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IAL 2023 Jan Q6

A Level / Edexcel / P3

IAL 2023 Jan Paper · Question 6

题目

Problem

Figure 2 shows a sketch of the graph with equation

y=3x5a2ay=\left|3x-5a\right|-2a

where aa is a positive constant.

The graph

• cuts the yy-axis at the point PP

• cuts the xx-axis at the points QQ and RR

• has a minimum point at SS

(a) Find, in simplest form in terms of aa, the coordinates of

(i) point PP

(ii) points QQ and RR

(iii) point SS

(4)

(b) Find, in simplest form in terms of aa, the values of xx for which

3x5a2a=x2a\left|3x-5a\right|-2a=\left|x-2a\right|
(4)
题目中文翻译

图 2 给出了图像

y=3x5a2ay=\left|3x-5a\right|-2a

的示意图,其中 aa 是正常数。

该图像

• 与 yy 轴交于点 PP

• 与 xx 轴交于点 QQRR

• 在点 SS 处取得最小值

(a) 用 aa 的最简形式求下列点的坐标:

(i) 点 PP

(ii) 点 QQ 和点 RR

(iii) 点 SS

(b) 用 aa 的最简形式求满足

3x5a2a=x2a\left|3x-5a\right|-2a=\left|x-2a\right|

xx 值。

解答

(a)

解法一

思路

展开

这是绝对值函数。与 yy 轴交点令 x=0x=0;与 xx 轴交点令 y=0y=0,也就是 3x5a=2a|3x-5a|=2a;最低点出现在绝对值内部等于 00 的位置。

答题过程

展开

The graph has equation

y=3x5a2a.y=|3x-5a|-2a.

For (i), find the yy-intercept.

For the yy-intercept, set x=0x=0:

y=3(0)5a2a.y=|3(0)-5a|-2a.

Since a>0a>0,

5a=5a.|-5a|=5a.

So

y=5a2a=3a.y=5a-2a=3a.

Therefore

P=(0,3a).\boxed{P=(0,3a)}.

For (ii), find the xx-intercepts.

For the xx-intercepts, set y=0y=0:

3x5a2a=0.|3x-5a|-2a=0.

So

3x5a=2a.|3x-5a|=2a.

Therefore

3x5a=2aor3x5a=2a.3x-5a=2a \qquad\text{or}\qquad 3x-5a=-2a.

From

3x5a=2a,3x-5a=2a,

we get

3x=7ax=73a.3x=7a \quad\Rightarrow\quad x=\frac73a.

From

3x5a=2a,3x-5a=-2a,

we get

3x=3ax=a.3x=3a \quad\Rightarrow\quad x=a.

Therefore the two xx-intercepts are

Q=(a,0),R=(73a,0).\boxed{Q=(a,0),\qquad R=\left(\frac73a,0\right)}.

For (iii), find the minimum point.

The minimum value of 3x5a|3x-5a| is 00, and this occurs when

3x5a=0.3x-5a=0.

So

x=53a.x=\frac53a.

At this value of xx,

y=02a=2a.y=0-2a=-2a.

Therefore

S=(53a,2a).\boxed{S=\left(\frac53a,-2a\right)}.

(b)

解法一

思路

展开

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 xx 值。

答题过程

展开

We need to solve

3x5a2a=x2a.|3x-5a|-2a=|x-2a|.

First take the case where both expressions inside the absolute values are non-negative. Then

3x5a=3x5a|3x-5a|=3x-5a

and

x2a=x2a.|x-2a|=x-2a.

So

3x5a2a=x2a.3x-5a-2a=x-2a.

Hence

3x7a=x2a.3x-7a=x-2a.

Therefore

2x=5a,2x=5a,

so

x=52a.x=\frac52a.

This value satisfies 3x5a03x-5a\ge0 and x2a0x-2a\ge0, so it is valid.

Now take the case where both expressions inside the absolute values are negative. Then

3x5a=(3x5a)|3x-5a|=-(3x-5a)

and

x2a=(x2a).|x-2a|=-(x-2a).

So

(3x5a)2a=(x2a).-(3x-5a)-2a=-(x-2a).

Simplify:

3x+5a2a=x+2a.-3x+5a-2a=-x+2a.

Thus

3x+3a=x+2a.-3x+3a=-x+2a.

Therefore

2x=a,-2x=-a,

so

x=12a.x=\frac12a.

This value satisfies 3x5a<03x-5a<0 and x2a<0x-2a<0, so it is valid.

The mixed-sign cases do not give additional valid solutions: if 3x5a03x-5a\ge0 and x2a<0x-2a<0, the equation gives x=94ax=\frac94a, which does not satisfy x<2ax<2a; the other mixed-sign condition 3x5a<03x-5a<0 and x2a0x-2a\ge0 is impossible because it would require x<53ax<\frac53a and x2ax\ge2a at the same time.

Hence the required values are

x=12a, 52a.\boxed{x=\frac12a,\ \frac52a}.