CAT Exam Model Paper 3 with solutions for free online practice

(a) ∵${\mathit{x}}_1,{\mathit{x}}_2,{\mathit{x}}_3$ are in AP.
⇒${\mathit{x}}_1=\mathit{a}-\mathit{d},{\mathit{x}}_2=\mathit{a},{\mathit{x}}_3=\mathit{a}+\mathit{d}$
Where, d is the common difference
Now, since ${\mathit{x}}_1,{\mathit{x}}_2,{\mathit{x}}_3$ are the roots of the given equation
${\mathit{x}}^3-{\mathit{x}}^2+\mathit{\beta }\mathit{x}+\mathit{\gamma }=0$
So, sum of roots = ${\mathit{x}}_1+{\mathit{x}}_2+{\mathit{x}}_3=1$
⇒ (a − d ) + (a) + (a + d ) =1 ... (i)
3a = 1 ⇒ a = $\frac{1}{3}$
$\mathit{\beta }={\mathit{x}}_1{\mathit{x}}_2+{\mathit{x}}_2+{\mathit{x}}_3+{\mathit{x}}_1{\mathit{x}}_3$
⇒ β = (a − d ) a + a(a + d ) + (a − d )(a + d ) ... (ii)
and product of roots = x${}_1$x${}_2$x${}_3$ = − $\mathit{\gamma }$ = (a − d )(a)(a + d ) ... (iii)
Hence from (i), we get a = $\frac{1}{3}$
and from eq. (ii), we get
$\mathit{\beta }=3{\mathit{a}}^2-{\mathit{d}}^2\Rightarrow \mathit{\beta }=\frac{1}{3}-{\mathit{d}}^2\left(\because \mathit{a}=\frac{1}{3}\right)$
Thus $\mathit{\beta }=\frac{1}{3}-{\mathit{d}}^2\le \frac{1}{3}$ [∵d${}^2$ ≥ 0]
⇒ β < =$\frac{1}{3}$
∴ β ∈ $\left(-\mathit{\infty },\frac{1}{3}\right]$
Again from eq. (iii), we get
$\mathit{a}\left({\mathit{a}}^2-{\mathit{d}}^2\right)=-\mathit{\gamma }\Rightarrow \left(\frac{1}{27}\right)+\left(-\frac{{\mathit{d}}^2}{3}\right)=-\mathit{\gamma }$
⇒$\mathit{\gamma }=\frac{{\mathit{d}}^2}{3}-\frac{1}{27}\Rightarrow \mathit{\gamma }\ge \frac{-1}{27}$ (∵d${}^2$ ≥ 0)
∴ y ∈ $\left[-\frac{1}{27},\mathit{\infty }\right]$
Hence option (a) is correct.