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

Transcript

Misc 19 The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that a : b = (m + โ(๐^2โ๐^2 )) : (m โ โ(๐^2โ๐^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 Misc 19 The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that a : b = (m + โ(๐^2โ๐^2 )) : (m โ โ(๐^2โ๐^2 ) ) Here, the two numbers be a and b. Arithmetic Mean =AM= (a+b)/2 & Geometric Mean=GM= โab According to the question, AM/( GM" " ) = ๐/๐ (๐ + ๐)/(2 โ๐๐ " " ) = ๐/๐ Applying componendo dividendo (๐+๐+2โ๐๐)/(๐+๐ โ2โ๐๐) = (๐ + ๐)/(๐ โ ๐ ) ((โ๐)2+(โ๐)2+2(โ๐รโ๐))/((โ๐)2+(โ๐)2โ2(โ๐รโ๐) ) =(๐ + ๐)/(๐ โ ๐ ) Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy (โ๐ + โ๐)2/(โ๐ โ โ๐)2 = (๐ + ๐)/(๐ โ ๐ ) ((โ๐ + โ๐)/(โ๐ โ โ๐))^2 = (๐ + ๐)/(๐ โ ๐ ) (โ๐ + โ๐)/(โ๐ โ โ๐) = โ((๐ + ๐)/(๐ โ ๐ )) (โ๐ + โ๐)/(โ๐ โ โ๐) = โ(๐ + ๐)/(โ(๐ โ ๐) ) Applying componendo dividendo ((โ๐ + โ๐) + (โ๐ โ โ๐))/((โ๐ + โ๐) โ (โ๐ โ โ๐) ) = (โ(๐ + ๐) + โ(๐ โ ๐))/(โ(๐ + ๐) โโ(๐ โ ๐)) (2โ๐)/(2โ๐) = (โ(๐ + ๐) + โ(๐ โ ๐))/(โ(๐ + ๐) โโ(๐ โ ๐)) โ๐/โ๐ = (โ(๐ + ๐) + โ(๐ โ ๐))/(โ(๐ + ๐) โโ(๐ โ ๐)) Squaring both sides (โ๐/โ๐)^2 = ((โ(๐ + ๐) + โ(๐ โ ๐))/(โ(๐ + ๐) โโ(๐ โ ๐)))^2 (โ๐)^2/(โ๐)^2 = (โ(๐ + ๐) + โ(๐ โ ๐))^2/(โ(๐ + ๐) โโ(๐ โ ๐))^2 Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy ๐/๐ = ((โ(๐ + ๐) )^2+(โ(๐ โ ๐) )^2+ 2(โ(๐+๐))(โ(๐โ๐)))/((โ(๐ + ๐) )^2+(โ(๐ โ ๐) )^2โ 2(โ(๐+๐))(โ(๐โ๐)) ) ๐/๐ = (๐ + ๐ + ๐ โ ๐ + 2โ((๐+๐)(๐โ๐) ))/(๐ + ๐ + ๐ โ ๐ โ 2โ((๐+๐)(๐โ๐) )) ๐/๐ = (๐ +๐ + ๐ โ ๐ + 2โ((๐^2โ ๐^2 ) ))/(๐ +๐ + ๐ โ ๐ โ 2โ((๐^2โ ๐^2 ) )) ๐/๐ = (2๐ + 2โ((๐^2โ ๐^2 ) ))/(2๐ โ 2โ((๐^2โ ๐^2 ) )) ๐/๐ = 2(๐ + โ((๐^2โ ๐^2 ) ))/2(๐ โ โ((๐^2โ ๐^2 ) )) ๐/๐ = (๐ + โ((๐^2โ ๐^2 ) ))/(๐ โ โ((๐^2โ ๐^2 ) )) Thus, a : b = (m + โ(๐^2โ๐^2 )) : (m โ โ(๐^2โ๐^2 ) ) Hence proved

Class 11
Important Question for exams Class 11