题目

Problem

Figure 3 shows a sketch of part of the graph with equation $y=f(x)$ , where

f(x)=|kx-10|+k\qquad x\in\mathbb{R}

and $k$ is a positive constant.

The graph

cuts the $y$ -axis at the point $A$
has a vertex at the point $P$

(a) Find, in simplest form in terms of $k$ ,

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

(ii) the coordinates of $P$

(3)

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

|kx-10|+k\ge 2k

(3)

Given that the line with equation $y=3x+1$ intersects the graph of $y=f(x)$ at 2 distinct points,

(4)

题目中文翻译

图 3 给出了图像 $y=f(x)$ 的部分草图，其中

f(x)=|kx-10|+k\qquad x\in\mathbb{R}

且 $k$ 为正常数。

该图像

与 $y$ 轴交于点 $A$
顶点为点 $P$

(a) 用 $k$ 的最简形式表示：

(i) 点 $A$ 的 $y$ 坐标；

(ii) 点 $P$ 的坐标。

(b) 用 $k$ 表示满足

|kx-10|+k\ge 2k

的 $x$ 的取值范围。

已知直线 $y=3x+1$ 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)-10|+k=10+k

Therefore the $y$ coordinate of $A$ is

\boxed{10+k}

(ii)

The vertex occurs when the expression inside the modulus is zero:

kx-10=0

Since $k>0$ ,

x=\frac{10}{k}

At the vertex,

y=0+k=k

Therefore

\boxed{P\left(\frac{10}{k},k\right)}

(b)

We need to solve

|kx-10|+k\ge 2k

|kx-10|\ge k

Since $k>0$ , this gives two cases.

First,

kx-10\ge k

kx\ge 10+k

and therefore

x\ge \frac{10+k}{k}

Second,

kx-10\le -k

kx\le 10-k

and therefore

x\le \frac{10-k}{k}

Hence

\boxed{x\le \frac{10-k}{k}\quad\text{or}\quad x\ge \frac{10+k}{k}}

(c)

The graph has two branches:

y=-kx+10+k

on the left of the vertex, and

y=kx-10+k

on the right of the vertex.

For the line

y=3x+1

to intersect the right-hand branch, the right-hand branch must have gradient greater than $3$ .

k>3

This gives the lower bound.

For the upper bound, consider the case where the line just passes through the vertex. Then the two intersections merge into one point.

At the vertex,

\left(x,y\right)=\left(\frac{10}{k},k\right)

Substitute this into

y=3x+1

to find the boundary value:

k=3\left(\frac{10}{k}\right)+1

Multiply by $k$ :

k^2=30+k

k^2-k-30=0

Factorise:

(k-6)(k+5)=0

Since $k$ is positive,

k=6

At $k=6$ , the line passes through the vertex, so there is only one intersection at the vertex rather than two distinct points. Therefore we need

k<6

Combining the two conditions,

\boxed{3<k<6}

IAL 2025 May Q10

题目

解答

(a)

(i)

(ii)

(b)

(c)