If Y = \x mid x is a positive factor of the number – Class 11 Mathematics

NCERT Exemplar · Q3

Y = \{x \mid x \text{ is a positive factor of the number } 2^{p-1}(2^p - 1),\ \text{where } 2^p - 1 \text{ is a prime number}\}

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Concept understanding — Relation Arrow Diagram

Relation Arrow Diagram

Imagine two groups of people at a party: one group of hosts, one group of guests. If you drew a string from every host to every guest they personally invited, you'd get a tangle of strings connecting the two groups. A relation arrow diagram does exactly this for sets — it draws an arrow from each element that is "related to" an element in the other set, so you can see a relation instead of just listing ordered pairs.

The Intuition First

A relation pairs up elements of one set with elements of another. Suppose:

Set $A = \{\text{Ravi}, \text{Anu}, \text{Kiran}\}$ (students)
Set $B = \{\text{Maths}, \text{Physics}, \text{Chemistry}\}$ (subjects)

and the relation is "studies": Ravi studies Maths and Physics; Anu studies only Chemistry; Kiran studies all three.

To draw the arrow diagram, list the elements of $A$ in one oval on the left, the elements of $B$ in another oval on the right, and draw one arrow for every pairing:

Ravi $\rightarrow$ Maths
Ravi $\rightarrow$ Physics
Anu $\rightarrow$ Chemistry
Kiran $\rightarrow$ Maths
Kiran $\rightarrow$ Physics
Kiran $\rightarrow$ Chemistry

Count the arrows leaving each name in $A$ : Ravi has 2, Anu has 1, Kiran has 3 — six arrows in total, one for each ordered pair in the relation. That is the whole diagram: two ovals of labelled points, joined by one arrow per related pair.

Note

The name comes from the arrows, not the ovals. What matters is: which element points to which, and in which direction.

The Precise Definition

A

Every arrow corresponds to exactly one ordered pair in $R$ . If a diagram has $k$ arrows, the relation has exactly $k$ ordered pairs — nothing more, nothing less.

Key Points to Remember

Direction matters. An arrow always starts at an element of $A$ and ends at an element of $B$ . An arrow "Ravi $\to$ Maths" is not the same statement as "Maths $\to$ Ravi" — the first element of the ordered pair is always where the arrow starts.
An element of $A$ can send multiple arrows. Kiran, above, sends three — one element can be related to many elements of $B$ .
An element of $A$ can send zero arrows. Nothing requires every element of $A$ to be related to something in $B$ . If a student studies none of the listed subjects, no arrow leaves their name.
An element of $B$ can receive multiple arrows. Both Ravi and Kiran point to Maths — an element of $B$ can be related to many elements of $A$ .

Watch out

A common mistake: assuming every element of $A$ must have at least one outgoing arrow, or that every element of $B$ must receive one. Neither is required. A relation can leave elements on either side completely unconnected.

A Relation on a Single Set

When a relation goes from a set to itself, both ovals contain the same elements, so they are usually drawn as one oval with arrows looping back into it.

Example. Let $A = \{1, 2, 3\}$ and let $R$ be "is less than" ( $<$ ). The pairs in $R$ are $(1,2), (1,3), (2,3)$ , so the arrow diagram has exactly these three arrows:

$1 \rightarrow 2$
$1 \rightarrow 3$
$2 \rightarrow 3$

Nothing points from $3$ , since no element of $A$ is greater than $3$ . Nothing points into $1$ , since no element of $A$ is less than $1$ .

Why This Matters

Arrow diagrams are the first visual bridge toward functions, one-one relations, onto relations, and equivalence relations. Once you can draw one, you can immediately see properties that are hard to spot from a bare list of ordered pairs — such as whether an element sends more than one arrow (which would rule the relation out as a function), or whether some element in $B$ is never hit by any arrow.

Tip

When a question asks you to "represent the relation using an arrow diagram," work systematically: list every ordered pair in the relation first, then draw exactly one arrow per pair, from the first coordinate to the second. Skipping the ordered-pair list is the most common source of missed or duplicated arrows.

Representing a relation using an arrow diagram is a visual technique taught early in the NCERT Class 11 Mathematics chapter on Relations and Functions, and "relation arrow diagram examples class 11" is a frequently searched revision topic for CBSE board preparation. This visualisation also makes it easier to identify one-one and onto relations, a distinction that is regularly tested in "relations and functions important questions".

The set $Y$ consists of all positive divisors of a perfect number of the form $2^{p-1}(2^p - 1)$ where $2^p - 1$ is prime. Since $2^p - 1$ is prime, the divisors are $1, 2, 4, \dots, 2^{p-1}$ and each multiplied by the prime $2^p - 1$ . So $Y = \{1, 2, 2^2, \dots, 2^{p-1}, (2^p - 1), 2(2^p - 1), 2^2(2^p - 1), \dots, 2^{p-1}(2^p - 1)\}$ .

The problem gives you a number of the form $N = 2^{p-1}(2^p - 1)$ , with the condition that $2^p - 1$ is prime. This is the classic form of an even perfect number — every even perfect number is of this shape, and conversely, whenever $2^p - 1$ is prime (a Mersenne prime), $N$ is perfect.

