题目

Problem

Figure 2 shows part of the graph with equation $y=f(x)$ , where

f(x)=|kx-9|-2\qquad x\in\mathbb{R}

and $k$ is a positive constant.

The graph intersects the $y$ -axis at the point $A$ and has a minimum point at $B$ as shown.

(a) (i) Find the $y$ coordinate of $A$

(ii) Find, in terms of $k$ , the $x$ coordinate of $B$

(2)

(b) Find, in terms of $k$ , the range of values of $x$ that satisfy the inequality

|kx-9|-2<0

(3)

Given that the line $y=3-2x$ intersects the graph $y=f(x)$ at two distinct points,

(3)

题目中文翻译

图 2 给出了图像 $y=f(x)$ 的一部分，其中

f(x)=|kx-9|-2\qquad x\in\mathbb{R}

且 $k$ 是正常数。

该图像与 $y$ 轴交于点 $A$ ，并在点 $B$ 处取得最小值，如图所示。

(a) (i) 求点 $A$ 的 $y$ 坐标。

(ii) 用 $k$ 表示点 $B$ 的 $x$ 坐标。

(b) 用 $k$ 表示满足不等式

|kx-9|-2<0

的 $x$ 的取值范围。

已知直线 $y=3-2x$ 与图像 $y=f(x)$ 有两个不同交点。

解答

(a)(i)

At the $y$ -axis,

x=0

f(0)=|k(0)-9|-2

Hence

f(0)=9-2=7

Therefore the $y$ coordinate of $A$ is

\boxed{7}

(a)(ii)

The minimum point occurs when the expression inside the modulus is zero:

kx-9=0

x=\frac9k

Therefore the $x$ coordinate of $B$ is

\boxed{\frac9k}

(b)

We need to solve

|kx-9|-2<0

|kx-9|<2

This gives

-2<kx-9<2

Add $9$ throughout:

7<kx<11

Since $k>0$ , divide by $k$ :

\boxed{\frac7k<x<\frac{11}{k}}

(c)

For the line

y=3-2x

to intersect the V-shaped graph at two distinct points, it must pass above the minimum point $B$ .

At $B$ ,

x=\frac9k,\qquad y=-2

The value of the line at $x=\dfrac9k$ is

3-2\left(\frac9k\right)

For two distinct intersections, this must be greater than $-2$ :

3-\frac{18}{k}>-2

5>\frac{18}{k}

Since $k>0$ ,

5k>18

Therefore

k>\frac{18}{5}

The limiting value is $k=\dfrac{18}{5}=3.6$ , and for two distinct intersections we need

\boxed{k>3.6}

IAL 2022 June Q5

题目

解答

(a)(i)

(a)(ii)

(b)

(c)