Question

If the polynomial $P\left(x\right)=24x^4+{\lambda }_1{x}^3+{\lambda }_2{x}^2+{\lambda }_{3}x+1$, where ${\lambda }_1,{\lambda }_2,{\lambda }_3\in R$ has four positive real roots $\alpha ,\beta ,\gamma ,\delta$ such that $\alpha +2\beta +3\gamma +4\delta =4$, then

• A. ${\lambda }_1+{\lambda }_2+{\lambda }_3=-25$
• B. ${\lambda }_1=-50$
• C. ${\lambda }_3=-10$
• D. ${\lambda }_3=25$

Answer 1

Answer: (A), (B), (C)

The correct options are
A ${\lambda }_1+{\lambda }_2+{\lambda }_3=-25$
B ${\lambda }_1=-50$
C ${\lambda }_3=-10$

$\alpha ,\beta ,\gamma ,\delta$ are roots of the polynomial $P\left(x\right)=24x^4+{\lambda }_1{x}^3+{\lambda }_2{x}^2+{\lambda }_{3}x+1$
$\therefore \alpha \beta \gamma \delta =\frac{1}{24}$ and $\alpha +2\beta +3\gamma +4\delta =4$
We know,
$A.M\ge G.M.\Rightarrow \frac{\alpha +2\beta +3\gamma +4\delta }{4}\ge (\alpha \cdot 2\beta \cdot 3\gamma \cdot 4\delta {)}^{1/4}$
But, $\frac{\alpha +2\beta +3\gamma +4\delta }{4}=(\alpha \cdot 2\beta \cdot 3\gamma \cdot 4\delta {)}^{1/4}$
Therfore, $\alpha =2\beta =3\gamma =4\delta =1$
$\Rightarrow \alpha =1,\beta =\frac{1}{2},\gamma =\frac{1}{3},\delta =\frac{1}{4}$
$\alpha +\beta +\gamma +\delta =\frac{-{\lambda }_1}{24}\Rightarrow {\lambda }_1=-50$
$\alpha \beta \gamma +\beta \gamma \delta +\gamma \delta \alpha +\delta \alpha \beta =\frac{-{\lambda }_3}{24}\Rightarrow \alpha \beta \gamma \delta (\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }+\frac{1}{\delta })=\frac{-{\lambda }_3}{24}\Rightarrow \frac{1}{24}(1+2+3+4)=\frac{-{\lambda }_3}{24}\Rightarrow {\lambda }_3=-10$
$\because \alpha$ is a root of the given polynomial.
$\therefore 24+{\lambda }_1+{\lambda }_2+{\lambda }_3+1=0\Rightarrow {\lambda }_1+{\lambda }_2+{\lambda }_3=-25$
$\therefore {\lambda }_2=35$