Relations and Functions: Class 12 Maths Chapter 1 Hinglish mein

Mathematics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

105 min
1 simulation
41 practice questions
17 MCQs

Is chapter mein aap kya seekhenge

Relation ko ordered pairs ke set ki tarah samjho, aur empty aur universal relations pehchaano.
Kisi relation ko reflexive, symmetric aur transitive properties ke liye test karo, aur jab bhi koi property fail ho to counterexample do.
Prove karo ki koi relation equivalence relation hai, aur uski equivalence classes ko set ke partition ki tarah nikaalo.
Algebra, graphs ya counting se tay karo ki koi function one-one, onto ya bijective hai.
Samjhao ki finite set par one-one function onto kyun hota hai, aur infinite set par yeh kyun fail ho sakta hai.
Do functions ka composition banao aur dikhao ki composition ka order matter karta hai.
Tay karo ki koi function invertible hai ya nahi, uska inverse nikaalo aur dono compositions se check karo.
Chhote finite sets par di gayi properties wale functions aur relations gino.

NCERT Mathematics Class 12, Chapter 1, pp. 1–17 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

1.1 Introduction

Class register har student ko ek roll number se jodta hai, railway chart har passenger ko ek berth se, aur scorecard har batter ko ek score se. Maths aisi pairings ka sirf dhaancha rakhta hai, yaani ordered pairs ka ek collection, aur dekhta hai ki woh collection kaun-kaun si shapes le sakta hai. Class XI mein tum relations, functions, domain, codomain aur range se mil chuke ho. Is chapter mein hum relations ko teen properties se sort karenge, functions ko is hisaab se sort karenge ki woh apne codomain ko kaise cover karte hain, aur phir functions ko ek ke baad ek jodenge aur unhe ulta chalayenge.

Yaad karo ki sets $A$ aur $B$ ke liye Cartesian product $A \times B$ un sabhi ordered pairs $(a, b)$ ka set hai jinmein $a \in A$ aur $b \in B$ . $A$ se $B$ tak ek relation (संबंध) bas $A \times B$ ka ek subset hai. Pairs kisi meaningful rule se chune ja sakte hain, jaise is a factor of, ya bina kisi rule ke; ek baar chun liye gaye, to relation bas wahi set of pairs hai. Jab $(a, b) \in R$ ho, to hum $a R b$ likhte hain aur kehte hain ki $a$ , $b$ se related hai.

Definition

Relation

$A$ se $B$ tak relation $R$ , $A \times B$ ka koi bhi subset hai. Jab $A = B$ ho, to hum $R$ ko $A$ mein relation kehte hain; tab yeh $A \times A$ ka subset hota hai.

Example $A = {2, 3, 5}$ aur $B = {4, 6, 9, 10}$ ke saath, rule ' $a$ divides $b$ ' yeh chunta hai: $R = {(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (5, 10)}$ .

2/3Rule $a$ divides $b$ inmein se 6 rakhta hai. Pairs ka wahi set relation $R$ hai.

Fig. 1.1Relation sabhi pairs ke grid mein se kuch cells ka chunaav hai.

Ek grid mein set A ke liye rows 2, 3 aur 5 hain aur set B ke liye columns 4, 6, 9 aur 10 hain, aur uske 12 cells mein se har ek mein ek chhota dot hai. Pehle step mein har cell plain hai: yeh A × B ke saare pairs hain, aur label kehta hai ki yeh 3 × 4 = 12 hain. Doosre step mein chhe cells gol ghere mein aa jaate hain, woh pairs jahan a divides b: (2, 4), (2, 6), (2, 10), (3, 6), (3, 9) aur (5, 10); baaki dots fade ho jaate hain aur label relation R ka naam batata hai. Teesre step mein grid left ki taraf khisakta hai aur wahi chhe pairs right mein circles 2, 3 aur 5 se circles 4, 6, 9 aur 10 tak arrows ki tarah dikhte hain.

Function ek khaas tarah ka relation hai: pehle set ke har element se exactly ek arrow nikalta hai. Fig. 1.1 mein number 2 se teen arrows nikalte hain, isliye yeh relation $A$ se $B$ tak function nahi hai. Dono pictures dimaag mein rakho, grid aur arrows, kyunki hum Section 1.2 mein relations aur Section 1.3 mein functions ko classify karenge.

Quick check

3 mein se 1

Set

A

mein 3 elements hain aur set

B

mein 4.

A

B

tak kitne alag relations hain?

A \times B

ka har subset gino, empty subset bhi.

Answer dekho

Answer 4096

A \times B

mein

3 \times 4 = 12

pairs hain, aur har pair ya to

R

mein hai ya bahar, isliye

2^{12} = 4096

relations hain.

1.2 Types of Relations

Ab se hamare zyadatar relations ek hi set ke andar rehte hain: $A$ mein relation, $A \times A$ ka subset hai. Do subsets hamesha milte hain, chahe $A$ kitna bhi ajeeb ho. Ek hai empty set $\emptyset$ , jo kisi ko kisi se relate nahi karta. Doosra hai poora $A \times A$ , jo har cheez ko har cheez se relate karta hai.

Definition

Empty relation

Relation $R = \emptyset$ : $A \times A$ ka woh subset jismein koi pair nahi hai, isliye $A$ ka koi element kisi se juda nahi hai.

Example $A = {1, 2, \dots, 9}$ mein rule $a + b = 20$ kabhi poora nahi hota, kyunki sabse bada possible sum $9 + 9 = 18$ hai.

Definition

Universal relation

Relation $R = A \times A$ , jismein $A$ ka har ordered pair hai; har element sabhi elements se juda hai, khud se bhi.

ExampleUsi $A$ mein rule $∣ a - b ∣ < 10$ sabhi $9 \times 9 = 81$ pairs ke liye sahi hai.

Yeh do extremes, empty (रिक्त) aur universal (सार्वत्रिक) relation, trivial relations kehlaate hain. Akele mein yeh boring hain, lekin aage aane wali properties ke liye yeh bahut acche test cases hain.

Example 1NCERT Example 1 · p. 2

Saal ke mahine

Maan lo $M$ ek non-leap year ke 12 mahinon ka set hai. $M$ mein, $a R_{1} b$ ka matlab hai ki mahine $a$ aur $b$ milkar 62 se zyada din rakhte hain, aur $a R_{2} b$ ka matlab hai ki $a$ aur $b$ ke dinon ki sankhya mein zyada se zyada 3 ka farak hai. Dikhao ki $R_{1}$ empty relation hai aur $R_{2}$ universal relation hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ek relation ko likhne ke teen tareeke. ${1, 2, 3, 4}$ mein $R$ lo jahan $b$ , $a$ ka double hai. Roster form mein, $R = {(1, 2), (2, 4)}$ . Set-builder form mein, $R = {(a, b) : b = 2 a}$ . Teesre, bahut common style mein, hum bas kehte hain ki $a R b$ if and only if $b = 2 a$ . Teeno pairs ke same set ko describe karte hain, isliye us waqt jo sabse clear ho wahi use karo.

Reflexive, symmetric aur transitive relations

Maths ke sabse important relations, jaise equality, triangles ki congruence aur 'same remainder chhodta hai', teen aadatein share karte hain. $A$ mein ek relation ko $A$ ke elements ke beech arrows ki tarah banao, aur jab koi element khud se related ho to ek chhota loop banao. Tab har aadat arrows ke baare mein ek rule ban jaati hai (Fig. 1.2).

Definition

Reflexive relation

$A$ mein ek relation $R$ jismein har $a \in A$ ke liye $(a, a) \in R$ : har element khud se related hai.

ExampleLogon ke set mein 'dono same saal mein paida hue'.

Definition

Symmetric relation

$A$ mein ek relation $R$ jismein sabhi $a, b \in A$ ke liye $(a, b) \in R$ ka matlab hai $(b, a) \in R$ .

ExampleIntegers ke set $𝐙$ mein $∣ a - b ∣ = 5$ .

Definition

Transitive relation

$A$ mein ek relation $R$ jismein sabhi $a, b, c \in A$ ke liye $(a, b) \in R$ aur $(b, c) \in R$ milkar $(a, c) \in R$ imply karte hain.

ExampleLogon ke set mein 'se umar mein bada hai'.

4/4Saath-saath. Jis relation mein teeno aadatein hon, woh equivalence relation hai.

Fig. 1.2Loops, wapsi ke arrows aur shortcuts: teen properties pictures mein.

Teen panels mein se har ek points 1, 2 aur 3 ko circles ki tarah dikhata hai. Pehle step mein left panel highlight hai: har point par khud tak wapas aata loop arrow hai, aur ek normal arrow 1 se 2 tak jaata hai, isliye relation reflexive hai. Doosre step mein beech wala panel highlight hai: arrows 1 aur 2 ko dono directions mein jodte hain, aur 2 aur 3 ko bhi dono directions mein, isliye har arrow ka ek wapas aane wala partner hai aur relation symmetric hai. Teesre step mein right panel highlight hai: arrows 1 se 2 aur 2 se 3 tak jaate hain, aur ek dashed shortcut arrow seedha 1 se 3 tak jaata hai, jaisa transitivity maangti hai. Aakhri step mein teeno panels poori strength ke saath saath-saath dikhte hain.

Do warnings bahut marks bachaati hain. Pehli, symmetry aur transitivity if–then statements hain. Agar kisi relation mein koi two-step path hai hi nahi, to transitivity kuch nahi maangti, aur relation apne aap transitive hai (hum kehte hain vacuously). Doosri, reflexivity har element ke baare mein claim hai, isliye ek bhi missing loop use bigaad deta hai: ${1, 2, 3}$ mein relation ${(1, 1), (2, 2)}$ reflexive nahi hai, kyunki $(3, 3)$ missing hai.

