jee-main 2020 Q52

jee-main · India · session2_03sep_shift1 Roots of polynomials Vieta's formulas: compute symmetric functions of roots
If $\alpha$ and $\beta$ are the roots of the equation $x ^ { 2 } + p x + 2 = 0$ and $\frac { 1 } { \alpha }$ and $\frac { 1 } { \beta }$ are the roots of the equation $2 x ^ { 2 } + 2 q x + 1 = 0$, then $\left( \alpha - \frac { 1 } { \alpha } \right) \left( \beta - \frac { 1 } { \beta } \right) \left( \alpha + \frac { 1 } { \beta } \right) \left( \beta + \frac { 1 } { \alpha } \right)$ is equal to:
(1) $\frac { 9 } { 4 } \left( 9 + q ^ { 2 } \right)$
(2) $\frac { 9 } { 4 } \left( 9 - q ^ { 2 } \right)$
(3) $\frac { 9 } { 4 } \left( 9 + p ^ { 2 } \right)$
(4) $\frac { 9 } { 4 } \left( 9 - p ^ { 2 } \right)$
If $\alpha$ and $\beta$ are the roots of the equation $x ^ { 2 } + p x + 2 = 0$ and $\frac { 1 } { \alpha }$ and $\frac { 1 } { \beta }$ are the roots of the equation $2 x ^ { 2 } + 2 q x + 1 = 0$, then $\left( \alpha - \frac { 1 } { \alpha } \right) \left( \beta - \frac { 1 } { \beta } \right) \left( \alpha + \frac { 1 } { \beta } \right) \left( \beta + \frac { 1 } { \alpha } \right)$ is equal to:

(1) $\frac { 9 } { 4 } \left( 9 + q ^ { 2 } \right)$\\
(2) $\frac { 9 } { 4 } \left( 9 - q ^ { 2 } \right)$\\
(3) $\frac { 9 } { 4 } \left( 9 + p ^ { 2 } \right)$\\
(4) $\frac { 9 } { 4 } \left( 9 - p ^ { 2 } \right)$