题目

Problem

Figure 2 shows a sketch of the curve with equation $y=f(x)$ , where

\begin{align*} f(x)=\cos 2x^\circ,\qquad 0\le x\le k. \end{align*}

Figure 2

The point $Q$ and the point $R(k,0)$ lie on the curve and are shown in Figure 2.

(a) State

(i) the coordinates of $Q$ ,

(ii) the value of $k$ .

(3)

(b) Given that there are exactly two solutions to the equation

\begin{align*} \cos 2x^\circ=p \end{align*}

in the region $0\le x\le k$ , find the range of possible values for $p$ .

(2)

解答

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$\cos 2x^\circ$ 的周期是 $180^\circ$ 。最小值 $-1$ 发生在 $2x=180^\circ$ ，即 $x=90^\circ$ 。右侧的 $x$ 轴交点满足 $\cos 2x^\circ=0$ 。

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For the minimum point $Q$ ,

\begin{align*} 2x=&\,180^\circ\\ x=&\,90. \end{align*}

\begin{align*} Q=(90,-1). \end{align*}

For $R(k,0)$ ,

\begin{align*} \cos 2k^\circ=&\,0. \end{align*}

The right-hand intercept shown after $x=180$ is when

\begin{align*} 2k=&\,450^\circ\\ k=&\,225. \end{align*}

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在 $0\le x\le225$ 上，图像从 $1$ 降到 $-1$ ，再升到 $1$ ，最后降到 $0$ 。水平线 $y=p$ 要刚好交两次：可以在 $-1<p<0$ ，也可以是 $p=1$ 。

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From the shape of $y=\cos2x^\circ$ on $0\le x\le225$ :

\begin{align*} -1<p<0 \end{align*}

gives exactly two intersections.

Also, when

\begin{align*} p=1, \end{align*}

the line $y=p$ meets the curve at $x=0$ and $x=180$ , so there are exactly two solutions.

Therefore

\begin{align*} -1<p<0 \quad\text{or}\quad p=1. \end{align*}