2nd_Year_Maths_Chapter1_Exercise1.3 -Graphical Representaion of Functions – New Book 2026-27

Question 1 — Logarithmic Function ka Inverse

f(x) = log₄(x−2)

(i) One-to-one hona: Horizontal line test ka matlab hai: agar koi bhi horizontal line graph ko sirf ek dafa cut kare to function one-to-one hai. Chunke log function hamesha strictly increasing hota hai, har horizontal line sirf ek point par milegi. Algebraically bhi dikhaya: agar f(x₁)=f(x₂) to x₁=x₂ nikalta hai — matlab do alag x kabhi same output nahi de sakte.

(ii) Inverse nikalna: y = log₄(x−2) liya, phir exponential form mein convert kiya (log ko exponent mein badalna): 4ʸ = x−2. Phir x aur y ko interchange kiya (ye hamesha inverse nikalne ka tareeqa hai): f⁻¹(x) = 4ˣ+2

(iii) Domain/Range: Original function ka domain (2,∞) hai (kyunke log ke andar positive number chahiye), aur range sab real numbers hai. Inverse mein ye ulat ho jate hain — ye hamesha hota hai: original ka domain = inverse ka range, aur original ka range = inverse ka domain.

(iv) Symmetry: f aur f⁻¹ ke graphs hamesha line y=x ke around mirror image hote hain — ye inverse functions ki sabse important property hai.

Question 2 — sin⁻¹x Function

Same concept jaise Q1, lekin trigonometric inverse ke saath:

• f(x) = sin⁻¹x ka domain [−1,1] hai (kyunke sin ki value hamesha is range mein hoti hai), aur range [−π/2, π/2]
• f⁻¹(x) = sin x (restricted) — matlab sin x ko sirf [−π/2,π/2] tak restrict kiya gaya hai (warna wo one-to-one nahi hota)
• Graph mein dikhaya ke sin⁻¹x strictly increasing hai, isliye one-to-one hai
• Dono graphs y=x ke around mirror image hain

Question 3 — Cubic Function ka Inverse

f(x) = (x−1)³, domain [0,3]

• y = (x−1)³ liya, phir cube root liya: x−1 = y^(1/3), isliye x = y^(1/3)+1
• x,y interchange kiya: f⁻¹(x) = x^(1/3)+1
• Endpoints check kiye: f(0)=−1, f(3)=8, isliye f, [0,3] ko [−1,8] mein map karta hai
• Isliye f⁻¹ ka domain [−1,8] aur range [0,3] hoga (ulta original ke)

Question 4, 5, 6, 7, 9, 10 — Transformations (Graph ko Shift/Stretch Karna)

Ye sab questions ek hi concept par based hain: function transformations. Yaad rakhne wale rules:

Important baat: Jab kai transformations ek saath ho rahi hon, to unhe order (tarteeb) mein apply karna zaroori hai — jis order mein diya gaya ho, wahi order follow karo, warna galat answer aayega.

Q4 (f(x)=√x): Pehle vertical stretch (2√x), phir horizontal shift (√(x−3)), phir dono mila diye (2√(x−3))

Q5 (f(x)=1/x): Char transformations diye gaye order mein apply kiye — har step mein pichle result par agla operation lagaya gaya, yahan tak ke final answer aa gaya: 3/(2(x+1)) + 1

Q6 (f(x)=ln x): Teen steps — right shift, phir vertical compression, phir upward shift — ek ke baad ek apply kiye.

Q7 (complicated fraction wala function): Same tareeqa, bas function thora complex hai (square root ke andar fraction). Har transformation ko x ki jagah substitute karke replace kiya gaya.

Q8 (f(x)=3ˣ vs g(x)=3^(x−2)+1): Dikhaya gaya ke shape same rehti hai sirf position badalti hai. Jab x ki jagah (x−2) aayi to graph 2 units right shift hua, aur +1 se 1 unit upar shift hua. Horizontal asymptote bhi y=0 se y=1 par chali gayi.

Q9 (ln x): Char transformations order mein apply hui — stretch, phir right shift, phir horizontal compression, phir upward shift.

Q10 (|x| — absolute value): Same tareeqa, sirf function |x| hai is dafa.

Sabse Zaroori Baat Yaad Rakhne Wali:

1. Transformation ka order matter karta hai — jis sequence mein diya jaye, usi mein karo
2. Horizontal transformations ulti feel hoti hain (x−k matlab right shift, x+k matlab left shift) — ye students ko confuse karta hai, isliye extra dhyan do
3. Vertical transformations seedhi feel hoti hain (+k matlab upar, k× matlab stretch)
4. Inverse function nikalne ka hamesha same 3-step tareeqa hai: (1) y likho, (2) x ke liye solve karo, (3) x aur y interchange karo