3 Marks Question - Maths STD 12 Science Questions - Vidyadip

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22 questions · timed · auto-graded

Question 13 Marks

If A = {1, 2, 3, 4} define relations on A which have properties of being:
Reflexive, transitive but not symmetric.

Answer

The relation on A having properties of being reflexive, transitive, but not symmetric is, R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)} Relation R satisfies reflexivity and transitivity. $ \Rightarrow(1, 1), (2, 2), (3, 3) \in\text{R}$$$and $(1, 1), (2, 1) \in \text{R}\Rightarrow(1, 1)\in \text{R}$
However, $(2,1)\in\text{R},$ but $(1,2)\notin\text{R}$

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Question 23 Marks

Test whether the following relations $R_2$ are:

Reflexive.
Symmetric.
Transitive.

$R_2$ on Z defined by $(\text{a, b})\in\text{R}_2\Leftrightarrow\ |\text{a}-\text{b}|\leq5$

Answer

Reflexivity: Let a be an arbitrary element of $R_2$. Then,$\text{a}\in\text{R}_2$
$\Rightarrow\ |\text{a}-\text{a}|=0\leq5$
So, $R_2$ is reflexive.
Symmetry: Let $(\text{a, b})\in\text{R}_2$
$\Rightarrow\ |\text{a}-\text{b}|\leq5$
$\Rightarrow\ |\text{b}-\text{a}|\leq5$ [Since, |a - b| = |b - a|]
$\Rightarrow\ (\text{b, a})\in\text{R}_2$
So, $R_2$ is symmetric.
Transitivity: Let $(1, 3)\in\text{R}_2$ and $(3,7)\in\text{R}_2$
$\Rightarrow\ |1-3|\leq5$ and $|3-7|\leq5$
But $|1-7|\nleq5$
$\Rightarrow\ (1,7)\notin\text{R}_2$
So, $R_2$ is transitive.

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Question 33 Marks

Let $A = {1, 2, 3},$ and let $R_1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}$. Find whether or not the relations $R_1$ on $A$ is:

Reflexive.
Symmetric.
Transitive.

Answer

We have $A=\{1,2,3\}$, and $R_1=\{(1,1),(1,3),(3,1),(2,2),(2,1),(3,3)\}$ $\therefore(1,1),(2,2)$ and $(3,3) \in R_1$
$\therefore R _1$ is not reflexive.
Now,
$\therefore(2,1) \in R_1 \text { but }(1,2) \notin R_1$
$\therefore R _1$ is not symmetric.
Again,
$\therefore(2,1) \in R _1$ and $(1,3) \in R _1$ but $(2,3) \notin R _1$
$\therefore R_1$ is not transitive.

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Question 43 Marks

The following relation are defined on the set of real numbers.
aRb if $1 + ab > 0$
Find whether these relation are reflexive, symmetric or transitive.

Answer

We have aRb if$ 1 + ab > 0$
Let R be the set of real numbers
Reflexive: Let $\text{a}\in\text{R}$
$\Rightarrow 1 + a^2 > 0$
$\Rightarrow aRa$
$\Rightarrow R $is reflexive.
Symmetric: Let aRb
$\Rightarrow 1 + ab > 0$
$\Rightarrow 1 + ba > 0$
$\Rightarrow bRa$
$\Rightarrow R$ is symmetric
Transitive: Let aRb and bRc
$\Rightarrow 1 + ab > 0$ and $1 + bc > 0$
$\Rightarrow 1 + ac > 0$
$\Rightarrow R$ is not transitive.

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Question 53 Marks

Let A$ = {1, 2, 3}, $and let $R_3 = {(1, 3), (3, 3)}$. Find whether or not the relations $R_3$ on $A$ is:

Reflexive.
Symmetric.
Transitive.

Answer

$R_3 = {(1, 3), (3, 3)}$
$\therefore\ (1,1)\notin\text{R}_3$
$\Rightarrow R_3$ is not reflexive.
Now, $(1,3)\in\text{R}_3$ but $(3,1)\in\text{R}_3$
$\therefore R _3$ is not symmetric.
Again, it is clear that $R_3$ is transitive.

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Question 63 Marks

Show that the relation $''\geq''$ on the set R of all real numbers is reflexive and transitive but not symmetric.

