MathematicsHard129×since 2002Q5144If α\alphaα and β\betaβ are the roots of the equation x² + px + 2 = 0 and 1α{1 \over \alpha }α1 and 1β{1 \over \beta }β1 are the roots of the equation 2x² + 2qx + 1 = 0, then (α−1α)(β−1β)(α+1β)(β+1α)\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right)(α−α1)(β−β1)(α+β1)(β+α1) is equal to :A94(9−q2){9 \over 4}\left( {9 - {q^2}} \right)49(9−q2)B94(9+q2){9 \over 4}\left( {9 + {q^2}} \right)49(9+q2)C94(9−p2){9 \over 4}\left( {9 - {p^2}} \right)49(9−p2)D94(9+p2){9 \over 4}\left( {9 + {p^2}} \right)49(9+p2)Check answerSkip