Write the following sets in the roster form: (i) D = \t mid t³ = t,t ∈ mathbbR\ (ii) E = \w mid (w-2)/(w+3) =

NCERT Exemplar · Q2

Q. Write the following sets in the roster form:

(i)

D = \{t \mid t^3 = t,\ t \in \mathbb{R}\}

(ii)

E = \left\{w \mid \dfrac{w-2}{w+3} = 3,\ w \in \mathbb{R}\right\}

(iii)

F = \{x \mid x^4 - 5x^2 + 6 = 0,\ x \in \mathbb{R}\}

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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".

Roster form lists all elements explicitly. Solve each defining equation, verify solutions lie in the given domain, then write the set as $\{\text{element}_1, \text{element}_2, \ldots\}$ .

When a set is given in set-builder notation (the form $\{x \mid \text{condition}\}$ ), it describes elements by a property they satisfy. The roster form simply lists those elements between braces. The task is straightforward: solve the equation or condition, find all real solutions, and enumerate them.

The key is to solve accurately and check that solutions belong to the specified universe (here $\mathbb{R}$ in all three cases).

(i) $D = \{t \mid t^3 = t,\ t \in \mathbb{R}\}$

We need all real numbers $t$ satisfying $t^3 = t$ .

Rearrange the equation:

t^3 - t = 0

Factor out $t$ :

t(t^2 - 1) = 0

Factor the quadratic:

t(t - 1)(t + 1) = 0

Read off the solutions:

The product is zero when any factor is zero, so $t = 0$ , $t = 1$ , or $t = -1$ .
Verify each solution:
- $0^3 = 0$ ✓
- $1^3 = 1$ ✓
- $(-1)^3 = -1$ ✓

All three are real numbers, so they all belong to $D$ .

Roster form: $D = \{-1, 0, 1\}$

(ii) $E = \left\{w \mid \dfrac{w-2}{w+3} = 3,\ w \in \mathbb{R}\right\}$

We solve the rational equation $\dfrac{w-2}{w+3} = 3$ .

Multiply both sides by $(w + 3)$ (noting that $w \neq -3$ to avoid division by zero):

w - 2 = 3(w + 3)

Expand the right side:

w - 2 = 3w + 9

Collect like terms:

w - 3w = 9 + 2

-2w = 11

Solve for $w$ :

w = -\frac{11}{2}

Check the solution is valid:

Substitute $w = -\frac{11}{2}$ into the original equation:

$\frac{-\frac{11}{2} - 2}{-\frac{11}{2} + 3} = \frac{-\frac{15}{2}}{-\frac{5}{2}} = \frac{15}{5} = 3$ ✓

Also, $-\frac{11}{2} \neq -3$ , so the denominator is non-zero.

Roster form: $E = \left\{-\dfrac{11}{2}\right\}$

(iii) $F = \{x \mid x^4 - 5x^2 + 6 = 0,\ x \in \mathbb{R}\}$

This is a biquadratic equation (quartic with only even powers). A substitution makes it quadratic.

Substitute $u = x^2$ (so $u \geq 0$ since $x$ is real):

u^2 - 5u + 6 = 0

Factor the quadratic:

(u - 2)(u - 3) = 0

Solve for $u$ :

$u = 2$ or $u = 3$
Back-substitute to find $x$ :
- If $u = x^2 = 2$ , then $x = \pm\sqrt{2}$
- If $u = x^2 = 3$ , then $x = \pm\sqrt{3}$
Verify each solution in the original equation:
- $(\pm\sqrt{2})^4 - 5(\pm\sqrt{2})^2 + 6 = 4 - 10 + 6 = 0$ ✓
- $(\pm\sqrt{3})^4 - 5(\pm\sqrt{3})^2 + 6 = 9 - 15 + 6 = 0$ ✓

All four values are real.

Roster form: $F = \{-\sqrt{3}, -\sqrt{2}, \sqrt{2}, \sqrt{3}\}$

Tip

For biquadratic equations $ax^4 + bx^2 + c = 0$ , always substitute $u = x^2$ to reduce to a quadratic. Don't forget both positive and negative square roots when back-substituting.

✓Final answer

D = \{-1, 0, 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
$y=3x$

←Previous questionQ1 — Write the following sets in the roster form: (i) A = \x : x ∈ mathbbR,…→Next questionQ3 — If Y = \x mid x is a positive factor of the number 2p-1(2p - 1),where …

Q.Write the following sets in the roster form:

(i) D={t∣t3=t, t∈R}D = \{t \mid t^3 = t,\ t \in \mathbb{R}\}D={t∣t3=t, t∈R}

(ii) E={w∣w−2w+3=3, w∈R}E = \left\{w \mid \dfrac{w-2}{w+3} = 3,\ w \in \mathbb{R}\right\}E={w∣w+3w−2​=3, w∈R}

(iii) F={x∣x4−5x2+6=0, x∈R}F = \{x \mid x^4 - 5x^2 + 6 = 0,\ x \in \mathbb{R}\}F={x∣x4−5x2+6=0, x∈R}

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(i) $D = \{t \mid t^3 = t,\ t \in \mathbb{R}\}$

(ii) $E = \left\{w \mid \dfrac{w-2}{w+3} = 3,\ w \in \mathbb{R}\right\}$

(iii) $F = \{x \mid x^4 - 5x^2 + 6 = 0,\ x \in \mathbb{R}\}$

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