Answer

We have,
relation $\text{R}=\ ''\geq''$ on the set R of all real numbers
Reflexivity: Let $\text{a}\in\text{R}$
$\Rightarrow\ \text{a}\geq\text{a}$
$\Rightarrow\ ''\geq''$ is reflexive.
Symmetric: Let $\text{a, b}\in\text{R}$
Such that $\text{a}\geq\text{b}\Rightarrow\ \text{b}\geq\text{a}$
$\therefore\ ''\geq''$ not symmetric.
Transitivity: Let $\text{a, b, c}\in\text{R}$
and $\text{a}\geq\text{b}\ \&\ \text{b}\geq\text{c}$
$\Rightarrow\ \text{a}\geq\text{c}$
$\Rightarrow\ ''\geq''$ is transitive.

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Question 73 Marks

Three relation$ R_1$is defined in set $A = {a, b, c}$ as follows:
$R_1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}$
Find whether or not the relation $R_1$on A is:

Reflexive.
Symmetric.
Transitive.

Answer

Consider R, Reflexive: Clearly, (a, a), (b, b) and (c, c) $\in\text{R}_1$Therefore, $R_1$ is reflexive.
Symmetric: We see that the ordered pairs obtained by interchanging the components of $R _1$ are also in $R _1$
Therefore,
$R _1$ is symmetric.
Transitive: Here, $\text{a, b}\in\text{R}_1,\ \text{b, c}\in\text{R}_1$ and also $\text{a, c}\in\text{R}_1$
Therefore, $R_1$_ is transitive.

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Question 83 Marks

The following relation are defined on the set of real numbers.
aRb if $|\text{a}|\leq\text{b}$
Find whether these relation are reflexive, symmetric or transitive.

Answer

We have aRb if $|\text{a}|\leq\text{b}$ Reflexive: Let $\text{a}\in\text{R}$$\Rightarrow\ |\text{a}|\nleq\text{a}$ $[\therefore|-2|=2>-2|]$
⇒ R is not reflexive. Symmetric: Let aRb $\Rightarrow\ |\text{a}|\leq\text{b}$ $\Rightarrow\ |\text{b}|\leq\text{a}$ $\begin{bmatrix}\therefore\ \ \ \text{Let a}=4, \text{b}=6 \\\ \ \ \ \ \ \ \ \ |4|\leq 8 \text{ but } |8|>4\end{bmatrix}$ ⇒ R is not symmetric. Transitive: Let aRb and bRc$\Rightarrow\ |\text{a}|\leq\text{b}$ and $|\text{b}|\leq\text{c}$
$\Rightarrow\ |\text{a}|\leq|\text{b}|\leq\text{c}$
$\Rightarrow\ |\text{a}|\leq\text{c}$
$\Rightarrow\ \text{aRc}$
⇒ R is transitive.

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Question 93 Marks

Give an example of a relation which is,
Reflexive and symmetric but not transitive.

Answer

Let A = {4, 6, 8} Define a relation R on A as: A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}Relation R is reflexive since for every $\text{a}\in\text{A},\ (\text{a, a})\in\text{R}$ i.e., (4, 4), (6, 6), (8, 8) $\in\text{R}$
Relation R is symmetric since $(\text{a, b})\in\text{R}\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{R.}$
Relation R is not transitive since (4, 6), (6, 8) $\in\text{R,}$ but $(4,8)\notin\text{R.}$
Hence, relation R is reflexive and symmetric but not transitive.

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Question 103 Marks

If R and S are transitive relations on a set A, then prove that $\text{R}\cup\text{S}$ may not be a transitive relation on A.

Answer

Let A = {a, b, c} and R and S be two relations on a, given by R = {(a, a), (a, b), (b, a), (b, b)}
And S = {(b, b), (b, c), (c, b), (c, c)}
Here, the relations R and S are transitive on A.
$\text{a, b}\in\text{R}\cup\text{S}$ and $\text{b, c}\in\text{R}\cup\text{S}$ But $\text{a, c}\notin\text{R}\cup\text{S}$
Hence, $\text{R}\cup\text{S}$ is not a transitive relation on A.

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Question 113 Marks

Three relation $R_2$ is defined in set $A=\{a, b, c\}$ as follows:
$R_2=\{(a, a)\}$
Find whether or not the relation $R _2$ on A is:

Reflexive.
Symmetric.
Transitive.

