1979 AMC 12 Problem 29

Attempt Problem 29 of the 1979 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1979 AMC 12 solutions, or check the answer key.

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29.

For each positive number

x,

let

f(x)= \frac{\begin{gathered} \left(x+\frac1x\right)^6\\ {}-\left(x^6+\frac1{x^6}\right)-2 \end{gathered}} {\begin{gathered} \left(x+\frac1x\right)^3\\ {}+\left(x^3+\frac1{x^3}\right) \end{gathered}}.

The minimum value of

f(x)

is

1

2

3

4

6

Answer: E

Concepts:algebraic manipulation difference of squares AM-GM Inequality

Difficulty rating: 2100

Small hint:

Set

t=x+\frac{1}{x}

and write

x^3+x^{-3}

in terms of

t

Big hint:

Factor the numerator as a difference of squares, then use

x+\frac{1}{x}\ge2

Solution:

Let

t=x+\frac{1}{x}

and

u=x^3+x^{-3}=t^3-3t.

Since

x^6+x^{-6}=u^2-2,

the numerator is

\begin{aligned} t^6-u^2&=(t^3-u)(t^3+u)\\ &=3t(t^3+u). \end{aligned}

The denominator is

t^3+u,

so

f(x)=3t=3(x+\frac{1}{x}).

For

x\gt0,

x+\frac{1}{x}\ge2,

with equality at

x=1.

Thus the minimum is

6.

Therefore, the correct answer is E.