Question

The real numbers $x_1,x_2,x_3$ satisfying the equation $x^3-x^2+\beta x+\gamma =0$ are in A.P. Find the intervals in which $\beta$ and $\gamma$ lie, respectively:

• A. $(-\infty ,\frac{1}{3}][\frac{-1}{27},\infty )$
• B. $(-\infty ,3)$
• C. $(\frac{-1}{2},\frac{1}{3})(\frac{-1}{27},\frac{1}{27})$
• D. none of these

Answer 1

The correct option is A $(-\infty ,\frac{1}{3}][\frac{-1}{27},\infty )$

$\because x_1,x_2,x_3$ are in AP.
$\Rightarrow x_1=a-d,x_2=a,x_3=a+d$
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, $\sum \alpha =x_1+x_2+x_3=1$
$\Rightarrow$ (a - d) + (a) + (a + d) = 1 . . . (i)
$\sum \alpha \beta =x_1{x}_2+x_2{x}_3+x_1{x}_3$
$\Rightarrow$ $\beta$ = (a - d) a + a(a +d) + (a - d) (a + d) . .. (ii)
and $\sum =\alpha \beta \gamma =x_1{x}_2{x}_3=-\gamma =(a-d)\left(a\right)(a+d)$ . . . (iii)
hence from (i) we get $a=\frac{1}{3}$
and from eq. (ii) we get
$\beta =3a^2-d^2\Rightarrow \beta =\frac{1}{3}-d^2(\because a=\frac{1}{3})Thus\beta =\frac{1}{3}-d^2\le \frac{1}{3}[\because d^2\ge 0]\Rightarrow \beta \le \frac{1}{3}\therefore \beta ϵ(-\infty ,\frac{1}{3}]$
Again from eq. (iii)
$a(a^2-d^2)=-\gamma \Rightarrow \left(\frac{1}{27}\right)+(-\frac{d^2}{3})=-\gamma \Rightarrow \gamma =\frac{d^2}{3}-\frac{1}{27}\Rightarrow \gamma \ge \frac{-1}{27}(\because d^2\ge 0)$
Hence option (a) is correct.