Question

All the roots of the equation $x^3-x^2+ax+b=0$ are real and distinct. If they are in A.P., then

• A. $a\in (-\infty ,\frac{1}{3})$
• B. $a\in (-\infty ,-\frac{1}{3})$
• C. $b\in (-\frac{1}{9},\infty )$
• D. $b\in (-\frac{1}{27},\infty )$

Answer 1

Answer: (A), (D)

The correct options are
A $a\in (-\infty ,\frac{1}{3})$
D $b\in (-\frac{1}{27},\infty )$

Let $x_1,x_2,x_3$ are the roots of $x^3-x^2+ax+b=0$
$x_1+x_2+x_3=1$
$\Rightarrow (A-d)+A+(A+d)=1,(d\ne 0)$
$\Rightarrow A=\frac{1}{3}$

$x_1{x}_2+x_2{x}_3+x_3{x}_1=a$
$\Rightarrow (A-d)A+A(A+d)+(A+d)(A-d)=a$
$\Rightarrow 3A^2-d^2=a$
$\Rightarrow 3{\left(\frac{1}{3}\right)}^2-d^2=a$
$\Rightarrow a=\frac{1}{3}-d^2<\frac{1}{3}$
$\Rightarrow a\in (-\infty ,\frac{1}{3})$

$x_1{x}_2{x}_3=-b$
$\Rightarrow (A-d)A(A+d)=-b$
$\Rightarrow \frac{1}{3}(\frac{1}{9}-d^2)=-b$
$\Rightarrow b=\frac{d^2}{3}-\frac{1}{27}>\frac{-1}{27}$
$\Rightarrow b\in (-\frac{1}{27},\infty )$