = \{x : x = y + 3, y \in N$ and $y > 3\}$
$x = y + 3$
When $y = 4$,
When $y = 5,$
When $y = 6$,
When $y = 7,$
When $y = 8,$
$...........................$
$...........................$
$x = 4+3 = 7$
$x = 5+3 = 8$
$x = 6+3 = 9$
$x = 7+3 = 10$
$x = 8+3 = 11$
$............................$
$............................$
$\therefore$ Roster form of the given set $A =\{7, 8, 9, 10, 11 ...........\}$
View full question & answer→Question 253 Marks
Is $ \{x : x^2 – 5x – 6 = 0\} = \{2, 3\}$ ?
Answer$x^5- 5x - 6 = 0$
$\Rightarrow x^2 - 6x + x - 6 = 0$
$\Rightarrow x (x - 6) + 1 (x - 6) = 0$
$\Rightarrow (x - 6) (x - 1) = 0$
$\therefore$ Either $x - 6 = 0$
i.e., $x = 6$
Or
$x + 1 = 0$
i.e. $x = -1$
$\therefore {x : x^2 - 5x - 6 = 0} \neq {2, 3}$
In other words ${x : x^2 - 5x -6 = 0}$
$= {2, 3}$ is not true.
View full question & answer→Question 263 Marks
Is $\{x : x^2 – 7x + 12 = 0\} = \{3, 4\}$?
Answer$x^2- 7x + 12 = 0$
$\Rightarrow x^2 - 4x - 3x + 12 = 0$
$\Rightarrow x (x - 4) - 3(x - 4) = 0$
$\Rightarrow (x - 4) (x - 3) = 0$
$\therefore$ Either $x - 4 = 0$
$\Rightarrow x = 4$ Or $x - 3 = 0$
$\Rightarrow x = 3$
$\therefore {x : x^2 - 7x + 12 = 0}$
$= {3, 4}$ is true.
View full question & answer→Question 273 Marks
Write the set of odd factors of $72$.
Answer$1 \times72 = 72$
$2 \times36 = 72$
$3 \times24 = 72$
$4 \times18 = 72$
$6 \times12 = 72$
$8 \times9 = 72$
Factors of $72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$
Set of odd factors of $72 = \{1, 3, 9\}$
View full question & answer→Question 283 Marks
Write the set of even factors of $124.$
Answer$1 \times 124= 1242 \times 62 = 124$
$4 \times 31 = 124$
Factors of $124 = 1, 2, 4, 31, 62, 124$
Set of even factors of $124 = \{2, 4, 62, 124\}$
View full question & answer→Question 293 Marks
Write the following sets in roster $($Tabular$)$ form :$A_5= \{x : x = 4n, n \in W$ and $n < 4\}$
Answer$A_5= \{x : x = 4n, n \in $ W and $n < 4\}$
$\because x = 4n$
$\therefore$ When $n = 0,$
$x = 4 \times 0$
$\Rightarrow x = 0$
When $n = 1,$
$x = 4 \times 1$
$\Rightarrow x = 4$
When $n = 2,$
$x = 4 \times 2$
$\Rightarrow x = 8$
When $n = 3,$
$x = 4 \times 3$
$\Rightarrow x = 12$
$\therefore $ Given set in roster $($Tabular$)$ form is
$A_5 = \{0, 4, 8, 12\}$
View full question & answer→Question 303 Marks
Write the following sets in roster $($Tabular$)$ form : $A_4 = \{x : x$ is a two digit number and sum of digits of $x$ is $7\}$
Answer$A_4 = \{x : x$ is a two digit number and sum of digits of $x$ is $7\}$
$\because x$ is a two digit number and sum of digits of $x$ is $7$
$\therefore x = 16, 25, 34, 43, 52, 61, 70$
$\therefore $ Given set in roster $($Tabular$)$ form is
$A_4 = \{16, 25, 34, 43, 52, 61, 70\}$
View full question & answer→Question 313 Marks
Write the following sets in roster $($Tabular$)$ form: $A_3 = {x : x \in Z, -3 \leq x <4}$
Answer$A_3 = \{x : x \in Z, -3 \leq x <4\}$
$\because -3 \leq x < 4$
$\therefore x = -3, -2, -1, 0, 1, 2, 3$
$\therefore$ Given set in roster $($Tabular$)$ form is
$A_3 = \{-3, -2, -1, 0, 1, 2, 3\}$
View full question & answer→Question 323 Marks
Write the following sets in roster $($Tabular$)$ form : $A_2 =\{x : x^2 - 4x -5 = 0\}$
Answer$A_2 =\{ x : x^2 - 4x - 5 = 0\}$
$\therefore x^2- 4x - 5 = 0$
$\Rightarrow x^2 - 5x + x - 5 = 0$
$\Rightarrow x (x - 5) + 1 (x - 5) = 0$
$\Rightarrow (x - 5) (x + 1) = 0$
$\therefore $ Either $x - 5 = 0$
$\Rightarrow x = 5$ or $x + 1 = 0$
$\Rightarrow x = -1$
$\therefore $ Given set in roster $($Tabular$)$ form is $A_2 = {5, -1}$
View full question & answer→Question 333 Marks
Write the following sets in roster $($Tabular$)$ form : $A_1 = \{x : 2x + 3 = 11\}$
Answer$A_1=\{x: 2 x+3=11\}$
$ \therefore 2 x+3=11$
$ \Rightarrow 2 x=11-3$
$ \Rightarrow 2 x=8$
$ \Rightarrow x=\frac{8}{2}$
$ \Rightarrow x=4$
$\therefore$ Given set in roster $($Tabular$)$ form is $\mathrm{A}_1=\{4\}$
View full question & answer→