Last updated at March 9, 2017 by Teachoo

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Ex9.3, 28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2 √2) :"(3 − 2 " √2) Introduction Componendo dividendo If 𝑥/𝑦 = 𝑎/𝑏 Applying componendo dividendo (𝑥 + 𝑦)/(𝑥 − 𝑦) = (𝑎 + 𝑏)/(𝑎 − 𝑏) Eg: Taking 1/2 = 4/8 (1+ 2)/(1 − 2) = (4 + 8)/(4 − 8) 3/(−1) = 12/(−4) -3 = -3 Ex 9.3, 28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2 √2) :"(3 − 2 " √2) Let a & b be the numbers We know that Geometric mean of two numbers a & b is √𝑎𝑏 i.e. GM of a & b = √𝑎𝑏 According to the question Sum of two numbers a and b is 6 times of their GM a + b = 6√𝑎𝑏 Solving, (𝑎 + 𝑏 )/(2√𝑎𝑏) = 3/1 Applying componendo & dividendo (𝑎+𝑏+2√𝑎𝑏)/(𝑎+𝑏 −2√𝑎𝑏) = (3 + 1)/(3 − 1 ) ((√𝑎)2+(√𝑏)2+2(√𝑎×√𝑏))/((√𝑎)2+(√𝑏)2−2(√𝑎×√𝑏) ) = 4/2 Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy (√𝑎 + √𝑏)2/(√𝑎 − √𝑏)2 = 2/1 ((√𝑎 + √𝑏)/(√𝑎 − √𝑏))^2 = 2/1 (√𝑎 + √𝑏)/(√𝑎 − √𝑏) = √2/( 1) Again applying componendo & dividendo ((√𝑎+ √𝑏)+(√𝑎 − √𝑏))/((√𝑎 + √𝑏) −(√𝑎 − √𝑏) ) = (√2 + 1)/(√2 − 1) (√𝑎 +√𝑎 + √𝑏 − √𝑏)/(√𝑎 + √𝑏 − √𝑎 + √𝑏) = (√2 + 1)/(√2 − 1) (2√𝑎 + 0)/(√𝑏 + √𝑏 + √𝑎 − √𝑎) = (√2 + 1)/(√2 − 1) (2√𝑎)/(2√𝑏 + 0) = (√2 + 1)/(√2 − 1) (2√𝑎)/(2√𝑏 ) = (√2 + 1)/(√2 − 1) √(𝑎/𝑏) = (√2 + 1)/(√2 − 1) Squaring both sides (√(𝑎/𝑏))^2 = ((√2 + 1)/(√2 − 1))^2 𝑎/𝑏 = ((√2 + 1)2)/((√2 − 1)2) 𝑎/𝑏 = ((√2)2 + (1)2 + 2√2 × 1)/((√2)2 + (1)2 − 2√2 × 1) 𝑎/𝑏 = (2 + 1 + 2√2)/(2 + 1 − 2√2) 𝑎/𝑏 = (3 + 2√2)/(3 − 2√2) Thus the ratio of a & b is 3 + 2√3: 3 – 2√2 Hence proved

Example 6
Important

Ex 9.2, 5 Important

Ex 9.2, 9 Important

Ex 9.2, 15 Important

Ex 9.2, 17 Important

Example 14 Important

Example 15 Important

Ex 9.3, 3 Important

Ex 9.3, 11 Important

Ex 9.3, 17 Important

Ex 9.3, 18 Important

Ex 9.3, 22 Important

Ex 9.3, 28 Important You are here

Ex 9.3, 29 Important

Ex 9.4.4 Important

Ex 9.4, 7 Important

Ex 9.4, 9 Important

Ex 9.4, 10 Important

Example 23 Important

Misc 16 Important

Misc 19 Important

Misc 25 Important

Misc 28 Important

Misc 32 Important

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .