题目
Problem
Figure 2 shows a sketch of the graph with equation
y=∣3x−5a∣−2a
where a is a positive constant.
The graph
• cuts the y-axis at the point P
• cuts the x-axis at the points Q and R
• has a minimum point at S
(a) Find, in simplest form in terms of a, the coordinates of
(i) point P
(ii) points Q and R
(iii) point S
(4)
(b) Find, in simplest form in terms of a, the values of x for which
∣3x−5a∣−2a=∣x−2a∣
(4)
题目中文翻译
图 2 给出了图像
y=∣3x−5a∣−2a
的示意图,其中 a 是正常数。
该图像
• 与 y 轴交于点 P
• 与 x 轴交于点 Q 和 R
• 在点 S 处取得最小值
(a) 用 a 的最简形式求下列点的坐标:
(i) 点 P
(ii) 点 Q 和点 R
(iii) 点 S
(b) 用 a 的最简形式求满足
∣3x−5a∣−2a=∣x−2a∣
的 x 值。
解答
(a)
解法一
思路
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这是绝对值函数。与 y 轴交点令 x=0;与 x 轴交点令 y=0,也就是 ∣3x−5a∣=2a;最低点出现在绝对值内部等于 0 的位置。
答题过程
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The graph has equation
y=∣3x−5a∣−2a.
For (i), find the y-intercept.
For the y-intercept, set x=0:
y=∣3(0)−5a∣−2a.
Since a>0,
∣−5a∣=5a.
So
y=5a−2a=3a.
Therefore
P=(0,3a).
For (ii), find the x-intercepts.
For the x-intercepts, set y=0:
∣3x−5a∣−2a=0.
So
∣3x−5a∣=2a.
Therefore
3x−5a=2aor3x−5a=−2a.
From
3x−5a=2a,
we get
3x=7a⇒x=37a.
From
3x−5a=−2a,
we get
3x=3a⇒x=a.
Therefore the two x-intercepts are
Q=(a,0),R=(37a,0).
For (iii), find the minimum point.
The minimum value of ∣3x−5a∣ is 0, and this occurs when
3x−5a=0.
So
x=35a.
At this value of x,
y=0−2a=−2a.
Therefore
S=(35a,−2a).
(b)
解法一
思路
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解绝对值方程时要按符号分情况。官方评分路线等价于分别取两边绝对值内部同为非负、同为负的情况,并解出两个有效的 x 值。
答题过程
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We need to solve
∣3x−5a∣−2a=∣x−2a∣.
First take the case where both expressions inside the absolute values are non-negative. Then
∣3x−5a∣=3x−5a
and
∣x−2a∣=x−2a.
So
3x−5a−2a=x−2a.
Hence
3x−7a=x−2a.
Therefore
2x=5a,
so
x=25a.
This value satisfies 3x−5a≥0 and x−2a≥0, so it is valid.
Now take the case where both expressions inside the absolute values are negative. Then
∣3x−5a∣=−(3x−5a)
and
∣x−2a∣=−(x−2a).
So
−(3x−5a)−2a=−(x−2a).
Simplify:
−3x+5a−2a=−x+2a.
Thus
−3x+3a=−x+2a.
Therefore
−2x=−a,
so
x=21a.
This value satisfies 3x−5a<0 and x−2a<0, so it is valid.
The mixed-sign cases do not give additional valid solutions: if
3x−5a≥0 and x−2a<0, the equation gives x=49a,
which does not satisfy x<2a; the other mixed-sign condition
3x−5a<0 and x−2a≥0 is impossible because it would require
x<35a and x≥2a at the same time.
Hence the required values are
x=21a, 25a.