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FP2 2021 Oct Q2

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 2

题目

Problem

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

x2−x⩽x+3x\begin{align*} \frac{x}{2-x}\leqslant\frac{x+3}{x} \end{align*}

(Solutions relying entirely on graphical methods are not acceptable.)

(8)
题目中文翻译

2. 用代数方法求使下列不等式成立的 xx 的取值集合：

x2−x⩽x+3x\begin{align*} \frac{x}{2-x}\leqslant\frac{x+3}{x} \end{align*}

(不接受完全依赖图像方法的解答。)

(8)

解答

解法一

思路

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这是分式不等式，不能直接乘以 x(2−x)x(2-x)，因为它的正负会随 xx 改变。比较稳妥的方法是先移到同一边，通分成一个分式，再用 critical values 分区间判断符号。

注意原式中分母不能为零，所以

x≠0,x≠2.\begin{align*} x\neq0,\qquad x\neq2. \end{align*}

答题过程

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The inequality is

x2−x⩽x+3x,\begin{align*} \frac{x}{2-x}\leqslant \frac{x+3}{x}, \end{align*}

where

x≠0,x≠2.\begin{align*} x\neq0,\qquad x\neq2. \end{align*}

Move all terms to the left:

x2−x−x+3x⩽0.\begin{align*} \frac{x}{2-x}-\frac{x+3}{x}\leqslant0. \end{align*}

Use the common denominator x(2−x)x(2-x):

 x2−x−x+3x= Fx(2−x),\begin{align*} &\,\frac{x}{2-x}-\frac{x+3}{x}\\[4mm] =&\,\frac{F}{x(2-x)}, \end{align*}

where the numerator is F=x2−(2−x)(x+3)F=x^2-(2-x)(x+3). Expanding,

F= x2−[2x+6−x2 −3x]= x2−(−x2−x+6)= 2x2+x−6= (2x−3)(x+2).\begin{align*} F=&\,x^2-\big[2x+6-x^2\\[4mm] &\,\hspace{2pt}-3x\big]\\[4mm] =&\,x^2-(-x^2-x+6)\\[4mm] =&\,2x^2+x-6\\[4mm] =&\,(2x-3)(x+2). \end{align*}

So the inequality becomes

(2x−3)(x+2)x(2−x)⩽0.\begin{align*} \frac{(2x-3)(x+2)} {x(2-x)} \leqslant0. \end{align*}

The critical values are

x=−2,0,32,2.\begin{align*} x=-2,\quad 0,\quad \frac32,\quad 2. \end{align*}

Now test the sign of

(2x−3)(x+2)x(2−x)\begin{align*} \frac{(2x-3)(x+2)} {x(2-x)} \end{align*}

on each interval:

xsign(−∞,−2)−−20(−2,0)+(0,32)−320(32,2)+(2,∞)−\begin{array}{c|c} x & \text{sign}\\ \hline (-\infty,-2) & -\\ -2 & 0\\ (-2,0) & +\\ (0,\frac32) & -\\ \frac32 & 0\\ (\frac32,2) & +\\ (2,\infty) & - \end{array}

Since we need the expression to be less than or equal to zero,

 x⩽−2, or 0<x⩽32, or x>2.\begin{align*} &\,x\leqslant-2,\\[4mm] &\,\text{or }0<x\leqslant\frac32,\\[4mm] &\,\text{or }x>2. \end{align*}

解法二

思路

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也可以乘以一个一定非负的表达式，避免不等号方向问题。这里可以乘以

x2(2−x)2.\begin{align*} x^2(2-x)^2. \end{align*}

因为它在定义域内一定为正，所以不等号方向不变。最后仍然要记得排除 x=0x=0 和 x=2x=2。

答题过程

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Start with

x2−x⩽x+3x,\begin{align*} \frac{x}{2-x}\leqslant \frac{x+3}{x}, \end{align*}

where

x≠0,x≠2.\begin{align*} x\neq0,\qquad x\neq2. \end{align*}

Multiply both sides by x2(2−x)2x^2(2-x)^2, which is positive for all allowed xx:

 x3(2−x)⩽ x(x+3)(2−x)2.\begin{align*} &\,x^3(2-x)\\[4mm] \leqslant&\,x(x+3)(2-x)^2. \end{align*}

Bring all terms to one side:

 x3(2−x) −x(x+3)(2−x)2⩽ 0.\begin{align*} &\,x^3(2-x)\\[4mm] &\,\hspace{2pt}-x(x+3)(2-x)^2\\[4mm] \leqslant&\,0. \end{align*}

Factor out x(2−x)x(2-x):

 x(2−x) ×[x2−(x+3)(2−x)]⩽ 0.\begin{align*} &\,x(2-x)\\[4mm] &\,\hspace{2pt}\times\big[x^2-(x+3)(2-x)\big]\\[4mm] \leqslant&\,0. \end{align*}

Now simplify the bracket:

 x2−(x+3)(2−x)= x2−(−x2−x+6)= 2x2+x−6= (2x−3)(x+2).\begin{align*} &\,x^2-(x+3)(2-x)\\[4mm] =&\,x^2-(-x^2-x+6)\\[4mm] =&\,2x^2+x-6\\[4mm] =&\,(2x-3)(x+2). \end{align*}

So

 x(2−x)(2x−3) ×(x+2)⩽ 0.\begin{align*} &\,x(2-x)(2x-3)\\[4mm] &\,\hspace{2pt}\times(x+2)\\[4mm] \leqslant&\,0. \end{align*}

Using the critical values

−2,0,32,2,\begin{align*} -2,\quad 0,\quad \frac32,\quad 2, \end{align*}

the product is negative on x<−2x<-2, 0<x<320<x<\frac32 and x>2x>2, and positive on the other intervals. It is zero at x=−2x=-2 and x=32x=\frac32. The values x=0x=0 and x=2x=2 are excluded from the original inequality.

Therefore the solution is

 x⩽−2, or 0<x⩽32, or x>2.\begin{align*} &\,x\leqslant-2,\\[4mm] &\,\text{or }0<x\leqslant\frac32,\\[4mm] &\,\text{or }x>2. \end{align*}