Skip to content
CalcGospel 國際數學圖譜
返回

P3 2020 Jan Q6

A Level / Edexcel / P3

IAL 2020 Jan Paper · Question 6

题目

Problem

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

f(x)=2∣2x−5∣+3x≥0f(x)=2|2x-5|+3\qquad x\ge 0

The vertex of the graph is at point PP as shown.

(a) State the coordinates of PP.

(2)

(b) Solve the equation f(x)=3x−2f(x)=3x-2.

(4)

Given that the equation

f(x)=kx+2f(x)=kx+2

where kk is a constant, has exactly two roots,

(c) find the range of values of kk.

(3)
题目中文翻译

图 2 给出了图像 y=f(x)y=f(x) 的一部分,其中

f(x)=2∣2x−5∣+3x≥0f(x)=2|2x-5|+3\qquad x\ge 0

图像的顶点为图中的点 PP。

(a) 写出点 PP 的坐标。

(b) 解方程 f(x)=3x−2f(x)=3x-2。

已知方程

f(x)=kx+2f(x)=kx+2

其中 kk 为常数,且该方程恰有两个根,

(c) 求 kk 的取值范围。

解答

(a)

The function is

f(x)=2∣2x−5∣+3f(x)=2|2x-5|+3

The vertex occurs when the expression inside the absolute value is zero:

2x−5=02x-5=0

So

x=52x=\frac52

At this value,

f(52)=2∣0∣+3=3f\left(\frac52\right)=2|0|+3=3

Therefore

P(52,3)\boxed{P\left(\frac52,3\right)}

(b)

解法一

思路

展开

按绝对值内的符号拆成两支,分别解方程并检查各自的区间。

答题过程

展开

We need to solve

f(x)=3x−2f(x)=3x-2

That is,

2∣2x−5∣+3=3x−22|2x-5|+3=3x-2

Case 1. x≥52x\geq \dfrac52

Then

∣2x−5∣=2x−5|2x-5|=2x-5

So

2(2x−5)+3=3x−22(2x-5)+3=3x-2

Hence

4x−7=3x−24x-7=3x-2

Therefore

x=5x=5

This is valid since 5≥525\geq \dfrac52.

Case 2. 0≤x<520\leq x<\dfrac52

Then

∣2x−5∣=5−2x|2x-5|=5-2x

So

2(5−2x)+3=3x−22(5-2x)+3=3x-2

Hence

13−4x=3x−213-4x=3x-2

Therefore

15=7x15=7x

so

x=157x=\frac{15}{7}

This is valid since

0≤157<520\leq \frac{15}{7}<\frac52

Thus the solutions are

x=157orx=5\boxed{x=\frac{15}{7}\quad \text{or}\quad x=5}

解法二

思路

展开

先孤立绝对值再平方,最后回验平方前右边非负的条件。

答题过程

展开

Rearrange to 2∣2x−5∣=3x−52|2x-5|=3x-5. Any solution must satisfy x≥5/3x\geq 5/3. Squaring gives

4(2x−5)2= (3x−5)2,7x2−50x+75= 0,(7x−15)(x−5)= 0.\begin{align*} 4(2x-5)^2=&\,(3x-5)^2,\\ 7x^2-50x+75=&\,0,\\ (7x-15)(x-5)=&\,0. \end{align*}

Both 157\frac{15}{7} and 55 are at least 53\frac53, so the two sides before squaring are non-negative. Thus both satisfy the original equation:

x=157orx=5.\boxed{x=\frac{15}{7}\quad\text{or}\quad x=5}.

(c)

We need the equation

f(x)=kx+2f(x)=kx+2

to have exactly two roots.

That is,

2∣2x−5∣+3=kx+22|2x-5|+3=kx+2

Left branch: 0≤x≤520\leq x\leq \dfrac52

On this branch,

f(x)=13−4xf(x)=13-4x

So

13−4x=kx+213-4x=kx+2

Hence

11=(k+4)x11=(k+4)x

The case k=−4k=-4 would give 11=011=0, so it yields no solution on this branch. Otherwise,

x=11k+4x=\frac{11}{k+4}

For this to lie on the left branch, we need

0≤11k+4≤520\leq \frac{11}{k+4}\leq \frac52

Since the numerator is positive, this gives

k+4>0k+4>0

and

11k+4≤52\frac{11}{k+4}\leq \frac52

So

22≤5(k+4)22\leq 5(k+4)

Hence

k≥25k\geq \frac25

Right branch: x≥52x\geq \dfrac52

On this branch,

f(x)=4x−7f(x)=4x-7

So

4x−7=kx+24x-7=kx+2

Hence

(4−k)x=9(4-k)x=9

The case k=4k=4 would give 0=90=9, so it yields no solution on this branch. Otherwise,

x=94−kx=\frac{9}{4-k}

For this to lie on the right branch, we need

94−k≥52\frac{9}{4-k}\geq \frac52

and also

4−k>04-k>0

So

k<4k<4

and

18≥5(4−k)18\geq 5(4-k)

Hence

k≥25k\geq \frac25

At k=25k=\dfrac25, the two branch solutions meet at the vertex, so there is only one root. Therefore, to have exactly two roots,

25<k<4\boxed{\frac25<k<4}