5 Marks Questions - Maths STD 12 Science Questions - Vidyadip

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5 Marks Questions

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34 questions · 6 auto-graded MCQ + 28 self-marked written.

Question 15 Marks

The relation S defined on the set R of all real number by the rule aSb iff a ≥ b is:

An equivalence relation.
Reflexive, transitive but not symmetric.
Symmetric, transitive but not reflexive.
Neither transitive nor reflexive but symmetric.

Answer

Reflexive, transitive but not symmetric.

Solution:

The relation S is reflexive, since for any $(\text{a, a})\in\text{S}$ the condition a2b holds,

The relation S is not symmetric since, for any $(\text{a, b}]\in\text{S}$ but $(\text{b, a})\notin\text{S}$

The relation S is transitive since, for any $(\text{a, b}]\in\text{S}$ and $(\text{b, c})\in\text{S}$

Therefore, $(\text{a, c})\notin\text{S}$

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MCQ 25 Marks

$S$ is a relation over the set $R$ of all real numbers and it is given by $(\text{a, b})\in\text{S}\Leftrightarrow\text{ab}\geq0.$ Then, $S$ is:

A
Symmetric and transitive only.
B
Reflexive and symmetric only.
C
Antisymmetric relation.
✓
An equivalence relation.

Answer

Correct option: D.

An equivalence relation.

Reflexivity: Let $\text{a}\in\text{R}$
Then,
$aa = a^2 > 0 \Rightarrow\ \text{a, }\forall$
So, $S$ is reflexive on $R.$
Symmetry: Let $(\text{a, b})\in\text{S}$
Then,
$\text{a, b}\in\text{S}\Rightarrow\ \text{ab}\geq0$
$\Rightarrow\ \text{ba}\geq0\Rightarrow\ \text{ba}\geq0$
$\Rightarrow\ \text{b, a}\in\text{S}\ \forall\ \text{a, b}\in\text{R}$
So, S is symmetric on $R.$
Transitivity: If $\text{a, b, b, c}\in\text{S}$
$\Rightarrow\ \text{ab}\geq0$ and $\text{bc}\geq0$
$\Rightarrow\ \text{ab}\times\text{bc}\geq0$
$\Rightarrow\ \text{ac}\geq0$
$\text{b}^2\geq0\Rightarrow\ \text{a, c}\in\text{S}$ for all $\text{a, b, c}\in\text{set R}$
Hence, $S$ is an equivalence relation on $R.$

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Question 35 Marks

The relation R = {(1, 1), (2, 2), (3, 3)} on the set {1, 2, 3} is:

Symmetric only.
Reflexive only.
An equivalence relation.
Transitive only.

Answer

An equivalence relation.

Solution:

R = {(a, b): a = b and a, b $\in\text{A}$}

Reflexivity: Let $\text{a}\in\text{A}$

Here,

a = a

$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is reflexive on A.

Symmetry: Let $\text{a, b}\in\text{A}$ such that $ (\text{a, b})\in\text{R}.$ Then,

$ (\text{a, b})\in\text{R}$

$\Rightarrow\ \text{a}=\text{b}$

$\Rightarrow\ \text{b}=\text{a}$

$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is symmetric on A.

Transitive: Let $\text{a, b, c}\in\text{A}$ such that $ (\text{a, b})\in\text{R}$ and $ (\text{b, c})\in\text{R}.$ Then,

$ (\text{a, b})\in\text{R}\Rightarrow\ \text{a}=\text{b}$

and $ (\text{b, c})\in\text{R}\Rightarrow\ \text{b}=\text{c}$

$\Rightarrow\ \text{a}=\text{c}$

$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is transitive on A.

Hence, R is an equivalence relation on A.

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Question 45 Marks

Let L denote the set of all straight lines in a plane. Let a relation R be defined by lRm if l is perpendicular to m for all l, m ∈ L. Then, R is:

Reflexive.
Symmetric.
Transitive.
None of these.

Answer

Symmetric.

Solution:

Given that L denote the set of all straight lines in a plane.

A relation R be defined by lRm if is perpendicular to m for all l, m ∈ L.

R is not reflexive. R is symmetric as we can say $\text{l}\bot\text{m}$ or $\text{m}\bot\text{l}.$

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Question 55 Marks

In the set Z of all integers, which of the following relation R is not an equivalence relation?

xRy : if $\text{x}\leq\text{y}$
xRy : if x = y
xRy : if x - y is an even integer
xRy : if $\text{x}\equiv\text{y}\ (\text{mod 3})$

Answer

xRy : if $\text{x}\leq\text{y}$

Solution:

In the set of Z of all integers xRy : if $\text{x}\leq\text{y}$ is not an equivalence relation.

