Question

设函数 $f$ 满足 $f\!\left(\dfrac{x}{3}\right) = x^2 + x + 1$ 。求所有满足 $f(3z) = 7$ 的 $z$ 的和。

Let $f$ be a function for which $f\!\left(\dfrac{x}{3}\right) = x^2 + x + 1.$ Find the sum of all values of $z$ for which $f(3z) = 7.$

$-\dfrac{1}{3}$

$-\dfrac{1}{9}$

$0$

$\dfrac{5}{9}$

$\dfrac{5}{3}$

令 $\dfrac{x}{3} = 3z$ 得到 $x = 9z$ ，所以 $\begin{aligned} f(3z) &= (9z)^2 + 9z + 1 \\ &= 81z^2 + 9z + 1 \\ &= 7. \end{aligned}$

整理得 $81z^2 + 9z - 6 = 0$ 。

由根和公式，所有 $z$ 的和为 $-\dfrac{9}{81} = -\dfrac{1}{9}$ 。

因此，正确答案是 B。

Setting $\dfrac{x}{3} = 3z$ gives $x = 9z,$ so $\begin{aligned} f(3z) &= (9z)^2 + 9z + 1 \\ &= 81z^2 + 9z + 1 \\ &= 7. \end{aligned}$

This rearranges to $81z^2 + 9z - 6 = 0.$

By the sum-of-roots formula, the sum of the values of $z$ is $-\dfrac{9}{81} = -\dfrac{1}{9}.$

Thus, the correct answer is B.