Class 12 Maths · Chapter 1
Short answer:
Is chapter me 2 exercises hain (Ex 1.1 aur Ex 1.2) jisme total karib 21 questions hain — relations ko reflexive/symmetric/transitive/equivalence check karna, aur functions ko one-one/onto/bijective check karna, composition of functions, aur invertible functions dhoondhna. Chapter 1 hone ki wajah se ye poore Class 12 Maths ka foundation hai — set theory pe based hai (Class 11) aur aage differentiation/integration ke domain-range wale concepts isi pe tike hain. JEE Main/Advanced me bhi bahut common hai.
Relations aur Functions Class 12 Maths ka pehla chapter hai, jo batata hai ki do sets ke beech elements kaise connect hote hain aur wo connection kis type ka hai. Board exam me is chapter se seedha 1-2 questions aate hain (reflexive/symmetric/transitive prove karna ya function ka one-one/onto check karna), lekin iska asli importance JEE me hai — composition of functions aur inverse function ke questions calculus (especially differentiation of composite/inverse functions) me directly use hote hain. Isliye is chapter ko sirf 'ratt' mat karo, concept clear karo — kyunki yehi foundation aage bhi kaam aayega.
Chapter 1 Summary — 5 Minute Revision
1. Relation kya hoti hai
Agar A aur B do sets hain, to A se B ki koi bhi relation R, A × B ka ek subset hoti hai. Simple bhasha me — R ek rule hai jo batata hai A ke kaunse elements B ke kaunse elements se "related" hain.
Jab R, set A se apne aap A me hi define hoti hai (matlab R ⊆ A × A), tab hum uski teen properties check karte hain:
- Reflexive: Har a ∈ A ke liye (a, a) ∈ R hona chahiye. Matlab har element khud se related hona chahiye.
- Symmetric: Agar (a, b) ∈ R hai to (b, a) bhi R me hona chahiye — sabke liye.
- Transitive: Agar (a, b) ∈ R aur (b, c) ∈ R hai to (a, c) bhi R me hona chahiye — sabke liye.
2. Equivalence Relation
Agar ek relation reflexive, symmetric, AUR transitive — teeno ek saath hai, tabhi wo equivalence relation kehlayegi. Ek bhi property miss ho gayi to nahi kehlayegi.
Equivalence relation set A ko disjoint "equivalence classes" me todh deti hai — ye idea board me directly nahi poocha jaata lekin concept samajhna zaroori hai.
3. Function kya hoti hai
Function f: A → B ek special relation hai jisme A ka har element B me exactly ek element se map hota hai — na zero, na do ya zyada.
4. Types of Functions
| Type | Definition | Test |
|---|---|---|
| One-one (Injective) | Alag-alag inputs ke liye alag-alag outputs | f(x₁) = f(x₂) ⟹ x₁ = x₂ |
| Onto (Surjective) | Codomain ka har element kisi na kisi input se cover hota hai | Range = Codomain |
| Bijective | One-one AUR onto dono | Dono tests pass hone chahiye |
| Many-one | Do ya zyada inputs same output dete hain | One-one test fail |
| Into | Codomain ka koi element uncovered reh jaata hai | Range ⊊ Codomain |
↔ Table ko side me swipe karein
5. Composition of Functions
Agar f: A → B aur g: B → C hain, to composite function gof: A → C aise define hoti hai: (gof)(x) = g(f(x)).
Important: fog aur gof generally same nahi hote. Order matter karta hai — pehle andar wale function ko apply karo, phir bahar wale ko.
6. Invertible Functions
Ek function f: A → B invertible tabhi hoti hai jab wo bijective ho (one-one + onto). Uska inverse f⁻¹: B → A hota hai jisme f(a) = b ⟺ f⁻¹(b) = a.
Yaad rakho: invertibility ki condition sirf "one-one" nahi, balki "bijective" hai — dono chahiye.
Poore Class 12 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q21)
Q1. Show that the relation R in the set R (real numbers) defined as R = {(a, b) : a ≤ b} is reflexive and transitive but not symmetric.
Reflexive: Har a ∈ R ke liye a ≤ a sach hai, isliye (a, a) ∈ R. Reflexive hai.