For the relation $\text{x}\leq\text{y}(\text{x, y})\in\text{R}$ but (y, x) not belongs to y as $\text{y}\geq\text{x}$ given.

Hence, it is not an equivalence relation.

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Question 65 Marks

If A = {1, 2, 3}, then a relation R = {(2, 3)} on A is:

Symmetric and transitive only.
Symmetric only.
Transitive only.
None of these.

Answer

Transitive only.

Solution:

The relation R is not reflexive because every element of A is not related to itself. Also, R is not symmetric since on interchanging the elements, the ordered pair in R is not contained in it.

R is transitive by default because there is only one element in it.

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MCQ 75 Marks

$R$ 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
$\{(10, 13), (8, 11)\}$
D
None of these.

Answer

Correct option: A.

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

Given that $R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\}$ defined by $y = x - 3.$
$R = \{(8, 11), (10, 13)\}$
$R^{-1} = \{(8, 11), (10, 13)\}$
As inverse function of $R$ is,
$y + 3 = x$
$\Rightarrow y = x + 3$

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Question 85 Marks

If R is a relation on the set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3)}, then R is:

Reflexive.
Symmetric.
Transitive.
All the three options.

Answer

All the three options.

Solution:

R = a, b : a = b and $\text{a, b}\in\text{A}$

Reflexivity: Let $\text{a}\in\text{A}$

Then, a = a

$\Rightarrow\ \text{a, a}\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is reflexive on A.

Symmetry: Let $\text{a, b}\in\text{A}$ such that $\text{a, b}\in\text{R.}$

Then, $\text{a, b}\in\text{R}$

⇒ a = b ⇒ b = a ⇒ b, $\text{a}\in\text{R}$ for all $\text{a}\in\text{A}$

So, R is symmetric on A.

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Question 95 Marks

If A = {a, b, c, d}, then a relation R = {(a, b), (b, a), (a, a)} on A is:

Symmetric and transitive only.
Reflexive and transitive only.
Symmetric only.
Transitive only.

Answer

Symmetric and transitive only.

Solution:

Given that A = {a, b, c, d} then a relation R = {(a, b), (b, a), (a, a)} on A.

(a, b), (b, a) $\in\text{R}$

⇒ R is symmetric.

Also for (a, a) R is symmetric.

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Question 105 Marks

If R is the largest equivalence relation on a set A and S is any relation on A, then:

$\text{R}\subset\text{S}$
$\text{S}\subset\text{R}$
$\text{R = S}$
None of these.

Answer

$\text{S}\subset\text{R}$

Solution:

Given that R is the largest relation on A and S is any relation on A.

We know that R is always subset of A × A.

Hence, $\text{S}\subset\text{R}.$

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MCQ 115 Marks

If a relation R is defined on the set Z of integers as follows: $(a, b) \in R ⇔ a^2 + b^2 = 25$. Then, domain (R) is:

A
$\{3, 4, 5\}$
B
$\{0, 3, 4, 5\}$
✓
$\{0,\pm3,\pm4,\pm5\}$
D
None of these.

Answer

Correct option: C.

$\{0,\pm3,\pm4,\pm5\}$

$\{0,\pm3,\pm4,\pm5\}$

Solution:
As aRb ⇔ a < b
does not satisfy reflexive and symmetric relation.

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Question 125 Marks

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:

{1, 4, 6, 9}
{4, 6, 9}
{1}
None of these.

Answer

{1}

Solution:

Here, $\text{R}=\text{x, y}:\text{x}\in\text{A}$ and $\text{y}\in\text{B}:\text{x}>\text{y}$ ⇒ R = 2, 1, 3, 1

Thus, Range of R = {1}

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MCQ 135 Marks

The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is:

A
$1$
B
$2$
C
$3$
✓
$5$

Answer

Correct option: D.

$5$

The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is,
$R_1 = \{(1, 1)\}$
$R_2 = \{(2, 2)\}$
$R_3 = \{(3, 3)\}$
$R_4 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
$R_5 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}$
The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is $5.$

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Question 145 Marks

Let R be the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$ Then, R is:

Symmetric.
Reflexive.
Transitive.
An equivalence relation.

Answer

Symmetric.

Solution:

Given R is the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$

It is symmetric relation as we can say either or $\text{l}_2\bot\text{l}_1.$

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Question 155 Marks

Let R be the relation on the set A = {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then,

R is reflexive and symmetric but not transitive.
R is reflexive and transitive but not symmetric.
R is symmetric and transitive but not reflexive.
R is an equivalence relation.

