Class 11 Maths · Chapter 2
Short answer:
Class 11 Maths ka Chapter 2 — Relations and Functions — seedha Chapter 1 Sets ke upar khada hota hai: ordered pairs se Cartesian product, Cartesian product se relation, aur relation ka ek special case — function. Yehi function ka idea aage Trigonometric Functions, Limits and Derivatives, sab jagah repeat hota hai, isliye is chapter ki definitions crystal-clear honi chahiye.
Agar class 11 maths ncert solutions dhoondh rahe ho aur Chapter 1 — Sets — already clear kar chuke ho, to Chapter 2 “Relations and Functions” natural agla step hai. Ye chapter do sets ke beech “link” banane ka tareeka sikhata hai: pehle ordered pair aur Cartesian product (A × B), phir uska subset — relation, aur relation ka ek strict version — function. Function ki definition (domain ke har x ka ek hi unique image) itni important hai ki ye Class 11 ke baaki chapters — jaise class 11 maths chapter 3 trigonometric functions ncert solutions aur aage Limits & Derivatives — sab isi base pe khade hain.
Neeche NCERT ke rationalised (2026–27 session wale) textbook ke Exercise 2.1, 2.2, 2.3 aur Miscellaneous Exercise ke saare questions step-by-step solve kiye gaye hain, saath mein ek formula table (apne class 11 maths formulas pdf download revision ke saath use karo), common mistakes, aur important-question style practice bhi di gayi hai. Agar syllabus poora dekhna ho to class 11 maths syllabus 2026-27 pdf NCERT ki official site se check kar sakte ho, aur class 11 maths ncert book pdf download karke saath mein rakh sakte ho.
Chapter 2 Summary — 5 Minute Revision
Is chapter mein humne dekha: Ordered pair → Cartesian product A × B (sab ordered pairs (a,b) jaha a∈A, b∈B) → Relation (A × B ka koi bhi subset) → Function (ek strict relation jisme domain ke har element ka bilkul ek hi image ho). Saath hi humne kuch standard real functions — identity, constant, polynomial, rational, modulus, signum, greatest integer — aur unke graphs dekhe, aur functions pe algebra (addition, subtraction, scalar multiplication, multiplication, quotient) seekhi.
Key takeaway: har function ek relation hai, lekin har relation function nahi hai. Agla step domain-range nikalna aur piecewise/modulus/greatest-integer functions ke graphs pehchanna practice karna hai — yeh skill Chapter 3 Trigonometric Functions aur baad mein Limits and Derivatives mein baar-baar kaam aayegi.
In-Text Questions — Solutions
(Ex 2.1, Q1) If (x/3 + 1, y − 2/3) = (5/3, 1/3), find the values of x and y.
Do ordered pairs equal hain, isliye corresponding elements equal honge:
x/3 + 1 = 5/3 aur y − 2/3 = 1/3
x/3 = 5/3 − 1 = 2/3 ⇒ x = 2
y = 1/3 + 2/3 = 1
Answer: x = 2, y = 1
(Ex 2.1, Q2) If set A has 3 elements and B = {3, 4, 5}, find the number of elements in (A × B).
n(A) = 3, n(B) = 3 (kyunki B mein 3, 4, 5 — 3 elements hain).
n(A × B) = n(A) × n(B) = 3 × 3 = 9
Answer: 9 elements
(Ex 2.1, Q3) If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
Definition se, har element of G ko har element of H se pair karo:
G × H = {(7,5), (7,4), (7,2), (8,5), (8,4), (8,2)}
H × G = {(5,7), (5,8), (4,7), (4,8), (2,7), (2,8)}
Note: G × H ≠ H × G, kyunki order matter karta hai.
(Ex 2.1, Q4) State whether true or false; agar false hai to correct karo: (i) P={m,n}, Q={n,m} ⇒ P×Q = {(m,n),(n,m)}. (ii) A,B non-empty ⇒ A×B non-empty set of pairs (x,y), x∈A, y∈B. (iii) A={1,2}, B={3,4} ⇒ A × (B ∩ φ) = φ.
(i) False. P = Q = {m,n} hai, isliye
P × Q = {(m,m), (m,n), (n,m), (n,n)}
(ii) True. Definition se hi A×B non-empty ordered pairs (x,y) ka set hai jaha x∈A, y∈B.