Reflexive: ∀a ∈ ℝ, a ≤ a ⟹ (a, a) ∈ R ✓
Symmetric nahi hai: Counter-example lo — a = 1, b = 2. (1, 2) ∈ R kyunki 1 ≤ 2. Lekin (2, 1) ∈ R nahi hai kyunki 2 ≤ 1 galat hai.
(1, 2) ∈ R lekin (2, 1) ∉ R ⟹ Not symmetric
Transitive: Maano (a, b) ∈ R aur (b, c) ∈ R, matlab a ≤ b aur b ≤ c. Real numbers ki property se a ≤ c bhi sach hai. Isliye (a, c) ∈ R.
a ≤ b aur b ≤ c ⟹ a ≤ c (transitivity of ≤) ⟹ Transitive ✓
Isliye R reflexive aur transitive hai, symmetric nahi.
Q2. Check whether the relation R in the set {1, 2, 3} defined as R = {(1, 2), (2, 1)} is reflexive, symmetric, and transitive.
Reflexive nahi: (1,1), (2,2), (3,3) — koi bhi R me nahi hai.
(1,1) ∉ R ⟹ Not reflexive
Symmetric hai: (1,2) ∈ R aur (2,1) ∈ R — dono present hain.
(1,2) ∈ R aur (2,1) ∈ R ⟹ Symmetric ✓
Transitive nahi: (1,2) ∈ R aur (2,1) ∈ R hai, to transitivity ke liye (1,1) ∈ R hona chahiye tha — lekin nahi hai.
(1,2) ∈ R aur (2,1) ∈ R lekin (1,1) ∉ R ⟹ Not transitive
Isliye R sirf symmetric hai, reflexive aur transitive nahi.
Q3. Show that the relation R in R defined as R = {(a, b) : a ≤ b²} is neither reflexive, nor symmetric, nor transitive.
Reflexive nahi: a = 1/2 lo. Kya 1/2 ≤ (1/2)² = 1/4? Nahi, 1/2 > 1/4.
a = 1/2: 1/2 ≤ 1/4 galat hai ⟹ Not reflexive
Symmetric nahi: a = 1, b = 4. 1 ≤ 4² = 16 sach hai, to (1,4) ∈ R. Lekin (4,1) ke liye 4 ≤ 1² = 1 galat hai.
(1,4) ∈ R lekin (4,1) ∉ R ⟹ Not symmetric
Transitive nahi: a = 3, b = 2, c = 1. 3 ≤ 2² = 4 sach (so (3,2) ∈ R). 2 ≤ 1² = 1 galat — is example se transitive chain nahi milti, better counter-example: a = 10, b = -4, c = 2. 10 ≤ (-4)² = 16 sach. -4 ≤ 2² = 4 sach. Par 10 ≤ 2² = 4 galat.
(10,-4) ∈ R aur (-4,2) ∈ R lekin (10,2) ∉ R ⟹ Not transitive
Isliye R teeno properties me se koi bhi satisfy nahi karti.
Q4. Show that the relation R in the set {1, 2, 3, 4, 5} given by R = {(a, b) : |a − b| is even} is an equivalence relation.
Reflexive: |a − a| = 0, jo even hai. Isliye (a, a) ∈ R har a ke liye.
|a − a| = 0 (even) ⟹ Reflexive ✓
Symmetric: Agar |a − b| even hai, to |b − a| = |a − b| bhi even hai (absolute value symmetric hota hai).
|a − b| = |b − a| ⟹ agar ek even to dusra bhi even ⟹ Symmetric ✓
Transitive: Maano |a − b| even aur |b − c| even hai. To a aur b ki parity (odd/even) same hai, aur b aur c ki parity bhi same hai. Isliye a aur c ki parity bhi same hai, matlab |a − c| bhi even hoga.
|a-b| even ⟹ a,b same parity. |b-c| even ⟹ b,c same parity. ⟹ a,c same parity ⟹ |a-c| even ⟹ Transitive ✓
Teeno properties satisfy ho rahi hain, isliye R equivalence relation hai. (Ye set ko do equivalence classes me todhti hai: {1,3,5} aur {2,4}.)
