jee-advanced 2007 Q50

jee-advanced · India · paper1 Roots of polynomials Vieta's formulas: compute symmetric functions of roots

Let $\alpha, \beta$ be the roots of the equation $x^2 - px + r = 0$ and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $x^2 - qx + r = 0$. Then the value of $r$ is
(A) $\frac{2}{9}(p-q)(2q-p)$
(B) $\frac{2}{9}(q-p)(2p-q)$
(C) $\frac{2}{9}(q-2p)(2q-p)$
(D) $\frac{2}{9}(2p-q)(2q-p)$

Let $\alpha, \beta$ be the roots of the equation $x^2 - px + r = 0$ and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $x^2 - qx + r = 0$. Then the value of $r$ is\\
(A) $\frac{2}{9}(p-q)(2q-p)$\\
(B) $\frac{2}{9}(q-p)(2p-q)$\\
(C) $\frac{2}{9}(q-2p)(2q-p)$\\
(D) $\frac{2}{9}(2p-q)(2q-p)$

Paper Questions

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