Property sahi hai yeh prove karne ke liye arbitrary elements se argue karo. Yeh dikhane ke liye ki woh fail hoti hai, ek specific counterexample kaafi hai, aur use naam se batao.

Definition

Equivalence relation

Aisa relation jo reflexive, symmetric aur transitive ho.

ExampleKisi bhi set mein equality, $a = b$ ; $𝐙$ mein '7 se divide karne par same remainder chhodta hai'.

InteractiveRelation lab

Switches ka ek 4 by 4 grid, har ordered pair (a, b) ke liye ek, jahan a aur b 1 se 4 tak hain, aur uske paas chaar points par usi relation ka arrow diagram. Switch tap karne se woh pair add ya remove hota hai, aur tum set ko teen elements tak chhota kar sakte ho. Ek panel batata hai ki relation reflexive, symmetric aur transitive hai ya nahi; jab koi property fail hoti hai to woh ek missing pair batata hai jo yeh prove karta hai, aur jab teeno sahi hon to equivalence classes list karta hai. Buttons ready-made relations load karte hain, grid clear karte hain, ya har property ko chahiye pairs add karte hain, aur ek challenge menu tumse di gayi properties wale relations banwata hai, jaise symmetric aur transitive lekin reflexive nahi.

Fig. 1.3

{1, 2, 3, 4}

par ek relation mein pairs ko on aur off karo aur teeno properties ko update hote dekho, har failure ke reason ke saath.

Example 2NCERT Example 2 · p. 3

Barabar area wale rectangles

Maan lo $S$ ek kaagaz par bane sabhi rectangles ka set hai, aur $P R Q$ ka matlab hai ki $P$ aur $Q$ ka area same hai. Dikhao ki $S$ mein $R$ ek equivalence relation hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 2 mein pattern dekho: relation tha 'ka area same hai', yaani kisi attribute ki equality. 'ka ___ same hai' form ka har relation teeno properties seedha equality se le leta hai. Example 21 yeh hamesha ke liye prove karta hai.

Example 3NCERT Example 3 · p. 3

Opposite signs

Non-zero real numbers ke set $𝐑^{*}$ mein $R = {(a, b) : a b < 0}$ lo, yaani $a R b$ jab $a$ aur $b$ ke signs opposite hon. Prove karo ki $R$ symmetric hai, lekin reflexive nahi hai aur transitive bhi nahi hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3 ke peeche ek simple reason hai. Do 'opposite sign' steps tumhe wapas original sign par le aate hain, isliye transitivity jo shortcut maangti hai woh hamesha same sign ke do numbers ko jodega, aur $R$ mein aisa pair kabhi nahi hota.

Example 4NCERT Example 4 · p. 3

List se properties padhna

Maan lo $A = {4, 5, 6}$ aur $R = {(4, 4), (5, 5), (6, 6), (5, 6), (6, 4)}$ . Tay karo ki $R$ mein teen mein se kaun si properties hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

List ke roop mein diye chhote relation ke liye transitivity ko systematically check karo: har pair $(a, b)$ ke liye, har woh pair $(b, c)$ dekho jo wahan se shuru hota hai jahan pehla khatam hota hai, aur confirm karo ki $(a, c)$ maujood hai. Loops kabhi problem nahi karte, kyunki $(a, a)$ ke baad $(a, c)$ sirf $(a, c)$ hi maangta hai.

Example 5NCERT Example 5 · p. 3

4 se divide karne par same remainder

Dikhao ki integers ke set $𝐙$ mein $R = {(a, b) : 4 divides a - b}$ ek equivalence relation hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Equivalence classes

Example 5 ke relation ko numbers 0 se 11 par chalao aur dekho kya hota hai (Fig. 1.4). Number 1, 5 aur 9 se related hai, jo usse 4 ke multiples se alag hain, aur list mein kisi aur se nahi. Number 2, 6 aur 10 se related hai. Baarah numbers chaar groups mein bant jaate hain, aur har number exactly ek group mein aata hai.

3/3Har number $[0], [1], [2], [3]$ mein se exactly ek mein hai, aur chaaron classes milkar poora set banati hain.

Fig. 1.4Equivalence relation ek set ko aisi classes mein sort karta hai jo overlap nahi karti.

0 se 11 tak numbered baarah circles. Pehle step mein woh bina kisi order ke bikhre hain, aur neeche rule likha hai: a R b agar 4, a minus b ko divide kare. Doosre step mein sirf 1, 5 aur 9 bright rehte hain aur gol ghere mein aate hain, jo 1 ki class banate hain, jabki baaki numbers fade ho jaate hain. Teesre step mein saare circles chaar dashed boxes mein khisak jaate hain jin par [0], [1], [2] aur [3] likha hai: box [0] mein 0, 4 aur 8 hain; [1] mein 1, 5 aur 9; [2] mein 2, 6 aur 10; aur [3] mein 3, 7 aur 11. Boxes overlap nahi karte aur milkar har number rakhte hain, jaisa unke neeche ka label kehta hai.

Remainder hi sab kuch kyun tay karta hai? $a = 4 q + r$ aur $b = 4 q^{'} + r^{'}$ likho, jahan remainders $r, r^{'} \in {0, 1, 2, 3}$ . Tab $a - b = 4 (q - q^{'}) + (r - r^{'})$ , isliye 4, $a - b$ ko tabhi divide karta hai jab 4, $r - r^{'}$ ko divide kare. Kyunki $- 3 \leq r - r^{'} \leq 3$ , 4 ka sirf ek multiple 0 hi iske barabar ho sakta hai. Isliye $a R b$ if and only if $r = r^{'}$ .

Definition

Equivalence class

$A$ mein equivalence relation $R$ aur element $a \in A$ ke liye, $a$ ki class $[a] = {x \in A : x R a}$ hai, yaani $a$ se related har cheez ka set.

Example $𝐙$ mein '4 se divide karne par same remainder' ke under, $[1] = {\dots, - 7, - 3, 1, 5, 9, \dots}$ , aur $[1] = [5] = [- 3]$ .

PropertyEquivalence relations aur partitions

Maan lo $R$ set $A$ mein ek equivalence relation hai. Tab (i) har element apni class mein hota hai, $a \in [a]$ ; (ii) koi bhi do classes ya to identical hoti hain ya disjoint; (iii) saari classes ka union $A$ hai. Ulta, agar $A$ ko aise non-empty, pairwise disjoint tukdon mein baanta jaaye jinka union $A$ ho ( $A$ ka ek partition), to 'usi tukde mein hai jismein' ek equivalence relation hai jiski classes exactly wahi tukde hain.

(i) sahi hai kyunki $R$ reflexive hai: $a R a$ . Isse (iii) bhi milta hai, kyunki har $a \in A$ kam se kam ek class mein hai, apni class mein.

(ii) Maan lo $[a]$ aur $[b]$ mein ek common element $c$ hai, yaani $c R a$ aur $c R b$ . Symmetry se $a R c$ , aur phir transitivity se $a R b$ . Ab koi bhi $x \in [a]$ lo: $x R a$ aur $a R b$ se transitivity $x R b$ deti hai, isliye $x \in [b]$ . Isliye $[a] \subseteq [b]$ . $a$ aur $b$ ke roles badalne se $[b] \subseteq [a]$ milta hai, isliye overlap karne wali classes barabar hain.

Converse ke liye, 'usi tukde mein hai jismein' reflexive hai (har element kisi tukde mein hai), symmetric hai (wording order ko ignore karti hai) aur transitive hai: agar $x, y$ ek tukde mein hain aur $y, z$ ek tukde mein, to dono tukdon mein $y$ hai, isliye woh same tukda hain kyunki alag tukde disjoint hote hain.

Yaani wahi information do tareekon se di ja sakti hai: equivalence relation ki tarah, ya set ke classes mein partition ki tarah. Integers 'same parity' ke under 2 classes mein bante hain, '4 se divide karne par same remainder' ke under 4 classes mein, aur ' $n$ se divide karne par same remainder' ke under $n$ classes $[0], [1], \dots, [n - 1]$ mein. Doosri taraf, kisi set ko non-overlapping groups mein sort karne ka har tareeka ek equivalence relation define karta hai: 'usi group mein hai jismein'.

Example 6NCERT Example 6 · p. 4

Digits ki sankhya se sort karna

Maan lo $A = {1, 2, \dots, 20}$ , aur $a R b$ ka matlab hai ki $a$ aur $b$ mein digits ki sankhya same hai. Prove karo ki $R$ ek equivalence relation hai, uski classes list karo, aur confirm karo ki $R$ alag-alag classes ke do numbers ko kabhi nahi jodta.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Classes ka finite list hona zaroori nahi. Plane mein do points ko related kaho jab woh origin $O$ se same distance par hon. Yeh phir se 'ka ___ same hai' hai, isliye yeh equivalence relation hai. Point $P (3, 4)$ ki class explore karne ke liye simulation use karo.

SimulationP se related points

Dono axes par minus 6 se 6 tak ka coordinate grid, origin O aur (3, 4) par ek fixed point P ke saath, jiski O se distance 5 bani aur naapi gayi hai. Doosre point Q ko kisi bhi grid point tak drag kiya ja sakta hai; uski O se distance banti hai aur chalne ke saath update hoti hai, taaki tum 'O se same distance' relation ke under P se related points dhoondh sako.

Fig. 1.5

Q

ko un grid points tak drag karo jinki

O

se distance

O P = 5

ke barabar hai.

Try karo

$Q$ ko $(4, 3)$ tak drag karo, phir $(0, - 5)$ tak. Kya har ek $P$ se related hai?
$O$ se distance 5 par saare 12 grid points dhoondho. Woh kis shape par hain?
$Q$ ko $(1, 1)$ tak drag karo. Kaunse grid points $(1, 1)$ ki class mein hain?

