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China Mathematical Competition

Algebra Difficulty 3.8 AMC 10/12 Find the answer China

The minimum value of $f(x) = \frac{5-4x+x^2}{2-x}$ for $x \in (-\infty, 2)$ is ( ).

This was a multiple-choice question, but the options didn't survive into the source we have. The answer given is C, and the solution below works it through.

Solution

Let $x < 2 \Rightarrow 2-x > 0$ . Then
$\begin{aligned} f(x) &= \frac{1 + (4 - 4x + x^2)}{2-x} = \frac{1}{2-x} + (2-x) \\ &\geq 2 \times \sqrt{\frac{1}{2-x} \times (2-x)} = 2. \end{aligned}$
The equality holds if and only if $\frac{1}{2-x} = 2-x$ , and it is so when $x = 1 \in (-\infty, 2)$ . This means that $f$ reaches the minimum value 2 at $x = 1$ .

Answer: C

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.