CalcGospel 國際數學圖譜
返回

FP2 2026 Jan A Q2

A Level / Edexcel / FP2

IAL 2026 Jan A Paper · Question 2

题目

Problem

2.

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

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

∣x2−9∣<∣1−2x∣\begin{align*} |x^2-9|<|1-2x| \end{align*}
(6)
题目中文翻译

本题必须展示所有解题步骤。 不接受完全依赖计算器技术的解答。

用代数方法求满足下列不等式的 xx 的取值集合：

∣x2−9∣<∣1−2x∣\begin{align*} |x^2-9|<|1-2x| \end{align*}
(6)

解答

解法一

思路

展开

绝对值相等的位置是临界点。分别解 x2−9=1−2xx^2-9=1-2x 和 x2−9=−(1−2x)x^2-9=-(1-2x)，再用数轴判断哪些区间满足左边距离更小。

答题过程

展开

The critical values occur when

x2−9=1−2x\begin{align*} x^2-9=1-2x \end{align*}

or

x2−9=−(1−2x)\begin{align*} x^2-9=-(1-2x) \end{align*}

First,

x2−9= 1−2xx2+2x−10= 0x= −2±22−4(1)(−10)2= −1±11\begin{align*} x^2-9 =&\, 1-2x\\[3mm] x^2+2x-10 =&\, 0\\[3mm] x =&\, \frac{-2\pm\sqrt{2^2-4(1)(-10)}}{2}\\[3mm] =&\, -1\pm\sqrt{11} \end{align*}

Next,

x2−9= −1+2xx2−2x−8= 0(x−4)(x+2)= 0\begin{align*} x^2-9 =&\, -1+2x\\[3mm] x^2-2x-8 =&\, 0\\[3mm] (x-4)(x+2) =&\, 0 \end{align*}

so

x=−2,4\begin{align*} x=-2,\quad 4 \end{align*}

The critical values in increasing order are

−1−11,−2,−1+11,4\begin{align*} -1-\sqrt{11},\quad -2,\quad -1+\sqrt{11},\quad 4 \end{align*}

Let α=−1−11\alpha=-1-\sqrt{11}, β=−1+11\beta=-1+\sqrt{11} and F(x)=∣x2−9∣−∣1−2x∣F(x)=|x^2-9|-|1-2x|. Testing one value in each interval gives

x(−∞,α)(α,−2)(−2,β)(β,4)(4,∞)test value−5−3035F(x)+−+−+\begin{align*} \begin{array}{c|ccccc} x&(-\infty,\alpha)&(\alpha,-2)&(-2,\beta)&(\beta,4)&(4,\infty)\\ \text{test value}&-5&-3&0&3&5\\\hline F(x)&+&-&+&-&+ \end{array} \end{align*}

Therefore F(x)<0F(x)<0 precisely when

−1−11<x<−2or−1+11<x<4\begin{align*} \boxed{ -1-\sqrt{11}<x<-2 \quad\text{or}\quad -1+\sqrt{11}<x<4 } \end{align*}

解法二

思路

展开

因为两边都是非负数，可以平方。平方后得到四次方程；但重点仍然是把它因式分解，再从数轴判断不等式的符号。

答题过程

展开

Since both sides are non-negative,

∣x2−9∣<∣1−2x∣\begin{align*} |x^2-9|<|1-2x| \end{align*}

is equivalent to

(x2−9)2<(1−2x)2\begin{align*} (x^2-9)^2<(1-2x)^2 \end{align*}

Bring all terms to one side:

(x2−9)2−(1−2x)2< 0(x2−9−(1−2x))(x2−9+(1−2x))< 0(x2+2x−10)(x2−2x−8)< 0(x+1−11)(x+1+11)(x−4)(x+2)< 0\begin{align*} (x^2-9)^2-(1-2x)^2 <&\,0\\[3mm] \Big(x^2-9-(1-2x)\Big) \Big(x^2-9+(1-2x)\Big) <&\,0\\[3mm] (x^2+2x-10)(x^2-2x-8) <&\,0\\[3mm] (x+1-\sqrt{11})(x+1+\sqrt{11})(x-4)(x+2) <&\,0 \end{align*}

The four roots are simple and the leading coefficient is positive. With α=−1−11\alpha=-1-\sqrt{11} and β=−1+11\beta=-1+\sqrt{11}, the sign alternates across the roots:

x(−∞,α)(α,−2)(−2,β)(β,4)(4,∞)product+−+−+\begin{align*} \begin{array}{c|ccccc} x&(-\infty,\alpha)&(\alpha,-2)&(-2,\beta)&(\beta,4)&(4,\infty)\\\hline \text{product}&+&-&+&-&+ \end{array} \end{align*}

Hence the product is negative on

−1−11<x<−2or−1+11<x<4\begin{align*} \boxed{ -1-\sqrt{11}<x<-2 \quad\text{or}\quad -1+\sqrt{11}<x<4 } \end{align*}