JEE Main 2020 (Online) 3rd September Morning ShiftQuestion 1
If $\alpha$ and $\beta$ are the roots of the equation $x^2+px+2=0$ and $\frac1\alpha$ and $\frac1\beta$ are the roots of the equation $2x^2+2qx+1=0$, then $\left(\alpha-\frac1\alpha\right)\left(\beta-\frac1\beta\right)\left(\alpha+\frac1\beta\right)\left(\beta+\frac1\alpha\right)$ is equal to :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. Pair the first two and the last two factors; each pair is symmetric in $\alpha,\beta$. Formulas used. (1) $(\alpha^2-1)(\beta^2-1)=(\alpha\beta)^2-(\alpha^2+\beta^2)+1$Step 1 - Basics. $\alpha\beta=2$, $\alpha+\beta=-p$, so $\alpha^2+\beta^2=p^2-4$. Step 2 - First pair. By (1), $$\left(\alpha-\frac1\alpha\right)\left(\beta-\frac1\beta\right)=\frac{(\alpha^2-1)(\beta^2-1)}{\alpha\beta}=\frac{4-(p^2-4)+1}{2}=\frac{9-p^2}{2}.$$ Step 3 - Second pair. $$\left(\alpha+\frac1\beta\right)\left(\beta+\frac1\alpha\right)=\alpha\beta+2+\frac{1}{\alpha\beta}=\frac92.$$ Step 4 - Product. $\dfrac{9-p^2}{2}\cdot\dfrac92=\dfrac94\left(9-p^2\right)$. Answer: (A)