题目

Problem

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 values of $x$ for which

\begin{align*} \left|\frac{x^2+5x-2}{x^2+1}\right|<2, \qquad x\in\mathbb{R} \end{align*}

(6)

解答

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因为 $x^2+1$ 永远为正，可以安全地乘过去，不会改变不等号方向。绝对值小于 $2$ 等价于同时满足小于 $2$ 和大于 $-2$ 。

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Since

\begin{align*} x^2+1>0 \qquad\text{for all real }x, \end{align*}

we can write

\begin{align*} -2 < \frac{x^2+5x-2}{x^2+1} < 2 \end{align*}

First,

\begin{align*} \frac{x^2+5x-2}{x^2+1}<2 \end{align*}

gives

\begin{align*} x^2+5x-2 &<2(x^2+1)\\[3mm] 0 &<x^2-5x+4\\[3mm] 0 &<(x-1)(x-4) \end{align*}

\begin{align*} x<1 \quad\text{or}\quad x>4 \end{align*}

Next,

\begin{align*} \frac{x^2+5x-2}{x^2+1}>-2 \end{align*}

gives

\begin{align*} x^2+5x-2 &>-2(x^2+1)\\[3mm] 3x^2+5x &>0\\[3mm] x(3x+5) &>0 \end{align*}

\begin{align*} x<-\frac53 \quad\text{or}\quad x>0 \end{align*}

Combining the two conditions,

\begin{align*} x<-\frac53, \qquad 0<x<1, \qquad x>4 \end{align*}

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也可以先找临界点，再用数轴判断符号。临界点来自分式等于 $2$ 或 $-2$ ，这比直接平方后解四次式更清楚。

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The critical values occur when

\begin{align*} \frac{x^2+5x-2}{x^2+1}=2 \end{align*}

\begin{align*} \frac{x^2+5x-2}{x^2+1}=-2 \end{align*}

Since $x^2+1>0$ , these give

\begin{align*} x^2+5x-2 ={}&2(x^2+1)\\[3mm] x^2-5x+4 ={}&0\\[3mm] (x-1)(x-4) ={}&0 \end{align*}

\begin{align*} x=1,\quad 4 \end{align*}

and

\begin{align*} x^2+5x-2 ={}&-2(x^2+1)\\[3mm] 3x^2+5x ={}&0\\[3mm] x(3x+5) ={}&0 \end{align*}

\begin{align*} x=0,\quad -\frac53 \end{align*}

Testing the intervals determined by

\begin{align*} -\frac53,\quad 0,\quad 1,\quad 4 \end{align*}

shows that the expression has modulus less than $2$ on

\begin{align*} \boxed{ x<-\frac53,\quad 0<x<1,\quad x>4 } \end{align*}