题目

Problem

Figure 1 shows a sketch of the curve $C_1$ with equation

\begin{align*} y = \frac{4x}{4 - |x|} \end{align*}

and the curve $C_2$ with equation

\begin{align*} y = x^2 - 8x \end{align*}

For $x > 0$ , $C_1$ has equation $y = \frac{4x}{4 - x}$

(a) Use algebra to show that $C_1$ touches $C_2$ at a point $P$ , stating the coordinates of $P$ .

(5)

(b) Hence or otherwise, using algebra, solve the inequality

\begin{align*} x^2 - 8x > \frac{4x}{4 - |x|} \end{align*}

(6)

题目中文翻译

3. 图 1 所示为曲线 $C_1$ 与曲线 $C_2$ 的示意图，其方程分别为

\begin{align*} C_1:\quad y=&\,\frac{4x}{4-|x|},\\[4mm] C_2:\quad y=&\,x^2-8x. \end{align*}

当 $x>0$ 时， $C_1$ 的方程为 $y=\frac{4x}{4-x}$ 。

(a) 用代数方法证明 $C_1$ 与 $C_2$ 在点 $P$ 处相切，并写出 $P$ 的坐标。

(5)

(b) 由此或用其他方法，使用代数解不等式

\begin{align*} x^2-8x>\frac{4x}{4-|x|}. \end{align*}

(6)

解答

(a)

解法一

思路

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题目已经说明当 $x>0$ 时，

\begin{align*} C_1:\quad y=\frac{4x}{4-x}. \end{align*}

令两条曲线相交，解方程。如果得到 repeated root，就说明它们在该点相切。

答题过程

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For $x>0$ ,

\begin{align*} C_1:\quad y=\frac{4x}{4-x}. \end{align*}

At an intersection with $C_2$ ,

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

Multiply by $4-x$ :

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

Expand the left hand side:

\begin{align*} (x^2-8x)(4-x) =&\,4x^2-32x-x^3+8x^2\\[2mm] =&\,-x^3+12x^2-32x. \end{align*}

\begin{align*} -x^3+12x^2-32x=4x. \end{align*}

Bring all terms to one side:

\begin{align*} -x^3+12x^2-36x=0. \end{align*}

Factorise:

\begin{align*} -x(x^2-12x+36)=0. \end{align*}

\begin{align*} x=0 \quad\text{or}\quad (x-6)^2=0. \end{align*}

For the point of contact on the positive branch,

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

Then

\begin{align*} y=6^2-8(6)=36-48=-12. \end{align*}

Since $x=6$ is a repeated root, the curves touch at

\begin{align*} P=(6,-12). \end{align*}

解法二

思路

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也可以比较两条曲线的 gradient。若它们在同一点相交且 gradient 相同，就说明它们相切。

答题过程

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For $x>0$ ,

\begin{align*} C_1:\quad y=\frac{4x}{4-x}. \end{align*}

Differentiate $C_1$ :

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \Big(\frac{4x}{4-x}\Big) =&\,\frac{4(4-x)-4x(-1)} {(4-x)^2}\\[2mm] =&\,\frac{16}{(4-x)^2}. \end{align*}

For $C_2$ ,

\begin{align*} y=x^2-8x, \end{align*}

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=2x-8. \end{align*}

At a point of tangency,

\begin{align*} 2x-8=\frac{16}{(4-x)^2}. \end{align*}

Since $x\ne4$ , multiplying by $(4-x)^2$ gives

\begin{align*} (2x-8)(4-x)^2=&\,16,\\[4mm] 2(x-4)(x-4)^2=&\,16,\\[4mm] (x-4)^3=&\,8,\\[4mm] x-4=&\,2,\\[4mm] x=&\,6. \end{align*}

Both gradients at $x=6$ are $4$ , since

\begin{align*} 2(6)-8=4 \end{align*}

and

\begin{align*} \frac{16}{(4-6)^2}=4. \end{align*}

At $x=6$ ,

\begin{align*} x^2-8x=-12, \end{align*}

and

\begin{align*} \frac{4x}{4-x} =&\,\frac{24}{-2}\\[2mm] =&\,-12. \end{align*}

So the curves meet with the same gradient at

\begin{align*} P=(6,-12). \end{align*}

(b)

