CAT Exam Model Paper 3 with solutions for free online practice

(a) ∵$x_1, x_2, x_3$ are in AP. ⇒$x_1 = a - d, x_2 = a, x_3 = a + d$ Where, d is the common difference Now, since $x_1, x_2, x_3$ are the roots of the given equation $x^{3} - x^{2} + \beta x + \gamma = 0$ So, sum of roots = $x_1 + x_2 + x_3 = 1$⇒ (a − d ) + (a) + (a + d ) =1 ... (i) 3a = 1 ⇒ a = $\frac{1}{3}$$\beta = x_1 x_2 + x_2 + x_3 + x_1 x_3$⇒ β = (a − d ) a + a(a + d ) + (a − d )(a + d ) ... (ii)and product of roots = x$_{1}$x$_{2}$x$_{3}$ = − $\gamma$ = (a − d )(a)(a + d ) ... (iii)Hence from (i), we get a = $\frac{1}{3}$and from eq. (ii), we get$\beta = 3 a^{2} - d^{2} \Rightarrow \beta = \frac{1}{3} - d^{2} \quad \left(\because a = \frac{1}{3}\right)$Thus $\beta = \frac{1}{3} - d^{2} \leq \frac{1}{3}$ [∵d$^{2}$ ≥ 0]⇒ β < =$\frac{1}{3}$∴ β ∈ $(-\infty, \frac{1}{3}]$Again from eq. (iii), we get$a (a^{2} - d^{2}) = -\gamma \Rightarrow \left( \frac{1}{27} \right) + \left( -\frac{d^{2}}{3} \right) = -\gamma$⇒$\gamma = \frac{d^{2}}{3} - \frac{1}{27} \Rightarrow \gamma \geq \frac{-1}{27}$ (∵d$^{2}$ ≥ 0)∴ y ∈ $\left[-\frac{1}{27}, \infty\right]$Hence option (a) is correct.