TS ICET 23 May 2018 Shift 2 Paper

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Question : 81 of 200

Marks: +1, -0

For k ≠ 0; if a =

√^3{\k} - 1/√^3{\k}

then

\a^3 + 3\a =

${\k-1}/{\k^2 +1}$
${\k^2-1}/\k$
${\k-1}/\k^2$
${\k}/{\k^2 -1}$

Solution:

Given

a=^3√{k}-\;{1}/{^3√{k}}

Cubing on both sides

a^{3}=(^3√{k})^{3}-(\;{1}/{^3√k})^{3}-3(^3√{k}-\;{1}/{^3√{k}})

⇒ a^{3}+3 a=k-\;{1}/{k}=\;{k^{2}-1}/{k}

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