1. Class 11
2. Important Question for exams Class 11

Transcript

Ex9.3, 28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2 โ2) :"(3 โ 2 " โ2) Introduction Componendo dividendo If ๐ฅ/๐ฆ = ๐/๐ Applying componendo dividendo (๐ฅ + ๐ฆ)/(๐ฅ โ ๐ฆ) = (๐ + ๐)/(๐ โ ๐) Eg: Taking 1/2 = 4/8 (1+ 2)/(1 โ 2) = (4 + 8)/(4 โ 8) 3/(โ1) = 12/(โ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 โ2) :"(3 โ 2 " โ2) Let a & b be the numbers We know that Geometric mean of two numbers a & b is โ๐๐ i.e. GM of a & b = โ๐๐ According to the question Sum of two numbers a and b is 6 times of their GM a + b = 6โ๐๐ Solving, (๐ + ๐ )/(2โ๐๐) = 3/1 Applying componendo & dividendo (๐+๐+2โ๐๐)/(๐+๐ โ2โ๐๐) = (3 + 1)/(3 โ 1 ) ((โ๐)2+(โ๐)2+2(โ๐รโ๐))/((โ๐)2+(โ๐)2โ2(โ๐รโ๐) ) = 4/2 Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy (โ๐ + โ๐)2/(โ๐ โ โ๐)2 = 2/1 ((โ๐ + โ๐)/(โ๐ โ โ๐))^2 = 2/1 (โ๐ + โ๐)/(โ๐ โ โ๐) = โ2/( 1) Again applying componendo & dividendo ((โ๐+ โ๐)+(โ๐ โ โ๐))/((โ๐ + โ๐) โ(โ๐ โ โ๐) ) = (โ2 + 1)/(โ2 โ 1) (โ๐ +โ๐ + โ๐ โ โ๐)/(โ๐ + โ๐ โ โ๐ + โ๐) = (โ2 + 1)/(โ2 โ 1) (2โ๐ + 0)/(โ๐ + โ๐ + โ๐ โ โ๐) = (โ2 + 1)/(โ2 โ 1) (2โ๐)/(2โ๐ + 0) = (โ2 + 1)/(โ2 โ 1) (2โ๐)/(2โ๐ ) = (โ2 + 1)/(โ2 โ 1) โ(๐/๐) = (โ2 + 1)/(โ2 โ 1) Squaring both sides (โ(๐/๐))^2 = ((โ2 + 1)/(โ2 โ 1))^2 ๐/๐ = ((โ2 + 1)2)/((โ2 โ 1)2) ๐/๐ = ((โ2)2 + (1)2 + 2โ2 ร 1)/((โ2)2 + (1)2 โ 2โ2 ร 1) ๐/๐ = (2 + 1 + 2โ2)/(2 + 1 โ 2โ2) ๐/๐ = (3 + 2โ2)/(3 โ 2โ2) Thus the ratio of a & b is 3 + 2โ3: 3 โ 2โ2 Hence proved

Class 11
Important Question for exams Class 11

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.