MCQ 11 Mark
If $R$ is a relation on a finite set having $n$ elements, then the number of relations on $A$ is:
- A
$2^{\text{n}}$
- ✓
$2^{\text{n}^2}$
- C
$\text{n}^2$
- D
$\text{n}^\text{n}$
AnswerCorrect option: B. $2^{\text{n}^2}$
Given, $A$ finite set with $n$ elements
Its Cartesian product with itself will have $n^2$ elements.
$\therefore$ Number of relations on $\text{A}=2^{\text{n}^2}$
View full question & answer→MCQ 21 Mark
If $R$ is a relation from a finite set $A$ having m elements of a finite set $B$ having $n$ elements, then the number of relations from $A$ to $B$ is:
- ✓
$2^{mn}$
- B
$2^{mn} - 1$
- C
$2mn$
- D
$m^n$
AnswerCorrect option: A. $2^{mn}$
Given, $n(A) = m$
$n(B) = n$
$\therefore$ $n(A \times B) = mn$
Then, the number of relations from $A$ to is $2^{mn}$
View full question & answer→MCQ 31 Mark
If the set $A$ has $p$ elements, $B$ has $q$ elements, then the number of elements in $A \times B$ is:
- A
$p + q$
- B
$p + q + 1$
- ✓
$pq$
- D
$p^2$
Answer$n(A \times B) = n(A) \times n(B)$
$n(A \times B) = p \times q = pq$
View full question & answer→MCQ 41 Mark
Let $R$ be a relation from a set $A$ to a set $B,$ then :
- A
$\text{R}=\text{A}\cup\text{B}$
- B
$\text{R}=\text{A}\cap\text{B}$
- ✓
$\text{R}\subseteq\text{A}\times\text{B}$
- D
$\text{R}\subseteq\text{B}\times\text{A}$
AnswerCorrect option: C. $\text{R}\subseteq\text{A}\times\text{B}$
If $R$ is a relation from set $A$ to set $B,$ then $R$ is always a subset of $A \times B$.
View full question & answer→MCQ 51 Mark
If $\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$ is a relation on $Z,$ then the domain of $R$ is:
- A
$\{0, 1, 2\}$
- B
$\{0, -1, -2\}$
- ✓
$\{-2, -1, 0, 1, 2\}$
- D
AnswerCorrect option: C. $\{-2, -1, 0, 1, 2\}$
$\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$
We know that,
$(-2)^2+0^2\leq4$
$\Rightarrow(2)^2+0^2\leq4$
$\Rightarrow(-1)^2+0^2\leq4$
$\Rightarrow(1)^2+0^2\leq4$
$\Rightarrow(-1)^2+(1)^2\leq4$
$\Rightarrow0^2+0^2\leq4$
$\Rightarrow(1)^2+(1)^2\leq4$
$\Rightarrow(-1)^2+(-1)^2\leq4$
Hence, domain $(R) = \{-2, -1, 0, 1, 2\}$
View full question & answer→MCQ 61 Mark
Let $A = \{1, 2, 3\}, B = \{1, 3, 5\}.$ If relation $R$ from $A$ to $B$ is given by $=\{(1, 3), (2, 5), (3, 3)\},$ Then $R^{-1}$ is:
- ✓
$\{(3, 3), (3, 1), (5, 2)\}$
- B
$\{(1, 3), (2, 5), (3, 3)\}$
- C
$\{(1, 3), (5, 2)\}$
- D
AnswerCorrect option: A. $\{(3, 3), (3, 1), (5, 2)\}$
$A =\{1, 2, 3\}, B = \{1, 3, 5\}$
$R = \{(1, 3), (2, 5), (3, 3)\}$
$\therefore R^{-1} = \{(3, 3), (3, 1), (5, 2)\}$
View full question & answer→MCQ 71 Mark
A relation $R$ is defined from $\{2, 3, 4, 5\}$ to $\{3, 6, 7, 10\}$ by : $x \ Ry \leftrightarrow x$ is relatively prime to $y$. Then, domain of $R$ is :
- A
$\{2, 3, 5\}$
- B
$\{3, 5\}$
- C
$\{2, 3, 4\}$
- ✓
$\{2, 3, 4, 5\}$
AnswerCorrect option: D. $\{2, 3, 4, 5\}$
Given,
From $\{2, 3, 4, 5\}$ to ${3, 6, 7, 10}, x \ Ry \leftrightarrow x$ is relatively prime to $y$
$2$ is relatively prime to $3, 7$
$3$ is relatively prime to $7, 10$
$4$ is relatively prime to $3, 7$
$5$ is relatively prime to $3, 6, 7$
So, domain of $R$ is $\{2, 3, 4, 5\}$
View full question & answer→MCQ 81 Mark
A relation $\phi$ from $C$ to $R$ is defined by $\text{x }\phi\text{ y}\Leftrightarrow|\text{x}|=\text{y}.$ Which one is correct?
