AMC12 2021 A

AMC12 2021 A · Q18

AMC12 2021 A · Q18. It mainly tests Functions basics, Primes & prime factorization.

Let $f$ be a function defined on the set of positive rational numbers with the property that $f(a\cdot b)=f(a)+f(b)$ for all positive rational numbers $a$ and $b$. Suppose that $f$ also has the property that $f(p)=p$ for every prime number $p$. For which of the following numbers $x$ is $f(x)<0$?

设 $f$ 是定义在正有理数集上的函数，具有性质 $f(a\cdot b)=f(a)+f(b)$，对所有正有理数 $a$ 和 $b$。假设 $f$ 还具有性质 $f(p)=p$，对每个质数 $p$。对于下列哪个数 $x$ 有 $f(x)<0$？

(A) \frac{17}{32} \frac{17}{32}

(B) \frac{11}{16} \frac{11}{16}

(D) \frac76 \frac76

(E) \frac{25}{11} \frac{25}{11}

Answer

Correct choice: (D)

正确答案：（D）

Solution

From the answer choices, note that \begin{align*} f(25)&=f\left(\frac{25}{11}\cdot11\right) \\ &=f\left(\frac{25}{11}\right)+f(11) \\ &=f\left(\frac{25}{11}\right)+11. \end{align*} On the other hand, we have \begin{align*} f(25)&=f(5\cdot5) \\ &=f(5)+f(5) \\ &=5+5 \\ &=10. \end{align*} Equating the expressions for $f(25)$ produces \[f\left(\frac{25}{11}\right)+11=10,\] from which $f\left(\frac{25}{11}\right)=-1.$ Therefore, the answer is $\boxed{\textbf{(E) }\frac{25}{11}}.$

从选项中，注意到 \begin{align*} f(25)&=f\left(\frac{25}{11}\cdot11\right) \\ &=f\left(\frac{25}{11}\right)+f(11) \\ &=f\left(\frac{25}{11}\right)+11. \end{align*} 另一方面，我们有 \begin{align*} f(25)&=f(5\cdot5) \\ &=f(5)+f(5) \\ &=5+5 \\ &=10. \end{align*} 将 $f(25)$ 的两个表达式等价，得 \[f\left(\frac{25}{11}\right)+11=10,\] 由此 $f\left(\frac{25}{11}\right)=-1$。因此答案是 $\boxed{\textbf{(E) }\frac{25}{11}}$。

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