Answer

R is reflexive and transitive but not symmetric.

Solution:

Reflexivity: Clearly, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$

So, R is reflexive on A.

Symmetry: Since, $1,2\in\text{R},$ but $2,1\notin\text{R,}$ R is not symmetric on A.

Transitivity: Since, $1,3,3,2\in\text{R}$ and $1,2\in\text{R},$ R is transitive on A.

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Question 165 Marks

Let R be a relation on the set N given by R = {(a, b): a = b - 2, b > 6}. Then,

(2, 4) ∈ R
(3, 8) ∈ R
(6, 8) ∈ R
(8, 7) ∈ R

Answer

(6, 8) ∈ R

Solution:

a = b - 2 ⇒ 6 = 8 - 2 and b = 8 > 6

Hence, (6, 8) ∈ R

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Question 175 Marks

Consider a non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then, R is:

Symmetric but not transitive.
Transitive but not symmetric.
Neither symmetric nor transitive.
Both symmetric and transitive.

Answer

Transitive but not symmetric.

Solution:

We have,

R = {(a, b): a is brother of b}

Let $(\text{a, b})\in\text{R}.$ Then,

a is brother of b.

but b is not necessary brother of a (As, b can be sister of a)

$\Rightarrow\ (\text{b, a})\notin\text{R}$

So, R is not symmetric.

Also,

Let $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$

⇒ a is brother of b and b is brother of c

⇒ a is brother of c

$\Rightarrow\ (\text{a, c})\in\text{R}$

So, R is transitive.

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Question 185 Marks

Let R be a relation on N defined by x + 2y = 8. The domain of R is:

{2, 4, 8}
{2, 4, 6, 8}
{2, 4, 6}
{1, 2, 3, 4}

Answer

{2, 4, 6}

Solution:

The relation R is defined as R = x, y: $\text{x, y}\in\text{N}$ and x + 2y = 8

⇒ R = x, y: $\text{x, y}\in\text{N}$ and $\text{y}=\frac{8-\text{x}}{2}$

Domain of R is all values of $\text{x}\in\text{N}$ satisfying the relation R.

Also, there are only three values of x that result in y, which is a natural number.

These are {2, 6, 4}.

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Question 195 Marks

The relation 'R' in N × N such that (a, b)R(c, d) ⇔ a + d = b + c is:

Reflexive but not symmetric.
Reflexive and transitive but not symmetric.
An equivalence relation.
None of the these.

Answer

An equivalence relation.

Solution:

We observe the following properties of relation R.

Reflexivity: Let $(\text{a, b})\in\text{N}\times\text{N}$

$\Rightarrow\ \text{a, b}\in\text{N}$

$\Rightarrow\ \text{a}+\text{b}=\text{b}+\text{a}$

$\Rightarrow\ (\text{a, b})\in\text{R}$

So, R is reflexive on N × N.

Symmetry: Let $(\text{a, b}),\ (\text{c, d})\in\text{N}\times\text{N}$ such that (a, b)R(c, d)

$\Rightarrow\ \text{a}+\text{d}=\text{b}+\text{c}$

$\Rightarrow\ \text{d}+\text{a}=\text{c}+\text{b}$

$\Rightarrow\ (\text{d, c}),\ (\text{b, a})\in\text{R}$

So, R is symmetric on N × N.

Transitivity: Let $(\text{a, b}),\ (\text{c, d}),\ (\text{e, f})\in\text{N}\times\text{N}$ such that (a, b)R(c, d) and (c, d)R(e, f)

⇒ a + d = b + c and c + f = d + e

⇒ a + d + c + f = b + c + d + e

⇒ a + f = b + e

⇒ (a, b)R(e, f)

So, R is transitive on N × N.

Hence, R is an equivalence relation on N.

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Question 205 Marks

Let A = {1, 2, 3} and B = {(1, 2), (2, 3), (1, 3)} be a relation on A. Then, R is:

Neither reflexive nor transitive.
Neither symmetric nor transitive.
Transitive.
None of these.

Answer

Transitive.

Solution:

Reflexivity: Since $(1, 1)\notin\text{B,}$ B is not reflexive on A.

Symmetry: Since $1,2\in\text{B}$ but $2,1\notin\text{B,}$ B is not symmetric on A.

Transitivity: Since $1,2\in\text{B},\ 2,3\in\text{B}$ and $1,3\in\text{B,}$ B is transitive on A.

