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FP2 2023 Jan Q5

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 5

题目

Problem

5.

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Use algebra to determine the set of values of xx for which

x2−9∣x+8∣> 6−2x\begin{align*} \frac{x^2 - 9}{|x + 8|} >\,& 6 - 2x \end{align*}
(6)
题目中文翻译

本题必须展示所有运算步骤。

完全依赖计算器技术的解答不予接受。

用代数方法求使下列不等式成立的 xx 的取值范围:

x2−9∣x+8∣> 6−2x. \begin{align*} \frac{x^2-9}{|x+8|}>&\,6-2x. \end{align*}
(6)

解答

解法一

思路

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因为分母是 ∣x+8∣|x+8|,所以要以 x=−8x=-8 为分界点分情况讨论。注意 x=−8x=-8 本身不能取。

在每个区间内把绝对值去掉,再解普通二次不等式。

答题过程

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We need

x2−9∣x+8∣>6−2x,\begin{align*} \frac{x^2-9}{|x+8|}>6-2x, \end{align*}

where

x≠−8.\begin{align*} x\neq -8. \end{align*}

If x>−8x>-8, then ∣x+8∣=x+8|x+8|=x+8. Since x+8>0x+8>0,

x2−9> (x+8)(6−2x)x2−9> 6x−2x2+48−16xx2−9> −2x2−10x+483x2+10x−57> 0.\begin{align*} x^2-9>&\,(x+8)(6-2x)\\[2mm] x^2-9>&\,6x-2x^2+48-16x\\[2mm] x^2-9>&\,-2x^2-10x+48\\[2mm] 3x^2+10x-57>&\,0. \end{align*}

Factorise:

3x2+10x−57= (3x+19)(x−3).\begin{align*} 3x^2+10x-57 =&\,(3x+19)(x-3). \end{align*}

So, in this case,

(3x+19)(x−3)>0.\begin{align*} (3x+19)(x-3)>0. \end{align*}

This gives

x<−193orx>3.\begin{align*} x<-\frac{19}{3} \quad\text{or}\quad x>3. \end{align*}

Combining with x>−8x>-8 gives

−8<x<−193orx>3.\begin{align*} -8<x<-\frac{19}{3} \quad\text{or}\quad x>3. \end{align*}

If x<−8x<-8, then ∣x+8∣=−(x+8)|x+8|=-(x+8). Since ∣x+8∣>0|x+8|>0, multiply by ∣x+8∣|x+8|:

x2−9> −(x+8)(6−2x)x2−9> (−x−8)(6−2x).\begin{align*} x^2-9>&\,-(x+8)(6-2x)\\[2mm] x^2-9>&\,(-x-8)(6-2x). \end{align*}

Compute the right hand side:

(−x−8)(6−2x)= −6x+2x2−48+16x= 2x2+10x−48.\begin{align*} (-x-8)(6-2x) =&\,-6x+2x^2-48+16x\\[2mm] =&\,2x^2+10x-48. \end{align*}

Therefore

x2−9> 2x2+10x−480> x2+10x−39x2+10x−39< 0.\begin{align*} x^2-9>&\,2x^2+10x-48\\[2mm] 0>&\,x^2+10x-39\\[2mm] x^2+10x-39<&\,0. \end{align*}

Factorise:

x2+10x−39= (x+13)(x−3).\begin{align*} x^2+10x-39 =&\,(x+13)(x-3). \end{align*}

Thus

(x+13)(x−3)<0,\begin{align*} (x+13)(x-3)<0, \end{align*}

so

−13<x<3.\begin{align*} -13<x<3. \end{align*}

Combining with x<−8x<-8 gives

−13<x<−8.\begin{align*} -13<x<-8. \end{align*}

Therefore the full solution is

−13<x<−8or−8<x<−193orx>3.\begin{align*} \boxed{ -13<x<-8 \quad\text{or}\quad -8<x<-\frac{19}{3} \quad\text{or}\quad x>3 }. \end{align*}

