Class 11 se aage — ab hum functions ke deeper types samjhenge aur composition explore karenge!
Socho ek school mein har student ko ek roll number diya gaya hai. Yeh ek function hai — har student (domain element) ka exactly ek roll number (codomain element) hai. Ab agar do alag students ka same roll number ho — toh yeh function nahi hai! Aur agar koi roll number kisi student ko assign nahi hua — toh yeh onto nahi hai. Class 12 mein hum in relationships ko formally define karte hain.
Ek set A par defined relation R ek subset hota hai A × A ka. Relation R ke kuch special types hote hain:
Har element apne aap se related ho: (a, a) ∈ R for all a ∈ A.
Example: "same age as" — Priya Priya ki same age ki hai ✓
Agar a R b, toh b R a bhi: (a,b)∈R ⟹ (b,a)∈R
Example: "is friend of" — Rahul Priya ka dost ⟹ Priya Rahul ki dost
a R b aur b R c ⟹ a R c: (a,b)∈R, (b,c)∈R ⟹ (a,c)∈R
Example: "is ancestor of" — Ram ka baap Shyam, Shyam ka baap Hari ⟹ Ram ka baap (ancestor) Hari
Jo teen teeno ho — Reflexive + Symmetric + Transitive.
Example: "same birthday as" — teeno conditions satisfy karta hai!
A = {1, 2, 3}, R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1)}
Reflexive? (1,1),(2,2),(3,3) ∈ R ✓
Symmetric? (1,2)∈R ⟹ (2,1)∈R ✓; (2,3)∈R ⟹ (3,2)∈R ✓
Transitive? (1,2)∈R, (2,3)∈R ⟹ (1,3)∈R ✓
∴ R ek equivalence relation hai!
Class 11 mein humne function ki basic definition seekhi thi. Ab Class 12 mein hum teen important types dekhenge jo board mein baar baar aate hain:
| Type | Condition | Example | Invertible? |
|---|---|---|---|
| One-One (Injective) | f(a)=f(b) ⟹ a=b | f(x)=2x | No (unless onto bhi ho) |
| Onto (Surjective) | Har y∈codomain ke liye ∃x s.t. f(x)=y | f(x)=x² on ℝ⁺ | No (unless one-one bhi ho) |
| Bijective | Both one-one AND onto | f(x)=x+3 | ✅ Yes! |
Agar f: A→B aur g: B→C hain, toh gof: A→C defined hai as: (gof)(x) = g(f(x))
Socho ek chain: pehle f apply karo, phir uske result par g apply karo. Yeh ek assembly line jaisa hai — pehla machine kaam karta hai, phir doosra.
f(x) = 2x + 1, g(x) = x²
gof(x) = g(f(x)) = g(2x+1) = (2x+1)² = 4x²+4x+1
fog(x) = f(g(x)) = f(x²) = 2x²+1
gof ≠ fog — clearly different!
Ek function f: A→B invertible hota hai agar bijective hai (one-one + onto). Tab uska inverse function f⁻¹: B→A exist karta hai jo satisfy karta hai:
f⁻¹(f(x)) = x for all x∈A, aur f(f⁻¹(y)) = y for all y∈B
f: ℝ→ℝ, f(x) = 3x - 5. Find f⁻¹.
Step 1: Let y = 3x - 5 ⟹ x = (y+5)/3
Step 2: f⁻¹(y) = (y+5)/3
Step 3: Verify: f⁻¹(f(x)) = f⁻¹(3x-5) = (3x-5+5)/3 = x ✓
Q1. Show that f: N→N defined by f(n) = n² is one-one but not onto.
One-one: f(a)=f(b) ⟹ a²=b² ⟹ a=b (N mein positive only) ✓. Not onto: 2∈N lekin koi n∈N nahi hai jiska n²=2 ✓
Q2. f(x) = (4x+3)/(6x-4), x≠2/3. Prove f is one-one and onto. Find f⁻¹.
f(a)=f(b) ⟹ (4a+3)/(6a-4)=(4b+3)/(6b-4) ⟹ cross multiply ⟹ a=b. For onto: y=(4x+3)/(6x-4) ⟹ x=(4y+3)/(6y-4). So f⁻¹(x)=(4x+3)/(6x-4) — interesting! f is its own inverse.
Q3. Let f(x)=x² aur g(x)=√x. Find gof(3) aur fog(9).
gof(3)=g(f(3))=g(9)=√9=3. fog(9)=f(g(9))=f(3)=9. Interesting — gof(3)=3 means gof=identity on this point!