2.2.2 · Maths › Functions
Intuition Big Idea kya hai?
Ek function ek machine ki tarah hai jo inputs ko outputs mein transform karta hai. Lekin har machine har input accept nahi karti (diesel wali gaadi mein petrol nahi daal sakte), aur har possible output actually nahi nikalta (ek vending machine mein 50 items ki slots ho sakti hain lekin sirf 20 hi stock ho sakti hain).
Domain : Woh SAARE inputs jo machine accept karti hai
Codomain : Woh SAARE outputs jo machine possibly produce kar sakti hai (the "target set")
Range : Woh outputs jo actually nikalte hain (codomain ka subset)
Ye kyun matter karta hai : Bina domain jaane, hum nahi bata sakte ki f ( x ) = x ka x = − 3 par kaam karega ya nahi. Bina range aur codomain mein fark kiye, hum nahi bata sakte ki function onto (surjective) hai ya nahi.
Definition Function Notation
Ek function f : A → B ka matlab hai:
A domain hai (saare valid inputs ka set)
B codomain hai (woh set jisme hum into map kar rahe hain)
Har x ∈ A ke liye, exactly ek y ∈ B exist karta hai jaise ki f ( x ) = y
Range (ya image ) hai Range ( f ) = { f ( x ) : x ∈ A } ⊆ B
Codomain ≠ Range kyun?
Codomain woh hai jo hum target set ke roop mein declare karte hain (jaise "mera function real numbers output karta hai")
Range woh hai jo actually hota hai (shayad sirf positive reals nikalte hain)
Range ⊆ Codomain hamesha hota hai, lekin equality tab hi hoti hai jab f onto ho
Scratch se Derivation :
Definition se shuru karo: ek value y range mein hai agar aur sirf agar equation f ( x ) = y ka domain mein kam se kam ek solution x ho
Isse hume existence condition milti hai ∃ x ∈ A
Isse ek solvability problem mein convert karo: function ko symbolically invert karo, phir check karo ki kaunse y ke liye x A mein aata hai
Worked example Example 1: Square Function
Maano f : R → R , jo f ( x ) = x 2 se define hota hai
Domain, codomain, range nikalo :
Domain : R (notation f : R → … mein diya gaya hai)
Kyun? Humne declare kiya hai ki function saare real numbers accept karta hai
Codomain : R (notation … → R mein diya gaya hai)
Range : Dekho ki kaunse y ∈ R achievable hain
y = x 2 rakho
x ke liye solve karo: x = ± y
Ye step kyun? Hume check karna hai ki diye gaye y ke liye, domain mein koi x exist karta hai ya nahi
Real x ke liye, humhe y ≥ 0 chahiye
Kyun? Negative numbers ka square root real nahi hota
Saath hi, kisi bhi y ≥ 0 ke liye, hum x = y ∈ R le sakte hain aur f ( x ) = y milega
Range = [ 0 , ∞ ) = { y ∈ R : y ≥ 0 }
Key insight : Yahan Range = Codomain! Function R par onto nahi hai kyunki negative numbers kabhi output nahi hote.
Worked example Example 2: Restricted Square Function
Maano g : [ 0 , 5 ] → R , jo g ( x ) = x 2 se define hota hai
Same formula, alag domain—kya badlega?
Domain : [ 0 , 5 ] (sirf non-negative inputs, 5 tak)
Codomain : R (ab bhi saare reals mein map ho raha hai)
Range :
x ∈ [ 0 , 5 ] ke liye, x 2 kya values leta hai?
Kyunki g ( x ) = x 2 [ 0 , 5 ] par increasing hai:
Minimum: g ( 0 ) = 0
Maximum: g ( 5 ) = 25
Ye step kyun? Closed interval par, continuous functions min aur max ke beech ki saari values achieve karte hain
Range = [ 0 , 25 ]
Notice karo : Domain ko restrict karne se range bhi restrict ho gayi !
Worked example Example 3: Rational Function
Maano h : R ∖ { 2 } → R , jo h ( x ) = x − 2 x + 1 se define hota hai
Domain : R ∖ { 2 } (2 ko chhodkar saare reals)
2 ko exclude kyun? x = 2 par zero se division undefined hai
Ye kaise nikaalein? Denominator = 0 set karo: x − 2 = 0 ⇒ x = 2
Codomain : R
Range : Kaunse y hit ho sakte hain?
y = x − 2 x + 1 set karo
x ke liye solve karo:
y ( x − 2 ) = x + 1
y x − 2 y = x + 1
y x − x = 2 y + 1
x ( y − 1 ) = 2 y + 1
Ye step kyun? Hum invert kar rahe hain yeh dekhne ke liye ki kaunse y ke paas corresponding x hai
Agar y = 1 : x = y − 1 2 y + 1
Ye x domain mein hona chahiye, isliye x = 2 :
y − 1 2 y + 1 = 2
2 y + 1 = 2 ( y − 1 )
2 y + 1 = 2 y − 2
1 = − 2 ✓
Ye check kyun? Ensure karne ke liye ki hamara inverted x domain restrictions violate na kare
Toh koi bhi y = 1 kaam karta hai!
Agar y = 1 ho? Toh x ( y − 1 ) = 0 force karta hai 0 = 2 y + 1 = 3 , contradiction
Range = R ∖ { 1 }
Key insight : Function codomain mein exactly ek value (y = 1 ) "miss" karta hai.
