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IAL 2026 Jan A Q10

A Level / Edexcel / P1

IAL 2026 Jan A Paper · Question 10

题目

Problem

Figure 4 shows a sketch of the curve C1C_1 with equation y=f(x)y=f(x) where f(x)f(x) is a cubic function in xx.

Figure 4

The curve

  • cuts the xx and yy axes at (−1,0)(-1,0) and (0,12)(0,12) respectively
  • has a minimum turning point at (4,0)(4,0)

as shown in Figure 4.

(a) Find f(x)f(x)

(4)

The curve C2C_2 has equation y=g(x)y=g(x) where g(x)=6(x+1)(x−2)g(x)=6(x+1)(x-2)

(b) Find the exact xx coordinates of the points of intersection of curves C1C_1 and C2C_2.

(Solutions relying on calculator technology are not acceptable.)

(4)

解答

(a)

解法一

思路

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三次函数在 x=−1x=-1 截 xx 轴，所以有因式 (x+1)(x+1)。又因为 (4,0)(4,0) 是最低转折点且在 xx 轴上，所以 x=4x=4 是重根，对应因式 (x−4)2(x-4)^2。最后用 yy 轴截距 (0,12)(0,12) 求常数倍。

答题过程

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Since the curve crosses the xx-axis at (−1,0)(-1,0) and has a turning point at (4,0)(4,0),

f(x)=a(x+1)(x−4)2.\begin{align*} f(x)=a(x+1)(x-4)^2. \end{align*}

Using the point (0,12)(0,12),

12= a(0+1)(0−4)212= 16aa= 34.\begin{align*} 12=&\,a(0+1)(0-4)^2\\ 12=&\,16a\\ a=&\,\frac34. \end{align*}

Therefore

f(x)=34(x+1)(x−4)2.\begin{align*} f(x)=\frac34(x+1)(x-4)^2. \end{align*}

解法二

思路

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也可以从一般三次函数 ax3+bx2+cx+dax^3+bx^2+cx+d 出发。四个条件分别来自 yy 轴截距、两个经过点，以及 x=4x=4 处斜率为 00。

答题过程

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Let

f(x)=ax3+bx2+cx+d.\begin{align*} f(x)=ax^3+bx^2+cx+d. \end{align*}

Since (0,12)(0,12) lies on the curve,

d=12.\begin{align*} d=12. \end{align*}

Since (−1,0)(-1,0) lies on the curve,

−a+b−c+12=0.\begin{align*} -a+b-c+12=0. \end{align*}

Since (4,0)(4,0) lies on the curve,

64a+16b+4c+12=0.\begin{align*} 64a+16b+4c+12=0. \end{align*}

Also,

f′(x)=3ax2+2bx+c.\begin{align*} f'(x)=3ax^2+2bx+c. \end{align*}

The curve has a turning point at x=4x=4, so

48a+8b+c=0.\begin{align*} 48a+8b+c=0. \end{align*}

Solving these simultaneous equations gives

a=34,b=−214,c=6,d=12.\begin{align*} a=\frac34,\qquad b=-\frac{21}{4},\qquad c=6,\qquad d=12. \end{align*}

Therefore

f(x)=34x3−214x2+6x+12.\begin{align*} f(x)=\frac34x^3-\frac{21}{4}x^2+6x+12. \end{align*}

This is equivalent to

f(x)=34(x+1)(x−4)2.\begin{align*} f(x)=\frac34(x+1)(x-4)^2. \end{align*}

(b)

解法一

思路

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交点满足 f(x)=g(x)f(x)=g(x)。因为两个式子都有因式 (x+1)(x+1)，所以 x=−1x=-1 直接给出一个交点；其余交点可以约去共同因式后解二次方程。

答题过程

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At intersections,

34(x+1)(x−4)2=6(x+1)(x−2).\begin{align*} \frac34(x+1)(x-4)^2=6(x+1)(x-2). \end{align*}

One solution is

x=−1.\begin{align*} x=-1. \end{align*}

For x≠−1x\neq-1, divide by (x+1)(x+1):

34(x−4)2= 6(x−2)3(x−4)2= 24(x−2)(x−4)2= 8(x−2).\begin{align*} \frac34(x-4)^2=&\,6(x-2)\\ 3(x-4)^2=&\,24(x-2)\\ (x-4)^2=&\,8(x-2). \end{align*}

Expand and rearrange:

x2−8x+16= 8x−16x2−16x+32= 0.\begin{align*} x^2-8x+16=&\,8x-16\\ x^2-16x+32=&\,0. \end{align*}

Complete the square:

x2−16x+32= 0(x−8)2−64+32= 0(x−8)2= 32x−8= ±42.\begin{align*} x^2-16x+32=&\,0\\ (x-8)^2-64+32=&\,0\\ (x-8)^2=&\,32\\ x-8=&\,\pm4\sqrt2. \end{align*}

Therefore the exact xx coordinates are

x=−1,x=8−42,x=8+42.\begin{align*} x=-1,\qquad x=8-4\sqrt2,\qquad x=8+4\sqrt2. \end{align*}