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IAL 2024 May R Q8

A Level / Edexcel / P1

IAL 2024 May (R) Paper · Question 8

题目

Problem

A curve CC with equation y=f(x)y=f(x) passes through the point R(4,13)R(4,13).

Given that

f′(x)=2(x−3)(3x+2)\begin{align*} f'(x)=2(x-3)(3x+2) \end{align*}

(a) use integration to find f(x)f(x), giving your answer in simplest form.

(5)

(b) Given that f(x)f(x) can be written in the form

(x−3)2(px+q)\begin{align*} (x-3)^2(px+q) \end{align*}

find the value of the constant pp and the value of the constant qq.

(2)

(c) Sketch the graph of y=f(2x)y=f(2x), showing the coordinates of any points where the curve touches or crosses the coordinate axes.

(4)

解答

(a)

解法一

思路

展开

先展开 f′(x)f'(x)，再积分得到 f(x)f(x)。积分常数用点 R(4,13)R(4,13) 求出。

答题过程

展开

Expand the derivative:

f′(x)= 2(x−3)(3x+2)= 2(3x2−7x−6)= 6x2−14x−12.\begin{align*} f'(x) =&\,2(x-3)(3x+2)\\ =&\,2(3x^2-7x-6)\\ =&\,6x^2-14x-12. \end{align*}

Integrate:

f(x)= ∫(6x2−14x−12) dx= 2x3−7x2−12x+c.\begin{align*} f(x) =&\,\int(6x^2-14x-12)\,dx\\ =&\,2x^3-7x^2-12x+c. \end{align*}

Use R(4,13)R(4,13):

13= 2(4)3−7(4)2−12(4)+c13= 128−112−48+c13= −32+cc= 45.\begin{align*} 13=&\,2(4)^3-7(4)^2-12(4)+c\\ 13=&\,128-112-48+c\\ 13=&\,-32+c\\ c=&\,45. \end{align*}

Therefore

f(x)=2x3−7x2−12x+45.\begin{align*} f(x)=2x^3-7x^2-12x+45. \end{align*}

(b)

解法一

思路

展开

把 (x−3)2(px+q)(x-3)^2(px+q) 展开，并与 (a) 中的三次式比较系数。也可以直接因式分解得到 (x−3)2(2x+5)(x-3)^2(2x+5)。

答题过程

展开

From part (a),

f(x)=2x3−7x2−12x+45.\begin{align*} f(x)=2x^3-7x^2-12x+45. \end{align*}

Factorise:

2x3−7x2−12x+45= (x−3)2(2x+5).\begin{align*} 2x^3-7x^2-12x+45 =&\,(x-3)^2(2x+5). \end{align*}

Therefore

p=2,q=5.\begin{align*} p=2,\qquad q=5. \end{align*}

(c)

解法一

思路

展开

由 (b) 得

f(x)=(x−3)2(2x+5).\begin{align*} f(x)=(x-3)^2(2x+5). \end{align*}

所以

f(2x)=(2x−3)2(4x+5).\begin{align*} f(2x)=(2x-3)^2(4x+5). \end{align*}

平方因子表示曲线在对应的 xx 截距处“触碰” xx 轴；一次因子表示曲线“穿过” xx 轴。

答题过程

展开

Using

f(x)=(x−3)2(2x+5),\begin{align*} f(x)=(x-3)^2(2x+5), \end{align*}

we have

f(2x)= (2x−3)2(2(2x)+5)= (2x−3)2(4x+5).\begin{align*} f(2x) =&\,(2x-3)^2(2(2x)+5)\\ =&\,(2x-3)^2(4x+5). \end{align*}

The xx-intercepts are found from

(2x−3)2(4x+5)=0.\begin{align*} (2x-3)^2(4x+5)=0. \end{align*}

So

2x−3=0or4x+5=0.\begin{align*} 2x-3=0 \quad\text{or}\quad 4x+5=0. \end{align*}

Hence

x=32orx=−54.\begin{align*} x=\frac32 \quad\text{or}\quad x=-\frac54. \end{align*}

At x=32x=\frac32, the repeated factor means the curve touches the xx-axis:

(32,0).\begin{align*} \left(\frac32,0\right). \end{align*}

At x=−54x=-\frac54, the curve crosses the xx-axis:

(−54,0).\begin{align*} \left(-\frac54,0\right). \end{align*}

The yy-intercept is

f(0)=45,\begin{align*} f(0)=45, \end{align*}

so for y=f(2x)y=f(2x) the yy-intercept is still

(0,45).\begin{align*} (0,45). \end{align*}

The sketch is a positive cubic curve, crossing the xx-axis at (−54,0)\left(-\frac54,0\right), touching the xx-axis at (32,0)\left(\frac32,0\right), and crossing the yy-axis at (0,45)(0,45).