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QJEEM15Maff32: JEE Main 2015 Mathematics - Class 11 JEEM15Maff32 Solution: If m is the A.M. of two distinct real numbers l and n and and…

If m is the A.M. of two distinct real numbers l and n (l,n > 1) and G1,G2{G_1},{G_2} and G3{G_3} are three geometric means between ll and n, then G14 +2G24 +G34G_1^4\, + 2G_2^4\, + G_3^4 equals:

Multiple Choice Options

  • A 4 lmn24\,lm{n^2}
  • B 4 l2m2n24\,{l^2}{m^2}{n^2}
  • C 4 l2m n4\,{l^2}m\,n
  • D 4 l m2n4\,l\,{m^2}n

Step-by-Step Verified Solution

m=l+n2m = {{l + n} \over 2} and common ratio of

G.P.G.P. =r=(nl)14= r = {\left( {{n \over l}} \right)^{{1 \over 4}}}

∴\therefore G1=l3/4 n1/4,{G_1} = {l^{3/4}}\,{n^{1/4}}, G2=l1/2n1/2, {G_2} = {l^{1/2}}{n^{1/2}},\,

              G3=l1/4n3/4\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}

G14+2G24+G34G_1^4 + 2G_2^4 + G_3^4

=l3n+2l2n2+ln⁡3= {l^3}n + 2{l^2}{n^2} + {\ln ^3}

=ln⁡(1+n)2= \ln {\left( {1 + n} \right)^2}

=ln⁡×2m2= \ln \times 2{m^2}

=4lm2n= 4l{m^2}n

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