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IAL 2025 May A Q8

A Level / Edexcel / P2

IAL 2025 May A Paper · Question 8

题目

Problem

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)

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

展开

曲线在 x=72x=\frac72 处切 xx-axis，所以 f(72)=0f\left(\frac72\right)=0，对应因式是 2x−72x-7。

答题过程

展开

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

f(72)=0.\begin{align*} f\left(\frac72\right)=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\left(\frac72\right)=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)\,dx\\ =&\,2x^3+13x^2-20x+c. \end{align*}

Since f(72)=0f\left(\frac72\right)=0,

2(72)3+13(72)2−20(72)+c= 03434+6374−70+c= 0245−70+c= 0c= −175.\begin{align*} 2\left(\frac72\right)^3 +13\left(\frac72\right)^2 -20\left(\frac72\right)+c=&\,0\\ \frac{343}{4}+\frac{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,

2x3+13x2−20x−175= (2x−7)(x2+10x+25)= (2x−7)(x+5)2.\begin{align*} 2x^3+13x^2-20x-175 =&\,(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

既然已知 2x−72x-7 是 f(x)f(x) 的因式，可以设 f(x)=(2x−7)(Ax2+Bx+C)f(x)=(2x-7)(Ax^2+Bx+C)。把它展开、求导，再和题目给出的 f′(x)f'(x) 比较系数。

答题过程

展开

Since 2x−72x-7 is a factor of f(x)f(x), let

f(x)=(2x−7)(Ax2+Bx+C).\begin{align*} f(x)=(2x-7)(Ax^2+Bx+C). \end{align*}

Expand:

f(x)= 2Ax3+2Bx2+2Cx −7Ax2−7Bx−7C= 2Ax3+(2B−7A)x2 +(2C−7B)x−7C.\begin{align*} f(x)=&\,2Ax^3+2Bx^2+2Cx\\ &\,\hspace{2pt}-7Ax^2-7Bx-7C\\ =&\,2Ax^3+(2B-7A)x^2\\ &\,\hspace{2pt}+(2C-7B)x-7C. \end{align*}

Differentiate:

f′(x)= 6Ax2+2(2B−7A)x +(2C−7B).\begin{align*} f'(x)=&\,6Ax^2+2(2B-7A)x\\ &\,\hspace{2pt}+(2C-7B). \end{align*}

But

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

Compare coefficients:

6A= 6⇒A=1,2(2B−7A)= 26⇒2B−7=13,⇒B=10,2C−7B= −20⇒2C−70=−20,⇒C=25.\begin{align*} 6A=&\,6 &&\Rightarrow A=1,\\ 2(2B-7A)=&\,26 &&\Rightarrow 2B-7=13,\\ &&&\Rightarrow B=10,\\ 2C-7B=&\,-20 &&\Rightarrow 2C-70=-20,\\ &&&\Rightarrow C=25. \end{align*}

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*}