$P (3, 4)$ ki class $O$ par centred radius 5 ka poora circle hai. Ismein infinitely many points hain, jinmein se sirf baarah ke coordinates whole numbers hain: $(\pm 5, 0)$ , $(0, \pm 5)$ , $(\pm 3, \pm 4)$ aur $(\pm 4, \pm 3)$ . Is relation ki classes $O$ par centred circles hain, saath mein ek-point class ${O}$ , aur yeh plane ka partition karti hain.

Practice · 16 questions

Practice 1.1: answers ke saath

1
Short answerMediumNCERT Exercise 1.1, Q1
Har relation ke liye tay karo ki woh reflexive, symmetric aur transitive hai ya nahi. Fail hone wali har property ke liye counterexample do.
1. $A = {1, 2, \dots, 10}$ mein $R = {(x, y) : x + y = 10}$
  Solution dikhao
  AnswerSirf symmetric.
  Reflexive nahi: $1 + 1 \neq 10$ , isliye $(1, 1) \notin R$ (sirf ek loop $(5, 5)$ hai). Symmetric: $x + y = 10$ se $y + x = 10$ milta hai. Transitive nahi: $(1, 9)$ aur $(9, 1)$ $R$ mein hain, lekin $(1, 1)$ nahi.
2. $𝐙$ mein $R = {(x, y) : x < 0 and y > 0}$
  Solution dikhao
  AnswerSirf transitive.
  Reflexive nahi: $(1, 1) \notin R$ . Symmetric nahi: $(- 1, 1) \in R$ lekin $(1, - 1) \notin R$ . Transitive: $(x, y), (y, z) \in R$ ke liye ek saath $y > 0$ aur $y < 0$ chahiye, jo kabhi nahi hota, isliye koi two-step path nahi hai aur condition vacuously sahi hai.
3. $A = {2, 3, \dots, 10}$ mein, $x R y$ jab $x$ aur $y$ ka 1 se bada koi common factor ho
  Solution dikhao
  AnswerReflexive aur symmetric, transitive nahi.
  Reflexive: har $x \geq 2$ ka khud ke saath common factor $x$ hai. Symmetric: $x$ aur $y$ ka common factor $y$ aur $x$ ka bhi common factor hai. Transitive nahi: $(2, 6)$ aur $(6, 3)$ $R$ mein hain, lekin 2 aur 3 ka 1 se bada koi common factor nahi hai.
4. $𝐑$ mein $R = {(x, y) : ⌊ x ⌋ = ⌊ y ⌋}$ , jahan $⌊ x ⌋$ woh greatest integer hai jo $x$ se zyada na ho
  Solution dikhao
  AnswerEk equivalence relation.
  Yeh 'ka greatest integer same hai' hai, yaani ek attribute ki equality, isliye yeh reflexive, symmetric aur transitive hai. Iski classes $n \in 𝐙$ ke liye intervals $[n, n + 1)$ hain.
5. School ke students ke set mein: ' $x$ aur $y$ same house mein hain'
  Solution dikhao
  AnswerEk equivalence relation.
  Ek attribute (house) ki equality, isliye teeno properties sahi hain.
6. Usi set mein: ' $x$ , $y$ se exactly ek din pehle paida hua'
  Solution dikhao
  AnswerTeeno mein se koi nahi.
  Reflexive nahi: koi bhi khud se ek din pehle paida nahi hota. Symmetric nahi: agar $x$ , $y$ se ek din pehle paida hua, to $y$ , $x$ se ek din baad paida hua. Transitive nahi: agar $x$ , $y$ se ek din bada hai aur $y$ , $z$ se ek din bada, to $x$ , $z$ se do din bada hai.
7. Usi set mein: ' $x$ , $y$ se chhota (umar mein) hai'
  Solution dikhao
  AnswerSirf transitive.
  Reflexive nahi aur symmetric nahi (agar $x$ , $y$ se chhota hai, to $y$ , $x$ se chhota nahi hai). Transitive: jo kisi aise insaan se chhota hai jo $z$ se chhota hai, woh $z$ se bhi chhota hai.
ProofMediumNCERT Exercise 1.1, Q2
Dikhao ki $𝐑$ mein $R = {(a, b) : a \geq 2 b}$ na reflexive hai, na symmetric, na transitive.
Solution dikhao
Reflexive nahi: $1 \geq 2 \times 1$ galat hai, isliye $(1, 1) \notin R$ .
Symmetric nahi: $(4, 1) \in R$ kyunki $4 \geq 2$ , lekin $(1, 4) \notin R$ kyunki $1 \geq 8$ galat hai.
Transitive nahi: $(- 8, - 4) \in R$ kyunki $- 8 \geq - 8$ , aur $(- 4, - 2) \in R$ kyunki $- 4 \geq - 4$ . Lekin $(- 8, - 2) \notin R$ , kyunki $- 8 \geq - 4$ galat hai.
Short answerEasyNCERT Exercise 1.1, Q3
$A = {1, 2, \dots, 10}$ mein relation $R = {(a, b) : b = 2 a}$ list karo aur check karo ki yeh reflexive, symmetric ya transitive hai.
Solution dikhao
Answer $R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)}$ ; teeno mein se koi property sahi nahi hai.
$R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)}$ . Reflexive nahi: $(1, 1) \notin R$ . Symmetric nahi: $(1, 2) \in R$ lekin $(2, 1) \notin R$ . Transitive nahi: $(1, 2), (2, 4) \in R$ lekin $(1, 4) \notin R$ .
ProofEasyNCERT Exercise 1.1, Q4
Logon ke ek group mein $x R y$ ka matlab hai ' $x$ , $y$ se lamba nahi hai'. Prove karo ki $R$ mein reflexive aur transitive properties hain lekin symmetry nahi hai (maan lo group mein alag-alag height ke do log hain).
Solution dikhao
$x R y$ ka matlab hai $h (x) \leq h (y)$ , jahan $h$ height hai. Reflexive: $h (x) \leq h (x)$ . Transitive: $h (x) \leq h (y)$ aur $h (y) \leq h (z)$ se $h (x) \leq h (z)$ milta hai.
Symmetric nahi: agar Asha 150 cm aur Ravi 165 cm lamba hai, to Asha $R$ Ravi, lekin Ravi, Asha se lamba hai, isliye Ravi, Asha se related nahi hai.
Short answerMediumNCERT Exercise 1.1, Q5
Check karo ki $𝐑$ mein $R = {(a, b) : a \leq b + 1}$ reflexive, symmetric ya transitive hai.
Solution dikhao
AnswerSirf reflexive.
Reflexive: $a \leq a + 1$ hamesha. Symmetric nahi: $(0, 5) \in R$ kyunki $0 \leq 6$ , lekin $(5, 0) \notin R$ kyunki $5 \leq 1$ galat hai.
Transitive nahi: $(2, 1) \in R$ kyunki $2 \leq 2$ , aur $(1, 0) \in R$ kyunki $1 \leq 1$ , lekin $(2, 0) \notin R$ kyunki $2 \leq 1$ galat hai.
ProofEasyNCERT Exercise 1.1, Q6
Dikhao ki $A = {4, 5, 6}$ mein $R = {(4, 6), (6, 4), (5, 5)}$ symmetric hai lekin na reflexive hai na transitive.
Solution dikhao
Symmetric: $(4, 6)$ aur $(6, 4)$ pair mein aate hain, aur loop $(5, 5)$ khud apna reverse hai.
Reflexive nahi: $(4, 4) \notin R$ . Transitive nahi: path $(4, 6), (6, 4)$ ko $(4, 4)$ chahiye, jo missing hai.
ProofEasyNCERT Exercise 1.1, Q7
Ek dukaan mein bikne wale mobile phones ke set mein dikhao ki ' $x$ aur $y$ ki price same hai' ek equivalence relation hai.
Solution dikhao
Maan lo $p (x)$ phone $x$ ki price hai. Reflexive: $p (x) = p (x)$ . Symmetric: $p (x) = p (y) \Rightarrow p (y) = p (x)$ . Transitive: $p (x) = p (y)$ aur $p (y) = p (z)$ se $p (x) = p (z)$ milta hai.
ProofMediumNCERT Exercise 1.1, Q8
Maan lo $A = {10, 11, \dots, 17}$ aur $R = {(a, b) : a + b is even}$ . Prove karo ki $R$ reflexive, symmetric aur transitive hai. Phir check karo ki ${10, 12, 14, 16}$ ke koi bhi do numbers $R$ se jude hain, ki ${11, 13, 15, 17}$ ke andar bhi yahi hota hai, aur ki $R$ pehle set ke kisi number ko doosre set ke kisi number se kabhi nahi jodta.
Solution dikhao
Reflexive: $a + a = 2 a$ even hai. Symmetric: $a + b = b + a$ . Transitive: agar $a + b$ aur $b + c$ even hain, to $a + c = (a + b) + (b + c) - 2 b$ even hai.
Do even numbers ka sum even hota hai, aur do odd numbers ka bhi, isliye dono sets apne andar related hain. Even number plus odd number odd hota hai, isliye ${10, 12, 14, 16}$ ka koi element ${11, 13, 15, 17}$ ke kisi element se related nahi hai. Yeh do sets hi equivalence classes hain.
