Question

If a polynomial $f\left(x\right)=4x^4-a{x}^3+b{x}^2-cx+5$, $(a,b,cϵR)$ has four positive real roots $r_1,r_2,r_3,r_4$ such that $\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}=1$ Then value of a will be

• A. 20
• B. 21
• C. 19
• D. 22

Answer 1

The correct option is C 19

From given information
$\prod r_1=\frac{5}{4}$
$A.M\ge G.M$
$\frac{\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}}{4}\ge {\left(\frac{r_1{r}_2{r}_3{r}_4}{\mathrm{2.4.5.8}}\right)}^{\frac{1}{4}}\ge {\left(\frac{1}{64\times 4}\right)}^{1/4}=\frac{1}{4}$
$A.M=G.M$
$\therefore$ Number are all equal
$\therefore \frac{r_{1}}{2}=\frac{r_{2}}{4}=\frac{r_{3}}{5}=\frac{r_{4}}{8}=\frac{1}
{4}$
$\therefore r_1=\frac{1}{2};r_2=1;r_3=\frac{5}{4};r_4=2;$
$\frac{a}{4}=\sum r_1\Rightarrow \frac{a}{4}=\frac{19}{4}\Rightarrow a=19$