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HMMT February

Algebra Difficulty 5.0 AIME Find the answer

Suppose f(x)f(x) is a rational function such that 3f(1x)+2f(x)x=x23f\left(\frac{1}{x}\right) + \frac{2f(x)}{x} = x^{2} for x≠0x \neq 0. Find f(−2)f(-2).

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let x=−12x = \frac{-1}{2}. Then 3f(−2)+2f(−12)−12=14⇒3f(−2)−4f(−12)=14\begin{align*} 3f(-2) + \frac{2f\left(\frac{-1}{2}\right)}{\frac{-1}{2}} = & \frac{1}{4} \\ \Rightarrow 3f(-2) - 4f\left(\frac{-1}{2}\right) & = \frac{1}{4} \tag{1} \end{align*} Let x=−2x = -2. Then 3f(−12)+2f(−2)−2=4⇒3f(−12)−f(−2)=4\begin{align*} & 3f\left(\frac{-1}{2}\right) + \frac{2f(-2)}{-2} = 4 \\ \Rightarrow 3f\left(\frac{-1}{2}\right) - f(-2) & = 4 \tag{2} \end{align*} Solving this system of equations {(1),(2)}\{(1),(2)\} for f(−2)f(-2) yields f(−2)=6720f(-2) = \frac{67}{20}.

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