9
Short answerMediumNCERT Exercise 1.1, Q9
Maan lo $A = {x \in 𝐙 : - 6 \leq x \leq 6}$ . Dikhao ki har relation $A$ mein ek equivalence relation hai, aur har case mein 1 ki class nikaalo.
1. $R_{1} = {(a, b) : 5 divides a - b}$
  Solution dikhao
  Answer $[1] = {- 4, 1, 6}$
  Reflexive: $a - a = 0 = 5 \times 0$ . Symmetric: agar $a - b = 5 k$ to $b - a = 5 (- k)$ . Transitive: agar $a - b = 5 k$ aur $b - c = 5 m$ , to $a - c = 5 (k + m)$ . $A$ ke jo elements 1 se 5 ke multiple se alag hain woh hain $1 - 5 = - 4$ , $1$ aur $1 + 5 = 6$ , isliye $[1] = {- 4, 1, 6}$ .
2. $R_{2} = {(a, b) : a^{2} = b^{2}}$
  Solution dikhao
  Answer $[1] = {- 1, 1}$
  Yeh 'ka square same hai' hai, yaani ek attribute ki equality, isliye yeh equivalence relation hai. $b^{2} = 1$ ka matlab hai $b = \pm 1$ , isliye $[1] = {- 1, 1}$ .
10
Short answerHardNCERT Exercise 1.1, Q10
$A = {1, 2, 3}$ use karke $A$ mein aise relation ka example do jo:
1. reflexive ho, lekin na symmetric ho na transitive;
  Solution dikhao
  AnswerJaise, pairs $(2, 1)$ aur $(1, 3)$ , saath mein teen loops $(1, 1)$ , $(2, 2)$ , $(3, 3)$ .
  Saare loops maujood hain. $(2, 1) \in R$ lekin $(1, 2) \notin R$ , isliye yeh symmetric nahi hai; $(2, 1), (1, 3)$ ko $(2, 3)$ chahiye, jo missing hai, isliye yeh transitive nahi hai.
2. symmetric aur transitive ho, lekin reflexive nahi;
  Solution dikhao
  AnswerJaise, ${(1, 2), (2, 1), (1, 1), (2, 2)}$ .
  1 aur 2 ke beech har pair maujood hai, isliye yeh symmetric aur transitive hai, lekin $(3, 3)$ missing hai.
3. transitive ho, lekin na reflexive ho na symmetric;
  Solution dikhao
  AnswerJaise ${(1, 2)}$ .
  Koi two-step path nahi hai, isliye yeh vacuously transitive hai; koi loop nahi hai, aur $(2, 1)$ missing hai.
4. reflexive aur symmetric ho, lekin transitive nahi;
  Solution dikhao
  AnswerJaise, pairs $(1, 2)$ , $(2, 1)$ , $(2, 3)$ , $(3, 2)$ , saath mein teeno loops.
  Saare loops maujood hain aur har pair ka reverse bhi hai, lekin $(1, 2), (2, 3)$ ko $(1, 3)$ chahiye, jo missing hai.
5. identity aur universal relations ke alawa koi equivalence relation ho.
  Solution dikhao
  AnswerJaise, pairs $(1, 3)$ aur $(3, 1)$ , saath mein teeno loops.
  Iski classes ${1, 3}$ aur ${2}$ hain, jo $A$ ka partition hai, isliye partition theorem se yeh equivalence relation hai.
ProofHardNCERT Exercise 1.1, Q11
Coordinate plane ke points ke set mein $P R Q$ lo jab $∣ x_{P} ∣ + ∣ y_{P} ∣ = ∣ x_{Q} ∣ + ∣ y_{Q} ∣$ . Dikhao ki $R$ ek equivalence relation hai, aur point $(2, 1)$ ki class describe karo.
Solution dikhao
AnswerSquare jiske vertices $(3, 0)$ , $(0, 3)$ , $(- 3, 0)$ aur $(0, - 3)$ hain.
$R$ hai ' $∣ x ∣ + ∣ y ∣$ ki value same hai', yaani ek attribute ki equality, isliye yeh equivalence relation hai.
$(2, 1)$ ke liye value 3 hai, isliye uski class ${(x, y) : ∣ x ∣ + ∣ y ∣ = 3}$ hai. First quadrant mein yeh $(3, 0)$ se $(0, 3)$ tak segment $x + y = 3$ hai; baaki quadrants iski mirror images dete hain. Milkar yeh ek square banate hain jiske vertices $(\pm 3, 0)$ aur $(0, \pm 3)$ hain, jo ek corner par khada hai.
Short answerMediumNCERT Exercise 1.1, Q12
Do rectangles similar hote hain jab lambi side aur chhoti side ka ratio same ho. Dikhao ki sabhi rectangles ke set mein similarity ek equivalence relation hai. Inmein se kaun related hain: $R_{1}$ hai $2 cm \times 3 cm$ , $R_{2}$ hai $4 cm \times 5 cm$ , $R_{3}$ hai $6 cm \times 10 cm$ , $R_{4}$ hai $4 cm \times 6 cm$ aur $R_{5}$ hai $3 cm \times 5 cm$ ?
Solution dikhao
Answer $R_{1}$ aur $R_{4}$ related hain; $R_{3}$ aur $R_{5}$ related hain; $R_{2}$ sirf khud se related hai.
Similarity ka matlab hai 'ka side ratio same hai', isliye yeh equivalence relation hai.
Ratios hain $R_{1} : \frac{3}{2}$ , $R_{2} : \frac{5}{4}$ , $R_{3} : \frac{10}{6} = \frac{5}{3}$ , $R_{4} : \frac{6}{4} = \frac{3}{2}$ aur $R_{5} : \frac{5}{3}$ . Isliye $R_{1} R R_{4}$ aur $R_{3} R R_{5}$ , jabki in paanch mein $R_{2}$ apni alag class mein hai.
Short answerMediumNCERT Exercise 1.1, Q13
Sabhi polyhedra (flat faces wale solids) ke set mein $P R Q$ lo jab $P$ aur $Q$ ke faces ki sankhya same ho. Dikhao ki $R$ ek equivalence relation hai. Kya square pyramid cube se related hai? Kya pentagonal pyramid hai?
Solution dikhao
AnswerSquare pyramid (5 faces) cube (6 faces) se related nahi hai; pentagonal pyramid (6 faces) related hai.
$R$ hai 'ke faces ki sankhya same hai', yaani ek attribute ki equality, isliye yeh equivalence relation hai.
Cube ke 6 faces hote hain. Square pyramid ke $1 + 4 = 5$ faces hain, isliye yeh cube se related nahi hai (yeh triangular prism se related hai, jiske bhi 5 faces hain). Pentagonal pyramid ke $1 + 5 = 6$ faces hain, isliye yeh cube ki class mein hai, har cuboid ke saath.
Short answerMediumNCERT Exercise 1.1, Q14
$x y$ -plane ke sabhi circles ke set mein $C_{1} R C_{2}$ lo jab $C_{1}$ aur $C_{2}$ ka centre same ho. Dikhao ki $R$ ek equivalence relation hai, aur circle $x^{2} + y^{2} - 4 x + 6 y - 3 = 0$ ki class describe karo.
Solution dikhao
Answer $r > 0$ ke saath saare circles $(x - 2)^{2} + (y + 3)^{2} = r^{2}$ .
'Ka centre same hai' ek attribute ki equality hai, isliye $R$ ek equivalence relation hai.
$x^{2} - 4 x + y^{2} + 6 y = 3$ se $(x - 2)^{2} + (y + 3)^{2} = 3 + 4 + 9 = 16$ milta hai, yaani centre $(2, - 3)$ aur radius 4 wala circle. Iski class $(2, - 3)$ par centred saare circles ki family hai: $(x - 2)^{2} + (y + 3)^{2} = r^{2}$ , $r > 0$ .
Quick answerMediumNCERT Exercise 1.1, Q15
${1, 2, 3, 4}$ mein $R$ mein chaaron loops hain aur saath mein $(2, 3), (3, 2), (1, 2), (2, 1)$ . Kaunsa statement sahi hai?
$R$ mein teeno properties hain. $R$ mein sirf symmetric aur transitive properties hain. $R$ mein sirf reflexive aur transitive properties hain. $R$ mein sirf reflexive aur symmetric properties hain.
Answer dekho
Answer (D) $R$ mein sirf reflexive aur symmetric properties hain.
Chaaron loops maujood hain aur har pair apne reverse ke saath hai, lekin $(1, 2), (2, 3) \in R$ jabki $(1, 3) \notin R$ .
Quick answerEasyNCERT Exercise 1.1, Q16
$𝐙$ mein $R = {(a, b) : a^{2} + b = 10, b < 0}$ lo. Kaunsa pair $R$ mein hai?
$(2, 6)$ $(- 3, 1)$ $(4, - 6)$ $(3, - 1)$
Answer dekho
Answer (C) $(4, - 6)$
$4^{2} + (- 6) = 10$ aur $- 6 < 0$ . Pairs $(2, 6)$ aur $(- 3, 1)$ equation ko satisfy karte hain lekin unmein $b > 0$ hai, aur $(3, - 1)$ se $9 - 1 = 8$ milta hai.

