Question

If a polynomial $f\left(x\right)=4x^4-a{x}^3+b{x}^2-cx+5$, $(a,b,c\in R)$ has four positive real zeros $r_1,r_2,r_3,r_4$ such that $\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}=1$, then value of $a$ is

• A. $20$
• B. $21$
• C. $19$
• D. $22$

Answer 1

Answer: (C)

$19$

We know that $A.M\ge G.M$
$\frac{\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}}{4}\ge {\left(\frac{r_1}{2}\frac{r_2}{4}\frac{r_3}{5}\frac{r_4}{8}\right)}^{\frac{1}{4}}$
We know from the given equation that $\left(r_1{r}_2{r}_3{r}_4\right)=\frac{5}{4}$
Substituting we get.
$G.M.={\left(\frac{1}{64\cdot \left(4\right)}\right)}^{\frac{1}{4}}=\frac{1}{4}$
Hence, $A.M=G.M(\because \frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}=1)$
$\therefore$ The numbers will be equal.
$\frac{r_1}{2}=\frac{r_2}{4}=\frac{r_3}{5}=\frac{r_4}{8}=\frac{1}{4}$
Hence,
$r_1=\frac{1}{2},r_2=1,r_3=\frac{5}{4},r_4=2\to \left(1\right)$
From the equation given , we get sum of the roots as
$r_1+r_2+r_3+r_4=\frac{a}{4}\to \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we get
$\frac{19}{4}=\frac{a}{4}$
$\Rightarrow a=19$