AMC12 2009 B

AMC12 2009 B · Q24

AMC12 2009 B · Q24. It mainly tests Functions basics, Trigonometry (basic).

For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$? Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.

在 $[0,\pi]$ 内，有多少个 $x$ 满足 $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$？注意：函数 $\sin^{ - 1} = \arcsin$ 与 $\cos^{ - 1} = \arccos$ 表示反三角函数。

(A) 3 3

(B) 4 4

(D) 6 6

(E) 7 7

Answer

Correct choice: (B)

正确答案：（B）

Solution

First of all, we have to agree on the range of $\sin^{-1}$ and $\cos^{-1}$. This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: $\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2$ and $\forall x: 0\leq \cos^{-1}(x) \leq \pi$. Hence we get that $\forall x\in[0,\pi]: \cos^{ - 1}(\cos x) = x$, thus our equation simplifies to $\sin^{ - 1}(\sin 6x) = x$. Consider the function $f(x) = \sin^{ - 1}(\sin 6x) - x$. We are looking for roots of $f$ on $[0,\pi]$. By analyzing properties of $\sin$ and $\sin^{-1}$ (or by computing the derivative of $f$) one can discover the following properties of $f$: - $f(0)=0$. - $f$ is increasing and then decreasing on $[0,\pi/6]$. - $f$ is decreasing and then increasing on $[\pi/6,2\pi/6]$. - $f$ is increasing and then decreasing on $[2\pi/6,3\pi/6]$. For $x=\pi/6$ we have $f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0$. Hence $f$ has exactly one root on $(0,\pi/6)$. For $x=2\pi/6$ we have $f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0$. Hence $f$ is negative on the entire interval $[\pi/6,2\pi/6]$. Now note that $\forall t: \sin^{-1}(t) \leq \pi/2$. Hence for $x > 3\pi/6$ we have $f(x) < 0$, and we can easily check that $f(3\pi/6)<0$ as well. Thus the only unknown part of $f$ is the interval $(2\pi/6,3\pi/6)$. On this interval, $f$ is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval. To prove that there are two roots, it is enough to find any $x$ from this interval such that $f(x)>0$. A good guess is its midpoint, $x=5\pi/12$, where the function $\sin^{-1}(\sin 6x)$ has its local maximum. We can evaluate: $f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0$. Summary: The function $f$ has $\boxed{\textbf{(B) }4}$ roots on $[0,\pi]$: the first one is $0$, the second one is in $(0,\pi/6)$, and the last two are in $(2\pi/6,3\pi/6)$. Actual solutions are $x=0$, $x=\pi/7$, $x=2\pi/5$, and $x=3\pi/7$.

首先需要约定 $\sin^{-1}$ 与 $\cos^{-1}$ 的值域。题目未写明，我们采用最常见的定义：$\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2$ 且 $\forall x: 0\leq \cos^{-1}(x) \leq \pi$。因此对任意 $x\in[0,\pi]$，有 $\cos^{ - 1}(\cos x) = x$，方程化为 $\sin^{ - 1}(\sin 6x) = x$。考虑函数 $f(x) = \sin^{ - 1}(\sin 6x) - x$，我们要找 $f$ 在 $[0,\pi]$ 上的零点。通过分析 $\sin$ 与 $\sin^{-1}$ 的性质（或计算 $f$ 的导数）可得到 $f$ 的如下性质： - $f(0)=0$。 - $f$ 在 $[0,\pi/6]$ 上先增后减。 - $f$ 在 $[\pi/6,2\pi/6]$ 上先减后增。 - $f$ 在 $[2\pi/6,3\pi/6]$ 上先增后减。当 $x=\pi/6$ 时，$f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0$，因此 $f$ 在 $(0,\pi/6)$ 上恰有一个零点。当 $x=2\pi/6$ 时，$f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0$，因此 $f$ 在整个区间 $[\pi/6,2\pi/6]$ 上为负。又注意到对任意 $t$，有 $\sin^{-1}(t) \leq \pi/2$。因此当 $x > 3\pi/6$ 时，$f(x) < 0$，并且也容易验证 $f(3\pi/6)<0$。因此 $f$ 唯一尚不确定的部分是区间 $(2\pi/6,3\pi/6)$。在该区间上，$f$ 在两端都为负，且先增后减，因此可能有 $0,1,$ 或 $2$ 个零点。要证明有两个零点，只需在该区间内找一个点使得 $f(x)>0$。取其中点 $x=5\pi/12$ 是一个好选择，此处 $\sin^{-1}(\sin 6x)$ 取得局部最大值。计算得 $f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0$。总结：函数 $f$ 在 $[0,\pi]$ 上共有 $\boxed{\textbf{(B) }4}$ 个零点：第一个是 $0$，第二个在 $(0,\pi/6)$，最后两个在 $(2\pi/6,3\pi/6)$。实际解为 $x=0$，$x=\pi/7$，$x=2\pi/5$，以及 $x=3\pi/7$。

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