1.3 Types of Functions

Class XI mein tum identity, constant, polynomial, rational, modulus aur signum functions se mile the, unke graphs banaye the aur functions ko add aur multiply karke combine kiya tha. Yahan hum function $f : X \to Y$ ke baare mein ek naya sawaal poochte hain: yeh codomain $Y$ ko kaise cover karta hai? Kya yeh kabhi do inputs ko same output par bhejta hai? Kya yeh $Y$ ke har element tak pahunchta hai?

Definition

Function

$X$ se $Y$ tak ek relation $f$ jismein har $x \in X$ , $Y$ ke exactly ek element se related hai, jise $f (x)$ likhte hain. $X$ domain hai, $Y$ codomain, aur ${f (x) : x \in X}$ range.

ExamplePost office ka PIN code: har post office ka exactly ek PIN code hota hai, halaanki ek PIN code kai post offices ko serve kar sakta hai.

Fig. 1.6 chaar inputs ke saath chaaron possibilities dikhata hai. Har picture mein do cheezein dekho: kya do arrows kabhi same square par girte hain, aur kya koi square bina arrow ke reh jaata hai.

1/4Do inputs output $a$ share karte hain, aur $b$ tak kabhi nahi pahuncha jaata.

Fig. 1.6Har function ke baare mein do sawaal: kya outputs kabhi share hote hain, aur kya codomain mein kuch chhoot jaata hai?NCERT Fig. 1.2 · p. 8

1 se 4 tak labelled chaar circles, yaani set X, squares ki ek row, yaani set Y, ke upar ek row mein hain, aur har circle ek arrow neeche ek square tak bhejta hai. Step one: 1 aur 2 dono a ki taraf point karte hain, 3, c ki taraf aur 4, d ki taraf, jabki b ko koi arrow nahi milta aur woh dashed bana hai; labels kehte hain one-one: no aur onto: no. Step two: paanch squares a se e tak hain, aur arrows a, b, d aur e tak jaate hain, sab alag, lekin c chhoot jaata hai aur dashed hai: one-one yes, onto no. Step three: teen squares hain; 1, a tak jaata hai, 2 aur 3 dono b tak, aur 4, c tak, isliye har square hit hota hai lekin b do baar: one-one no, onto yes. Step four: chaar squares, arrows 1 se b, 2 se d, 3 se a aur 4 se c, isliye har square exactly ek baar hit hota hai: one-one yes, onto yes.

Definition

One-one (injective) function

Function $f : X \to Y$ jismein alag inputs alag outputs dete hain. Doosre shabdon mein, jab bhi $f (x_{1}) = f (x_{2})$ ho, tab $x_{1} = x_{2}$ hona hi chahiye. Jo function one-one nahi hota use many-one kehte hain.

Example $𝐍$ par $f (x) = x + 3$ one-one hai; $𝐙$ par $g (x) = x^{2}$ many-one hai, kyunki $g (- 2) = g (2)$ .

Proofs mein ' $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ ' wala form use karo: maan lo do outputs barabar hain aur nikaalo ki inputs barabar the. Yeh dikhane ke liye ki $f$ one-one nahi hai, same output wale do alag inputs dikhao.

Definition

Onto (surjective) function

Function $f : X \to Y$ jismein har $y \in Y$ kam se kam ek $x \in X$ ki image hai; yaani har $y \in Y$ ke liye equation $f (x) = y$ ka $X$ mein solution hai.

Example $f : 𝐑 \to 𝐑$ , $f (x) = x^{3}$ onto hai, kyunki $y$ , $\sqrt[3]{y}$ ki image hai.

Doosre shabdon mein, $f$ tabhi onto hai jab uski range uske codomain ke barabar ho. Isliye 'onto' is par depend karta hai ki hum kaunsa codomain declare karte hain: $x \mapsto x^{2}$ map $𝐑 \to 𝐑$ ki tarah onto nahi hai, kyunki $- 1$ tak kuch nahi jaata, lekin map $𝐑 \to [0, \infty)$ ki tarah onto hai. Judge karne se pehle hamesha arrow $f : X \to Y$ padho.

Definition

Bijective function

Aisa function jo one-one bhi ho aur onto bhi. Tab codomain ka har element domain ke exactly ek element ki image hota hai.

Example $f : 𝐑 \to 𝐑$ , $f (x) = 2 x - 7$ .

InteractiveMapping lab

Points ke do columns: domain X jismein 1 se numbered 2 se 5 elements hain, aur codomain Y jismein a se lettered 2 se 5 elements hain, aur dono sizes set karne ke liye menus. X ke har element ke liye ek drop-down menu Y mein uski image chunta hai, aur arrow diagram update hota hai. Ek status panel batata hai ki function one-one hai ya nahi, aur agar nahi to output share karne wale do inputs ka naam leta hai, aur onto hai ya nahi, aur agar nahi to chhoote hue output ka naam leta hai. Ek Reverse arrows switch har arrow ko palat deta hai aur batata hai ki palti hui picture Y se X tak function hai ya nahi. Challenge buttons di gayi properties wala function maangte hain aur samjhate hain jab chune gaye sizes ke liye aisa function impossible ho, aur ek counter dikhata hai ki un sizes ke liye kitne functions aur kitne one-one functions hain.

Fig. 1.7Chuno ki

X

ka har element kahan jaata hai aur turant dekho ki tumhara function one-one, onto ya bijective hai, aur kya uske arrows ulta karne se function milta hai.

Example 7NCERT Example 7 · p. 8

Library mein accession numbers

Ek school library mein 1850 books hain, aur har book par 1 se 1850 tak ek accession number laga hai, koi do books same number share nahi karti. Maan lo $B$ in books ka set hai aur $f : B \to 𝐍$ , $f (b) =$ $b$ ka accession number. Prove karo ki $f$ one-one hai aur onto nahi hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 8NCERT Example 8 · p. 8

Natural numbers ko shift karna

Dikhao ki $f : 𝐍 \to 𝐍$ , $f (x) = x + 3$ , one-one hai lekin onto nahi.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9NCERT Example 9 · p. 9

Real line ka ek bijection

Prove karo ki $f : 𝐑 \to 𝐑$ , $f (x) = \frac{x + 4}{3}$ , one-one aur onto hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Graphs dono properties ki quick picture dete hain (Fig. 1.8). Function $f : 𝐑 \to 𝐑$ ke graph ke aar-paar horizontal lines $y = c$ khincho. Agar koi line graph ko do baar milti hai, to do inputs output $c$ share karte hain, aur $f$ one-one nahi hai. Agar koi line graph ko miss karti hai, to $c$ tak kabhi nahi pahuncha jaata, aur $f$ onto nahi hai. $𝐑$ ka bijection aisa graph hai jise har horizontal line exactly ek baar milti hai; Example 9 ki straight line, slope $\frac{1}{3}$ wali, yeh test pass karti hai.

1/3Line $y = 5$ , $y = x^{2} - 4 x$ ko $x = - 1$ aur $x = 5$ par milti hai: one-one nahi.

Fig. 1.8

𝐑

𝐑

tak function ke liye horizontal-line test.NCERT Fig. 1.4 · p. 9

Axes jinmein x minus 2 se 6 tak aur y minus 7 se 9 tak hai. Step one: y = x squared minus 4x ke U-shaped graph ko, jo point (2, minus 4) par sabse neeche hai, dashed horizontal line y = 5 do marked points, x = minus 1 aur x = 5, par kaatti hai. Step two: dashed line neeche y = minus 6 par aati hai, lowest point se bhi neeche, aur graph ko kahin nahi milti. Step three: U-shaped graph fade hota hai aur ek S-shaped cubic, y equals one half of (x minus 2) cubed plus 1, dikhta hai; y = minus 5, minus 1 aur 5 par teen dashed lines mein se har ek use exactly ek marked point par milti hai.

Example 10NCERT Example 10 · p. 9

Digits ginna

Maan lo $f : 𝐍 \to 𝐍$ , $n$ ke digits ki sankhya deta hai, yaani $f (7) = 1$ , $f (42) = 2$ aur $f (2026) = 4$ . Dikhao ki $f$ onto hai lekin one-one nahi.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 11NCERT Example 11 · p. 9

Ek parabola jo do baar fail hota hai

Dikhao ki $f : 𝐑 \to 𝐑$ , $f (x) = x^{2} - 4 x$ , na one-one hai na onto.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 12NCERT Example 12 · p. 9

Padosiyon ko swap karna

$f : 𝐍 \to 𝐍$ ko aise define karo: $f (n) = n + 1$ agar 3 se divide karne par $n$ remainder 1 chhode, $f (n) = n - 1$ agar remainder 2 chhode, aur $f (n) = n$ agar $n$ , 3 ka multiple ho. Yaani $f$ 1 aur 2 ko swap karta hai, 4 aur 5 ko swap karta hai, 7 aur 8 ko swap karta hai, aur 3, 6, 9, … ko fix rakhta hai. Dikhao ki $f$ one-one aur onto hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Finite set par dono properties aapas mein bandhi hain. Socho chaar letters chaar letter-boxes mein daale ja rahe hain, har letter exactly ek box mein.

Example 13NCERT Example 13 · p. 10

Finite set par onto hone se one-one hona padta hai

Maan lo $S = {a, b, c, d}$ . Dikhao ki har onto function $f : S \to S$ one-one hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14NCERT Example 14 · p. 10

Finite set par one-one hone se onto hona padta hai

Phir se $S = {a, b, c, d}$ ke saath, dikhao ki har one-one function $f : S \to S$ onto hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

PropertyFinite set par one-one aur onto

Maan lo $X$ ek finite set hai aur $f$ , $X$ se khud tak ek map hai. Tab $f$ tabhi one-one hai jab woh onto ho, isliye akeli koi bhi property $f$ ko bijection bana deti hai.

1.4 Composition of Functions aur Invertible Function

Taxi ka kiraya chali gayi distance par depend karta hai, aur distance us destination par jo tum chunte ho. Destination ko pehle rule mein daalo, result ko doosre rule mein daalo, aur tumhe seedha kiraya mil jaata hai. Is tarah do functions ko jodne se ek naya function banta hai, jise unka composition kehte hain.

Definition

Composition of functions

$f : A \to B$ aur $g : B \to C$ ke liye composite $g \circ f : A \to C$ har $x \in A$ ke liye $(g \circ f) (x) = g (f (x))$ se define hota hai. Ise ' $g$ after $f$ ' padho. NCERT ise $g o f$ likhta hai.

ExampleAgar $𝐑$ par $f (x) = x + 2$ aur $g (x) = x^{2}$ , to $(g \circ f) (3) = g (5) = 25$ .

Do details important hain. Right mein likha function, $f$ , pehle kaam karta hai. Aur composite tabhi meaningful hai jab $f$ ka har output $g$ ke liye valid input ho, isiliye definition mein $f$ ka codomain $g$ ka domain hai (Fig. 1.9).

4/4Bahar se dekho to do machines milkar ek function banati hain, $(g \circ f) (x) = (x + 2)^{2}$ .