(iii) True. B ∩ φ = φ, aur A × φ = φ (koi bhi set with empty set ka product empty hota hai).
(Ex 2.1, Q5) If A = {−1, 1}, find A × A × A.
Har position mein −1 ya 1 aa sakta hai, total 2³ = 8 triplets:
A × A × A = {(−1,−1,−1), (−1,−1,1), (−1,1,−1), (−1,1,1), (1,−1,−1), (1,−1,1), (1,1,−1), (1,1,1)}
(Ex 2.1, Q6) If A × B = {(a,x),(a,y),(b,x),(b,y)}, find A and B.
A = set of all first elements, B = set of all second elements:
A = {a, b}, B = {x, y}
(Ex 2.1, Q7) A={1,2}, B={1,2,3,4}, C={5,6}, D={5,6,7,8}. Verify: (i) A × (B ∩ C) = (A × B) ∩ (A × C). (ii) A × C is a subset of B × D.
(i) B ∩ C = φ (B aur C mein koi common element nahi), isliye A × (B ∩ C) = φ.
A × B = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}
A × C = {(1,5),(1,6),(2,5),(2,6)}
In dono sets mein koi common pair nahi (second elements alag range mein hain), isliye (A×B) ∩ (A×C) = φ = A × (B ∩ C). Verified.
(ii) A × C = {(1,5),(1,6),(2,5),(2,6)}. Har pair ka first element A ⊆ B mein hai aur second element C ⊆ D mein hai, isliye har (a,c) ∈ B × D bhi hai. A × C ⊆ B × D verified.
(Ex 2.1, Q8) A={1,2}, B={3,4}. Write A × B. A × B ke kitne subsets honge? List them.
A × B = {(1,3), (1,4), (2,3), (2,4)}
n(A × B) = 4, isliye subsets = 2⁴ = 16:
φ, {(1,3)}, {(1,4)}, {(2,3)}, {(2,4)}, {(1,3),(1,4)}, {(1,3),(2,3)}, {(1,3),(2,4)}, {(1,4),(2,3)}, {(1,4),(2,4)}, {(2,3),(2,4)}, {(1,3),(1,4),(2,3)}, {(1,3),(1,4),(2,4)}, {(1,3),(2,3),(2,4)}, {(1,4),(2,3),(2,4)}, {(1,3),(1,4),(2,3),(2,4)}
(Ex 2.1, Q9) n(A)=3, n(B)=2. (x,1),(y,2),(z,1) ∈ A × B, jaha x, y, z distinct hain. Find A and B.
Second elements sirf 1 aur 2 hain aur n(B)=2, isliye B = {1, 2}.
First elements x, y, z distinct hain aur n(A)=3, isliye A = {x, y, z}.
(Ex 2.1, Q10) A × A ke 9 elements hain, jinme (−1,0) aur (0,1) hain. Find set A aur A × A ke remaining elements.
n(A × A) = 9 ⇒ n(A)² = 9 ⇒ n(A) = 3.
(−1,0) aur (0,1) ∈ A × A ⇒ −1, 0, 1 ∈ A. Chunki n(A)=3, A = {−1, 0, 1}.
A × A = {(−1,−1),(−1,0),(−1,1),(0,−1),(0,0),(0,1),(1,−1),(1,0),(1,1)}
Given (−1,0) aur (0,1) ko hata ke remaining 7 elements:
(−1,−1), (−1,1), (0,−1), (0,0), (1,−1), (1,0), (1,1)
(Ex 2.2, Q1) A={1,2,...,14}. R = {(x,y): 3x − y = 0, x,y∈A} se defined relation ka domain, codomain aur range likho.
3x − y = 0 ⇒ y = 3x. x = 1,2,3,4 ke liye y = 3,6,9,12 (A mein hai); x=5 pe y=15 jo A se bahar hai, ruk jao.
R = {(1,3), (2,6), (3,9), (4,12)}
Domain = {1,2,3,4}; Codomain = A = {1,2,...,14}; Range = {3,6,9,12}
(Ex 2.2, Q2) N par relation R = {(x,y): y = x+5, x is a natural number less than 4}. Roster form mein likho, domain aur range batao.
x < 4 ke natural numbers: x = 1, 2, 3.