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Question 215 Marks

R is a relation on the set Z of integers and it is given by (x, y) ∈ R ⇔ | x - y | ≤ 1. Then, R is:

Reflexive and transitive.
Reflexive and symmetric.
Symmetric and transitive.
An equivalence relation.

Answer

Reflexive and symmetric.

Solution:

Reflexivity: Let $\text{x}\in\text{R.}$ Then,

$\text{x}-\text{x}=0<1$

$\Rightarrow\ |\text{x}-\text{x}|\leq1$

$\Rightarrow\ (\text{x, x})\in\text{R}$ for all $\text{x}\in\text{Z}$

So, R is reflexive on Z.

Symmetry: Let $\text{x, y}\in\text{R.}$ Then,

$|\text{x}-\text{y}|\leq0$

$\Rightarrow\ |-(\text{y}-\text{x})|\leq1$

$\Rightarrow\ |(\text{y}-\text{x})|\leq1$ [Since |x - y| = |y - x|]

$\Rightarrow\ (\text{y, x})\in\text{R}$ for all $\text{x, y}\in\text{Z}$

So, R is symmetric on Z.

Transitivity: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}.$ Then,

$|\text{x}-\text{y}|\leq1$ and $|\text{y}-\text{z}|\leq1$

⇒ It is not always true that $|\text{x}-\text{y}|\leq1.$

$\Rightarrow\ (\text{x, z})\notin\text{R}$

So, R is not transitive on Z.

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Question 225 Marks

If A = {a, b, c}, then the relation R = {(b, c)} on A is:

Reflexive only.
Symmetric only.
Transitive only.
Reflexive and transitive only.

Answer

Transitive only.

Solution:

The relation R = {(b, c)} is neither reflexive nor symmetric because every element of A is not related to itself. Also, the ordered pair of R obtained by interchanging its elements is not contained in R.

We observe that R is transitive on A because there is only one pair.

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Question 235 Marks

For real numbers x and y, define xRy if $\text{x}-\text{y}+\sqrt{2}$ is an irrational number. Then the relation R is:

Reflexive.
Symmetric.
Transitive.
None of these.

Answer

Reflexive.

Solution:

We have,

$\text{R} = \big\{(\text{x, y}):\text{x}-\text{y}+\sqrt{2}$ $$ is an irrational number, $\text{x, y}\in\text{R}\big\}$

As, $\text{x}-\text{x}+\sqrt{2}=\sqrt{2},$ which is an irrational number

$\Rightarrow\ (\text{x, x})\in\text{R}$

So, R is reflexive relation.

Since, $\Big(\sqrt{2},2\Big)\in\text{R}$

i.e. $\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2,$ which is an irrational number

but $2-\sqrt{2}+\sqrt{2}=2,$ which is a rational number

$\Rightarrow\ \Big(2,\sqrt{2}\Big)\notin\text{R}$

So, R is not symmetric relation.

Also, $\Big(\sqrt{2},2\Big)\in\text{R}$ and $\Big(2,2\sqrt{2}\Big)\in\text{R}$

$\Rightarrow\ \Big(\sqrt{2},2\sqrt{2}\Big)\notin\text{R}$

So, R is not transitive relation.

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Question 245 Marks

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b for all a, b ∈ T. Then, R is:

Reflexive but not symmetric.
Transitive but not symmetric.
Equivalence.
None of these.

Answer

Equivalence.

Solution:

Given that R is T be the set of all triangle in the Euclidean plane, and a relation R on T be defined as aRb if a is congruent to b for all a, b ∈ T.

Here, congruency of triangles follows reflexive, symmetric and transitive property.

Hence, it is an equivalence relation.

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Question 255 Marks

Let R = {(a, a), (b, b), (c, c), (a, b)} be a relation on set A = a, b, c. Then, R is:

Identify relation.
Reflexive.
Symmetric.
Antisymmetric.

Answer

Reflexive.

Solution:

Reflexivity: Since $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A},$ R is reflexive on A.

Symmetry: Since $(\text{a, b})\in\text{R}$ but $(\text{b, a})\notin\text{R,}$ is not symmetric on A.

⇒ R is not antisymmetric on A.

Also, R is not an identity relation on A.

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MCQ 265 Marks

Let $A = \{1, 2, 3\}$. Then, the number of equivalence relations containing $(1, 2)$ is:

A
$1$
✓
$2$
C
$3$
D
$4$

Answer

Correct option: B.

$2$

Given that $A = \{1, 2, 3\}.$ Then, the number of equivalence relation containing $(1, 2)$ is,
$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
$R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)\}$
Then, the number of equivalence relation containing $(1, 2)$ is $2.$

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Question 275 Marks

If R is a relation on the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} given by xRy ⇔ y = 3x, then R =

{(3, 1), (6, 2), (8, 2), (9, 3)}
{(3, 1), (6, 2), (9, 3)}
{(3, 1), (2, 6), (3, 9)}
None of these.

