Question

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

• A. $b$
• B. $c$
• C. $d$
• D. $e$

Answer 1

Answer: (C)

Given $f\left(x\right)=x^{6}-12x^{5}+bx^{4}+cx^{3}+dx^{2}+ex+64=0$ .
Let the roots of the equation be $x_{1},x_{2},x_{3},x_{4},x_{5}\&x_{6}$ .
So, sum of roots $\left(x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6}\right)=12...\left(1\right)$
and product of roots $\left(x_{1} . x_{2} . x_{3} . x_{4} . x_{5} . x_{6}\right)=64...\left(2\right)$
Since, all the roots are positive, so applying $AM-GM$ inequality, we get
$\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6}}{6}\geq \left(x_{1} . x_{2} . x_{3} . x_{4} . x_{5} . x_{6}\right)^{\frac{1}{6}}$
$\Rightarrow \frac{12}{6}\geq \left(64\right)^{\frac{1}{6}}$ (from equations $\left(1\right)\&\left(2\right)$
$\Rightarrow 2\geq 2$
$\Rightarrow AM=GM$
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. $240$ .