题目

Problem

(i) Figure 3 shows part of the graph of the trigonometric function with equation $y=f(x)$ .

Figure 3

(a) Write down an expression for $f(x)$ .

(2)

On a separate diagram,

(b) sketch, for $-2\pi<x<2\pi$ , the graph of the curve with equation

\begin{align*} y=f\left(x+\frac{\pi}{4}\right). \end{align*}

Show clearly the coordinates of all the points where the curve intersects the coordinate axes.

(3)

(ii) Figure 4 shows part of the graph of the trigonometric function with equation $y=g(x)$ .

Figure 4

(a) Write down an expression for $g(x)$ .

(2)

On a separate diagram,

(b) sketch, for $-2\pi<x<2\pi$ , the graph of the curve with equation $y=g(x)-2$ .

Show clearly the coordinates of the $y$ intercept.

(2)

解答

(i)(a)

解法一

思路

展开

图像最大值是 $3$ ，最小值是 $-3$ ，且在 $x=0$ 处达到最大值，所以是 $3\cos x$ 。

答题过程

展开

\begin{align*} f(x)=3\cos x. \end{align*}

(i)(b)

解法一

思路

展开

$f\left(x+\frac\pi4\right)=3\cos\left(x+\frac\pi4\right)$ ，表示原图向左平移 $\frac\pi4$ 。截距可直接令 $y=0$ 或 $x=0$ 。

答题过程

展开

The curve is

\begin{align*} y=3\cos\left(x+\frac{\pi}{4}\right). \end{align*}

For $x$ -intercepts,

\begin{align*} \cos\left(x+\frac{\pi}{4}\right)=0. \end{align*}

\begin{align*} x+\frac{\pi}{4}=\frac{\pi}{2}+n\pi, \end{align*}

and hence

\begin{align*} x=\frac{\pi}{4}+n\pi. \end{align*}

For $-2\pi<x<2\pi$ , the $x$ -intercepts are

\begin{align*} \left(-\frac{7\pi}{4},0\right),\quad \left(-\frac{3\pi}{4},0\right),\quad \left(\frac{\pi}{4},0\right),\quad \left(\frac{5\pi}{4},0\right). \end{align*}

The $y$ -intercept is found by setting $x=0$ :

\begin{align*} y =&\,3\cos\frac{\pi}{4}\\ =&\,\frac{3\sqrt2}{2}. \end{align*}

So the $y$ -intercept is

\begin{align*} \left(0,\frac{3\sqrt2}{2}\right). \end{align*}

The sketch should be the graph of $3\cos x$ shifted left by $\frac{\pi}{4}$ .

(ii)(a)

解法一

思路

展开

图像是振幅为 $1$ 的 sine curve，并且在 $-2\pi$ 到 $2\pi$ 内完成更多周期；对应 $g(x)=\sin(2x)$ 。

答题过程

展开

\begin{align*} g(x)=\sin(2x). \end{align*}

(ii)(b)

解法一

思路

展开

$y=g(x)-2$ 是把 $y=g(x)$ 整体向下平移 $2$ 。因此 $y$ 截距也比原来低 $2$ 。

答题过程

展开

The curve is

\begin{align*} y=\sin(2x)-2. \end{align*}

At $x=0$ ,

\begin{align*} y=\sin0-2=-2. \end{align*}

So the $y$ -intercept is

\begin{align*} (0,-2). \end{align*}

The sketch should have the same shape as $y=\sin(2x)$ , translated $2$ units down.

IAL 2023 May Q9

题目

解答

(i)(a)

解法一

思路

答题过程

(i)(b)

解法一

思路

答题过程

(ii)(a)

解法一

思路

答题过程

(ii)(b)

解法一

思路

答题过程