题目

Problem

Given that $k$ is a positive constant,

(a) on separate diagrams, sketch the graph with equation

(i) $y=k-2|x|$

(ii) $y=\left|2x-\dfrac{k}{3}\right|$

Show on each sketch the coordinates, in terms of $k$ , of each point where the graph meets or cuts the axes.

(4)

(b) Hence find, in terms of $k$ , the values of $x$ for which

\left|2x-\frac{k}{3}\right|=k-2|x|

giving your answers in simplest form.

(4)

题目中文翻译

已知 $k$ 为正常数。

(a) 在分别的图上，画出下列方程所表示图像的草图：

(i) $y=k-2|x|$

(ii) $y=\left|2x-\dfrac{k}{3}\right|$

在每幅图上用 $k$ 表示图像与坐标轴相交或相切的各点坐标。

(b) 由此求满足

\left|2x-\frac{k}{3}\right|=k-2|x|

的 $x$ 值，并将答案化为最简形式。

解答

(a)

(i)

For

y=k-2|x|

the graph is an upside-down V shape.

The maximum occurs when

x=0

so the $y$ -intercept is

(0,k)

For the $x$ -intercepts, set $y=0$ :

k-2|x|=0

|x|=\frac{k}{2}

Hence

x=\pm \frac{k}{2}

So the $x$ -intercepts are

\left(-\frac{k}{2},0\right),\qquad \left(\frac{k}{2},0\right)

(ii)

For

y=\left|2x-\frac{k}{3}\right|

the graph is a V shape.

The minimum occurs when

2x-\frac{k}{3}=0

x=\frac{k}{6}

Thus the $x$ -intercept is

\left(\frac{k}{6},0\right)

For the $y$ -intercept, set $x=0$ :

y=\left|-\frac{k}{3}\right|=\frac{k}{3}

So the $y$ -intercept is

\left(0,\frac{k}{3}\right)

(b)

We need to solve

\left|2x-\frac{k}{3}\right|=k-2|x|

The intersections can be found by considering the relevant branches.

Left intersection

For $x<0$ ,

|x|=-x

and

\left|2x-\frac{k}{3}\right|=\frac{k}{3}-2x

So the equation becomes

\frac{k}{3}-2x=k+2x

Hence

-4x=\frac{2k}{3}

Therefore

x=-\frac{k}{6}

Right intersection

For $x\geq \dfrac{k}{6}$ ,

\left|2x-\frac{k}{3}\right|=2x-\frac{k}{3}

and

|x|=x

So the equation becomes

2x-\frac{k}{3}=k-2x

Hence

4x=\frac{4k}{3}

Therefore

x=\frac{k}{3}

Thus the values of $x$ are

\boxed{x=-\frac{k}{6},\qquad x=\frac{k}{3}}

IAL 2021 June Q6

题目

解答

(a)

(i)

(ii)

(b)

Left intersection

Right intersection