Answer

$R_2$ is Reflexive: Clearly $a, a \in R_2$
Therefore, $R_2$ is reflexive.
Symmetric: Clearly, $a, a \in R \Rightarrow a, a \in R$.
Therefore, $R _2$ is symmetric.
Transitive: $R _2$ is clearly a transitive relation, since there is only one element in it.

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Question 123 Marks

Let $A = {1, 2, 3},$ and let $R_2 = {(2, 2), (3, 1), (1, 3)}$. Find whether or not the relations $R_2$ on$ A$ is:

Reflexive.
Symmetric.
Transitive.

Answer

$R_2 = {(2, 2), (3, 1), (1, 3)}$
$\therefore\ (1,1)\notin\text{R}_2$
$\Rightarrow R_1$ is not reflexive.
Now, $(1,3) \in R _2$
$\Rightarrow(3,1) \in R_2$
$\therefore R _2$ is symmetric.
Again,
$(3,1) \in R_2 \text { and }(1,3) \in R_2 \text { but }(3,3) \notin R_2$
$\therefore R _2$ is not transitive.

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Question 133 Marks

Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

Answer

Reflexivity: Let a be an arbitrary element of R. Then, a = a + 1 cannot be true for all $\text{a}\in\text{A.}$ $\Rightarrow\ (\text{a, a})\notin\text{R}$ So, R is not reflexive on A. Symmetry: Let $(\text{a, b})\in\text{R}$ ⇒ b = a + 1 ⇒ -a = -b + 1 ⇒ a = b - 1 Thus, $(\text{b, a})\notin\text{R}$ So, R is not symmetric on A.Transitivity: Let (1, 2) and (2, 3) $\in\text{R}$
⇒ 2 = 1 + 1 and 3 = 2 + 1 is true. But $3\neq1+1$ $\Rightarrow\ (1,3)\notin\text{R}$ So, R is not transitive on A.

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Question 143 Marks

Check whether the relation $R$ on $R$ defined by $R = \{(a, b): a \leq b^3\}$ is reflexive, symmetric or transitive.

Answer

The relation R on R is defined by R. We observe that $(-2)\leq(-2)^3$ is not true. Therefore, R is not reflexive. Since $1\leq\Big(3^{\frac{1}{3}}\Big)^3$ but $3^\frac{1}{3}\leq1$ i.e. $\Big(1,3^\frac{1}{3}\Big)\in\text{R.}$
Therefore, R is not symmetric.
Hence, R is not transitive because $(5,2)\in\text{R}$ and $\Big(2,2^\frac{1}{3}\Big)\in\text{R}$ but $\Big(5,2^\frac{1}{3}\Big)\notin\text{R.}$

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Question 153 Marks

Three relation $R_3$ is defined in set $A=\{a, b, c\}$ as follows:
$R_3=\{(b, c)\}$
Find whether or not the relation $R_3$ on $A$ is:

Reflexive.
Symmetric.
Transitive.

Answer

$R_3$ is Reflexive: Here $b, b \notin R_3$ neither $c, c \notin R_3$ Therefore, $R_3$ is not reflexive.
Symmetric: Here, b, c $\notin R _3$, but c, c $\notin R _3$
So, $R_3$ is not symmetric.
Transitive: Here, $R_3$ has only two elements.
Hence, $R_3$ is transitive.

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Question 163 Marks

Give an example of a relation which is,Symmetric but neither reflexive nor transitive.

Answer

Let A = {5, 6, 7}.
Define a relation R on A as R = {(5, 6), (6, 5)}.
Relation R is not reflexive as $(5, 5), (6, 6), (7, 7)\notin\text{R.}$ $$
Now, as $(5, 6)\in\text{R}$ and also $(6,5)\in\text{R,}$ R is symmetric.
$\Rightarrow(5, 6), (6, 5)\in\text{R,}\text{but}(5, 5)\notin\text{R}$ $$ $$ $$
Therefore, R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.

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Question 173 Marks

Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.

Answer

Let A be a set. Then $\text{I}_\text{A}=\{(\text{a, a});\text{ a}\in\text{A}\}$ is the identity relation on A. Hence, every identity relation on a set is reflexive by defination. Converse: Let A = {(a, b, c)} be a set. Let R = {(a, a), (b, b), (c, c), (a, b)} be a relation defined on A.Clearly R is reflexive on set A, but it is not identity relation on set A as $(\text{a, b})\in\text{R}$
Hence, a reflexive relation need not be identity relation.

