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Question

If α and β are the roots of the equation x2+px+2=0 and 1α and 1β are the roots of the equation 2x2+2qx+1=0, then (α1α)(β1β)(α+1β)(β+1α) is equal to

A
94(9+p2)
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B
94(9p2)
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C
94(9+q2)
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D
94(9q2)
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Solution

The correct option is B 94(9p2)
α and β are the roots of the equation x2+px+2=0.
α+β=p, αβ=2
1α and 1β are the roots of the equation 2x2+2qx+1=0
1α+1β=q,1αβ=12
α+βαβ=qp2=q
p=2q
Now
(α+1β)(β+1α)=αβ+1αβ+2
=2+12+2=92
and
(α1α)(β1β)=αβ+1αβαββα
=2+12[α2+β2αβ]
=52[(α+β)22αβαβ]
=52[p242]
=9p22

(α1α)(β1β)(α+1β)(β+1α)=(9p22)(92)
=94(9p2)

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