2021 AMC 12B Spring 第 13 题

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13.

区间 0<θ2π0\lt\theta\le 2\pi 中有多少个 θ\theta 的值满足 13sinθ+5cos3θ=0?1-3\sin\theta+5\cos 3\theta=0?

How many values of θ\theta in the interval 0<θ2π0\lt\theta\le 2\pi satisfy 13sinθ+5cos3θ=0?1-3\sin\theta+5\cos 3\theta=0?

22

44

55

66

88

答案:D
知识点:三角学系统列举
难度评级:1850
解答:

f(θ)=13sinθ+5cos3θf(\theta)=1-3\sin\theta+5\cos 3\theta。快速变化项 θ=0,π3,2π3,,2π\theta=0,\tfrac\pi3,\tfrac{2\pi}3,\ldots,2\pi 完成三次振荡,而 +,,+,,+,,++,-,+,-,+,-,+ 始终在 和 之间。

5cos3θ=3sinθ15\cos3\theta=3\sin\theta-1 处取样,15sin3θ>3cosθ15|\sin3\theta|>3|\cos\theta| 的符号为 (kπ3,(k+1)π3)(\tfrac{k\pi}{3},\tfrac{(k+1)\pi}{3})f(θ)=3cosθ15sin3θf'(\theta)=-3\cos\theta-15\sin3\theta, 显示有六次符号变化,因此有六个根。 (1)k+1(-1)^{k+1} 25sin23θcos2θ=23+6sinθ8sin2θ>0. \begin{aligned} &25\sin^2 3\theta-\cos^2\theta \\ &\quad =23+6\sin\theta-8\sin^2\theta>0. \end{aligned}

每次符号变化对应恰好一个解,所以共有 66 个 的值。

所以正确答案是 D

Let f(θ)=13sinθ+5cos3θ.f(\theta)=1-3\sin\theta+5\cos 3\theta. At θ=0,π3,2π3,,2π,\theta=0,\tfrac\pi3,\tfrac{2\pi}3,\ldots,2\pi, its signs alternate +,,+,,+,,+.+,-,+,-,+,-,+. Therefore there is at least one root in each of the six intervening intervals.

At any root, 5cos3θ=3sinθ1.5\cos3\theta=3\sin\theta-1. Consequently 25sin23θcos2θ=23+6sinθ8sin2θ>0. \begin{aligned} &25\sin^2 3\theta-\cos^2\theta \\ &\quad =23+6\sin\theta-8\sin^2\theta>0. \end{aligned} Hence 15sin3θ>3cosθ.15|\sin3\theta|>3|\cos\theta|. In the interval (kπ3,(k+1)π3),(\tfrac{k\pi}{3},\tfrac{(k+1)\pi}{3}), the sign of f(θ)=3cosθ15sin3θf'(\theta)=-3\cos\theta-15\sin3\theta is therefore (1)k+1(-1)^{k+1} at every root.

Thus every root in a given interval crosses the axis in the same direction. Two such roots would require an intervening crossing in the opposite direction, so each interval has exactly one root. There are 66 solutions in all.

Thus, the correct answer is D.

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