AnswerCorrect option: D. $\text{i}\ \phi\ 1$
We have,
$|\text{i}|=\sqrt{1^2+0^2}=1$
Thus, $\text{i }\phi\ 1$ satisfies $\text{x}\ \phi\text{ y}\Leftrightarrow|\text{x}|=\text{y}$
View full question & answer→MCQ 91 Mark
Let $R$ be a relation on $N$ defined by $x + 2y = 8$. The domain of $R$ is :
- A
$\{2, 4, 8\}$
- B
$\{2, 4, 6, 8\}$
- ✓
$\{2, 4, 6\}$
- D
$\{1, 2, 3, 4\}$
AnswerCorrect option: C. $\{2, 4, 6\}$
$x + 2y = 8$
$\Rightarrow x = 8 - 2y$
For $y = 1, x = 6$
$y = 2, x = 4$
$y = 3, x = 2$
Then $R = \{(2, 3), (4, 2), (6, 1)\}$
$\therefore$ Domain of $R = \{2, 4, 6\}$
View full question & answer→MCQ 101 Mark
If $R$ is a relation on the set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ given by $x \ Ry \leftrightarrow y = 3x$, then $R =$
- A
$[(3, 1), (6, 2), (8, 2), (9, 3)]$
- B
$[(3, 1), (6, 2), (9, 3)]$
- C
$[(3, 1), (2, 6), (3, 9)]$
- ✓
Answer$A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$
$x \ R y \leftrightarrow y = 3x$
For $x = 1, y = 3$
For $x = 2, y = 6$
For $x = 3, y = 9$
Thus, $R = \{(1, 3), (2, 6), (3, 9)\}$
View full question & answer→MCQ 111 Mark
$R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\}$ defined by $y = x - 3.$ Then, $R^{-1} $ is:
- ✓
- B
- C
$\{(10, 13), (8, 11), (12, 10)\}$
- D
Answer$R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\},$ defined by $y = x - 3$
Now, we have,
$11 - 3 = 8$
$13 - 3 = 10$
So, $R = \{(13, 10), (11, 8)\}$
$\therefore R^{-1} = \{(10, 13), (8, 11)\}$
View full question & answer→MCQ 121 Mark
If $A = \{1, 2, 4\}, B = \{2, 4, 5\}, C = \{2, 5\},$ then $(A - B) \times (B - C)$ is :
- A
$\{(1, 2), (1, 5), (2, 5)\}$
- ✓
$\{(1, 4)\}$
- C
$(1, 4)$
- D
AnswerCorrect option: B. $\{(1, 4)\}$
$A = \{1, 2, 4\}, B = \{2, 4, 5\}$ and $C = \{2, 5\}$
$(A - B) = {1}$
$(B - C) = {4}$
So $, (A - B) \times (B - C) = \{(1, 4)\}$
View full question & answer→MCQ 131 Mark
If $A = \{1, 2, 3\}, B = \{1, 4, 6, 9\}$ and $R$ is a relation from $A$ to $B$ defined by $'x\ '$ is greater than $y$. The range of $R$ is
- A
$\{1, 4, 6, 9\}$
- B
$\{4, 6, 9\}$
- ✓
$\{1\}$
- D
AnswerCorrect option: C. $\{1\}$
$A = \{1, 2, 3\}$ and $B = \{1, 4, 6, 9\}$
$R$ is a relation from $A$ to $B$ defined by: $x$ is greater than $y$.
Then $R = \{(2, 1), (3, 1)\}$
$\therefore$ Range $(R) = \{1\}$
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