M.C.Q (1 Marks)

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
$\text{n}^\text{n}$

Answer

Correct 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 $\text{A}=2^{\text{n}^2}$

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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
$m^n$

Answer

Correct 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 $2^{mn}$

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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

Correct option: C.

$pq$

$n(A \times B) = n(A) \times n(B)$
$n(A \times B) = p \times q = pq$

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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
$\text{R}\subseteq\text{B}\times\text{A}$

Answer

Correct 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 × B.

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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
none of these.

Answer

Correct 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}

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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
none of these.

Answer

Correct option: A.

$\{(3,3),(3,1),(5,2)\}$

$A=\{1,2,3\}, B=\{1,3,5\}$
$R=\{(1,3),(2,5),(3,3)\}$
$\therefore R^{-1}=\{(3,3),(3,1),(5,2)\}$

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MCQ 71 Mark

A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : x R y ⇔ 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}

Answer

Correct option: D.

{2, 3, 4, 5}

Given,
From {2, 3, 4, 5} to {3, 6, 7, 10}, x R y ⇔ 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}

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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?

A
$(2+3\text{i})\ \phi\ 13$
B
$3\phi\ (-3)$
C
$(1+\text{i})\ \phi\ 2$
✓
$\text{i}\ \phi\ 1$

Answer

Correct option: D.

$\text{i}\ \phi\ 1$

We have,
$|\text{i}|=\sqrt{1^2+0^2}=1$
Thus, $\text{i }\phi\ 1$ satisfies $\text{x}\ \phi\text{ y}\Leftrightarrow|\text{x}|=\text{y}$

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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}

Answer

Correct option: C.

{2, 4, 6}

x + 2y = 8
⇒ 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}

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MCQ 101 Mark

If R is a relation on the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} given by x R y ⇔ 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)]
✓
none of these.

Answer

Correct option: D.

none of these.

A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
x R y ⇔ 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)}

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MCQ 111 Mark

$R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\}$ defined by $y = x - 3.$ Then, $R^{-1}$ is:

✓
$\{(8, 11), (10, 13)\}$
B
$\{(11, 8), (13, 10)\}$
C
$\{(10, 13), (8, 11), (12, 10)\}$
D
none of these.

Answer

Correct option: A.

$\{(8, 11), (10, 13)\}$

$R$ 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, $R = \{(13, 10), (11, 8)\}$
$\therefore R^{-1} = \{(10, 13), (8, 11)\}$

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MCQ 121 Mark

If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A - B) × (B - C) is:

✓
{(1, 2), (1, 5), (2, 5)}
B
{(1, 4)}
C
(1, 4)
D
none of these.

Answer

Correct option: A.

{(1, 2), (1, 5), (2, 5)}

A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}
(A - B) = {1}
(B - C) = {4}
So, (A - B) × (B - C) = {(1, 4)}

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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
none of these.

Answer

Correct 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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Maths M.C.Q (1 Marks)

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