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This question was basically targeted to further goers who have been taught sequences and series and have some basic knowledge of the natural log (log with base e-exponent function ... Which is denoted as ln)
a)
ReplyDeleteEvery1 is requested to provide with them the sol/method/reason along with their ans
ReplyDeleteif not it will be considered as an educated guess
The and to this question is c)
ReplyDeleteThis question was basically targeted to further goers who have been taught sequences and series and have some basic knowledge of the natural log (log with base e-exponent function ... Which is denoted as ln)
ReplyDeleteFurther*---FIITJEE
ReplyDeleteLet's see. In this question we require a concept of arithmetic mean and geometric mean
ReplyDeleteWe should know that
AM > GM
(x^lny-lnz + y^lny-lnz + z^lnx-lny) /3 >(x^lny-lnz * y^lny-lnz * z^lnx-lny)^1/3
Let's solve rhs
Let a=lnx, b=lny, c=lnz
x=e^a, y=e^b, z=e^c
Replacing in this we get...
ReplyDeleteReplacing in rhs we get...
(e^a(b-c) * e^b(c-a) * e^c(a-b)) ^1/3
Opening brackets we get
=(e^ab-ac+bc-ab+ac-bc) ^1/3
=(e^0)^1/3
≠(1)^1/3
=1
Hence we get...
Hence by replacing in the original eqn. We get...
ReplyDelete(x^lny-lnz + y^lny-lnz + z^lnx-lny)/3 > 1
Which implies...
(x^lny-lnz + y^lny-lnz + z^lnx-lny) > 3
Which implies...
(x^lny-lnz + y^lny-lnz + z^lnx-lny) € (3,infinity)
Hence ans is (c)
Hope you in understand the solution B-)
ReplyDelete