AMC10 2025 A

AMC10 2025 A · Q9

AMC10 2025 A · Q9. It mainly tests Quadratic equations, Algebra misc.

Let $f(x) = 100x^3 - 300x^2 + 200x$. For how many real numbers $a$ does the graph of $y = f(x - a)$ pass through the point $(1, 25)$?

设$f(x) = 100x^3 - 300x^2 + 200x$。有几个实数$a$使得$y = f(x - a)$的图像经过点$(1, 25)$？

(A) $1 $1

(B) $2 $2

(D) $4 $4

(E) more than $4 more than $4

Answer

Correct choice: (C)

正确答案：（C）

Solution

Substitute $1 - a$ for $x$ and set this expression equal to $25.$ The problem boils down to finding how many real roots \[100(1-a)^3 - 300(1-a)^2 + 200(1-a) = 25\] has. We further simplify this expression and create a function $f(x):$ \[f(x) = -100a^3 + 100a - 25\] Using Descarte's Rule of Signs we get: Sign changes for $f(x)$ (possible number of positive roots): 2 \[f(-x) = +100a^3 - 100a - 25\] Sign changes for $f(-x)$ (possible number of negative roots): 1 Possibilities for roots: 1) $2$ positive roots, $1$ negative root 2) $0$ positive roots, $1$ negative root, $2$ imaginary roots So which one is it? We know if the function changes sign between an interval, then a root exists in that interval (Intermediate Value Theorem). From $a = 0$ to $\frac{1}{2},$ the function changes sign because $f(0) = -25$ while $f(\frac{1}{2}) = +\frac{25}{2}$, so a positive root exists. This eliminates the second possibility, implying that there must be $2$ positive and $1$ negative roots. So the answer is $2 + 1 = \boxed{\textbf{(C) } 3}.$

将$x$替换为$1 - a$并设此表达式等于$25$。问题归结为求方程\[100(1-a)^3 - 300(1-a)^2 + 200(1-a) = 25\]有多少实根。我们进一步简化此表达式并创建函数$f(x)$： \[f(x) = -100a^3 + 100a - 25\] 使用笛卡尔符号法则我们得到： $f(x)$的符号变化（可能的正根数）：2 \[f(-x) = +100a^3 - 100a - 25\] $f(-x)$的符号变化（可能的负根数）：1 根的可能性： 1) $2$个正根，$1$个负根 2) $0$个正根，$1$个负根，$2$个虚根那么是哪一种？我们知道如果函数在区间内改变符号，则该区间内存在根（中间值定理）。从$a = 0$到$\frac{1}{2}$，函数改变符号因为$f(0) = -25$而$f(\frac{1}{2}) = +\frac{25}{2}$，所以存在一个正根。这排除了第二种可能性，意味着必须有$2$个正根和$1$个负根。所以答案是$2 + 1 = \boxed{\textbf{(C) } 3}$。

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