Fig. 1.9

g \circ f

pehle

f

chalata hai, phir

g

.NCERT Fig. 1.5 · p. 12

Ek dashed conveyor belt left se right do square machines ke beech se chalti hai, left mein f aur right mein g, f ke neeche f(x) = x + 2 aur g ke neeche g(x) = x squared likha hai. Step one: 3 wala ek chip input end par intezaar karta hai. Step two: 5 wala chip machines ke beech khisakta hai, kyunki f(3) = 5, aur input chip fade ho jaata hai. Step three: 25 wala chip output end par khisakta hai, kyunki g(5) = 25. Step four: ek dashed box dono machines ko gher leta hai aur us par g composed with f likha hai, neeche (g composed with f)(x) = (x + 2) squared likha hai, yaani yeh jodi ek machine ki tarah kaam karti hai jo 3 ko 25 bana deti hai.

Example 15NCERT Example 15 · p. 12

Finite functions ka composition

Maan lo $f : {0, 1, 2, 3} \to {1, 2, 5, 10}$ , $f (x) = x^{2} + 1$ hai, aur $g : {1, 2, 5, 10} \to {0, 1, 2}$ woh remainder deta hai jo $y$ ko 3 se divide karne par bachta hai. $g \circ f$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 16NCERT Example 16 · p. 12

Order matter karta hai

Maan lo $f, g : 𝐑 \to 𝐑$ , $f (x) = x + 2$ aur $g (x) = x^{2}$ hain. $g \circ f$ aur $f \circ g$ nikaalo, aur dikhao ki yeh alag functions hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Isliye composition commutative nahi hai. Lekin yeh associative hai: $f : A \to B$ , $g : B \to C$ aur $h : C \to D$ ke liye $h \circ (g \circ f)$ aur $(h \circ g) \circ f$ dono $x$ ko $h (g (f (x)))$ tak bhejte hain. Isiliye hum bina brackets ke $h \circ g \circ f$ likh sakte hain.

Invertible functions

Kuch functions ko ulta chalaya ja sakta hai. Agar $f (x) = x + 2$ , 3 ko 5 tak bhejta hai, to $y \mapsto y - 2$ , 5 ko wapas 3 tak bhejta hai, aur yeh har input ke liye $f$ ko undo karta hai. Ise precisely kehne ke liye hum identity function $I_{X} : X \to X$ , $I_{X} (x) = x$ use karte hain, jo har element ko wahin chhod deta hai jahan woh hai.

Definition

Invertible function

Function $f : X \to Y$ invertible hai agar aisa function $g : Y \to X$ ho jiske liye $g \circ f = I_{X}$ aur $f \circ g = I_{Y}$ . Tab hum $g$ ko $f$ ka inverse kehte hain aur ise $f^{- 1}$ likhte hain.

Example $f : 𝐑 \to 𝐑$ , $f (x) = x + 2$ , invertible hai aur $f^{- 1} (y) = y - 2$ .

Dono conditions zaroori hain. $g \circ f = I_{X}$ kehta hai ki $g$ , $f$ ko undo karta hai: $X$ se shuru karo, $Y$ tak jao aur wapas wahin aao jahan se chale the. $f \circ g = I_{Y}$ kehta hai ki $f$ , $g$ ko undo karta hai. Kin functions ka inverse hota hai? $f$ ke arrows ulte karo aur poocho ki result ab bhi function hai ya nahi (Fig. 1.10).

2/3Ulta karne par $B$ ke har element se exactly ek arrow nikalta hai: yeh $f^{- 1}$ hai.

Fig. 1.10Arrows ulte karne se function tabhi milta hai jab

B

ka har element exactly ek baar hit hua ho.

Circles ke do columns: left mein A jismein 1, 2 aur 3 hain aur right mein B jismein p, q aur r hain. Step one: f ke arrows 1 se q, 2 se r aur 3 se p tak jaate hain, aur neeche ki line yeh values list karti hai. Step two: arrows ulte aur highlighted ho jaate hain, q se 1, r se 2 aur p se 3 tak; B ke har element se exactly ek arrow nikalta hai, isliye ulti picture function f inverse hai. Step three ek alag function h dikhata hai jismein h(1) = h(2) = p aur h(3) = q, ulta kiya hua: ab p se do arrows nikalte hain, jis par two arrows likha hai, aur r, jo dashed bana hai, se koi nahi, jis par no arrow likha hai, isliye ulti picture function nahi hai.

TheoremInvertible matlab bijective

Function $f : X \to Y$ invertible hai if and only if woh one-one aur onto ho.

Invertible se bijective. Maan lo $g$ ke liye $g \circ f = I_{X}$ aur $f \circ g = I_{Y}$ . Agar $f (x_{1}) = f (x_{2})$ , to dono sides par $g$ lagao: $x_{1} = g (f (x_{1})) = g (f (x_{2})) = x_{2}$ , isliye $f$ one-one hai. Kisi bhi $y \in Y$ ke liye element $x = g (y)$ se $f (x) = f (g (y)) = y$ , isliye $f$ onto hai.

Bijective se invertible. Agar $f$ onto hai, to har $y \in Y$ ki kam se kam ek preimage hai; agar $f$ one-one hai, to zyada se zyada ek. Isliye hum $g (y)$ ko woh unique $x$ define kar sakte hain jiske liye $f (x) = y$ . Tab $g (f (x)) = x$ , kyunki $x$ , $f (x)$ ki unique preimage hai, aur $g (y)$ ke choice se $f (g (y)) = y$ .

Isse do practical raaste milte hain. $f^{- 1}$ nikaale bina yeh dikhane ke liye ki $f$ invertible hai, prove karo ki yeh one-one aur onto hai. $f^{- 1}$ nikaalne ke liye $y = f (x)$ ko $x$ ke liye solve karo: jo expression milta hai woh $f^{- 1} (y)$ hai, aur tum dono compositions check karke use confirm kar sakte ho.

Example 17NCERT Example 17 · p. 12

Inverse nikaalna aur check karna

Maan lo $Y = {8, 11, 14, 17, \dots} = {y \in 𝐍 : y = 3 x + 5 for some x \in 𝐍}$ aur $f : 𝐍 \to Y$ , $f (x) = 3 x + 5$ hai. Dikhao ki $f$ invertible hai aur $f^{- 1}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Domain ko restrict karna

Rule $x \mapsto x^{2} + 4$ , $𝐑$ se $𝐑$ tak map ki tarah invertible nahi hai. Dikhao ki $f : [0, \infty) \to [4, \infty)$ , $f (x) = x^{2} + 4$ , invertible hai aur $f^{- 1}$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Miscellaneous Examples

Yeh examples chapter ke ideas ko mix karte hain. Kai examples general relations ke baare mein proofs hain; baaki chhote sets par relations ya functions ginte hain, jahan cases ko dhyaan se list karna hi poora skill hai.

Example 18NCERT Example 18 · p. 13

Do equivalence relations ko combine karna

$𝐙$ mein $R_{1}$ ko 'same parity' (2, $a - b$ ko divide kare) aur $R_{2}$ ko '3 se divide karne par same remainder' (3, $a - b$ ko divide kare) lo. (a) Dikhao ki agar $R_{1}$ aur $R_{2}$ set $A$ mein koi bhi do equivalence relations hain, to $R_{1} \cap R_{2}$ ek equivalence relation hai. (b) Upar ke do relations ke liye $R_{1} \cap R_{2}$ describe karo. (c) Dikhao ki yahan $R_{1} \cup R_{2}$ equivalence relation nahi hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 19NCERT Example 19 · p. 13

Barabar differences wale pairs

Natural numbers ke ordered pairs ke set $A = 𝐍 \times 𝐍$ mein $(a, b) R (c, d)$ define karo if and only if $a + d = b + c$ . Dikhao ki $R$ ek equivalence relation hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 20NCERT Example 20 · p. 13

Ek relation, do descriptions

Maan lo $X = {1, 2, \dots, 10}$ , $R_{1} = {(x, y) : 4 divides x - y}$ , aur $x R_{2} y$ ka matlab hai ki $x$ aur $y$ dono sets ${1, 5, 9}$ , ${2, 6, 10}$ , ${3, 7}$ aur ${4, 8}$ mein se same set mein hain. Dikhao ki $R_{1} = R_{2}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 21NCERT Example 21 · p. 14

Same output, yaani same class

$𝐑$ mein do numbers ko related kaho jab unka square same ho: $a \sim b$ agar $a^{2} = b^{2}$ . (a) Dikhao ki $\sim$ ek equivalence relation hai aur iski classes nikaalo. (b) Samjhao ki har function $f : X \to Y$ ke liye ' $f$ ke under same output' $X$ mein equivalence relation kyun hai, aur iski classes tumhe $f$ ke baare mein kya batati hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 22NCERT Example 22 · p. 14

One-one functions ginna

$A = {1, 2, 3}$ se $B = {p, q, r, s, t}$ tak kitne one-one functions hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 23NCERT Example 23 · p. 14

Di gayi properties wale relations ginna

Maan lo $A = {1, 2, 3}$ . $A$ mein kitne aise relations hain jismein $(2, 1)$ aur $(3, 1)$ hon aur jo reflexive aur transitive hon, lekin symmetric nahi?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Equivalence relations ginna wahi hai jo elements ko groups mein sort karne ke tareeke ginna, aur ise picture karna aasaan hai (Fig. 1.11).

4/4 $1 + 3 + 1 = 5$ partitions, isliye 5 equivalence relations.

Fig. 1.11

{a, b, c}

ko classes mein sort karne ka har tareeka ek equivalence relation hai.

