If α,β are the roots of the equation x2+px+2=0 and 1αand 1β are the roots of the equation 2x2+2qx+1=0 then (α−1α)(β−1β)(α+1β)(β+1α) is equal to

α+β=−p,α β=2, 1α+1β=−q given expression =(αβ−αβ−βα+1αβ)(αβ+1+1+1αβ)=(52−α2+β2αβ)(92)=94(9−p2)

If α,β are the roots of the equation x2+px+2=0 and 1αand 1β are the roots of the equation 2x2+2qx+1=0 then (α−1α)(β−1β)(α+1β)(β+1α) is equal to

α+β=−p,α β=2, 1α+1β=−q given expression =(αβ−αβ−βα+1αβ)(αβ+1+1+1αβ)=(52−α2+β2αβ)(92)=94(9−p2)

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If $α, β$ are the roots of the equation $x^{2} + p x + 2 = 0$ and $\frac{1}{α} and \frac{1}{β}$ are the roots of the equation $2 x^{2} + 2 q x + 1 = 0$ then $(α - \frac{1}{α}) (β - \frac{1}{β}) (α + \frac{1}{β}) (β + \frac{1}{α})$ is equal to

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$\frac{9}{4} (9 - p^{2})$

$\frac{9}{4} (9 + p^{2})$

$\frac{9}{4} (9 - q^{2})$

$\frac{9}{4} (9 + q^{2})$

answer is B.

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Detailed Solution

$α + β = - p, α β = 2, \frac{1}{α} + \frac{1}{β} = - q$
given expression
$= (α β - \frac{α}{β} - \frac{β}{α} + \frac{1}{α β}) (α β + 1 + 1 + \frac{1}{α β})$
$= (\frac{5}{2} - \frac{α^{2} + β^{2}}{α β}) (\frac{9}{2}) = \frac{9}{4} (9 - p^{2})$