题目

Problem

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

Solutions relying on calculator technology are not acceptable.

Given that

\begin{align*} \frac{x + 2}{x + 4} \leqslant{}& \frac{x}{k(x - 1)} \end{align*}

where $k$ is a positive constant,

(a) show that

\begin{align*} (x + 4)(x - 1)(px^2 + qx + r) \leqslant{}& 0 \end{align*}

where $p$ , $q$ and $r$ are expressions in terms of $k$ to be determined.

(3)

(b) Hence, or otherwise, determine the values for $x$ for which

\begin{align*} \frac{x + 2}{x + 4} \leqslant{}& \frac{x}{3(x - 1)} \end{align*}

(4)

解答

(a)

解法一

思路

展开

不等式中有分母，不能直接乘以 $(x+4)(x-1)$ ，因为它的正负会变。比较稳的做法是先移到同一边，再乘以 $k(x+4)^2(x-1)^2$ 。由于 $k>0$ ，而平方项非负且在定义域内不为 $0$ ，不等号方向不变。

答题过程

展开

Start from

\begin{align*} \frac{x+2}{x+4} \leqslant{}& \frac{x}{k(x-1)} \end{align*}

Bring all terms to the left-hand side:

\begin{align*} \frac{x+2}{x+4} -\frac{x}{k(x-1)} \leqslant{}&0\\[4mm] \frac{k(x+2)(x-1)-x(x+4)} {k(x+4)(x-1)} \leqslant{}&0 \end{align*}

Since $k>0$ , multiply by $k(x+4)^2(x-1)^2$ , which is positive for $x\ne -4,1$ :

\begin{align*} (x+4)(x-1) \left[k(x+2)(x-1)-x(x+4)\right] \leqslant{}&0 \end{align*}

Now expand the bracket:

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

Therefore

\begin{align*} (x+4)(x-1) \left((k-1)x^2+(k-4)x-2k\right) \leqslant{}&0 \end{align*}

This is in the required form, where

\begin{align*} p=k-1,\qquad q=k-4,\qquad r=-2k \end{align*}

(b)

解法一

思路

展开

这里用 (a) 的结果最直接。把 $k=3$ 代入后，得到一个四个一次因式相乘的不等式。最后一定要记得原分母不能为 $0$ ，所以 $x=-4$ 和 $x=1$ 不能取。

答题过程

展开

For $k=3$ ,

\begin{align*} (k-1)x^2+(k-4)x-2k ={}& 2x^2-x-6\\[4mm] ={}& (2x+3)(x-2) \end{align*}

So the inequality becomes

\begin{align*} (x+4)(x-1)(2x+3)(x-2)\leqslant 0 \end{align*}

The critical values are

\begin{align*} x=-4,\qquad x=-\frac32,\qquad x=1,\qquad x=2 \end{align*}

Check the signs on the intervals:

\begin{array}{c|ccccc} \text{interval} & x<-4 & -4<x<-\frac32 & -\frac32<x<1 & 1<x<2 & x>2\\ \hline \text{sign} & + & - & + & - & + \end{array}

We need the product to be non-positive. The zeros from the quadratic can be included, but the original denominator excludes $x=-4$ and $x=1$ . Therefore,

\begin{align*} \boxed{ -4<x\leqslant -\frac32 \quad\text{or}\quad 1<x\leqslant 2 } \end{align*}

IAL 2023 June Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程