Let $\alpha, \beta$ be the roots of the equation $x^{2}-p x+r=0$ and $\frac{\alpha}{2}, 2 \beta$ be the roots of the equation $x^{2}-q x+r=0$. Then the value of $r$ is

(D) $\frac{2}{9}(2 p-q)(2 q-p)$

(A) $\frac{2}{9}(p-q)(2 q-p)$

(B) $\frac{2}{9}(q-p)(2 p-q)$

(C) $\frac{2}{9}(q-2 p)(2 q-p)$

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Quadratic Equations and Inequalities

Let

α, β

be the roots of the equation

x^{2} - p x + r = 0

and

\frac{α}{2}, 2 β

be the roots of the equation

x^{2} - q x + r = 0

. Then the value of

r

Answer: (D)

From the first equation,

α + β = p

and

α β = r

.

From the second equation,

\frac{α}{2} + 2 β = q

and

\frac{α}{2} \cdot 2 β = α β = r

.

From

α + β = p

and

\frac{α}{2} + 2 β = q

, we get:

β = \frac{2 q - p}{3}

and

α = \frac{2 ( 2 p - q )}{3}

Therefore,

r = α β = \frac{2 ( 2 p - q )}{3} \cdot \frac{2 q - p}{3} = \frac{2}{9} (2 p - q) (2 q - p)

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Single correct option