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Question 183 Marks

Three relation $R_4$ is defined in set$ A = {a, b, c}$ as follows:
$R_4 = {(a, b), (b, c), (c, a)}$
Find whether or not the relation $R_4$ on $A$ is:

Reflexive.
Symmetric.
Transitive.

Answer

$R_4$is reflexive: Here, $\text{a}\notin\text{R}_4,\ \text{b},\ \text{b}\notin\text{R}_4,\ \text{c},\ \text{c}\notin\text{R}_4$
Therefore, $R_4$ is not reflexive.
Symmetric: Here, $\text{a, b}\notin\text{R}_4,$ but $\text{b, b}\notin\text{R}_4$
Therefore, $R_4$is not symmetric.
Transitive: Here, $\text{a, b}\notin\text{R}_4,\ \text{b, c}\in\text{R}_4$
But $\text{a, c}\notin\text{R}_4$
Therefore,$R_4$ is not transitive.

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Question 193 Marks

The following relation are defined on the set of real numbers.
aRb if a - b > 0
Find whether these relations are reflexive, symmetric or transitive.

Answer

aRb if a - b > 0 Let R be the set of real numbers. Reflexivity: Let $\text{a}\in\text{R}$ $\Rightarrow\ \text{a}-\text{a}=0$ $\Rightarrow\ (\text{a, a})\notin\text{R}$ $\therefore$ R is not reflexive. Symmetric: Let aRb ⇒ a - a > 0 ⇒ b - a < 0$\therefore$ R is not symmetric.
Transitive: Let aRb and bRc
⇒ a - a > and b - c > 0 ⇒ a - c > 0 ⇒ aRc$\therefore$ R is transitive.

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Question 203 Marks

An integer m is said to be related to another integer n if m is a multiple of n.Check if the relation is symmetric, reflexive and transitive.

Answer

$\text{R} = \big\{{(\text{m, n})}(1, 1), (2, 1)\Rightarrow(1, 1): \text{m, n} \in\text{Z,}\text{m} \\=\text{kn},\text{where}\text{k}\in\text{N}\big\}$$$$$$$ $$ $$
$$Reflexivity: Let m be an arbitrary element of R. Then, m = km is true for k = 1$\Rightarrow\ (\text{m, m})\in\text{R}$
Thus, R is reflexive. Symmetry: Let $(\text{m, n})\in\text{R}$ ⇒ m = kn for some $\text{k}\in\text{N}$ $\Rightarrow\ \text{n}=\frac{1}{\text{k}}\text{m}$ $\Rightarrow\ (\text{n, m})\notin\text{R}$ Thus, R is not symmetric.Transitivity: Let (m, n) and (n, o) $\in\text{R}$
⇒ m = kn and n = lo for some $\text{k, l}\in\text{N}$ ⇒ m = (kl)o Here, $\text{kl}\in\text{N}$ $\Rightarrow\ (\text{m, o})\in\text{R}$ Thus, R is transitive.

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Question 213 Marks

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, add a minimum number of ordered pairs so that the enlarged relation is symmeteric, transitive and reflexive.

Answer

We have,
R = {(1, 2), (2, 3)}
R can be a transitive only when the elements (1, 3) is added
R can be a reflexive only when the elements (1, 1), (2, 2), (3, 3) are added
R can be a symmetric only when the elements (2, 1), (3, 1) and (3, 2) are added
So, the required enlarged relation, R' = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} = A × A

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Question 223 Marks

Let A be the set of all human beings in a town at a particular time. Determine whether the following relations are reflexive, symmetric and transitive:
R = {(x, y): x is father of and y}

Answer

A be the set of human beings
R = {(x, y): x is father of and y}
Reflexive: Since x can not be father of x
$\therefore\ (\text{x, x})\notin\text{R}$
⇒ R is not reflexive.
Symmetric: Let $(\text{x, y})\in\text{R}$
⇒ x is father of y
⇒ y can not be father of x
$\Rightarrow\ (\text{y, x})\notin\text{R}$
⇒ R is not Symmetric.
Transitive: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}$
⇒ x is father of y and y is father of z
⇒ x is grandfather of z
$\Rightarrow\ (\text{x, z})\notin\text{R}$
⇒ R is not transitive.

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3 Marks Question - Maths STD 12 Science Questions - Vidyadip