Q5. Check if f: R → R defined by f(x) = 3 − 4x is one-one and onto.
One-one check: Maano f(x₁) = f(x₂).
3 − 4x₁ = 3 − 4x₂ ⟹ −4x₁ = −4x₂ ⟹ x₁ = x₂
Isliye f one-one hai.
Onto check: Koi bhi y ∈ R (codomain) lo. Dekhna hai kya koi x ∈ R milega jisse f(x) = y ho.
y = 3 − 4x ⟹ x = (3 − y)/4
x = (3−y)/4 hamesha real number hai kisi bhi real y ke liye, isliye har y ka ek preimage exist karta hai. Isliye f onto hai.
Result: f one-one aur onto dono hai, matlab f bijective hai.
Q6. Show that f: N → N given by f(1) = f(2) = 1 and f(x) = x − 1 for every x > 2, is onto but not one-one.
One-one nahi: f(1) = 1 aur f(2) = 1. Dono different inputs (1 aur 2) ka output same (1) hai.
f(1) = 1 = f(2) lekin 1 ≠ 2 ⟹ Not one-one
Onto hai: Codomain N ka har element check karte hain. y = 1 ke liye x = 1 (ya 2) available hai. y > 1 kisi bhi natural number ke liye, x = y + 1 lo (jo 2 se bada hai), to f(x) = (y+1) − 1 = y.
y > 1 ke liye x = y + 1 (x > 2) ⟹ f(x) = x − 1 = y
Isliye N ka har element kisi na kisi x se cover ho raha hai — f onto hai.
Q7. Let f: {1, 2, 3} → {a, b, c} be defined as f(1) = a, f(2) = b, f(3) = c. Show that f is invertible and find f⁻¹.
Pehle check karte hain f bijective hai ya nahi.
One-one: Teeno inputs (1,2,3) ke outputs (a,b,c) sab alag hain. One-one hai.
Onto: Codomain {a,b,c} ka har element kisi input se cover ho raha hai. Onto hai.
f bijective hai (one-one + onto) ⟹ f invertible hai
Ab inverse define karte hain — bas mapping ulti kar do:
f⁻¹: {a,b,c} → {1,2,3} defined as f⁻¹(a) = 1, f⁻¹(b) = 2, f⁻¹(c) = 3
Q8. If f: R → R is given by f(x) = (3 − x³)^(1/3), find fof(x).
fof(x) ka matlab hai f(f(x)) — pehle andar wala f(x) nikalo, phir usi expression pe f dobara apply karo.
f(x) = (3 − x³)^(1/3)
f(f(x)) = [3 − (f(x))³]^(1/3) = [3 − ((3 − x³)^(1/3))³]^(1/3)
Cube aur cube-root cancel ho jaate hain andar wale term me:
= [3 − (3 − x³)]^(1/3) = [3 − 3 + x³]^(1/3) = (x³)^(1/3) = x
Isliye fof(x) = x. (Ye batata hai ki f apna khud ka inverse hai, kyunki fof = identity function.)
Q9. Let f: R → R be defined as f(x) = x⁴. Choose the correct answer regarding whether f is one-one, onto, both, or neither.
One-one check: x = 1 aur x = −1 lo.
f(1) = 1⁴ = 1, f(−1) = (−1)⁴ = 1 ⟹ f(1) = f(−1) lekin 1 ≠ −1
Isliye f one-one nahi hai.
Onto check: f(x) = x⁴ hamesha ≥ 0 hoga, kabhi negative nahi ho sakta. Lekin codomain R me negative numbers bhi hain (jaise −1), jinka koi preimage nahi hai.
y = −1 ke liye koi x ∈ R nahi milega jisse x⁴ = −1 ho ⟹ Not onto
Result: f na one-one hai na onto hai.
Q10. Let f: R → R be defined as f(x) = 3x. Choose the correct answer regarding one-one and onto.
One-one: f(x₁) = f(x₂) ⟹ 3x₁ = 3x₂ ⟹ x₁ = x₂. One-one hai.
