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IAL 2025 June A Q1

A Level / Edexcel / FP2

IAL 2025 June A Paper · Question 1

题目

Problem

Figure 1
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.

Figure 1 shows a sketch of the curve CC with equation

y=10xx−6x≠6\begin{align*} y=\frac{10x}{x-6}\qquad x\ne 6 \end{align*}

and the line ll with equation y=2x+12y=2x+12

(a) Use algebra to determine the xx coordinates of the points of intersection of CC and ll

(2)

(b) Determine the range of values of xx for which

(i) 2x+12>10xx−62x+12>\dfrac{10x}{x-6}

(ii) ∣2x+12∣>∣10xx−6∣\left|2x+12\right|>\left|\dfrac{10x}{x-6}\right|

(4)

解答

(a)

解法一

思路

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交点的 xx 坐标来自两条曲线的 yy 值相等。题目强调用代数，所以要把分式方程化成二次方程并因式分解，不能只从图上读。

答题过程

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At the points of intersection,

10xx−6= 2x+1210x= (2x+12)(x−6)= 2x2−722x2−10x−72= 0x2−5x−36= 0(x+4)(x−9)= 0\begin{align*} \frac{10x}{x-6} =&\,2x+12\\[4mm] 10x =&\,(2x+12)(x-6)\\[4mm] =&\,2x^2-72\\[4mm] 2x^2-10x-72 =&\,0\\[4mm] x^2-5x-36 =&\,0\\[4mm] (x+4)(x-9) =&\,0 \end{align*}

Therefore

x=−4orx=9\begin{align*} x=-4 \quad \text{or} \quad x=9 \end{align*}

(b)(i)

解法一

思路

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把不等式移到一边，通分后用临界点分区间。注意分母 x−6x-6 会改变符号，所以不能直接两边乘以 x−6x-6 后忘记讨论符号。

答题过程

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Consider

2x+12−10xx−6>0(2x+12)(x−6)−10xx−6>02x2−72−10xx−6>02(x+4)(x−9)x−6>0\begin{align*} 2x+12-\frac{10x}{x-6} &>0\\[4mm] \frac{(2x+12)(x-6)-10x}{x-6} &>0\\[4mm] \frac{2x^2-72-10x}{x-6} &>0\\[4mm] \frac{2(x+4)(x-9)}{x-6} &>0 \end{align*}

The critical values are

x=−4,x=6,x=9\begin{align*} x=-4,\qquad x=6,\qquad x=9 \end{align*}

Testing the intervals gives

−4<x<6orx>9\begin{align*} -4<x<6 \quad \text{or} \quad x>9 \end{align*}

(b)(ii)

解法一

思路

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两个函数都取绝对值，所以交界点来自 2x+12=10xx−62x+12=\dfrac{10x}{x-6} 和 2x+12=−10xx−62x+12=-\dfrac{10x}{x-6}。找出四个临界值后，用数轴分区间判断。

答题过程

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The first pair of boundary points comes from part (a):

x=−4,x=9\begin{align*} x=-4,\qquad x=9 \end{align*}

The other pair of boundary points is found by solving

−(2x+12)>10xx−6−2x−12−10xx−6>0(−2x−12)(x−6)−10xx−6>0−2x2+72−10xx−6>0−2(x+9)(x−4)x−6>0\begin{align*} -(2x+12) &>\frac{10x}{x-6}\\[4mm] -2x-12-\frac{10x}{x-6} &>0\\[4mm] \frac{(-2x-12)(x-6)-10x}{x-6} &>0\\[4mm] \frac{-2x^2+72-10x}{x-6} &>0\\[4mm] \frac{-2(x+9)(x-4)}{x-6} &>0 \end{align*}

So

(x+9)(x−4)= 0x= −9, 4\begin{align*} (x+9)(x-4)=&\,0\\[4mm] x=&\,-9,\ 4 \end{align*}

The critical values are

−9,−4,4,9\begin{align*} -9,\quad -4,\quad 4,\quad 9 \end{align*}

Testing intervals in ∣2x+12∣>∣10xx−6∣\left|2x+12\right|>\left|\dfrac{10x}{x-6}\right| gives

x<−9,−4<x<4,x>9\begin{align*} x<-9,\qquad -4<x<4,\qquad x>9 \end{align*}