Paanch panels mein se har ek points a, b aur c dikhata hai, classes ke around shaded outlines ke saath. Pehle step mein sirf pehla panel highlight hai: ek outline mein teeno points hain, jisse universal relation milta hai. Doosre step mein doosra panel highlight hai: har point ki apni outline hai, teen classes, jisse identity relation milta hai. Teesre step mein aakhri teen panels highlight hain: har ek mein do points ek outline share karte hain aur teesra akela hai, akele point ke har choice ke liye ek baar. Aakhri step mein paanchon panels dikhte hain aur neeche count 1 + 3 + 1 = 5 likha hai.

Example 24NCERT Example 24 · p. 14

Teen elements par saare equivalence relations

$A = {a, b, c}$ mein saare equivalence relations nikaalo aur unhe gino.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 25NCERT Example 25 · p. 14

Onto functions ka sum

Maan lo $f, g : 𝐙 \to 𝐙$ , $f (n) = n$ aur $g (n) = - n$ hain. Dikhao ki $f$ aur $g$ dono onto hain, lekin $f + g$ nahi.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 26NCERT Example 26 · p. 15

One-one functions ka sum

Prove karo ki $𝐑$ ke maps $x \mapsto x^{3}$ aur $x \mapsto - 3 x$ dono one-one hain, aur unka sum $x \mapsto x^{3} - 3 x$ nahi hai. Maps ko $f$ aur $g$ kaho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

{1, 2, 3, 4}

{0, 1}

tak kitne functions hain?

functions

Answer dekho

Answer 16 functions

4 inputs mein se har ek independently 2 outputs mein se ek chunta hai:

2^{4} = 16

Recap

Summary aur key points

$A$ se $B$ tak relation $A \times B$ ka koi bhi subset hai; $A$ mein relation $A \times A$ ka subset hai (Section 1.1).
Empty relation $\emptyset$ aur universal relation $A \times A$ do trivial relations hain (Section 1.2).
Reflexive: har $a$ ke liye $(a, a) \in R$ . Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$ . Transitive: $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$ (Reflexive, symmetric aur transitive relations).
Equivalence relation mein teeno properties hoti hain; 'ka ___ same hai' iska typical example hai (Example 21).
Class $[a]$ , $a$ se related elements ka set hai. Classes set ka partition karti hain, aur har partition ek equivalence relation se aata hai (Property).
$f : X \to Y$ one-one hai agar $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ , onto hai agar uski range $Y$ ho, aur bijective hai agar dono ho (Section 1.3).
$f : 𝐑 \to 𝐑$ ke liye: one-one ka matlab hai koi horizontal line graph ko do baar nahi milti; onto ka matlab hai har horizontal line use milti hai (Fig. 1.8).
Finite set se khud tak maps ke liye one-one aur onto equivalent hain; infinite sets ke liye nahi (Property).
$(g \circ f) (x) = g (f (x))$ , isliye $f$ pehle kaam karta hai. Composition associative hai, lekin general mein $g \circ f \neq f \circ g$ (Section 1.4).
$f$ invertible hai jab kisi $g$ ke liye $g \circ f = I_{X}$ aur $f \circ g = I_{Y}$ ; yeh tabhi hota hai jab $f$ bijective ho (Theorem).
$f^{- 1}$ nikaalne ke liye $y = f (x)$ ko $x$ ke liye solve karo, phir dono compositions check karo (Example 17).
$k$ -element set se $n$ -element set tak $n^{k}$ functions hain, aur inmein se $n (n - 1) \dots (n - k + 1)$ one-one hain (Example 22).

Timeline

Function ka idea kaise bada hua

1637
Curves equations ban jaate hain
René Descartes ki La Géométrie curves ko do unknowns wali equations se describe karti hai, taaki ek quantity doosri par depend kare.
1673
Leibniz 'function' shabd use karte hain
Curves par apne notes mein Gottfried Leibniz ise kisi point par slope ya tangent jaisi quantities ke liye use karte hain.
1718
Ek variable ka function
Johann Bernoulli function ko ek variable aur constants se bani quantity kehte hain, aur ise $φ x$ likhte hain.
1734
Euler $f (x)$ likhte hain
Leonhard Euler notation $f (x)$ introduce karte hain, aur baad mein functions ko analysis ke central objects bana dete hain.
1797
Lagrange ke analytic functions
Joseph-Louis Lagrange Théorie des fonctions analytiques mein calculus ko functions par khada karte hain aur derivative ke liye $f^{'} (x)$ likhte hain.
1837
Dirichlet: koi bhi rule chalega
Peter Gustav Lejeune Dirichlet $y$ ko $x$ par kisi bhi rule se depend karne dete hain, zaroori nahi ki woh formula ho.
1900s ki shuruaat
Functions pairs ke sets ki tarah
Georg Cantor ki set theory aane ke baad function ordered pairs ka aisa set ban jaata hai jismein koi pehli entry repeat nahi hoti: yahi definition is chapter mein use hui hai.

Practice · 13 questions

Practice questions answers ke saath (NCERT “Miscellaneous exercises”)

