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P2 2025 May A Q8

A Level / Edexcel / P2

IAL 2025 May A Paper · Question 8

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

A curve CC has equation y=f(x)y=f(x) where f(x)f(x) is a polynomial in xx.

Given

f′(x)=2(3x−2)(x+5)f'(x)=2(3x-2)(x+5)

(a) deduce the range of values of xx for which yy is decreasing.

(2)

Given further that CC cuts the xx-axis at 72\frac72

(b) state a factor of f(x)f(x), giving the answer in the form (Ax+B)(Ax+B) where AA and BB are integers.

(1)

(c) Hence find an expression for f(x)f(x) in a fully factorised form.

(6)
题目中文翻译

本题必须展示计算过程的所有步骤。

完全依赖计算器技术的解答不予接受。

曲线 CC 的方程为 y=f(x)y=f(x),其中 f(x)f(x) 是关于 xx 的多项式。

已知

f′(x)=2(3x−2)(x+5)f'(x)=2(3x-2)(x+5)

(a) 推出 yy 递减时 xx 的取值范围。

(2)

还已知 CC 在 72\frac72 处与 xx 轴相交。

(b) 写出 f(x)f(x) 的一个因式,答案写成 (Ax+B)(Ax+B) 的形式,其中 AA、BB 是整数。

(1)

(c) 由此求 f(x)f(x) 的表达式,写成完全因式分解的形式。

(6)

解答

(a)

解法一

思路

展开

yy decreasing 表示 f′(x)<0f'(x)<0。先找导数的两个零点,再判断符号。

答题过程

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The critical values are found from

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

So

x=23orx=−5.x=\frac23 \quad\text{or}\quad x=-5.

Since the quadratic 2(3x−2)(x+5)2(3x-2)(x+5) has positive leading coefficient, it is negative between its two roots.

Therefore, yy is decreasing when

−5<x<23.\begin{align*} -5<x<\frac23. \end{align*}

(b)

解法一

思路

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曲线在 x=72x=\frac72 处与 xx 轴相交,所以 f(72)=0f\bigg(\frac72\bigg)=0,对应因式是 2x−72x-7。

答题过程

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Since CC cuts the xx-axis at x=72x=\frac72,

f(72)=0.\begin{align*} f\bigg(\frac72\bigg)=0. \end{align*}

Therefore, a factor of f(x)f(x) is

2x−7.\begin{align*} 2x-7. \end{align*}

(c)

解法一

思路

展开

先把 f′(x)f'(x) 展开并积分,得到 f(x)f(x),但会有常数 cc。再用 f(72)=0f\bigg(\frac72\bigg)=0 求 cc,最后因式分解。

答题过程

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First expand f′(x)f'(x):

f′(x)= 2(3x−2)(x+5)= 2(3x2+15x−2x−10)= 6x2+26x−20.\begin{align*} f'(x)=&\,2(3x-2)(x+5)\\ =&\,2(3x^2+15x-2x-10)\\ =&\,6x^2+26x-20. \end{align*}

Integrate:

f(x)= ∫(6x2+26x−20) dx= 2x3+13x2−20x+c.\begin{align*} f(x)=&\,\int(6x^2+26x-20)\,\mathrm{d}x\\ =&\,2x^3+13x^2-20x+c. \end{align*}

Since f(72)=0f\bigg(\frac72\bigg)=0,

 2(72)3+13(72)2 −20(72)+c= 0.\begin{align*} &\,2\bigg(\frac72\bigg)^3+13\bigg(\frac72\bigg)^2\\ &\,\hspace{2pt}-20\bigg(\frac72\bigg)+c\\ =&\,0. \end{align*}

Simplifying,

343+6374−70+c= 0245−70+c= 0c= −175.\begin{align*} \frac{343+637}{4}-70+c=&\,0\\ 245-70+c=&\,0\\ c=&\,-175. \end{align*}

So

f(x)=2x3+13x2−20x−175.\begin{align*} f(x)=2x^3+13x^2-20x-175. \end{align*}

Using the factor 2x−72x-7, divide the polynomial. The first quotient term is x2x^2:

 2x3+13x2−20x−175 −(2x3−7x2)= 20x2−20x−175.\begin{align*} &\,2x^3+13x^2-20x-175\\ &\,\hspace{2pt}-(2x^3-7x^2)\\ =&\,20x^2-20x-175. \end{align*}

The next quotient term is 10x10x:

 20x2−20x−175−(20x2−70x)= 50x−175.\begin{align*} &\,20x^2-20x-175-(20x^2-70x)\\ =&\,50x-175. \end{align*}

The final quotient term is 2525:

50x−175−(50x−175)=0.50x-175-(50x-175)=0.

So

f(x)= (2x−7)(x2+10x+25)= (2x−7)(x+5)2.\begin{align*} f(x)=&\,(2x-7)(x^2+10x+25)\\ =&\,(2x-7)(x+5)^2. \end{align*}

Therefore,

f(x)=(2x−7)(x+5)2.\begin{align*} f(x)=(2x-7)(x+5)^2. \end{align*}

解法二

思路

展开

先把导数展开并积分,保留积分常数 cc。由(b)的因式设 f(x)=(2x−7)(Ax2+Bx+C)f(x)=(2x-7)(Ax^2+Bx+C),与积分结果比较系数,求出二次因式和常数。这条路线也明确展示了所需积分过程。

答题过程

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First expand and integrate:

f′(x)= 2(3x−2)(x+5)= 6x2+26x−20,f(x)= ∫(6x2+26x−20) dx= 2x3+13x2−20x+c.\begin{align*} f'(x)=&\,2(3x-2)(x+5)\\ =&\,6x^2+26x-20,\\ f(x)=&\,\int(6x^2+26x-20)\,\mathrm{d}x\\ =&\,2x^3+13x^2-20x+c. \end{align*}

From part (b), 2x−72x-7 is a factor. Let

f(x)=(2x−7)(Ax2+Bx+C).f(x)=(2x-7)(Ax^2+Bx+C).

Expanding this gives

f(x)= 2Ax3+(2B−7A)x2 +(2C−7B)x−7C.\begin{align*} f(x)=&\,2Ax^3+(2B-7A)x^2\\ &\,\hspace{2pt}+(2C-7B)x-7C. \end{align*}

Compare with the integrated expression:

2A= 2A= 1,2B−7A= 132B−7= 13B= 10,2C−7B= −202C−70= −20C= 25.\begin{align*} 2A=&\,2\\ A=&\,1,\\[2mm] 2B-7A=&\,13\\ 2B-7=&\,13\\ B=&\,10,\\[2mm] 2C-7B=&\,-20\\ 2C-70=&\,-20\\ C=&\,25. \end{align*}

The constant term is c=−7C=−175c=-7C=-175. Therefore,

f(x)= (2x−7)(x2+10x+25)= (2x−7)(x+5)2.\begin{align*} f(x)=&\,(2x-7)(x^2+10x+25)\\ =&\,(2x-7)(x+5)^2. \end{align*}

Equivalently, dividing the integrated expression before finding cc gives quotient x2+10x+25x^2+10x+25 and remainder c+175c+175. Since 2x−72x-7 is a factor, that remainder is zero, again giving c=−175c=-175.