题目
Problem
Figure 1 shows a sketch of part of the graph with equation y=f(x), where
f(x)=∣3x−13∣+5x∈R
The vertex of the graph is at point P, as shown in Figure 1.
(a) State the coordinates of P.
(2)
(b) (i) State the range of f.
(ii) Find the value of ff(4)
(2)
(c) Solve, using algebra and showing your working,
16−2x>∣3x−13∣+5
(4)
The graph with equation y=f(x) is transformed onto the graph with equation y=af(x+b).
The vertex of the graph with equation y=af(x+b) is (4,20).
Given that a and b are constants,
(d) find the value of a and the value of b.
(2)
题目中文翻译
图 1 给出了图像 y=f(x) 的一部分，其中
f(x)=∣3x−13∣+5x∈R
图中的点 P 是图像的顶点。
(a) 写出点 P 的坐标。
(b) (i) 写出 f 的值域。
(ii) 求 ff(4) 的值。
(c) 用代数方法并写出过程，解不等式
16−2x>∣3x−13∣+5
图像 y=f(x) 经变换得到图像 y=af(x+b)。
已知图像 y=af(x+b) 的顶点是 (4,20)。
设 a、b 为常数，
(d) 求 a 与 b 的值。
解答
(a)
The vertex occurs when the expression inside the modulus is zero:
3x−13=0
So
x=313
At this point,
f(x)=0+5=5
Therefore
P=(313,5)
(b)(i)
Since
∣3x−13∣≥0
we have
f(x)=∣3x−13∣+5≥5
So the range is
f(x)≥5
(b)(ii)
First find f(4):
f(4)=∣12−13∣+5=6
Then
ff(4)=f(6)
Now
f(6)=∣18−13∣+5=10
Therefore
ff(4)=10
(c)
We need to solve
16−2x>∣3x−13∣+5
Subtract 5 from both sides:
11−2x>∣3x−13∣
This requires
11−2x>0
Now use
∣3x−13∣<11−2x
which gives
−(11−2x)<3x−13<11−2x
Solve the left inequality:
−11+2x<3x−13
so
x>2
Solve the right inequality:
3x−13<11−2x
so
5x<24
and hence
x<524
Therefore
2<x<524
(d)
The vertex of y=f(x) is
(313,5)
For
y=af(x+b)
the vertex occurs when
x+b=313
The new vertex has x coordinate 4, so
4+b=313
Thus
b=31
The new vertex has y coordinate 20, so
5a=20
Hence
a=4
Therefore
a=4,b=31 