Question

Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1\lt a\lt b\lt c.$ What is $g(1)$ in terms of $a, b,$ and $c?$

$\dfrac{1+a+b+c}{c}$

$1+a+b+c$

$\dfrac{1+a+b+c}{c^2}$

$\dfrac{a+b+c}{c^2}$

$\dfrac{1+a+b+c}{a+b+c}$

Answer: A

Concepts:Vieta’s Formulas polynomial

Difficulty rating: 1720

Solution:

Let $f$ have roots $r,s,t.$ Since $g$ is monic with roots $\tfrac1r,\tfrac1s,\tfrac1t,$ $\begin{aligned} g(1) &= \left(1-\tfrac1r\right)\left(1-\tfrac1s\right) \\ &\quad {}\cdot \left(1-\tfrac1t\right) \\ &= \dfrac{(r-1)(s-1)(t-1)}{rst}. \end{aligned}$

Now $f(1)=(1-r)(1-s)(1-t)$ $=1+a+b+c,$ so $(r-1)(s-1)(t-1)$ $=-(1+a+b+c).$ Also $rst=-c.$

Therefore $g(1)$ $=\dfrac{-(1+a+b+c)}{-c}$ $=\dfrac{1+a+b+c}{c}.$

Thus, the correct answer is A.

2021 AMC 12B Spring Problem 16

Answer: A

Difficulty rating: 1720

Solution:

Problem 16 in Other Years