R = {(1,6), (2,7), (3,8)}
Domain = {1,2,3}; Range = {6,7,8}
(Ex 2.2, Q3) A={1,2,3,5}, B={4,6,9}. R = {(x,y): x aur y ka difference odd hai}. Roster form mein likho.
Har x∈A ko har y∈B se check karo: |x−y| odd hona chahiye.
x=1: |1−4|=3(odd), |1−6|=5(odd), |1−9|=8(even)
x=2: |2−4|=2(even), |2−6|=4(even), |2−9|=7(odd)
x=3: |3−4|=1(odd), |3−6|=3(odd), |3−9|=6(even)
x=5: |5−4|=1(odd), |5−6|=1(odd), |5−9|=4(even)
R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}
(Ex 2.2, Q4) Fig 2.7 mein P={5,6,7} se Q={3,4,5} tak relation dikhaya gaya hai jaha 5→3, 6→4, 7→5. (i) set-builder form, (ii) roster form likho. Domain aur range batao.
Pattern: har y, x se 2 kam hai (y = x−2).
(i) Set-builder: R = {(x, y): y = x − 2, x ∈ P}
(ii) Roster: R = {(5,3), (6,4), (7,5)}
Domain = {5,6,7}; Range = {3,4,5}
(Ex 2.2, Q5) A={1,2,3,4,6}. R = {(a,b): a,b∈A, b exactly divisible by a}. (i) Roster form, (ii) domain, (iii) range likho.
Har a ke liye A mein wo b dhoondo jo a se divisible ho:
a=1: b=1,2,3,4,6; a=2: b=2,4,6; a=3: b=3,6; a=4: b=4; a=6: b=6
R = {(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)}
Domain = {1,2,3,4,6}; Range = {1,2,3,4,6}
(Ex 2.2, Q6) R = {(x, x+5): x ∈ {0,1,2,3,4,5}} ka domain aur range nikalo.
R = {(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)}
Domain = {0,1,2,3,4,5}; Range = {5,6,7,8,9,10}
(Ex 2.2, Q7) Relation R = {(x, x³): x is a prime number less than 10} ko roster form mein likho.
10 se chote prime numbers: 2, 3, 5, 7
R = {(2,8), (3,27), (5,125), (7,343)}
(Ex 2.2, Q8) A={x,y,z}, B={1,2}. A se B tak kitne relations ban sakte hain?
n(A)=3, n(B)=2 ⇒ n(A × B) = 6.
Relations ki total sankhya = subsets of A×B = 2⁶ = 64
(Ex 2.2, Q9) Z par R = {(a,b): a,b∈Z, a−b is an integer}. Domain aur range find karo.
a aur b dono integers hain, isliye a−b hamesha ek integer hoga — ye condition har (a,b) pair ke liye true hai.
Isliye Domain = Z, Range = Z (poora Z hi relation mein hai).
(Ex 2.3, Q1) Kaunse relations functions hain? Reason do; agar function hai to domain-range batao. (i) {(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)} (ii) {(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)} (iii) {(1,3),(1,5),(2,5)}
(i) Function hai. Saare first elements (2,5,8,11,14,17) distinct hain, har ek ka unique image hai. Domain={2,5,8,11,14,17}; Range={1}.
(ii) Function hai. Saare first elements distinct. Domain={2,4,6,8,10,12,14}; Range={1,2,3,4,5,6,7}.
(iii) Function nahi hai. First element 1, do alag images (3 aur 5) de raha hai — single-valued condition fail.
(Ex 2.3, Q2) Domain aur range nikalo: (i) f(x) = −|x| (ii) f(x) = √(9 − x²)
(i) |x| har real x ke liye defined hai, isliye Domain = R. Chunki |x| ≥ 0, −|x| ≤ 0. Range = (−∞, 0].
(ii) Square root ke andar non-negative hona chahiye:
9 − x² ≥ 0 ⇒ x² ≤ 9 ⇒ −3 ≤ x ≤ 3
Domain = [−3, 3]. x=0 pe max value √9=3, x=±3 pe min value 0. Range = [0, 3].
