Question

Suppose $a$ and $b$ are real no. such that the roots of the cubic equation $a{x}^3-x^2+bx+1=0$ are all positive real no. then
$0<3ab\le 1$

• A. True
• B. False

Answer 1

The correct option is A True

The given equation is $a{x}^3-x^2+bx+1=0$

Let $\alpha ,\beta ,\gamma$ be the roots of the given equation.

We have

$\alpha +\beta +\gamma =\frac{1}{a}$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{b}{a}$

$\alpha \beta \gamma =\frac{1}{a}$

It follows that $a,b$ are positive. we obtain

$\frac{3b}{a}=3(\alpha \beta +\beta \gamma +\gamma \alpha )\le (\alpha +\beta +\gamma {)}^2=\frac{1}{a^2}$

Which gives, $0<3ab\le 1.$