Question

If $\lambda ,\mu$ be a real numbers such that $x^3-\lambda x^2+\mu x-6=0$ has its roots real and positive then the minimum value of $\mu$ is

• A. $3{\times }^3\sqrt{36}$
• B. $11$
• C. $0$
• D. none of these

Answer 1

The correct option is A $3{\times }^3\sqrt{36}$

Let $\alpha ,\beta ,\gamma$ are the roots of $x^3-\lambda x^2+\mu x-6=0$
given that roots are real and positive.
$\alpha +\beta +\gamma =\lambda$
$\alpha \beta +\beta \gamma +\gamma \alpha =\mu$
$\alpha \beta \gamma =6$
using the result $A.M\ge G.M$
$\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{3}\ge {\left({\alpha }^2{\beta }^2{\gamma }^2\right)}^{\frac{1}{3}}$
$\Rightarrow \frac{\mu }{3}\ge {\left(36\right)}^{\frac{1}{3}}$
$\Rightarrow \mu \ge 3\times {\left(36\right)}^{\frac{1}{3}}$
$\therefore$The minimum value of $\mu$ is $3\times {\left(36\right)}^{\frac{1}{3}}$