Answer

None of these.

Solution:

The relation R is defined as,

$\text{R}=\{(\text{x, y}):\ \text{x, y}\in\text{A}:\text{y}=3\text{x}\}$

⇒ R = {(1, 3), (2, 6), (3, 9)}

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Question 285 Marks

A relation $\phi$ from C to R is defined by $\text{x }\phi\text{ y}\Leftrightarrow|\text{x}|=\text{y.}$ Which one is correct?

$(2+3\text{i})\phi13$
$3\phi(-3)$
$(1+\text{i})\phi2$
$\text{i}\phi1$

Answer

$\text{i}\phi1$

Solution:

$\because\ |2+3\text{i}|=\sqrt{13}\neq13$

$|3|\neq-3$

$|1+\text{i}|=\sqrt{2}\neq2$

and $|\text{i}|=1$

So, $(\text{i, }1)\in\phi$

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MCQ 295 Marks

The relation $R$ defined on the set $A=\{1,2,3,4,5\}$ by $R=\left\{(a, b):\left|a^2-b^2\right|<16\right\}$ is given by:

A
{(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)}
B
{(2, 2), (3, 2), (4, 2), (2, 4)}
C
{(3, 3), (4, 3), (5, 4), (3, 4)}
✓
None of these.

Answer

Correct option: D.

None of these.

None of these.

Solution:
R is given by {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4),(1, 3), (3, 1), (1, 4), (4, 1), (2, 4), (4, 2)} which is not mentioned in (a), (b) or (c).

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Question 305 Marks

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then, R is:

Reflexive but not symmetric.
Reflexive but not transitive.
Symmetric and transitive.
Neither symmetric nor transitive.

Answer

Reflexive but not symmetric.

Solution:

We have,

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

As, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$

So, R is reflexive relation.

Also, $(1,2)\in\text{R}$ but $(2,1)\notin\text{R}$

So, R is not symmetric relation.

And, $(1,2)\in\text{R},\ (2,3)\in\text{R}$ and $(1,3)\in\text{R}$

So, R is transitive relation.

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Question 315 Marks

Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then, R is:

Reflexive and symmetric.
Transitive and symmetric.
Equivalence.
Reflexive, transitive but not symmetric.

Answer

Reflexive, transitive but not symmetric.

Solution:

We have,

R = {(m, n): n divides m; m, n ∈ N}

As, m divides m

$\Rightarrow\ (\text{m, m})\in\text{R}\ \forall\ \text{m}\in\text{N}$

So, R is reflexive.

Since, $(2,1)\in\text{R}$ i.e. 1 divides 2

but 2 cannot divide 1 i.e. $(2,1)\notin\text{R}$

So, R is not symmetric.

Let $(\text{m, n})\in\text{R}$ and $(\text{n, p})\in\text{R.}$ Then,

n divides m and p divides n

⇒ p divides m

$\Rightarrow\ (\text{m, p})\in\text{R}$

So, R is transitive.

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Question 325 Marks

Let A = {2, 3, 4, 5, ..., 17, 18}. Let $'\simeq'$ be the equivalence relation on A × A, cartesian product of A with itself, defined by $(\text{a, b})\simeq(\text{c, d)}$ if ad = bc. Then, the number of ordered pairs of the equivalence class of (3, 2) is:

4
5
6
7

Answer

6

Solution:

The ordered pairs of the equivalence class of (3, 2) are {(3, 2), (6, 4), (9, 6), (12, 8), (15, 10), (18, 12)}.

We observe that these are 6 pairs.

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Question 335 Marks

A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. Then, domain of R is:

{2, 3, 5}
{3, 5}
{2, 3, 4}
{2, 3, 4, 5}

Answer

{2, 3, 4, 5}

Solution:

Given that relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. R can be written as,

{(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}

Here we can see that domain means x element which is $2\leq\text{x}\leq5.$

Hence, {2, 3, 4, 5}

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Question 345 Marks

Let A = {1, 2, 3}. Then, the number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is:

1
2
3
4

Answer

1

Solution:

Given that A = {1, 2, 3}

To find the number of relations containing (1, 2) and (1, 3) then R can be written as {(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)}

Here, we can see that

(3, 1) and (1, 2) ⇒ (3, 2) which is not belongs to R.

The number of relations containing (1, 2) and (1, 3)

Which are reflexive and symmetric but not transitive is 1.

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