TS ICET 23 May 2018 Shift 1 Paper

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Question : 81 of 200

Marks: +1, -0

If a =

{√3 - 1}/{√3 + 1}

and b =

{√3 + 1}/{√3 - 1}

then the value of

{a^2-ab+b^2}/{a^2+ab+b^2}

is

$5/4$
$13/15$
$13/14$
$4/5$

Solution:

Given

a={√{3}-1}/{√{3}+1}={√{3}-1}/{√{3}+1} × {√{3}-1}/{√{3}-1}=2-√{3}

Similarly,

b={√{3}+1}/{√{3}-1}={√{3}+1}/{√{3}-1}× {√{3}+1}/{√{3}+1}=2+√{3}

ab=(2-√{3}) ×(2+√{3})=4-3=1

Required answer

\;{a^{2}-a b+b^{2}}/{a^{2}+a b+b^{2}}=\;{(a+b)^{2}-3 a b}/{(a+b)^{2}-a b}

={4^2-3×1}/{4^2-1}=13/15

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