解法二

思路

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另一种思路是先找出所有临界点(即等式成立的边界值以及无定义点),然后在数轴上划分区间进行代值测试。此法可免去在不等式中处理负号和不等号方向变号的困扰。

等式方程为 x2−9∣x+8∣=6−2x\dfrac{x^2-9}{|x+8|} = 6-2x,因为绝对值恒为非负,故可直接变形为 x2−9=(6−2x)∣x+8∣x^2-9 = (6-2x)|x+8|,再依 x>−8x > -8 和 x<−8x < -8 分类找出所有解。

答题过程

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First, identify all critical values. The vertical asymptote (where the expression is undefined) is:

x= −8\begin{align*} x =&\, -8 \end{align*}

Now solve the equation for the boundary points of intersection:

x2−9∣x+8∣= 6−2x\begin{align*} \frac{x^2-9}{|x+8|} =&\, 6-2x \end{align*}

Since ∣x+8∣>0|x+8| > 0 for all x≠−8x \ne -8, this is equivalent to:

x2−9= (6−2x)∣x+8∣\begin{align*} x^2-9 =&\, (6-2x)|x+8| \end{align*}

Case 1: x>−8x > -8, which means ∣x+8∣=x+8|x+8| = x+8:

x2−9= (6−2x)(x+8)x2−9= −2x2−10x+483x2+10x−57= 0(3x+19)(x−3)= 0x=−193orx=3\begin{align*} x^2-9 =&\, (6-2x)(x+8)\\[3mm] x^2-9 =&\, -2x^2-10x+48\\[3mm] 3x^2+10x-57 =&\, 0\\[3mm] (3x+19)(x-3) =&\, 0\\[3mm] x = -\frac{19}{3} &\quad\text{or}\quad x = 3 \end{align*}

Both values are greater than −8-8, so they are valid critical values.

Case 2: x<−8x < -8, which means ∣x+8∣=−(x+8)|x+8| = -(x+8):

x2−9= −(6−2x)(x+8)x2−9= 2x2+10x−48x2+10x−39= 0(x+13)(x−3)= 0x=−13orx=3\begin{align*} x^2-9 =&\, -(6-2x)(x+8)\\[3mm] x^2-9 =&\, 2x^2+10x-48\\[3mm] x^2+10x-39 =&\, 0\\[3mm] (x+13)(x-3) =&\, 0\\[3mm] x = -13 &\quad\text{or}\quad x = 3 \end{align*}

Since we require x<−8x < -8, the only valid root in this case is:

x= −13\begin{align*} x =&\, -13 \end{align*}

Thus, the critical values are:

x=−13,x=−8,x=−193,x=3\begin{align*} x = -13,\quad x = -8,\quad x = -\frac{19}{3},\quad x = 3 \end{align*}

These values divide the real number line into five intervals. Testing a value from each interval in the inequality x2−9∣x+8∣>6−2x\dfrac{x^2-9}{|x+8|} > 6-2x yields:

  • For x<−13x<-13, take x=−14x=-14: the left-hand side is 187/6<34187/6<34, so the inequality is false.
  • For −13<x<−8-13<x<-8, take x=−10x=-10: the left-hand side is 91/2>2691/2>26, so the inequality is true.
  • For −8<x<−19/3-8<x<-19/3, take x=−7x=-7: the left-hand side is 40>2040>20, so the inequality is true.
  • For −19/3<x<3-19/3<x<3, take x=0x=0: the left-hand side is −9/8<6-9/8<6, so the inequality is false.
  • For x>3x>3, take x=4x=4: the left-hand side is 7/12>−27/12>-2, so the inequality is true.

Therefore, the inequality holds on:

−13<x<−8or−8<x<−193orx>3\begin{align*} \boxed{ -13<x<-8 \quad\text{or}\quad -8<x<-\frac{19}{3} \quad\text{or}\quad x>3 } \end{align*}