题目

Problem

(i) Find the range of values of $z$ for which

\begin{align*} \frac4z<8 \end{align*}

giving the answer in simplest form.

(2)

(ii)

Figure 1

Figure 1 shows a plot of

the line with equation $y=x+5$
the curve with equation $y=x(x-3)$

On the opposite page there is a copy of Figure 1 labelled Diagram 1.

Diagram 1

On Diagram 1 represent the following inequality graphically.

\begin{align*} x+5\leqslant y\leqslant x(x-3) \quad\text{for }x\geqslant-2 \end{align*}

(3)

解答

(i)

解法一

思路

展开

含有 $\dfrac{1}{z}$ 的不等式不能直接乘 $z$ ，因为 $z$ 可能为正也可能为负。分成 $z>0$ 和 $z<0$ 两种情况最稳。

答题过程

展开

Consider two cases.

If $z>0$ , then multiplying by $z$ does not change the inequality:

\begin{align*} \frac4z&<8\\[3mm] 4&<8z\\[3mm] z&>\frac12 \end{align*}

If $z<0$ , then

\begin{align*} \frac4z \end{align*}

is negative, so it is always less than $8$ .

Therefore

\begin{align*} \boxed{ z<0 \quad\text{or}\quad z>\frac12 } \end{align*}

(ii)

解法一

思路

展开

要表示的是夹在直线 $y=x+5$ 和抛物线 $y=x(x-3)$ 之间，并且满足 $x\geqslant-2$ 的区域。边界都是实线，因为不等号包含等号。

答题过程

展开

The required region satisfies

\begin{align*} y\geqslant x+5 \end{align*}

and

\begin{align*} y\leqslant x(x-3) \end{align*}

with

\begin{align*} x\geqslant-2 \end{align*}

So draw the vertical boundary line

\begin{align*} x=-2 \end{align*}

as a solid line, and shade the region above the line, below the curve, and to the right of $x=-2$ .

A completed diagram is:

IAL 2026 Jan Q3

题目

解答

(i)

解法一

思路

答题过程

(ii)

解法一

思路

答题过程