3x₁ = 3x₂ ⟹ x₁ = x₂ ⟹ One-one ✓
Onto: Koi bhi y ∈ R lo, x = y/3 lene se f(x) = 3(y/3) = y milta hai, jo hamesha real hai.
y ∈ R ⟹ x = y/3 ∈ R ⟹ f(x) = y ⟹ Onto ✓
Result: f one-one aur onto dono hai, isliye bijective hai.
Q11. Show that the Signum Function f: R → R given by f(x) = 1 if x > 0, 0 if x = 0, −1 if x < 0, is neither one-one nor onto.
One-one nahi: x = 2 aur x = 5 dono positive hain, dono ka f(x) = 1.
f(2) = 1 = f(5) lekin 2 ≠ 5 ⟹ Not one-one
Onto nahi: Range of f sirf {−1, 0, 1} hai, jabki codomain poora R hai. Koi bhi y = 5 (ya koi bhi number {−1,0,1} ke alawa) ka preimage exist nahi karta.
Range = {−1, 0, 1} ≠ R = Codomain ⟹ Not onto
Isliye Signum function na one-one hai na onto.
Q12. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1,4), (2,5), (3,6)} be a function from A to B. Show that f is one-one.
Har input ka output alag hai — 1→4, 2→5, 3→6. Koi bhi do alag inputs same output nahi de rahe.
f(1)=4, f(2)=5, f(3)=6 — sab distinct outputs ⟹ f(x₁)=f(x₂) ⟹ x₁=x₂ hamesha sach hai
Isliye f one-one hai. (Note: f onto nahi hai kyunki codomain B me 7 ka koi preimage nahi hai.)
Q13. Let f: {2, 3, 4, 5} → {3, 4, 5, 9} and g: {3, 4, 5, 9} → {7, 11, 15} be functions defined as f(2)=3, f(3)=4, f(4)=5, f(5)=9 and g(3)=7, g(4)=7, g(5)=11, g(9)=11. Find gof.
gof matlab pehle f apply karo, phir uske output pe g apply karo.
(gof)(2) = g(f(2)) = g(3) = 7
(gof)(3) = g(f(3)) = g(4) = 7
(gof)(4) = g(f(4)) = g(5) = 11
(gof)(5) = g(f(5)) = g(9) = 11
Isliye gof = {(2,7), (3,7), (4,11), (5,11)}.
Q14. Show that if f: R − {7/5} → R − {3/5} is defined by f(x) = (3x + 4)/(5x − 7) and g: R − {3/5} → R − {7/5} is defined by g(x) = (7x + 4)/(5x − 3), then fog = I_A and gof = I_B, where A, B are appropriate domains.
Pehle gof(x) nikalte hain, matlab f(x) me x ki jagah g(x) daalte hain.
g(x) = (7x+4)/(5x−3)
f(g(x)) = [3g(x) + 4] / [5g(x) − 7]
Numerator aur denominator alag-alag simplify karte hain:
Numerator = 3·(7x+4)/(5x−3) + 4 = [3(7x+4) + 4(5x−3)]/(5x−3) = [21x+12+20x−12]/(5x−3) = 41x/(5x−3)
Denominator = 5·(7x+4)/(5x−3) − 7 = [5(7x+4) − 7(5x−3)]/(5x−3) = [35x+20−35x+21]/(5x−3) = 41/(5x−3)
f(g(x)) = [41x/(5x−3)] ÷ [41/(5x−3)] = 41x/41 = x
Isliye fog(x) = x, matlab fog = identity function.
Ab isi tarah gof(x) nikalte hain — g(x) me x ki jagah f(x) daalte hain.
f(x) = (3x+4)/(5x−7)
g(f(x)) = [7f(x) + 4] / [5f(x) − 3]
Numerator = 7·(3x+4)/(5x−7) + 4 = [7(3x+4) + 4(5x−7)]/(5x−7) = [21x+28+20x−28]/(5x−7) = 41x/(5x−7)
Denominator = 5·(3x+4)/(5x−7) − 3 = [5(3x+4) − 3(5x−7)]/(5x−7) = [15x+20−15x+21]/(5x−7) = 41/(5x−7)
g(f(x)) = [41x/(5x−7)] ÷ [41/(5x−7)] = 41x/41 = x
Isliye gof(x) = x bhi. Dono fog = I aur gof = I mile, jo prove karta hai ki f aur g ek-dusre ke inverse hain.
