Q.$A = \{1, 2, 3, 5\}$ and $B = \{4, 6, 9\}$. Define a relation from A to B by $R = \{(x, y) : \text{the difference between } x \text{ and } y \text{ is odd}, x \in A, y \in B\}$. Write R in roaster form.

R only relates people who BOTH voted; since Y did not cast a vote, no pair involving Y as either element can be in R.

The relation is $R=\{(V_1,V_2): V_1,V_2\in A \text{ and both } V_1 \text{ and } V_2 \text{ casted their votes}\}$.

We're told $X$ casted his vote, but $Y$ did not cast his vote.

For the ordered pair $(X,Y)$ to belong to $R$, both $X$ and $Y$ must have voted. Since $Y$ did not vote, this condition fails regardless of $X$'s status.

So $(X,Y) \notin R$.

…

Since both X and W (X's wife) casted their votes, the pair condition of R is satisfied in both orders.

We're told $X$ casted his vote, and $W$ (his wife) also casted her vote.

Since the relation $R$ requires only that both elements of the pair voted — with no directional/order-dependent condition — and both $X$ and $W$ voted:

$(X,W)\in R \quad\text{and}\quad (W,X)\in R$

…

F1, F2, F3 all casted their votes, so every pair drawn from among them satisfies R's "both voted" condition.

We're told $F_1, F_2, F_3$ are all voters who all casted their votes.

…

Testing each property: R fails reflexivity (since only ~65% of voters actually voted, not everyone in A), but R IS symmetric and IS transitive whenever it holds — a case the listed 4 options don't fully capture.

Let's test each property of $R = \{(V_1,V_2): V_1,V_2\in A,\ V_1\text{ and }V_2\text{ both voted}\}$ against the full set $A$ (all eligible voters, not just those who voted):

Reflexive? This requires $(a,a)\in R$ for every $a\in A$, i.e. every eligible voter must have voted. Since the turnout is only about 65% (and we're explicitly told $Y$ did not vote), this fails for people like $Y$: $(Y,Y)\notin R$. R is NOT reflexive.

Symmetric? If $(a,b)\in R$, then both $a$ and $b$ voted — which automatically means both $b$ and $a$ voted too, so $(b,a)\in R$. R IS symmetric.

Transitive? If $(a,b)\in R$ and $(b,c)\in R$, then $a,b$ both voted and $b,c$ both voted — so $a$ and $c$ both voted, giving $(a,c)\in R$. R IS transitive.

Conclusion: $R$ is both symmetric and transitive, but not reflexive — so it is not an equivalence relation on the full set $A$ (it would be an equivalence relation only on the subset of people who actually voted).

…

$R$ is built from $x\in\{0,1,2,3,4,5\}$, so its domain is exactly that set: $\{0,1,2,3,4,5\}$.

The relation is $R=\{(x,x+5): x\in\{0,1,2,3,4,5\}\}$. Writing out the pairs: $(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)$.

…

Only (b) is a mapping from $A$ to $B$.

A function from $A=\{1,2,3\}$ to $B$ must pair every element of $A$ with exactly one element of $B$ (NCERT Class 11 Relations and Functions).

• (a) has $(2,a)$ and $(2,b)$ — $2$ has two images ✗. …

Only $x = 1, 2, 3, 4$ give a $y = 3x$ that stays within $A = \{1, \ldots, 14\}$.

$A = \{1, 2, 3, \ldots, 14\}$ and $R = \{(x, y) : y = 3x,\ x, y \in A\}$.

We need $y = 3x \le 14$, so $x \le 14/3 \approx 4.67$, meaning $x \in \{1, 2, 3, 4\}$ (and $x \ge 1$ since $x \in A$).

…

Domain $=\{0,1,2,3,4,5\}$; Range $=\{5,6,7,8,9,10\}$.

$R=\{(x,x+5): x\in\{0,1,2,3,4,5\}\}$ gives the pairs $(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)$.

…

Number of relations from $A$ to $B$ is $2^{mn}$.

…

Only $x = 0, 1, 2$ give $y = x+5$ inside A, so the range is {5, 6, 7}.

$A = \{0,1,2,3,4,5,6,7\}$ and $R = \{(x,y) : y = x+5,\ x,y \in A\}$.

For each $x \in A$, we need $y = x+5$ to also be a member of A, i.e. $x + 5 \le 7$, so $x \le 2$.

• $x=0 \Rightarrow y=5$
• $x=1 \Rightarrow y=6$
• $x=2 \Rightarrow y=7$ …

The domain of $R$ is $\{1,2,3,4\}$.

$R=\{(x,y): 3x-y=0,\ x,y\in A\}$ means $y=3x$. For $(x,y)$ to be a valid pair, both $x$ and $y=3x$ must lie in $A=\{1,2,\ldots,14\}$.

…