But we don't need the perfection property here. What matters is the factor structure. Since $2^p - 1$ is prime, call it $q$ . Then $N = 2^{p-1} \cdot q$ , where $q$ is an odd prime and $2^{p-1}$ is a power of 2. The two parts are coprime (one is a power of 2, the other is an odd prime), so the divisors of $N$ are simply all possible products of a divisor of $2^{p-1}$ and a divisor of $q$ .

Let's list them systematically.

Divisors of $2^{p-1}$ : These are $1, 2, 2^2, 2^3, \dots, 2^{p-1}$ . That's $p$ numbers.
Divisors of $q$ (where $q = 2^p - 1$ is prime): Only $1$ and $q$ itself.
All divisors of $N$ : Take each divisor of $2^{p-1}$ and multiply it by each divisor of $q$ . That gives:
- Multiply by $1$ : $1, 2, 2^2, \dots, 2^{p-1}$
- Multiply by $q$ : $q, 2q, 2^2 q, \dots, 2^{p-1} q$
No other combinations exist because $q$ has no other factors.
Total count: There are $p$ divisors from the first row and $p$ from the second, so $2p$ divisors in all. This matches the divisor-count formula: if $N = 2^{p-1} \cdot q^1$ , then $\tau(N) = (p-1+1)(1+1) = p \cdot 2 = 2p$ .

Watch out

A common mistake is to forget that $1$ is a divisor, or to think that $2^p - 1$ itself might factor further. The problem explicitly states $2^p - 1$ is prime, so it has exactly two divisors: $1$ and itself.

Tip

Notice that the divisors come in natural pairs: each divisor $d$ from the first row pairs with $N/d$ from the second row. For example, $1$ pairs with $2^{p-1}q$ , $2$ pairs with $2^{p-2}q$ , and so on. This is a hallmark of perfect numbers — the sum of all divisors equals $2N$ .

So the roster form of $Y$ is simply the list of all these $2p$ numbers.

✓Final answer

Y = \{1, 2, 2^2, \dots, 2^{p-1}, (2^p - 1), 2(2^p - 1), 2^2(2^p - 1), \dots, 2^{p-1}(2^p - 1)\}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2022Set ANNUAL2 marks
$R = \{(x, x+5) : x \in \{0,1,2,3,4,5\}\}$
›Reveal solutionSolution

Substituting each $x\in\{0,1,2,3,4,5\}$ into $x+5$ gives the relation's pairs directly, from which the domain and range are read off.

$R=\{(x,x+5): x\in\{0,1,2,3,4,5\}\}$

Computing each pair:

$(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)$

The domain is the set of all first components (the $x$ -values):

$\{0,1,2,3,4,5\}$

The range is the set of all second components (the $x+5$ values):

$\{5,6,7,8,9,10\}$

✓Final answer
$=\{0,1,2,3,4,5\}$
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL2 marks
$A = \{1, 2, 3, ..., 14\}$
›Reveal solutionSolution

⚠️ This question's printed relation definition is genuinely unclear in the scan — it shows " $(x,-1)$ " as the pair template alongside the rule " $3x-y=0$ ", which is inconsistent (that rule needs a variable $y$ as the second coordinate, not the fixed value $-1$ ). Read honestly, the legible and mathematically sensible part is the rule $3x-y=0$ (i.e. $y=3x$ ) applied over $A=\{1,2,\dots,14\}$ — the standard textbook form of this exact problem — so that is what is solved here, restricted to pairs where both $x$ and $y$ lie in $A$ .

Note on the ambiguity: the printed stem reads " $R=\{(x,-1):3x-y=0,\text{ where }x,y\in R\}$ ", which cannot be read literally as written (a fixed " $-1$ " second coordinate contradicts a rule that defines $y$ via $3x-y=0$ , and " $x,y\in R$ " should say " $x,y\in A$ " since $R$ is the relation being defined, not a domain). The clearly legible and internally consistent core of the question is the rule $3x-y=0$ , i.e. $y=3x$ , with $x,y$ drawn from $A=\{1,2,\dots,14\}$ — this matches the standard NCERT-style problem this question is based on, so it is solved on that basis rather than guessed at literally.

Given $A=\{1,2,\dots,14\}$ and $R=\{(x,y): y=3x,\ x,y\in A\}$ , we need $y=3x$ to also lie in $A$ , i.e. $3x\le14$ , so $x\le4.67$ , giving $x=1,2,3,4$ (for $x=5$ , $y=15\notin A$ ).

$x=1\Rightarrow y=3,\quad x=2\Rightarrow y=6,\quad x=3\Rightarrow y=9,\quad x=4\Rightarrow y=12.$

So $R=\{(1,3),(2,6),(3,9),(4,12)\}$ .

Domain of $R$ = set of first coordinates = $\{1,2,3,4\}$ .

Range of $R$ = set of second coordinates = $\{3,6,9,12\}$ .

✓Final answer
$y=3x$

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