Q15. State whether the function f: N → N given by f(x) = 5x is injective, surjective or both.
Injective (one-one): f(x₁) = f(x₂) ⟹ 5x₁ = 5x₂ ⟹ x₁ = x₂.
5x₁ = 5x₂ ⟹ x₁ = x₂ ⟹ Injective ✓
Surjective nahi: Codomain N ka har element multiple of 5 nahi hai. Jaise y = 2, iske liye x = 2/5 chahiye jo natural number nahi hai.
y = 2: x = 2/5 ∉ N ⟹ koi preimage nahi ⟹ Not surjective
Isliye f sirf injective hai, surjective nahi.
Q16. Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by f(x) = (x − 2)/(x − 3). Show that f is one-one and onto.
One-one: f(x₁) = f(x₂) maano.
(x₁−2)/(x₁−3) = (x₂−2)/(x₂−3)
(x₁−2)(x₂−3) = (x₂−2)(x₁−3)
x₁x₂ −3x₁ −2x₂ +6 = x₁x₂ −3x₂ −2x₁ +6
−3x₁ −2x₂ = −3x₂ −2x₁ ⟹ −3x₁ +2x₁ = −3x₂ +2x₂ ⟹ −x₁ = −x₂ ⟹ x₁ = x₂
Isliye f one-one hai.
Onto: y ∈ B (matlab y ≠ 1) lo, x nikalte hain.
y = (x−2)/(x−3) ⟹ y(x−3) = x−2 ⟹ xy − 3y = x − 2 ⟹ x(y−1) = 3y − 2 ⟹ x = (3y−2)/(y−1)
Kyunki y ≠ 1 (given), x hamesha well-defined hai aur x ≠ 3 bhi verify hota hai. Isliye har y ∈ B ka preimage A me milta hai — f onto hai.
Q17. Let f: N → R be a function defined as f(x) = 4x² + 12x + 15. Show that f: N → S, where S is the range of f, is invertible. Find the inverse.
Pehle f(x) ko complete-the-square form me likhte hain.
f(x) = 4x² + 12x + 15 = (2x+3)² + 6
One-one (N pe): N pe x badhne ke saath (2x+3)² strictly badhta hai, isliye f strictly increasing hai, matlab one-one hai.
x₁ ≠ x₂ ⟹ (2x₁+3)² ≠ (2x₂+3)² (kyunki N pe 2x+3 > 0 hamesha) ⟹ f(x₁) ≠ f(x₂)
Onto (S pe): S ko range hi define kiya gaya hai (S = f(N)), isliye definition se f: N → S onto hai.
Isliye f: N → S bijective hai, matlab invertible hai. Ab inverse nikalte hain — y = f(x) se x ko y ke terms me likhte hain.
y = (2x+3)² + 6 ⟹ y − 6 = (2x+3)² ⟹ √(y−6) = 2x+3 (positive root, kyunki x ∈ N)
x = [√(y−6) − 3]/2
f⁻¹(y) = [√(y−6) − 3]/2
Q18. Let f: X → Y be a function. Define a relation R in X given by R = {(a, b) : f(a) = f(b)}. Show that R is an equivalence relation in X.
Reflexive: Har a ∈ X ke liye f(a) = f(a) obviously sach hai, isliye (a,a) ∈ R.
f(a) = f(a) ⟹ (a,a) ∈ R ⟹ Reflexive ✓
Symmetric: Agar (a,b) ∈ R, matlab f(a) = f(b). To f(b) = f(a) bhi sach hai (equality symmetric hoti hai), isliye (b,a) ∈ R.
f(a) = f(b) ⟹ f(b) = f(a) ⟹ (b,a) ∈ R ⟹ Symmetric ✓
Transitive: Agar (a,b) ∈ R aur (b,c) ∈ R hai, matlab f(a) = f(b) aur f(b) = f(c). To f(a) = f(c) bhi sach hai (transitivity of equality).
f(a) = f(b) aur f(b) = f(c) ⟹ f(a) = f(c) ⟹ (a,c) ∈ R ⟹ Transitive ✓
Teeno properties satisfy ho rahi hain, isliye R equivalence relation hai.
