题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Figure 2

Figure 2 shows a sketch of the curve $C$ with equation

\begin{align*} y=\frac{15x}{|x|+4} \end{align*}

and the line $l$ with equation $y=x-2$

(a) Use algebra to determine the exact values of $x$ for which

\begin{align*} x-2>\frac{15x}{|x|+4} \end{align*}

(6)

(b) Hence use algebra to determine the exact values of $x$ for which

\begin{align*} |x-2|>\frac{|15x|}{|x|+4} \end{align*}

(4)

解答

(a)

解法一

思路

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绝对值在 $x=0$ 处分段。分别处理 $x>0$ 和 $x<0$ ，因为 $|x|+4$ 始终为正，所以分母可以安全乘过去。最后根据每段的条件保留合适区间。

答题过程

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For $x>0$ ,

\begin{align*} |x|=x \end{align*}

\begin{align*} x-2 >& \frac{15x}{x+4} \end{align*}

Since $x+4>0$ ,

\begin{align*} (x-2)(x+4) >& 15x\\[2mm] x^2+2x-8 >& 15x\\[2mm] x^2-13x-8 >& 0 \end{align*}

The roots of $x^2-13x-8=0$ are

\begin{align*} x ={}& \frac{13\pm\sqrt{13^2+32}}{2}\\[2mm] ={}& \frac{13\pm\sqrt{201}}{2} \end{align*}

Since $x>0$ , this gives

\begin{align*} x>\frac{13+\sqrt{201}}{2} \end{align*}

For $x<0$ ,

\begin{align*} |x|=-x \end{align*}

\begin{align*} x-2 >& \frac{15x}{-x+4} \end{align*}

Since $-x+4>0$ ,

\begin{align*} (x-2)(-x+4) >& 15x\\[2mm] -x^2+6x-8 >& 15x\\[2mm] x^2+9x+8 <& 0\\[2mm] (x+8)(x+1) <& 0 \end{align*}

Thus

\begin{align*} -8<x<-1 \end{align*}

Therefore

\begin{align*} \boxed{ -8<x<-1 \quad \text{or} \quad x>\frac{13+\sqrt{201}}{2} } \end{align*}

(b)

解法一

思路

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左边变成 $|x-2|$ ，右边变成 $\frac{|15x|}{|x|+4}$ 。当 $x<0$ 时，图像对应到 (a) 中已经得到的左外侧边界；当 $x>0$ 时，还要处理 $x-2$ 的正负，新的交点来自 $\frac{15x}{x+4}=-x+2$ 。

答题过程

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From part (a), the regions outside the intersections include

\begin{align*} x<-8 \quad \text{and} \quad x>\frac{13+\sqrt{201}}{2} \end{align*}

We also need the positive solution where

\begin{align*} \frac{15x}{x+4} ={}& -x+2 \end{align*}

because this corresponds to the reflected line $y=|x-2|$ for $0<x<2$ .

Solve:

\begin{align*} 15x ={}& (-x+2)(x+4)\\[2mm] ={}& -x^2-2x+8\\[2mm] x^2+17x-8 ={}& 0 \end{align*}

Thus

\begin{align*} x ={}& \frac{-17\pm\sqrt{17^2+32}}{2}\\[2mm] ={}& \frac{-17\pm\sqrt{321}}{2} \end{align*}

The positive root is

\begin{align*} \frac{-17+\sqrt{321}}{2} \end{align*}

Therefore the solution is

\begin{align*} \boxed{ x<-8 \quad \text{or} \quad -1<x<\frac{-17+\sqrt{321}}{2} \quad \text{or} \quad x>\frac{13+\sqrt{201}}{2} } \end{align*}

IAL 2025 June Q8

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程