Common mistake Mistake 1: Codomain aur Range ko Confuse Karna
Galat soch : "Agar f : A → B hai, toh range B ke barabar honi chahiye"
Ye sahi kyun lagta hai : Notation → B suggest karta hai ki hum B ko poora hit kar rahe hain
Fix :
Codomain target set hai (jahan hum land kar sakte hain )
Range woh hai jahan hum actually land karte hain
Range ⊆ Codomain, lekin equality ke liye proof chahiye (surjectivity)
Example : f : R → R , f ( x ) = x 2 ka codomain R hai lekin range [ 0 , ∞ ) hai
Common mistake Mistake 2: Range Nikalte Waqt Domain Restrictions Bhool Jaana
Galat approach : f ( x ) = 4 − x 2 ke liye, sirf y = 4 − x 2 solve karo aur x = ± 4 − y 2 paao, toh "koi bhi y ≥ 0 kaam karega"
Ye sahi kyun lagta hai : Humne function ko sahi se invert kiya
Fix : Invert karne ke baad, check karo ki x original domain mein rehta hai ya nahi!
Domain: 4 − x 2 ≥ 0 ⇒ x 2 ≤ 4 ⇒ x ∈ [ − 2 , 2 ]
y = 4 − x 2 aur x ∈ [ − 2 , 2 ] se:
Max y : jab x = 0 , y = 2
Min y : jab x = ± 2 , y = 0
Sahi range : [ 0 , 2 ] , [ 0 , ∞ ) nahi
Common mistake Mistake 3: Domain ko Single Number Likhna
Galat : "f ( x ) = x 1 ka domain x = 0 hai"
Ye sahi kyun lagta hai : Hum excluded point identify kar rahe hain
Fix : Domain ek set hai, condition nahi
Sahi : Domain = R ∖ { 0 } ya ( − ∞ , 0 ) ∪ ( 0 , ∞ )
Condition x = 0 domain ko describe karta hai, khud domain nahi hai
Recall Ek 12-Saal ke Bacche ko Samjhao
Socho tumhare paas ek juice machine hai.
Domain = kaunse fruits tum daalne ki permission rakhte ho (apples, oranges, lekin NOT pathar ya khilone)
Codomain = woh saari tarah ki drinks jo machine bana sakti hai (apple juice, orange juice, grape juice, mango juice...)
Range = woh drinks jo actually nikalti hain jab tum sirf apne allowed fruits use karo. Agar tum sirf apples aur oranges daalte ho, toh grape juice kabhi nahi aayega, chahe machine bana sakti ho agar tumhare paas angoor hote.
Toh: Range (actual outputs) ⊆ Codomain (possible outputs), aur Domain (allowed inputs) batata hai ki tum safely kya use kar sakte ho.
Mnemonic Memory Aid: DCR = "Doctor"
D omain: D ata andar jaata hai
C odomain: C ould come out (aa sakta hai)
R ange: R eally comes out (actually aata hai)
Ya: D on't C onfuse R ange! (Domain → Codomain ⊇ Range)
Function Basics - prerequisite: kya cheez kisi ko function banati hai
Injective Functions - range use karta hai: one-to-one matlab alag inputs → alag outputs
Surjective Functions - directly use karta hai: onto ⟺ range equals codomain
Inverse Functions - f ki range = codomain chahiye taaki f − 1 ka domain ban sake
Composite Functions - g ∘ f ke liye f ki range g ke domain mein fit honi chahiye
Graphing Functions - domain x-axis extent determine karta hai, range y-axis extent determine karta hai
#flashcards/maths
Ek function ka domain kya hota hai? Saare valid input values ka set jo function accept karta hai (woh set jisme se hum x values choose karte hain).
Ek function ka codomain kya hota hai? Woh target set jisme function map karta hai; notation f: A → B mein set B. Ismein woh saare possible outputs hote hain jo hum declare karte hain.
Ek function ka range kya hota hai? Woh saare actual output values ka set jo function produce karta hai; range = {f(x) : x ∈ Domain}. Hamesha codomain ka subset hota hai.
Range aur codomain mein kya relation hai? Range ⊆ Codomain hamesha hota hai. Dono equal hote hain agar aur sirf agar function onto (surjective) ho.
f(x) = x², domain = ℝ ke liye, range kya hai? [0, ∞), kyunki real x ke liye x² kabhi negative nahi hota, aur har y ≥ 0, x = √y se achieve hota hai.
f(x) = 1/(x-3) ka domain kya hai? ℝ \ {3}, kyunki denominator x - 3 zero nahi ho sakta (zero se division undefined hai).
f: [0, 4] → ℝ, f(x) = x² ke liye, range kya hai? [0,16], kyunki x² [0,4] par increase karta hai, minimum x=0 par 0 deta hai, maximum x=4 par 16 deta hai.
Algebraically function ka range kaise nikaalein? (1) y = f(x) likho, (2) x ko y ke terms mein solve karo, (3) x par domain constraints lagao aur dekho kaunse y values valid hain.
f(x) = (2x+1)/(x-1), x ≠ 1 ke liye, range mein kaunsi value NAHI hai? y = 2. y = (2x+1)/(x-1) set karne aur solve karne par x = (y+1)/(y-2) milta hai. Jab y=2, denominator zero hai (koi solution exist nahi karta).
Sach ya Jhooth: Notation f: A → B ka matlab range = B hai Jhooth. Iska matlab codomain = B hai. Range woh subset of B hai jo actually achieve hota hai aur B se chhota ho sakta hai.
Exactly one output per input