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IAL 2026 Jan Q7

A Level / Edexcel / P3

IAL 2026 Jan Paper · Question 7

题目

Problem

Figure 1 shows a sketch of part of the graph with equation y=f(x)y=f(x), where

f(x)=3x114f(x)=|3x-11|-4

The vertex of the graph is at the point PP, as shown in Figure 1.

(a) State the coordinates of PP.

(2)

(b) Solve the equation f(x)=8f(x)=8.

(3)

The line ll has equation y=mxy=mx, where mm is a constant.

Given that ll intersects the graph of y=f(x)y=f(x) at exactly one point,

(c) find the possible values of mm.

(3)

The graph with equation y=f(x)y=f(x) is transformed onto the graph with equation y=af(xb)y=af(x-b), where aa and bb are constants.

Given that the vertex of the graph with equation y=af(xb)y=af(x-b) is (5,16)(5,16),

(d) find the value of aa and the value of bb.

(2)
题目中文翻译

图 1 给出了曲线 y=f(x)y=f(x) 的部分草图,其中

f(x)=3x114f(x)=|3x-11|-4

图像顶点为点 PP

(a) 写出点 PP 的坐标;

(b) 解方程 f(x)=8f(x)=8

直线 ll 的方程为 y=mxy=mx,其中 mm 为常数。

若直线 ll 与图像 y=f(x)y=f(x) 恰有一个交点,

(c) 求 mm 的可能取值;

图像 y=f(x)y=f(x) 经过变换得到图像 y=af(xb)y=af(x-b),其中 a,ba,b 为常数。

若图像 y=af(xb)y=af(x-b) 的顶点为 (5,16)(5,16)

(d) 求 aabb 的值。

解答

(a)

解法一

思路

展开

绝对值图像

y=3x114y=|3x-11|-4

的顶点出现在绝对值内部等于 00 的位置,也就是 3x11=03x-11=0。此时函数值为 4-4

答题过程

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The vertex occurs when

3x11=03x-11=0

So

x=113x=\frac{11}{3}

At this value of xx,

f(x)=04=4f(x)=|0|-4=-4

Therefore

P(113,4)\boxed{P\left(\frac{11}{3},-4\right)}

(b)

解法一

思路

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解绝对值方程时,要把

3x11=12|3x-11|=12

拆成

3x11=12or3x11=123x-11=12 \qquad\text{or}\qquad 3x-11=-12

两种情况。

答题过程

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We need to solve

f(x)=8f(x)=8

So

3x114=83x11=12\begin{align*} |3x-11|-4=&\,8 \\[2mm] |3x-11|=&\,12 \end{align*}

Hence

3x11=12or3x11=123x-11=12 \qquad\text{or}\qquad 3x-11=-12

For the first equation,

3x11=123x=23x=233\begin{align*} 3x-11=&\,12 \\[2mm] 3x=&\,23 \\[2mm] x=&\,\frac{23}{3} \end{align*}

For the second equation,

3x11=123x=1x=13\begin{align*} 3x-11=&\,-12 \\[2mm] 3x=&\,-1 \\[2mm] x=&\,-\frac13 \end{align*}

Therefore

x=13, 233\boxed{x=-\frac13,\ \frac{23}{3}}

(c)

解法一

思路

展开

把绝对值图像分成左右两条直线:

f(x)={3x+7,x<113,3x15,x113.f(x)= \begin{cases} -3x+7, & x<\frac{11}{3},\\ 3x-15, & x\geq\frac{11}{3}. \end{cases}

直线 y=mxy=mx 要与整个 V 形图像恰有一个交点。可能发生在三种情况:

  • 只与右边分支相交;
  • 只与左边分支相交;
  • 正好经过顶点,此时两条分支的交点是同一个点。

答题过程

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For the right-hand branch,

f(x)=3x15(x113)f(x)=3x-15 \qquad \left(x\geq\frac{11}{3}\right)

Intersect with y=mxy=mx:

mx=3x15(3m)x=15x=153m\begin{align*} mx=&\,3x-15 \\[2mm] (3-m)x=&\,15 \\[2mm] x=&\,\frac{15}{3-m} \end{align*}

For this to lie on the right-hand branch,

153m113\frac{15}{3-m}\geq\frac{11}{3}

This gives an intersection on the right-hand branch when

1211m<3-\frac{12}{11}\leq m<3

For the left-hand branch,

f(x)=3x+7(x<113)f(x)=-3x+7 \qquad \left(x<\frac{11}{3}\right)

Intersect with y=mxy=mx:

mx=3x+7(m+3)x=7x=7m+3\begin{align*} mx=&\,-3x+7 \\[2mm] (m+3)x=&\,7 \\[2mm] x=&\,\frac{7}{m+3} \end{align*}

This gives an intersection on the left-hand branch when

m1211orm<3m\geq-\frac{12}{11} \quad\text{or}\quad m<-3

Now consider exactly one intersection with the whole graph.

If m3m\geq3, the line intersects only the left-hand branch, so this works.

If m<3m<-3, the line intersects only the left-hand branch, so this also works.

If m=1211m=-\dfrac{12}{11}, the line passes through the vertex

(113,4)\left(\frac{11}{3},-4\right)

so the two branch intersections coincide as one point.

Therefore the possible values of mm are

m3,m=1211,m<3\boxed{m\geq3,\qquad m=-\frac{12}{11},\qquad m<-3}

(d)

解法一

思路

展开

原图像 y=f(x)y=f(x) 的顶点是

(113,4)\left(\frac{11}{3},-4\right)

变换 y=af(xb)y=af(x-b) 中,xbx-b 表示图像向右平移 bb,所以顶点的 xx 坐标变成

113+b\frac{11}{3}+b

而外面的 aa 会把 yy 坐标乘以 aa

答题过程

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The vertex of y=f(x)y=f(x) is

(113,4)\left(\frac{11}{3},-4\right)

For

y=af(xb)y=af(x-b)

the vertex becomes

(113+b, 4a)\left(\frac{11}{3}+b,\ -4a\right)

We are given that this vertex is (5,16)(5,16). Therefore

113+b=5\frac{11}{3}+b=5

and

4a=16-4a=16

Hence

b=5113=43a=4\begin{align*} b=&\,5-\frac{11}{3}=\frac{4}{3} \\[2mm] a=&\,-4 \end{align*}

Therefore

a=4,b=43\boxed{a=-4,\qquad b=\frac43}