(Ex 2.3, Q3) f(x) = 2x − 5. Find (i) f(0), (ii) f(7), (iii) f(−3).
f(0) = 2(0) − 5 = −5
f(7) = 2(7) − 5 = 14 − 5 = 9
f(−3) = 2(−3) − 5 = −6 − 5 = −11
(Ex 2.3, Q4) t(C) = 9C/5 + 32 (Celsius se Fahrenheit). Find (i) t(0) (ii) t(28) (iii) t(−10) (iv) C jab t(C) = 212.
t(0) = 9(0)/5 + 32 = 32
t(28) = 9(28)/5 + 32 = 252/5 + 32 = 50.4 + 32 = 82.4
t(−10) = 9(−10)/5 + 32 = −18 + 32 = 14
(iv) t(C) = 212 rakh ke solve karo:
9C/5 + 32 = 212 ⇒ 9C/5 = 180 ⇒ C = 100
(Ex 2.3, Q5) Range nikalo: (i) f(x)=2−3x, x∈R, x>0 (ii) f(x)=x²+2, x real (iii) f(x)=x, x real.
(i) x>0 hai, jaise-jaise x badhta hai 2−3x ghatta hai. x→0⁺ pe f→2 (nahi pahunchta), x→∞ pe f→−∞. Range = (−∞, 2).
(ii) x² ≥ 0 hamesha, isliye x²+2 ≥ 2. Range = [2, ∞).
(iii) f(x)=x identity function hai jiska x koi bhi real number ho sakta hai. Range = R.
Poore Class 11 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–Q12)
(Misc, Q1) f(x) = x² for 0≤x≤3, 3x for 3≤x≤10. g(x) = x² for 0≤x≤2, 3x for 2≤x≤10. Show ki f function hai aur g function nahi hai.
f ke liye: overlap point x=3 pe dono pieces ki value check karo — x²=9 aur 3x=9. Dono same hain, isliye x=3 pe f ki ek hi value hai. Har x ka ek hi image hai ⇒ f ek function hai.
g ke liye: overlap point x=2 pe check karo — x²=4 (pehle piece se) aur 3x=6 (dusre piece se). x=2 pe do alag values (4 aur 6) mil rahi hain, matlab ek hi x ka do images ban raha hai ⇒ g function nahi hai.
(Misc, Q2) f(x)=x². [f(1.1) − f(1)] / (1.1 − 1) nikalo.
f(1.1) = (1.1)² = 1.21, f(1) = 1² = 1
[f(1.1) − f(1)] / (1.1 − 1) = (1.21 − 1) / 0.1 = 0.21 / 0.1 = 2.1
(Misc, Q3) f(x) = (x²+2x+1) / (x²−8x+12) ka domain nikalo.
Denominator zero nahi hona chahiye. Factorise karo:
x² − 8x + 12 = (x − 6)(x − 2)
Ye zero hota hai x = 6 aur x = 2 pe. Inko domain se hata do:
Domain = R − {2, 6}
(Misc, Q4) f(x) = √(x − 1) ka domain aur range nikalo.
Square root ke andar non-negative hona chahiye:
x − 1 ≥ 0 ⇒ x ≥ 1
Domain = [1, ∞). Chunki square root ki value hamesha ≥ 0 hoti hai, Range = [0, ∞).
(Misc, Q5) f(x) = |x − 1| ka domain aur range nikalo.
Modulus har real number ke liye defined hai, isliye Domain = R.
|x−1| ki value kabhi negative nahi ho sakti, isliye Range = [0, ∞).
(Misc, Q6) f = {(x, x²/(1+x²)) : x∈R} ka range nikalo.
Let y = x²/(1+x²). Isko is form mein likho:
y = x²/(1+x²) = 1 − 1/(1+x²)
Chunki x² ≥ 0, 1+x² ≥ 1, isliye 1/(1+x²) ki range (0, 1] hai (x=0 pe max value 1, x→∞ pe 0 ke paas). Isliye y = 1 − (kuch value between 0 aur 1] ka range [0, 1) banta hai.
Range = [0, 1)
(Misc, Q7) f(x)=x+1, g(x)=2x−3. f+g, f−g aur f/g nikalo.
(f+g)(x) = (x+1) + (2x−3) = 3x − 2
(f−g)(x) = (x+1) − (2x−3) = −x + 4
(f/g)(x) = (x+1) / (2x−3), x ≠ 3/2
(Misc, Q8) f = {(1,1),(2,3),(0,−1),(−1,−3)} ek linear function f(x)=ax+b hai. a aur b nikalo.
