2003 AMC 12B 第 20 题

先试着解答 2003 AMC 12B 第 20 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2003 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

20.

图中给出了 f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d 的部分图像。bb 是多少?
Part of the graph of f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d is shown. What is b?b?
−4-4
−2-2
00
22
44
答案:B
知识点:多项式换元法
小提示:
从图像读出 f(−1)=0f(-1) = 0,f(1)=0f(1) = 0,和 f(0)=2f(0) = 2
Read off f(−1)=0,f(-1) = 0, f(1)=0,f(1) = 0, and f(0)=2f(0) = 2 from the graph
大提示:
把 f(1)+f(−1)f(1) + f(-1) 相加会消去 aa 和 cc,留下 2b+2d=02b + 2d = 0
Adding f(1)+f(−1)f(1) + f(-1) eliminates aa and c,c, leaving 2b+2d=02b + 2d = 0
解答:
图像经过 (−1,0)(-1, 0),(1,0)(1, 0),和 (0,2)(0, 2)。所以 f(0)=d=2f(0) = d = 2。
将两式相加:f(1)+f(−1)=(a+b+c+d)+(−a+b−c+d)=2b+2d=0, \begin{aligned} &f(1) + f(-1) \\ &= (a + b + c + d) \\ &\quad {}+ (-a + b - c + d) \\ &= 2b + 2d = 0\text{,} \end{aligned} 因此 b=−d=−2b = -d = -2。
因此,正确答案是 B。
The graph passes through (−1,0),(-1, 0), (1,0),(1, 0), and (0,2).(0, 2). So f(0)=d=2.f(0) = d = 2.
Adding f(1)+f(−1)=(a+b+c+d)+(−a+b−c+d)=2b+2d=0, \begin{aligned} &f(1) + f(-1) \\ &= (a + b + c + d) \\ &\quad {}+ (-a + b - c + d) \\ &= 2b + 2d = 0, \end{aligned} so b=−d=−2.b = -d = -2.
Thus, the correct answer is B.
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