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

• A. $\beta \in (-\infty ,\frac{1}{3}]$ and $\gamma \in [-\frac{1}{27},\infty )$
• B. $\beta \in (-\infty ,\frac{1}{3})$ and $\gamma \in [-\frac{1}{27},\infty )$
  
• C. $\beta \in (-\infty ,-\frac{1}{3}]$ and $\gamma \in (-\frac{1}{27},\infty )$
• D. None of these

Answer 1

Answer: (A)

$\beta \in (-\infty ,\frac{1}{3}]$ and $\gamma \in [-\frac{1}{27},\infty )$

As $x_1,x_2,x_3$ are in A.P, $x_1=x_2-d$ and $x_3=x_2+d$
$x_1+x_2+x_3=1\Rightarrow 3x_2=1\Rightarrow x_2=\frac{1}{3}$
Now $\beta =x_2(x_1+x_3)+x_1{x}_3=2x_2^2+x_1{x}_3$
$\Rightarrow \beta -\frac{2}{9}=(\frac{1}{3}-d)(\frac{1}{3}+d)$
where $d$ is the common difference of A.P
$\Rightarrow d^2=\frac{1}{9}+\frac{2}{9}-\beta$
$\Rightarrow \beta \le \frac{1}{3}$ and $d^2\ge 0$
Next $-\gamma =x_1{x}_2{x}_3=\frac{1}{3}(\frac{1}{9}-d^2)\le \frac{1}{27}\Rightarrow \gamma \ge -\frac{1}{27}$
Thus $\beta \in (-\infty ,\frac{1}{3}]$ and $\gamma \in [-\frac{1}{27},\infty )$