题目

Problem

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

Figure 2

The curve passes through the points $(-5,0)$ and $(0,-3)$ , and touches the $x$ -axis at the point $(2,0)$ .

(i) On separate diagrams, sketch the curve with equation

(a) $y=f(x+2)$ ,

(b) $y=f(-x)$ .

On each diagram, show clearly the coordinates of each point at which the curve crosses or touches the coordinate axes.

(6)

Figure 3 shows a sketch of the curve with equation

Figure 3

\begin{align*} y=k\cos\left(x+\frac{\pi}{6}\right),\qquad 0\leq x\leq 2\pi. \end{align*}

The curve meets the $y$ -axis at the point $(0,\sqrt3)$ and passes through the points $(p,0)$ and $(q,0)$ .

(ii) Find the value of $k$ and the exact values of $p$ and $q$ .

(3)

解答

(i)(a)

解法一

思路

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$y=f(x+2)$ 表示原图向左平移 $2$ 个单位。每个点的 $x$ 坐标减去 $2$ ， $y$ 坐标不变。

答题过程

展开

The transformation $y=f(x+2)$ is a translation 2 units to the left.

Therefore

\begin{align*} (-5,0)&\mapsto(-7,0),\\ (0,-3)&\mapsto(-2,-3),\\ (2,0)&\mapsto(0,0). \end{align*}

The sketch should have the same shape as the original curve, touch the $x$ -axis at $(0,0)$ , cross the $x$ -axis at $(-7,0)$ , and pass through $(-2,-3)$ .

(i)(b)

解法一

思路

展开

$y=f(-x)$ 是关于 $y$ 轴的反射。每个点的 $x$ 坐标变号， $y$ 坐标不变。

答题过程

展开

The transformation $y=f(-x)$ is a reflection in the $y$ -axis.

Therefore

\begin{align*} (-5,0)&\mapsto(5,0),\\ (0,-3)&\mapsto(0,-3),\\ (2,0)&\mapsto(-2,0). \end{align*}

The sketch should touch the $x$ -axis at $(-2,0)$ , cross the $x$ -axis at $(5,0)$ , and pass through $(0,-3)$ .

(ii)

解法一

思路

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先把 $x=0$ 和 $y=\sqrt3$ 代入，求出 $k$ 。再令 $y=0$ ，也就是令余弦部分等于 $0$ ，找出区间 $0\leq x\leq2\pi$ 内的两个解。

答题过程

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At $x=0$ ,

\begin{align*} \sqrt3=&\,k\cos\left(\frac{\pi}{6}\right)\\ =&\,k\cdot\frac{\sqrt3}{2}. \end{align*}

\begin{align*} k=2. \end{align*}

For the $x$ -intercepts,

\begin{align*} 2\cos\left(x+\frac{\pi}{6}\right)=&\,0\\ \cos\left(x+\frac{\pi}{6}\right)=&\,0. \end{align*}

In $0\leq x\leq2\pi$ ,

\begin{align*} x+\frac{\pi}{6} =\frac{\pi}{2} \quad\text{or}\quad x+\frac{\pi}{6} =\frac{3\pi}{2}. \end{align*}

Hence

\begin{align*} x=\frac{\pi}{3} \quad\text{or}\quad x=\frac{4\pi}{3}. \end{align*}

Therefore

\begin{align*} k=2,\qquad p=\frac{\pi}{3},\qquad q=\frac{4\pi}{3}. \end{align*}

IAL 2020 Oct Q5

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)

解法一

思路

答题过程