ProofHardNCERT Miscellaneous Exercise, Q1
Dikhao ki $f : (- 1, 1) \to 𝐑$ , $f (x) = \frac{x}{1 - ∣ x ∣}$ , ek bijection hai.
Solution dikhao
$0 \leq x < 1$ ke liye $f (x) = \frac{x}{1 - x} \geq 0$ ; $- 1 < x < 0$ ke liye $f (x) = \frac{x}{1 + x} < 0$ . Isliye $f (x)$ ka sign $x$ jaisa hai, aur $f (x) = 0$ sirf $x = 0$ par.
One-one: agar $f (x_{1}) = f (x_{2})$ , to $x_{1}$ aur $x_{2}$ , 0 ki same side par hain. Agar dono $\geq 0$ hain, to $\frac{x_{1}}{1 - x_{1}} = \frac{x_{2}}{1 - x_{2}}$ se $x_{1} - x_{1} x_{2} = x_{2} - x_{1} x_{2}$ milta hai, isliye $x_{1} = x_{2}$ . Agar dono negative hain, to $\frac{x_{1}}{1 + x_{1}} = \frac{x_{2}}{1 + x_{2}}$ se isi tarah $x_{1} = x_{2}$ milta hai.
Onto: $y \geq 0$ ke liye $x = \frac{y}{1 + y} \in [0, 1)$ lo; tab $1 - x = \frac{1}{1 + y}$ aur $f (x) = y$ . $y < 0$ ke liye $x = \frac{y}{1 - y} \in (- 1, 0)$ lo; tab $1 + x = \frac{1}{1 - y}$ aur $f (x) = y$ . Dono cases mein $x = \frac{y}{1 + ∣ y ∣}$ , jo asal mein $f^{- 1}$ ka formula hai.
ProofMediumNCERT Miscellaneous Exercise, Q2
Dikhao ki $f : 𝐑 \to 𝐑$ , $f (x) = x^{3} + x$ , one-one hai.
Solution dikhao
Agar $x_{1}^{3} + x_{1} = x_{2}^{3} + x_{2}$ , to $(x_{1} - x_{2}) (x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} + 1) = 0$ .
Doosra factor ${(x_{1} + \frac{1}{2} x_{2})}^{2} + \frac{3}{4} x_{2}^{2} + 1 \geq 1$ ke barabar hai, isliye yeh kabhi zero nahi hota. Isliye $x_{1} - x_{2} = 0$ , aur $f$ one-one hai.
3
Short answerMediumNCERT Miscellaneous Exercise, Q3
Maan lo $X$ ek non-empty finite set hai aur $P (X)$ uske saare subsets ka set hai. $A, B \in P (X)$ ke liye $A R B$ lo jab $A$ aur $B$ mein elements ki sankhya same ho.
1. Kya $P (X)$ mein $R$ ek equivalence relation hai? Apna answer justify karo.
  Solution dikhao
  AnswerHaan.
  $R$ hai 'ka size same hai', yaani ek attribute ki equality, isliye yeh reflexive, symmetric aur transitive hai.
2. $X = {1, 2, 3}$ ke liye equivalence classes list karo. Kitni hain?
  Solution dikhao
  Answer4 classes.
  Classes hain ${\emptyset}$ ; teen one-element subsets; teen two-element subsets; aur ${X}$ . Isliye 4 classes hain, jinmein 1, 3, 3 aur 1 members hain (kul 8 subsets).
4
NumericalMediumNCERT Miscellaneous Exercise, Q4
Maan lo $S = {1, 2, 3, 4, 5}$ .
1. $S \to S$ kitne functions hain?
  Answer dekho
  Answer 3125
  5 inputs mein se har ek ke paas 5 choices hain: $5^{5} = 3125$ .
2. Inmein se kitne one-one hain?
  Answer dekho
  Answer 120
  $5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$ .
3. Inmein se kitne onto hain?
  Answer dekho
  Answer 120
  Finite set par onto aur one-one ek hi baat hai, isliye phir se $5! = 120$ .
4. Kitne na one-one hain na onto?
  Answer dekho
  Answer 3005
  120 one-one functions exactly wahi hain jo onto hain, isliye $3125 - 120 = 3005$ functions dono mein se kuch nahi hain.
5
Short answerMediumNCERT Miscellaneous Exercise, Q5
Maan lo $A = {1, 2, 3}$ , $B = {1, 8, 27}$ , aur $f, g : A \to B$ , $f (x) = x^{3}$ aur $g (x) = 6 x^{2} - 11 x + 6$ hain. Same domain aur codomain wale do functions equal hote hain jab woh domain ke har point par barabar hon.
1. Kya $f$ aur $g$ equal hain?
  Solution dikhao
  AnswerHaan.
  $g (1) = 6 - 11 + 6 = 1$ , $g (2) = 24 - 22 + 6 = 8$ aur $g (3) = 54 - 33 + 6 = 27$ , jo $f (1) = 1$ , $f (2) = 8$ aur $f (3) = 27$ se match karte hain. Isliye $f = g$ , chahe formulas alag dikhte hon.
2. Kya domain ${1, 2, 3, 4}$ par (kaafi bade codomain ke saath) $f$ aur $g$ ab bhi equal honge?
  Solution dikhao
  AnswerNahi.
  $f (4) = 64$ lekin $g (4) = 96 - 44 + 6 = 58$ . Asal mein $f (x) - g (x) = (x - 1) (x - 2) (x - 3)$ , jo sirf 1, 2 aur 3 par zero hota hai.
Quick answerMediumNCERT Miscellaneous Exercise, Q6
Maan lo $A = {1, 2, 3}$ . $A$ mein kitne relations $(1, 2)$ contain karte hain aur reflexive aur symmetric hain, lekin transitive nahi?
Answer dekho
Answer 2
Link ${1, 2}$ hona hi chahiye, isliye 4 candidates hain: kuch add mat karo, ${1, 3}$ add karo, ${2, 3}$ add karo, ya dono add karo. Kuch add na karne se equivalence relation milta hai, aur dono add karne se universal relation, dono transitive. Exactly ek link add karne se transitivity toot jaati hai (jaise $(3, 1), (1, 2)$ ko $(3, 2)$ chahiye), isliye answer 2 hai.
Quick answerMediumNCERT Miscellaneous Exercise, Q7
${1, 2, 3, 4}$ mein kitne equivalence relations pair $(1, 2)$ contain karte hain?
Answer dekho
Answer 5
Jin partitions mein 1 aur 2 saath hain woh hain ${12} {3} {4}$ , ${123} {4}$ , ${124} {3}$ , ${12} {34}$ aur ${1234}$ : kul 5.
Quick answerEasy
3 elements wale set mein kitne relations reflexive hain?
Answer dekho
Answer 64
9 pairs mein se 3 loops andar hone hi chahiye, aur baaki 6 mein se har ek andar ya bahar ho sakta hai: $2^{6} = 64$ .
Quick answerMedium
${1, 2, 3}$ mein kitne relations reflexive aur symmetric dono hain?
Answer dekho
Answer 8
Loops forced hain, aur teen two-way links ${1, 2}$ , ${1, 3}$ , ${2, 3}$ mein se har ek andar ya bahar hai: $2^{3} = 8$ .
Quick answerMedium
Agar $f : A \to B$ aur $g : B \to C$ dono one-one hain, to $g \circ f$ hai
hamesha one-onehamesha ontoone-one sirf tab jab $A = C$ kabhi one-one nahi
Answer dekho
Answer (A) hamesha one-one
$g (f (x_{1})) = g (f (x_{2}))$ se $f (x_{1}) = f (x_{2})$ milta hai kyunki $g$ one-one hai, aur phir $x_{1} = x_{2}$ kyunki $f$ one-one hai.
11
Quick answerMedium
Har function ko uske type se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
$𝐑$ par $x \mapsto x^{3}$ : bijective
$𝐍$ par $x \mapsto x + 3$ : one-one, onto nahi
$𝐑$ par $x \mapsto ∣ x ∣$ : dono nahi
digits ki sankhya, $𝐍 \to 𝐍$ : onto, one-one nahi
$x^{3}$ ka inverse $\sqrt[3]{y}$ hai; $x + 3$ , 1, 2, 3 ko miss karta hai; $∣ x ∣$ mein $∣ - 1 ∣ = ∣ 1 ∣$ hai aur yeh $- 1$ ko miss karta hai; digit count onto hai aur $f (1) = f (2)$ .
Quick answerEasy
Function $f : X \to Y$ invertible hai if and only if woh aur ho.
Answer dekho
Answer (1) one-one, (2) onto
Invertible matlab bijective: one-one aur onto (Theorem).
13
Quick answerEasy
Bijection $f$ ka inverse nikaalne ke steps ko order mein lagao.
1. $f^{- 1} (f (x)) = x$ aur $f (f^{- 1} (y)) = y$ check karo
2. Result ko $f^{- 1} (y)$ kaho
3. $x$ ko $y$ ke terms mein solve karo
4. $y = f (x)$ likho
Answer dekho
Answer
$y = f (x)$ likho
$x$ ko $y$ ke terms mein solve karo
Result ko $f^{- 1} (y)$ kaho
$f^{- 1} (f (x)) = x$ aur $f (f^{- 1} (y)) = y$ check karo
$y = f (x)$ ko $x$ ke liye solve karne se inverse formula milta hai, aur dono compositions use confirm karte hain.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 1.1, Q1	Practice 1.1 · Q1
NCERT Exercise 1.1, Q2	Practice 1.1 · Q2
NCERT Exercise 1.1, Q3	Practice 1.1 · Q3
NCERT Exercise 1.1, Q4	Practice 1.1 · Q4
NCERT Exercise 1.1, Q5	Practice 1.1 · Q5
NCERT Exercise 1.1, Q6	Practice 1.1 · Q6
NCERT Exercise 1.1, Q7	Practice 1.1 · Q7
NCERT Exercise 1.1, Q8	Practice 1.1 · Q8
NCERT Exercise 1.1, Q9	Practice 1.1 · Q9
NCERT Exercise 1.1, Q10	Practice 1.1 · Q10
NCERT Exercise 1.1, Q11	Practice 1.1 · Q11
NCERT Exercise 1.1, Q12	Practice 1.1 · Q12
NCERT Exercise 1.1, Q13	Practice 1.1 · Q13
NCERT Exercise 1.1, Q14	Practice 1.1 · Q14
NCERT Exercise 1.1, Q15	Practice 1.1 · Q15
NCERT Exercise 1.1, Q16	Practice 1.1 · Q16
NCERT Exercise 1.2, Q1	Practice 1.2 · Q1
NCERT Exercise 1.2, Q2	Practice 1.2 · Q2
NCERT Exercise 1.2, Q3	Practice 1.2 · Q3
NCERT Exercise 1.2, Q4	Practice 1.2 · Q4
NCERT Exercise 1.2, Q5	Practice 1.2 · Q5
NCERT Exercise 1.2, Q6	Practice 1.2 · Q6
NCERT Exercise 1.2, Q7	Practice 1.2 · Q7
NCERT Exercise 1.2, Q8	Practice 1.2 · Q8
NCERT Exercise 1.2, Q9	Practice 1.2 · Q9
NCERT Exercise 1.2, Q10	Practice 1.2 · Q10
NCERT Exercise 1.2, Q11	Practice 1.2 · Q11
NCERT Exercise 1.2, Q12	Practice 1.2 · Q12
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercises · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous exercises · Q7

Quick check: MCQs

1.1 Introduction3 questions
Equivalence classes4 questions
1.3 Types of Functions4 questions
Invertible functions4 questions
Miscellaneous Examples2 questions

Glossary · 15

Key terms (English + Hindi)

bijective functionएकैकी आच्छादक फलनekaiki aachchhaadak phalan: Aisa function jo one-one bhi ho aur onto bhi.

compositionफलनों का संयोजनphalanon ka sanyojan: $f : A \to B$ aur $g : B \to C$ ke liye function $g \circ f : A \to C$ , $(g \circ f) (x) = g (f (x))$ .Chapter mein pehli baar yahan aaya →

empty relationरिक्त संबंधrikt sambandh: Relation $\emptyset$ , jismein koi element kisi element se related nahi hai.
equivalence classतुल्यता वर्गtulyata varg: Equivalence relation ke liye, $a$ se related saare elements ka set $[a]$ .
equivalence relationतुल्यता संबंधtulyata sambandh: Aisa relation jo reflexive, symmetric aur transitive ho.

functionफलनphalan: $X$ se $Y$ tak relation jo $X$ ke har element ko $Y$ ke exactly ek element se relate karta hai.

identity functionतत्समक फलनtatsamak phalan: Function $I_{X} : X \to X$ jismein har $x$ ke liye $I_{X} (x) = x$ .Chapter mein pehli baar yahan aaya →
invertible functionव्युत्क्रमणीय फलनvyutkramaniya phalan: Function $f : X \to Y$ jiske liye aisa function $f^{- 1} : Y \to X$ ho ki $f^{- 1} \circ f = I_{X}$ aur $f \circ f^{- 1} = I_{Y}$ .

one-one functionएकैकी फलनekaiki phalan: Function jismein $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ : alag inputs alag outputs dete hain.
onto functionआच्छादक फलनaachchhaadak phalan: Function jiski range uske codomain ke barabar hai: codomain ka har element kisi ki image hai.

reflexive relationस्वतुल्य संबंधsvatulya sambandh: Aisa relation jismein set ke har $a$ ke liye $(a, a) \in R$ .
relationसंबंधsambandh: $A \times B$ ka subset; $A$ mein relation $A \times A$ ka subset hai.Chapter mein pehli baar yahan aaya →

symmetric relationसममित संबंधsammit sambandh: Aisa relation jismein $(a, b) \in R$ hamesha $(b, a) \in R$ imply karta hai.

transitive relationसंक्रामक संबंधsankraamak sambandh: Aisa relation jismein $(a, b) \in R$ aur $(b, c) \in R$ hamesha $(a, c) \in R$ imply karte hain.

universal relationसार्वत्रिक संबंधsaarvatrik sambandh: Relation $A \times A$ , jismein har element har element se related hai.

Relations and Functions: Class 12 Maths Chapter 1 Hinglish mein

Is chapter mein aap kya seekhenge

1.1 Introduction

1.2 Types of Relations

Reflexive, symmetric aur transitive relations

Equivalence classes

Practice 1.1: answers ke saath

1.3 Types of Functions

Practice 1.2: answers ke saath

1.4 Composition of Functions aur Invertible Function

Invertible functions

Miscellaneous Examples

Summary aur key points

Practice questions answers ke saath (NCERT “Miscellaneous exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye