CAT 2024

Slot 2

QA

Question & Solution

From the sum and product of roots, we get: $\alpha + \beta = -\frac{\lambda}{3}$ and $\alpha \beta = -\frac{1}{3}$

Simplifying the expression given in the question, we get: $\frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = 15$
and substituting the denominator's value as 1/9, we get: $\alpha^2 + \beta^2 = \frac{15}{9}$

We want the expression $\alpha^3 + \beta^3$, so multiplying both sides by $\alpha+\beta$, we get:
$\alpha^3 + \beta^3 + \alpha \beta(\alpha + \beta) = \frac{15}{9}(\alpha + \beta)$
$\alpha^3 + \beta^3 + \frac{\lambda}{9} = \frac{15}{9}\left(-\frac{\lambda}{3}\right)$
$\alpha^3 + \beta^3 + \frac{\lambda}{9} = -\frac{5\lambda}{9} - \frac{\lambda}{9} = -\frac{2\lambda}{3}$

We would still need to find the value of $\lambda$
This we can do from the initial relation we had:

$\alpha^2 + \beta^2 = \frac{15}{9}$
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = \frac{15}{9}$
$\frac{\lambda^2}{9} + \frac{2}{3} = \frac{15}{9}$
$\frac{\lambda^2}{9} = \frac{15-6}{9} = \frac{9}{9} = 1$
This would finally give us $\lambda^2=9$

Using this in our required expression, we get:

$(\alpha^3 + \beta^3)^2 = \left(-\frac{2\lambda}{3}\right)^2 = \frac{4 \times 9}{9} = 4$

Therefore, Option B is the correct answer.