Question

The real number $x_{1}, x_{2}, x_{3}$ satisfying the equation $x^{3}-x^{2}+\beta x+\lambda=0$ are in A.P.
All possible values of $\beta$ are:

• A. $\left(-\infty, \frac{1}{3}\right)$
• B. $\left(-\infty,-\frac{1}{3}\right)$
• C. $\left(\frac{1}{3}, \infty\right)$
• D. $\left(-\frac{1}{3}, \infty\right)$

Answer 1

Answer: (A)

From the equation, the real roots of $x^{3}-x^{2}+\beta x+\gamma=0$ are $x_{1}, x_{2}, x_{3}$ and they are in A.P., As $x_{1} x_{2}, x_{3}$ are in A.P. let $x_{1}=a$ $-d, x_{2}=a, x_{3}=a+d$. Now,
$ x_{1}+x_{2}+x_{3}=-\frac{-1}{1}=1 $
$\Rightarrow a-d+a+a+d=1 $
$ \Rightarrow a=\frac{1}{3} $
$x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=\frac{\beta}{1}=\beta$
$\Rightarrow (a-d) a+a(a+d)+(a+d)(a-d)=\beta$
$x_{1} x_{2} x_{3}=-\frac{\gamma}{1}=-\gamma$
$\Rightarrow (a-d) a(a+d)=-\gamma$
From (1) and (2), we get
$3 a ^{2}- d ^{2}=\beta $
$3 \frac{1}{9}- d ^{2}=\beta, $
so $ \beta=\frac{1}{3}- d ^{2}<\frac{1}{3}$
From (1) and (3), we get
$ \frac{1}{3}\left(\frac{1}{9}- d ^{2}\right)=-\gamma $
$\Rightarrow \gamma=\frac{1}{3}\left( d ^{2}-\frac{1}{9}\right)>\frac{1}{3}\left(-\frac{1}{9}\right)=-\frac{1}{27}$
$ \gamma \in\left(-\frac{1}{27},+\infty\right)$