Question

If all roots of the equation $f\left(x\right)=x^6-12x^5+b{x}^4+c{x}^3+d{x}^2+ex+64=0$ are ;positive, then which has the greatest numerical (absolute) value

• A. $b$
• B. $c$
• C. $d$
• D. $e$

Answer 1

Answer: (C)

Given $f\left(x\right)=x^6-12x^5+b{x}^4+c{x}^3+d{x}^2+ex+64=0$ .
Let the roots of the equation be $x_1,x_2,x_3,x_4,x_5\&x_6$ .
So, sum of roots $\left(x_1+x_2+x_3+x_4+x_5+x_6\right)=\mathrm{12...}\left(1\right)$
and product of roots $\left(x_1.x_2.x_3.x_4.x_5.x_6\right)=\mathrm{64...}\left(2\right)$
Since, all the roots are positive, so applying $AM-GM$ inequality, we get
$\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}\ge {\left(x_1.x_2.x_3.x_4.x_5.x_6\right)}^{\frac{1}{6}}$
$\Rightarrow \frac{12}{6}\ge {\left(64\right)}^{\frac{1}{6}}$ (from equations $\left(1\right)\&\left(2\right)$
$\Rightarrow 2\ge 2$
$\Rightarrow AM=GM$
This is possible only when $x_1=x_2=x_3=x_4=x_5=x_6$ .
So, from equation $\left(2\right)$ , we get
${\left(x_1\right)}^6=64=2^6\Rightarrow x_1=2$ .
So, the given equation is equivalent to $f\left(x\right)={\left(x-2\right)}^6$ .
$\Rightarrow f\left(x\right)=\_{}^6{C}_0^{}{x}^6{\left(-2\right)}^0+\_{}^6{C}_1^{}{x}^5{\left(-2\right)}^1+\_{}^6{C}_2^{}{x}^4{\left(-2\right)}^2+\_{}^6{C}_3^{}{x}^3{\left(-2\right)}^3$
$+\_{}^6{C}_4^{}{x}^2{\left(-2\right)}^4+\_{}^6{C}_5^{}{x}^1{\left(-2\right)}^5+\_{}^6{C}_6^{}{x}^0{\left(-2\right)}^6=0$
(using binomial expansion)
$\Rightarrow f\left(x\right)=x^6-12x^5+60x^4-160x^3+240x^2-192x+64=0$
Comparing with the given function, we get maximum absolute value for $d$ i.e. $240$ .