题目
Problem
Figure 2 shows a sketch of part of the graph with equation y=f(x) where
f(x)=21−2∣2−x∣x≥0
(a) Find f(f(6)).
(2)
(b) Solve the equation f(x)=5x.
(2)
Given that the equation f(x)=k, where k is a constant, has exactly two roots,
(c) state the set of possible values of k.
(2)
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 (6,3).
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 2 给出了方程 y=f(x) 所表示图像一部分的草图，其中
f(x)=21−2∣2−x∣x≥0
(a) 求 f(f(6))。
(b) 解方程 f(x)=5x。
已知方程 f(x)=k（其中 k 为常数）恰有两个根，
(c) 写出 k 的可能取值集合。
图像 y=f(x) 经变换得到图像 y=af(x−b)。
已知图像 y=af(x−b) 的顶点为 (6,3)。
其中 a 和 b 为常数，
(d) 求 a 和 b 的值。
解答
(a)
We have
f(x)=21−2∣2−x∣
First find f(6):
f(6)=21−2∣2−6∣
So
f(6)=21−2(4)=13
Now find f(f(6)):
f(f(6))=f(13)
Therefore
f(13)=21−2∣2−13∣
So
f(13)=21−2(11)=−1
Hence
f(f(6))=−1
(b)
We need to solve
f(x)=5x
That is,
21−2∣2−x∣=5x
Since x≥0, consider the two branches.
Case 1: 0≤x≤2
Then
∣2−x∣=2−x
So
21−2(2−x)=5x
Hence
17+2x=5x
Therefore
x=317
But this is not in the interval 0≤x≤2, so reject it.
Case 2: x>2
Then
∣2−x∣=x−2
So
21−2(x−2)=5x
Hence
25−2x=5x
Therefore
x=725
This is valid since
725>2
So
x=725
(c)
The graph has vertex at
(2,21)
The domain is
x≥0
At the left endpoint,
f(0)=21−2∣2∣=17
For a horizontal line
y=k
to meet the graph in exactly two points, it must lie below the maximum value 21 but at or above the endpoint value 17.
Therefore
17≤k<21
(d)
The original graph
y=f(x)
has vertex
(2,21)
For
y=af(x−b)
the replacement x↦x−b translates the graph b units to the right. So the vertex moves from
(2,21)
to
(2+b,21a)
We are told that the new vertex is
(6,3)
So
2+b=6
and
21a=3
Therefore
b=4
and
a=71
Hence
a=71,b=4 