题目

Problem

(a) On the axes in the answer book sketch and clearly label the graphs of

(i) $y=x(a-x)$

(ii) $y=x^2(b-x)$

where $a$ and $b$ are positive constants $b>a$ .

Show clearly the coordinates of all the points where the curves cross or meet the coordinate axes.

(5)

(b) Show that the $x$ -coordinates of the points of intersection of

\begin{align*} y=x(4-x) \qquad\text{and}\qquad y=x^2(7-x) \end{align*}

are given by the solutions to the equation

\begin{align*} x(x^2-8x+4)=0 \end{align*}

(2)

The point $A$ lies on both of the curves and the $x$ and $y$ coordinates of $A$ are both positive.

(c) Find the exact coordinates of $A$ , leaving the answer in the form $(p+q\sqrt3,r+s\sqrt3)$ , where $p$ , $q$ , $r$ and $s$ are integers.

(Solutions relying on calculator technology are not acceptable.)

(5)

解答

(a)

解法一

思路

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$y=x(a-x)$ 是开口向下的二次曲线，根为 $x=0$ 和 $x=a$ 。

$y=x^2(b-x)$ 是三次曲线，根为 $x=0$ 和 $x=b$ ，其中 $x=0$ 是重根，所以曲线在原点与 $x$ 轴相切。

因为 $b>a>0$ ，三次曲线的另一个截距在二次曲线截距的右边。

答题过程

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For

\begin{align*} y=x(a-x), \end{align*}

the $x$ -intercepts are found from

\begin{align*} x(a-x)=0. \end{align*}

So the curve crosses the $x$ -axis at

\begin{align*} (0,0)\quad\text{and}\quad(a,0). \end{align*}

It is a negative quadratic curve.

For

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

the $x$ -intercepts are found from

\begin{align*} x^2(b-x)=0. \end{align*}

So the curve meets the $x$ -axis at

\begin{align*} (0,0)\quad\text{and}\quad(b,0). \end{align*}

The factor $x^2$ means the curve touches the $x$ -axis at the origin. Since $b>a$ , the point $(b,0)$ is to the right of $(a,0)$ .

A suitable sketch is:

(b)

解法一

思路

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交点的 $x$ 坐标来自两条曲线的 $y$ 值相等。把两个表达式相等后展开并整理，就要自然得到题目给出的形式。

答题过程

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At intersections,

\begin{align*} x(4-x)=&\,x^2(7-x). \end{align*}

Expanding both sides gives

\begin{align*} 4x-x^2=&\,7x^2-x^3. \end{align*}

Bring all terms to one side:

\begin{align*} x^3-8x^2+4x=&\,0. \end{align*}

Factor out $x$ :

\begin{align*} x(x^2-8x+4)=0. \end{align*}

This is the required equation.

(c)

解法一

思路

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由 (b) 得到 $x=0$ 或 $x^2-8x+4=0$ 。点 $A$ 的 $x,y$ 坐标都为正，所以要选使 $y=x(4-x)$ 为正的那个根，即 $0<x<4$ 的根。

答题过程

展开

From part (b),

\begin{align*} x(x^2-8x+4)=0. \end{align*}

Since point $A$ has positive $x$ and $y$ coordinates, $x\neq0$ .

Solve

\begin{align*} x^2-8x+4=0. \end{align*}

Using the quadratic formula,

\begin{align*} x=&\,\frac{8\pm\sqrt{(-8)^2-4(1)(4)}}{2}\\ =&\,\frac{8\pm\sqrt{48}}{2}\\ =&\,4\pm2\sqrt3. \end{align*}

For $y=x(4-x)$ to be positive, we need $0<x<4$ , so

\begin{align*} x=4-2\sqrt3. \end{align*}

Now substitute into $y=x(4-x)$ :

\begin{align*} y =&\,(4-2\sqrt3)\bigl(4-(4-2\sqrt3)\bigr)\\ =&\,(4-2\sqrt3)(2\sqrt3)\\ =&\,8\sqrt3-12. \end{align*}

Therefore

\begin{align*} A=(4-2\sqrt3,\,-12+8\sqrt3). \end{align*}

IAL 2025 Oct A Q10

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