Do points use karo — (0,−1) aur (1,1):
f(0) = a(0) + b = b = −1 ⇒ b = −1
f(1) = a(1) + b = a + b = 1 ⇒ a − 1 = 1 ⇒ a = 2
Check: f(2) = 2(2)−1 = 3 ✓, f(−1) = 2(−1)−1 = −3 ✓
a = 2, b = −1
(Misc, Q9) R = {(a,b): a,b∈N, a=b²} N se N tak. Check karo: (i) (a,a)∈R sabhi a ke liye (ii) (a,b)∈R ⇒ (b,a)∈R (iii) (a,b)∈R, (b,c)∈R ⇒ (a,c)∈R.
(i) False. (a,a)∈R ke liye a=a² chahiye, jo sirf a=1 ke liye true hai, har a ke liye nahi. Counter-example: a=2, 2 ≠ 2²=4.
(ii) False. (4,2)∈R kyunki 4=2². Lekin (2,4)∈R ke liye 2=4²=16 chahiye, jo false hai.
(iii) False. Example: a=16, b=4, c=2. (16,4)∈R kyunki 16=4², (4,2)∈R kyunki 4=2². Lekin (16,2)∈R ke liye 16=2²=4 chahiye, jo false hai.
(Misc, Q10) A={1,2,3,4}, B={1,5,9,11,15,16}, f={(1,5),(2,9),(3,1),(4,5),(2,11)}. (i) Kya f, A se B tak ek relation hai? (ii) Kya f, A se B tak ek function hai?
(i) True. f ke saare ordered pairs A × B ke elements hain (har first element A mein hai, har second element B mein hai), isliye f ⊆ A×B hai — ye ek valid relation hai.
(ii) False. Element 2, do alag images (9 aur 11) de raha hai — (2,9) aur (2,11) dono f mein hain. Single-valued condition fail, isliye f function nahi hai.
(Misc, Q11) f = {(ab, a+b): a,b∈Z} ⊆ Z × Z. Kya f, Z se Z tak ek function hai?
Check karo ki ek hi first element (ab) alag-alag second elements (a+b) de sakta hai ya nahi. Lo ab = 6:
(a,b)=(1,6) ⇒ a+b=7; (a,b)=(2,3) ⇒ a+b=5
Same first element 6, do alag images (7 aur 5) — single-valued condition fail.
f, Z se Z tak function nahi hai.
(Misc, Q12) A={9,10,11,12,13}, f:A→N, f(n) = highest prime factor of n. Range of f nikalo.
Har n ko factorise karke highest prime factor nikalo:
9 = 3² ⇒ f(9)=3; 10 = 2×5 ⇒ f(10)=5; 11 (prime) ⇒ f(11)=11; 12 = 2²×3 ⇒ f(12)=3; 13 (prime) ⇒ f(13)=13
Range of f = {3, 5, 11, 13}
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| Ordered pair equality | (a, b) = (c, d) ⇔ a = c aur b = d |
| Cartesian product | A × B = {(a, b) : a ∈ A, b ∈ B} |
| Number of elements in A × B | n(A) = p, n(B) = q ⇒ n(A × B) = pq |
| Empty set property | A × φ = φ; agar A ya B infinite hai to A × B bhi infinite |
| Order matters | General mein A × B ≠ B × A |
| Relation | R ⊆ A × B; Domain = first elements ka set, Range = second elements ka set, Range ⊆ Codomain |
| Total relations from A to B | n(A)=p, n(B)=q ⇒ total relations = 2pq |
| Function (definition) | f: A → B function hai agar A ke har element ka B mein bilkul ek unique image ho |
| Identity function | f(x) = x, Domain = Range = R |
| Constant function | f(x) = c, Domain = R, Range = {c} |
| Polynomial function | f(x) = a0 + a1x + a2x2 + ... + anxn |
| Rational function | f(x) = p(x)/q(x), q(x) ≠ 0 |
| Modulus function | f(x) = |x| = x agar x≥0, −x agar x<0 |
| Signum function | f(x) = 1 (x>0), 0 (x=0), −1 (x<0); Range = {−1,0,1} |
| Greatest integer function | f(x) = [x] = greatest integer ≤ x |
| Algebra of functions | (f+g)(x)=f(x)+g(x); (f−g)(x)=f(x)−g(x); (fg)(x)=f(x)g(x); (f/g)(x)=f(x)/g(x) agar g(x)≠0; (αf)(x)=αf(x) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Relation aur function ko same maan lena — har function relation hai, lekin har relation function nahi hoti (agar koi x do alag y de raha ho to woh function nahi hai).
