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
$3a^2-d^2=\beta$
$3\frac{1}{9}-d^2=\beta ,$
so $\beta =\frac{1}{3}-d^2<\frac{1}{3}$
From (1) and (3), we get
$\frac{1}{3}\left(\frac{1}{9}-d^2\right)=-\gamma$
$\Rightarrow \gamma =\frac{1}{3}\left(d^2-\frac{1}{9}\right)>\frac{1}{3}\left(-\frac{1}{9}\right)=-\frac{1}{27}$
$\gamma \in \left(-\frac{1}{27},+\infty \right)$