解法一

思路

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因为有 $|x|$ ，要分成 $x>0$ 和 $x<0$ 两个分支。

当 $x>0$ 时，右边是 $\dfrac{4x}{4-x}$ ，而 (a) 已经告诉我们 $x=6$ 是相切点。还要注意 $x=4$ 是 vertical asymptote。

当 $x<0$ 时，右边是 $\dfrac{4x}{4+x}$ ，需要另外求交点。

把左边减去右边后因式分解，判断差值的正负，而不是直接乘上正负未定的分母。注意 $x=0$ 、交点处都是等号，不符合严格大于。

答题过程

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We solve

\begin{align*} x^2-8x>\frac{4x}{4-|x|}. \end{align*}

The original expression is undefined at $x=\pm4$ . At $x=0$ the two sides are equal, so $x=0$ is not a solution.

Let

\begin{align*} F(x)=x^2-8x-\frac{4x}{4-|x|}. \end{align*}

We require $F(x)>0$ .

First consider $x>0$ . Then $|x|=x$ , so

\begin{align*} x^2-8x>\frac{4x}{4-x}. \end{align*}

From part (a), the curves touch at $x=6$ . Also, this branch has a vertical asymptote at

\begin{align*} x=4. \end{align*}

Using the factorisation from part (a),

\begin{align*} F(x)=&\,\frac{(x^2-8x)(4-x)-4x}{4-x}\\[4mm] =&\,\frac{-x(x-6)^2}{4-x}\\[4mm] =&\,\frac{x(x-6)^2}{x-4}. \end{align*}

For $x>0$ and $x\ne6$ , the numerator is positive. Thus the sign is determined by $x-4$ :

\begin{array}{c|c} \text{Interval}&\text{Sign of }F(x)\\\hline 0<x<4&-\\ 4<x<6&+\\ x>6&+ \end{array}

At $x=6$ , $F(x)=0$ , so the positive branch gives

\begin{align*} 4<x<6 \quad\text{or}\quad x>6. \end{align*}

Now consider $x<0$ . Then $|x|=-x$ , so

\begin{align*} x^2-8x>\frac{4x}{4+x}. \end{align*}

Find the intersections:

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

Multiply by $4+x$ :

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

Expand:

\begin{align*} 4x^2+x^3-32x-8x^2=4x. \end{align*}

\begin{align*} x^3-4x^2-36x=0. \end{align*}

Factorise:

\begin{align*} x(x^2-4x-36)=0. \end{align*}

Thus

\begin{align*} x=0 \quad\text{or}\quad x=2\pm2\sqrt{10}. \end{align*}

For $x<0$ , the relevant intersection is

\begin{align*} x=2-2\sqrt{10}. \end{align*}

This branch also has a vertical asymptote at

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

On this branch,

\begin{align*} F(x)=&\,\frac{(x^2-8x)(x+4)-4x}{x+4}\\[4mm] =&\,\frac{x(x^2-4x-36)}{x+4}. \end{align*}

Let

\begin{align*} \alpha=&\,2-2\sqrt{10},\\[4mm] \beta=&\,2+2\sqrt{10}. \end{align*}

Then $\alpha<-4<0<\beta$ , and

\begin{align*} F(x)=\frac{x(x-\alpha)(x-\beta)}{x+4}. \end{align*}

For $x<0$ , both $x$ and $x-\beta$ are negative. Their product is positive, so the sign of $F(x)$ is the sign of $(x-\alpha)/(x+4)$ :

\begin{array}{c|c} \text{Interval}&\text{Sign of }F(x)\\\hline x<\alpha&+\\ \alpha<x<-4&-\\ -4<x<0&+ \end{array}

Exclude $x=\alpha$ , where $F(x)=0$ , and $x=-4$ , where it is undefined. Hence the negative branch gives

\begin{align*} x<2-2\sqrt{10} \quad\text{or}\quad -4<x<0. \end{align*}

Combining the positive and negative branches, the solution is

\begin{align*} x<2-2\sqrt{10}, &\quad\text{or}\\[4mm] -4<x<0, &\quad\text{or}\\[4mm] 4<x<6, &\quad\text{or}\\[4mm] x>6.& \end{align*}

FP2 2022 Jan Q3

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程