- Range aur Codomain ko same samajhna — Range hamesha Codomain ka subset hoti hai, jaruri nahi ki dono equal ho.
- A × B mein order bhool jaana — (a,b) ≠ (b,a) jab tak a=b na ho, aur general mein A × B ≠ B × A.
- n(A × B) ko n(A) + n(B) samajh lena, jabki sahi formula n(A × B) = n(A) × n(B) hai.
- Modulus, Signum aur Greatest Integer jaise piecewise functions mein negative x ke liye sign ya interval galat laga dena.
- √g(x) ya f(x)/g(x) type functions ka domain nikalte waqt g(x) ≥ 0 (square root ke liye) ya g(x) ≠ 0 (denominator ke liye) ki condition check karna bhool jaana.
Board-Style Important Questions
- Practice Question 1: Agar n(A) = 3 aur n(B) = 4 hai, to n(A × B) kitna hoga?
- Practice Question 2: f(x) = √(9 − x²) ka domain aur range nikalo.
- Practice Question 3: A = {1,2}, B = {3,4} ke liye A × B likho aur A × B ke subsets ki total sankhya batao.
- Practice Question 4: f = {(1,1),(2,3),(0,−1),(−1,−3)} ek linear function f(x) = ax + b hai. a aur b ki value nikalo.
- Practice Question 5: f(x) = x²/(1+x²), x ∈ R, ka range nikalo.
- Practice Question 6: f(x) = x² for 0≤x≤3, 3x for 3≤x≤10 diya gaya hai; g(x) = x² for 0≤x≤2, 3x for 2≤x≤10. Sabit karo ki f function hai lekin g function nahi hai.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 2 — Relations and Functions — ka is subject mein kya role hai?
Ye chapter Sets (Chapter 1) ko aage badhata hai aur ‘function’ ka wo strict definition deta hai jo poore Class 11-12 Maths — Trigonometric Functions, Complex Numbers, aur khaaskar Limits and Derivatives — mein baar-baar use hota hai. Isliye ye ek foundational chapter hai, sirf isolated topic nahi.
Class 11 Maths chapter wise weightage 2026-27 ke liye kahan dekhein?
Exact marks-weightage numbers is article mein jaanbujh kar nahi diye gaye kyunki alag-alag sources ke figures aapas mein conflict karte hain. Sahi, verified weightage ke liye CBSE ki official curriculum document (class 11 maths chapter wise weightage 2026-27 pdf) hi check karo.
Is chapter ke formulas ka pdf kahan milega?
Upar diya gaya formula table Chapter 2 ke saare key rules cover karta hai. Agar poore Class 11 Maths ka combined class 11 maths formulas pdf download chahiye, to NCERT/CBSE ki official study material use karna best rahega.
Ye chapter Class 11 Maths Chapter 1 Sets NCERT solutions se kaise juda hai?
Cartesian product aur relation dono concepts seedhe set theory pe based hain — agar Chapter 1 ke Union, Intersection jaise operations clear nahi hain, to Chapter 2 ke A × (B ∩ C) jaise questions confusing lagenge. Isliye Chapter 1 pehle revise karna helpful hai.
Relations and Functions ke baad Class 11 Maths mein kaunsa chapter aata hai?
Rationalised syllabus mein iske baad Chapter 3 — Trigonometric Functions — aata hai, jaha function ka concept naye tareeke se (angles aur ratios ke saath) use hota hai.
Practice ke liye NCERT Exercise ke alawa aur kya solve karna chahiye?
NCERT Exercise 2.1–2.3 aur Miscellaneous Exercise clear karne ke baad class 11 maths exemplar solutions aur class 11 maths important questions with solutions try karo — ye thoda tricky domain-range aur piecewise-function variations dete hain jo textbook se ek level upar hote hain.
Class 11 Maths — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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