Q19. Determine whether the relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x − y = 0} is reflexive, symmetric, or transitive.
Pehle R ke elements likhte hain: y = 3x, jaha dono x aur y set A me hone chahiye (1 se 14).
R = {(1,3), (2,6), (3,9), (4,12)}
Reflexive nahi: (1,1) ke liye 3(1) − 1 = 2 ≠ 0, isliye (1,1) ∉ R.
(1,1) ∉ R ⟹ Not reflexive
Symmetric nahi: (1,3) ∈ R hai lekin (3,1) ke liye 3(3) − 1 = 8 ≠ 0, isliye (3,1) ∉ R.
(1,3) ∈ R lekin (3,1) ∉ R ⟹ Not symmetric
Transitive nahi: (1,3) ∈ R aur (3,9) ∈ R hai, lekin (1,9) ke liye 3(1) − 9 = −6 ≠ 0, isliye (1,9) ∉ R.
(1,3) ∈ R, (3,9) ∈ R lekin (1,9) ∉ R ⟹ Not transitive
Isliye R koi bhi property satisfy nahi karti.
Q20. Show that the function f: R → R defined by f(x) = x² is neither one-one nor onto.
One-one nahi: x = 2 aur x = −2 lo.
f(2) = 4, f(−2) = 4 ⟹ f(2) = f(−2) lekin 2 ≠ −2 ⟹ Not one-one
Onto nahi: f(x) = x² kabhi negative nahi hota, lekin codomain R me negative numbers bhi hain.
y = −4 ke liye koi x ∈ R nahi milega jisse x² = −4 ho (real solution exist nahi karta) ⟹ Not onto
Isliye f na one-one hai na onto.
Q21. Let A and B be sets. Show that f: A × B → B × A defined by f(a, b) = (b, a) is bijective.
One-one: Maano f(a₁,b₁) = f(a₂,b₂).
(b₁,a₁) = (b₂,a₂) ⟹ b₁=b₂ aur a₁=a₂ ⟹ (a₁,b₁) = (a₂,b₂)
Isliye f one-one hai.
Onto: Koi bhi (b,a) ∈ B × A lo. (a,b) ∈ A × B lene se f(a,b) = (b,a) milta hai — matlab har element ka preimage exist karta hai.
(b,a) ∈ B×A ⟹ preimage (a,b) ∈ A×B hai ⟹ f(a,b) = (b,a) ⟹ Onto ✓
Isliye f bijective hai (one-one + onto dono).
Important Equations — Ek Nazar Me
| Concept | Definition / Test |
|---|---|
| Reflexive | ∀a ∈ A, (a, a) ∈ R |
| Symmetric | (a, b) ∈ R ⟹ (b, a) ∈ R, sabke liye |
| Transitive | (a, b) ∈ R aur (b, c) ∈ R ⟹ (a, c) ∈ R, sabke liye |
| Equivalence relation | Reflexive + Symmetric + Transitive teeno ek saath |
| One-one (injective) | f(x₁) = f(x₂) ⟹ x₁ = x₂ (koi do alag inputs same output nahi denge) |
| Onto (surjective) | Range(f) = Codomain (har codomain element ka preimage exist karta hai) |
| Bijective | One-one AUR onto dono ek saath |
| Composition of functions | (gof)(x) = g(f(x)) — pehle f, phir g. fog ≠ gof generally. |
| Invertibility condition | f invertible ⟺ f bijective (one-one aur onto) |
| Inverse function property | f(a) = b ⟺ f⁻¹(b) = a; fof⁻¹ = I aur f⁻¹of = I |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Reflexive property ko specific values pe check karna. Students sirf ek-do numbers daal ke bolte hain 'reflexive hai' — lekin reflexive property SABHI elements ke liye prove karni padti hai, general 'a' lekar. Ek specific value se conclude karna galat hai.
- One-one aur onto ko confuse karna. Bahut students sochte hain agar function 'increasing' dikh raha hai to wo onto bhi hoga — ye galat hai. One-one ka matlab hai alag inputs ke alag outputs; onto ka matlab hai range = codomain. Dono independent properties hain, ek dusre se guarantee nahi hoti.
