The ascending order of \sqrt^4 10,\sqrt^3 6,\sqrt^2 3,\sqrt^3 9 \and \sqrt^6 8 is

Correct Answer is : 86,32,104,63,93

\sqrt^6 8,\sqrt^4 10,\sqrt^3 6,\sqrt^3 9,\sqrt^2 3

\sqrt^6 8,\sqrt^2 3,\sqrt^3 6,\sqrt^3 9,\sqrt^4 10

\sqrt^6 8,\sqrt^2 3,\sqrt^4 10,\sqrt^3 6,\sqrt^3 9

\sqrt^2 3,\sqrt^6 8,\sqrt^4 10,\sqrt^3 6,\sqrt^3 9

TS ICET 24 May 2018 Shift 1 Paper

Question : 89 of 200

Marks: +1, -0

The ascending order of

√^4 10,√^3 6,√^2 3,√^3 9 \and √^6 8

Solution:

^4{√10}=10^{\;{1}/{4}}=10^{\;{1}/{4}} x^{\;{3}/{3}}=10^{\;{3}/{12}}=(10^{3})^{\;{1}/{12}}=1,000^{\;{1}/{12}}

^3{√6}=6^{\;{1}/{3}}=6^{\;{1}/{3} × \;{4}/{4}}=6^{\;{4}/{12}}=(6^{4})^{\;{1}/{12}}=1,296^{\;{1}/{12}}

^2{√3}=3^{\;{1}/{2}}=3^{\;{1}/{2} × \;{6}/{6}}=3^{\;{6}/{12}}=(3^{6})^{\;{1}/{12}}=729^{\;{1}/{12}}

^3{√9}=9^{\;{1}/{3}}=9^{\;{1}/{3} × \;{4}/{4}}=9^{\;{4}/{12}}=(9^{4})^{\;{1}/{12}}=6,561^{\;{1}/{12}}

^6{√8}=8^{\;{1}/{6}}=8^{\;{1}/{6} × \;{3}/{3}}=8^{\;{3}/{12}}=(8^{3})^{\;{1}/{12}}=512^{\;{1}/{12}}

Since,

512 < 729 < 1,000 < 1,296 < 6,561

Hence

^6{√8}<\;^2{√3}<\;^4{√10}<^5{√6}<\;^3{√9}

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