If a =[\sqrt^3{11}, \b = \sqrt^4{23} \and \c =\sqrt^6{119} then the decreasing order of a,b,c is

TS ICET 23 May 2018 Shift 2 Paper

Question : 89 of 200

Marks: +1, -0

If a =

[√^3{11}, \b = √^4{23} \and \c =√^6{119}

then the decreasing order of a,b,c is

Solution:

LCM of 3, 4 and 6 is 12

a=^3√{11}=11^{{1}/{3}}=11^{{4}/{12}}=(11^{4})^{{1}/{12}}=14641^{{1}/{12}}

b=^4√{23}=23^{{1}/{4}}=23{3}/{12}=(23^{3})^{{1}/{12}}=12167^{{1}/{12}}

c=^6√{119}=119^{{1}/{6}}=119^{{2}/{12}}=(119^{2})^{{1}/{12}}=14161^{{1}/{12}}

Clearly, a > c > b

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