If alpha and beta are the roots of the equati | vedclass

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(C) Given $\alpha, \beta$ are roots of $x^{2}+px+2=0$,so $\alpha+\beta = -p$ and $\alpha\beta = 2$.The roots of $2x^{2}+2qx+1=0$ are $\frac{1}{\alpha}

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

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(C) Given $\alpha, \beta$ are roots of $x^{2}+px+2=0$,so $\alpha+\beta = -p$ and $\alpha\beta = 2$.The roots of $2x^{2}+2qx+1=0$ are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.Sum of roots: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-p}{2} = -q \Rightarrow p = 2q$.Product of roots: $\frac{1}{\alpha\beta} = \frac{1}{2}$,which is consistent.We need to evaluate $E = \left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$.$E = \left(\frac{\alpha^{2}-1}{\alpha}\right)\left(\frac{\beta^{2}-1}{\beta}\right)\left(\frac{\alpha\beta+1}{\beta}\right)\left(\frac{\alpha\beta+1}{\alpha}\right) = \frac{(\alpha^{2}-1)(\beta^{2}-1)(\alpha\beta+1)^{2}}{(\alpha\beta)^{2}}$.Since $\alpha^{2} = -p\alpha-2$ and $\beta^{2} = -p\beta-2$,then $\alpha^{2}-1 = -p\alpha-3$ and $\beta^{2}-1 = -p\beta-3$.$E = \frac{(-p\alpha-3)(-p\beta-3)(2+1)^{2}}{2^{2}} = \frac{(p\alpha+3)(p\beta+3)(9)}{4} = \frac{9}{4}(p^{2}\alpha\beta + 3p(\alpha+\beta) + 9)$.Substituting $\alpha\beta=2$ and $\alpha+\beta=-p$:$E = \frac{9}{4}(2p^{2} - 3p^{2} + 9) = \frac{9}{4}(9-p^{2})$.

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