题目

Problem

Figure 2

(a) Express $6x-\dfrac{27}{4}-x^2$ in the form $a+b(x+c)^2$ where $a$ , $b$ and $c$ are constants to be found.

(3)

Figure 2 shows part of a sketch of curve $C_1$ with equation

\begin{align*} y=6x-\frac{27}{4}-x^2. \end{align*}

Given that the point $P$ is the maximum point on $C_1$ ,

(b) state the coordinates of $P$ .

(2)

Figure 2 also shows part of a sketch of curve $C_2$ with equation

\begin{align*} y=\cos(kx) \end{align*}

where $k$ is a constant and $x$ is measured in radians.

Given that $C_1$ and $C_2$ intersect on the $x$ -axis at point $A$ and at point $B$ , as shown in Figure 2,

(ii) state the value of $k$

(iii) state the period of $C_2$ .

(3)

The line segment $L$ joins $P$ and $B$ .

The region $R$ , shown shaded in Figure 2, is bounded by $L$ , $C_1$ and $C_2$ .

(d) Use inequalities to define $R$ .

(5)

解答

(a)

解法一

思路

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先把二次项提负号，再配方。目标是把它写成顶点形式，从而直接读出最大点。

答题过程

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\begin{align*} 6x-\frac{27}{4}-x^2 =&\,-(x^2-6x)-\frac{27}{4}\\ =&\,-\left((x-3)^2-9\right)-\frac{27}{4}\\ =&\,-(x-3)^2+9-\frac{27}{4}\\ =&\,\frac94-(x-3)^2. \end{align*}

\begin{align*} a=\frac94,\qquad b=-1,\qquad c=-3. \end{align*}

(b)

解法一

思路

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由 (a) 的顶点形式

\begin{align*} y=\frac94-(x-3)^2 \end{align*}

可知最大值在 $x=3$ 时取得，此时 $y=\frac94$ 。

答题过程

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From

\begin{align*} y=\frac94-(x-3)^2, \end{align*}

the maximum occurs when

\begin{align*} x=3. \end{align*}

Then

\begin{align*} y=\frac94. \end{align*}

\begin{align*} P=\left(3,\frac94\right). \end{align*}

(c)

解法一

思路

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点 $A,B$ 是两条曲线在 $x$ 轴上的交点，所以先找 $C_1$ 的两个 $x$ 截距。由顶点形式可知截距关于 $x=3$ 对称。

对 $C_2:y=\cos(kx)$ ，它在 $x$ 轴上满足 $\cos(kx)=0$ 。图中 $A$ 和 $B$ 是相邻的两个零点，用它们之间的距离可以确定周期和 $k$ 。

答题过程

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For $C_1$ on the $x$ -axis,

\begin{align*} \frac94-(x-3)^2=&\,0\\ (x-3)^2=&\,\frac94\\ x-3=&\,\pm\frac32. \end{align*}

So the two $x$ -intercepts are

\begin{align*} x=\frac32,\qquad x=\frac92. \end{align*}

Point $B$ is the right-hand intercept, so

\begin{align*} x_B=\frac92. \end{align*}

The adjacent zeros of $C_2$ shown are at

\begin{align*} x=\frac32 \quad\text{and}\quad x=\frac92. \end{align*}

The distance between adjacent zeros is half a period, so

\begin{align*} \frac{\text{period}}{2} =&\,\frac92-\frac32\\ =&\,3. \end{align*}

Thus the period is

\begin{align*} 6. \end{align*}

For $y=\cos(kx)$ , the period is $\frac{2\pi}{k}$ , so

\begin{align*} \frac{2\pi}{k}=&\,6\\ k=&\,\frac{\pi}{3}. \end{align*}

Therefore

\begin{align*} x_B=\frac92,\qquad k=\frac{\pi}{3},\qquad \text{period}=6. \end{align*}

(d)

解法一

思路

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区域 $R$ 的下边界是 $C_2$ ，上边界分成两段：左侧是 $C_1$ ，右侧是直线段 $L$ 。所以需要先求 $L$ 的方程，再用不等式表达 $y$ 位于这些曲线之间。

答题过程

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The lower boundary is

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

The upper curve $C_1$ is

\begin{align*} y=6x-\frac{27}{4}-x^2. \end{align*}

Now find the equation of $L$ , which passes through

\begin{align*} P\left(3,\frac94\right) \quad\text{and}\quad B\left(\frac92,0\right). \end{align*}

Its gradient is

\begin{align*} m =&\,\frac{0-\frac94}{\frac92-3}\\ =&\,\frac{-\frac94}{\frac32}\\ =&\,-\frac32. \end{align*}

Using point $B$ ,

\begin{align*} y-0=&\,-\frac32\left(x-\frac92\right)\\ y=&\,-\frac32x+\frac{27}{4}. \end{align*}

Therefore the region $R$ is defined by

\begin{align*} y&>\cos\left(\frac{\pi x}{3}\right),\\ y&<6x-\frac{27}{4}-x^2,\\ y&<-\frac32x+\frac{27}{4}. \end{align*}

IAL 2025 Jan Q9

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

(d)

解法一

思路

答题过程