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IAL 2025 May Q7

A Level / Edexcel / P1

IAL 2025 May Paper · Question 7

题目

Problem

Figure 3 shows a sketch of part of the curve C1C_1.

Figure 3

Given that C1C_1

  • has equation y=f(x)y=f(x) where f(x)f(x) is a cubic function
  • touches the xx-axis at the origin and cuts the xx-axis at x=4x=4
  • passes through the point (10,120)(10,120)

(a) find f(x)f(x).

(3)

The curve C2C_2 has equation y=1.2x(8−x)y=1.2x(8-x).

On the following page there is a copy of Figure 3 called Diagram 1.

Diagram 1

(b) On Diagram 1 sketch a graph of the curve C2C_2.

(2)

(c) Use algebra to find the coordinates of the points where C1C_1 and C2C_2 intersect. Show each stage of your working.

(5)

解答

(a)

解法一

思路

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曲线在原点与 xx 轴相切，说明 x=0x=0 是重根；又在 x=4x=4 处穿过 xx 轴，所以另一个因子是 (x−4)(x-4)。因此设 f(x)=λx2(x−4)f(x)=\lambda x^2(x-4)，再代入点 (10,120)(10,120) 求 λ\lambda。

答题过程

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Since the curve touches the xx-axis at the origin and cuts the xx-axis at x=4x=4,

f(x)=λx2(x−4).\begin{align*} f(x)=\lambda x^2(x-4). \end{align*}

Use the point (10,120)(10,120):

120= λ(10)2(10−4)120= 600λλ= 0.2.\begin{align*} 120=&\,\lambda(10)^2(10-4)\\ 120=&\,600\lambda\\ \lambda=&\,0.2. \end{align*}

Therefore

f(x)=0.2x2(x−4).\begin{align*} f(x)=0.2x^2(x-4). \end{align*}

解法二

思路

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也可以从一般三次式出发。原点在曲线上给出常数项；在原点相切说明导数在 00 处为 00；再用 x=4x=4 和 (10,120)(10,120) 求剩下常数。

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 the curve passes through the origin,

d=0.\begin{align*} d=0. \end{align*}

Since it touches the xx-axis at the origin, the gradient there is 00:

f′(x)= 3ax2+2bx+c,f′(0)= c=0.\begin{align*} f'(x)=&\,3ax^2+2bx+c,\\ f'(0)=&\,c=0. \end{align*}

So

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

Using f(4)=0f(4)=0,

64a+16b= 0b= −4a.\begin{align*} 64a+16b=&\,0\\ b=&\,-4a. \end{align*}

Using f(10)=120f(10)=120,

1000a+100b= 1201000a−400a= 120600a= 120a= 0.2.\begin{align*} 1000a+100b=&\,120\\ 1000a-400a=&\,120\\ 600a=&\,120\\ a=&\,0.2. \end{align*}

So b=−0.8b=-0.8, and

f(x)=0.2x3−0.8x2=0.2x2(x−4).\begin{align*} f(x)=0.2x^3-0.8x^2=0.2x^2(x-4). \end{align*}

(b)

解法一

思路

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y=1.2x(8−x)y=1.2x(8-x) 是开口向下的二次曲线，根为 x=0x=0 和 x=8x=8。所以图像经过原点，并在 x=4x=4 右侧再次穿过 xx 轴。

答题过程

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The curve

y=1.2x(8−x)\begin{align*} y=1.2x(8-x) \end{align*}

is a downward-opening parabola.

Its xx-intercepts are found from

1.2x(8−x)=0,\begin{align*} 1.2x(8-x)=0, \end{align*}

so it passes through

(0,0)and(8,0).\begin{align*} (0,0)\quad\text{and}\quad(8,0). \end{align*}

The sketch should show a downward-opening parabola through the origin and through a positive xx-intercept to the right of x=4x=4.

A completed sketch is:

(c)

解法一

思路

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交点处两条曲线的 yy 值相等。建立方程后提取公因式 xx，再解二次方程。最后要把每个 xx 代回任一曲线求 yy。

答题过程

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At intersections,

1.2x(8−x)= 0.2x2(x−4).\begin{align*} 1.2x(8-x)=&\,0.2x^2(x-4). \end{align*}

Multiply by 55:

6x(8−x)= x2(x−4).\begin{align*} 6x(8-x)=&\,x^2(x-4). \end{align*}

Expand and collect terms:

48x−6x2= x3−4x2x3+2x2−48x= 0x(x2+2x−48)= 0x(x+8)(x−6)= 0.\begin{align*} 48x-6x^2=&\,x^3-4x^2\\ x^3+2x^2-48x=&\,0\\ x(x^2+2x-48)=&\,0\\ x(x+8)(x-6)=&\,0. \end{align*}

Thus

x=0,x=−8,x=6.\begin{align*} x=0,\qquad x=-8,\qquad x=6. \end{align*}

Using y=1.2x(8−x)y=1.2x(8-x):

x=0⇒y=0,x=−8⇒y=1.2(−8)(16)=−153.6,x=6⇒y=1.2(6)(2)=14.4.\begin{align*} x=0&\Rightarrow y=0,\\ x=-8&\Rightarrow y=1.2(-8)(16)=-153.6,\\ x=6&\Rightarrow y=1.2(6)(2)=14.4. \end{align*}

Therefore the points of intersection are

(0,0),(−8,−153.6),(6,14.4).\begin{align*} (0,0),\qquad (-8,-153.6),\qquad (6,14.4). \end{align*}