- fog aur gof me order ulta karna. (fog)(x) = f(g(x)) hota hai, na ki g(f(x)). Exam me students jaldi me andar-bahar ulta likh dete hain — hamesha yaad rakho: jo function bahar (left side) likha hai wo LAST apply hota hai.
- Bina domain-codomain match kiye invertibility bol dena. Function invertible tabhi hai jab wo poori tarah bijective ho — sirf 'monotonic dikh raha hai' bolna kaafi nahi. Domain aur codomain dono specify karke check karna padta hai, especially jab function R → R jaisa likha ho lekin actual range chhota ho.
- Symmetric property me sirf ek pair check karke reflexive/transitive assume kar lena. Teeno properties ALAG-ALAG independently check karni padti hain. Ek property hone se doosri automatically nahi aati — jaise symmetric hona transitive hone ki guarantee nahi deta.
- Composite function ka domain ignore karna. gof define hone ke liye f ka range, g ke domain ka subset hona chahiye. Students sirf formula substitute kar dete hain bina check kiye ki composition valid bhi hai ya nahi — especially rational functions me jaha domain restrictions hoti hain.
Board-Style Important Questions
- 1 mark: Define an equivalence relation.
- 1 mark: Give an example of a relation which is symmetric but neither reflexive nor transitive.
- 2 marks: Check whether the relation R = {(a, b) : a = b} in the set of real numbers is reflexive, symmetric, and transitive.
- 3 marks: Show that the function f: R → R defined by f(x) = 2x + 3 is one-one and onto.
- 5 marks: Let A = R − {3} and B = R − {1}. Consider f: A → B defined by f(x) = (x−2)/(x−3). Show f is one-one and onto, hence bijective.
- 5 marks: Prove that the relation R in the set of integers Z defined by R = {(a, b) : 2 divides (a − b)} is an equivalence relation.
Aksar Poochhe Jaane Wale Sawaal
Equivalence relation aur simple relation me kya farak hai?
Har relation equivalence nahi hoti. Relation sirf koi bhi rule hai jo elements ko connect karta hai. Equivalence relation banne ke liye teeno properties — reflexive, symmetric, aur transitive — ek saath hona zaroori hai. Ek bhi missing ho to wo sirf 'relation' hai, 'equivalence relation' nahi.
One-one function ka matlab kya hota hai simple bhasha me?
One-one (injective) ka matlab hai ki koi bhi do alag-alag inputs same output nahi de sakte. Har input ka apna unique output hota hai — jaise Aadhar number, har person ka alag number hota hai, do logo ka same nahi ho sakta.
fog aur gof kabhi equal ho sakte hain kya?
Haan, kabhi-kabhi ho sakte hain (jaise Q8 me fof(x) = x aaya), lekin general rule ye nahi hai — most cases me fog ≠ gof. Isliye exam me hamesha calculate karke hi batao, assume mat karo.
Kya har function invertible hoti hai?
Nahi. Sirf bijective functions (jo one-one AUR onto dono hon) invertible hoti hain. Agar function many-one hai ya into hai, to uska proper inverse exist nahi karta.
Reflexive property prove karne ke liye kya specific numbers use kar sakte hain?
Nahi, reflexive property prove karne ke liye general element 'a' lekar dikhana padta hai ki (a,a) hamesha relation me hoga — sirf ek ya do specific numbers check karna proof nahi maana jaata. Lekin property DISPROVE karne ke liye ek counter-example kaafi hai.
Composition of functions kis order me likhi jaati hai? Aur is chapter ka JEE me kya importance hai?
(gof)(x) = g(f(x)) — pehle andar wala function f apply hota hai, uska output phir g me jaata hai. Naam se ulta lagta hai isliye confusion hoti hai, hamesha formula yaad rakhna: right se left apply karo. Composition aur inverse functions ke concepts differentiation ke chain rule aur inverse trigonometric functions ki foundation hain, isliye ye chapter sirf board ke liye nahi, JEE ke aage ke calculus ke liye bhi zaroori hai.
Class 12 Maths — Saare Chapters
Board exam tak sirf revision karna hai?
Class 12 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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