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P2 2022 May Q7

A Level / Edexcel / P2

IAL 2022 May Paper · Question 7

题目

Problem

f(x)=Ax3+6x2−4x+Bf(x) = Ax^3 + 6x^2 - 4x + B

where AA and BB are constants.

Given that

  • (x+2)(x + 2) is a factor of f(x)f(x)
  • ∫35f(x) dx=176\displaystyle\int_{3}^{5} f(x) \, \mathrm{d}x = 176

find the value of AA and the value of BB.

(7)
题目中文翻译

已知多项式:

f(x)=Ax3+6x2−4x+Bf(x) = Ax^3 + 6x^2 - 4x + B

其中 AA 和 BB 是常数。

已知:

  • (x+2)(x + 2) 是 f(x)f(x) 的一个因式
  • ∫35f(x) dx=176\displaystyle\int_{3}^{5} f(x) \, \mathrm{d}x = 176

求 AA 的值和 BB 的值。 (7)

解答

解法一

思路

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我们要确定两个未知常数 AA 和 BB,需要根据已知条件列出两个独立的方程进行联立求解:

  1. 利用因式定理(Factor Theorem): 因为 (x+2)(x + 2) 是 f(x)f(x) 的一个因式,所以当 x=−2x = -2 时,函数值必须为 00,即 f(−2)=0f(-2) = 0。
f(−2)=A(−2)3+6(−2)2−4(−2)+B=0f(-2) = A(-2)^3 + 6(-2)^2 - 4(-2) + B = 0

由此可得关于 AA 和 BB 的第一个线性方程。

  1. 利用定积分条件: 计算 f(x)f(x) 在区间 [3,5][3, 5] 上的定积分。根据不定积分公式:
∫(Ax3+6x2−4x+B) dx=A4x4+2x3−2x2+Bx+c\int (Ax^3 + 6x^2 - 4x + B) \, \mathrm{d}x = \frac{A}{4}x^4 + 2x^3 - 2x^2 + Bx + c

将上限 55 和下限 33 分别代入,并相减,令其等于 176176。 这将给出关于 AA 和 BB 的第二个线性方程。

  1. 联立求解方程组: 解这个二元一次方程组,求出常数 AA 和 BB。

答题过程

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Step 1: Apply the factor theorem

Since (x+2)(x + 2) is a factor of f(x)f(x), we have f(−2)=0f(-2) = 0:

A(−2)3+6(−2)2−4(−2)+B=0A(-2)^3 + 6(-2)^2 - 4(-2) + B = 0 −8A+6(4)+8+B=0-8A + 6(4) + 8 + B = 0 −8A+24+8+B=0-8A + 24 + 8 + B = 0 −8A+B+32=0  ⟹  B=8A−32(1)-8A + B + 32 = 0 \implies B = 8A - 32 \tag{1}

Step 2: Integrate f(x)f(x) from 33 to 55

Perform the integration term-by-term:

∫35(Ax3+6x2−4x+B) dx= [A4x4+63x3−42x2+Bx]35= [A4x4+2x3−2x2+Bx]35\begin{align*} \int_{3}^{5} \Big( Ax^3 + 6x^2 - 4x + B \Big) \, \mathrm{d}x =&\, \Big[ \frac{A}{4}x^4 + \frac{6}{3}x^3 - \frac{4}{2}x^2 + Bx \Big]_{3}^{5} \\ =&\, \Big[ \frac{A}{4}x^4 + 2x^3 - 2x^2 + Bx \Big]_{3}^{5} \end{align*}

Substitute the upper limit 55 and the lower limit 33:

  • At x=5x = 5:
A4(5)4+2(5)3−2(5)2+B(5)= 6254A+2(125)−2(25)+5B= 156.25A+250−50+5B= 156.25A+200+5B\begin{align*} \frac{A}{4}(5)^4 + 2(5)^3 - 2(5)^2 + B(5) =&\, \frac{625}{4}A + 2(125) - 2(25) + 5B \\ =&\, 156.25A + 250 - 50 + 5B \\ =&\, 156.25A + 200 + 5B \end{align*}
  • At x=3x = 3:
A4(3)4+2(3)3−2(3)2+B(3)= 814A+2(27)−2(9)+3B= 20.25A+54−18+3B= 20.25A+36+3B\begin{align*} \frac{A}{4}(3)^4 + 2(3)^3 - 2(3)^2 + B(3) =&\, \frac{81}{4}A + 2(27) - 2(9) + 3B \\ =&\, 20.25A + 54 - 18 + 3B \\ =&\, 20.25A + 36 + 3B \end{align*}

Subtract the lower limit evaluation from the upper limit evaluation:

 (156.25A+200+5B)−(20.25A+36+3B)= (156.25−20.25)A+(200−36)+(5B−3B)= 136A+164+2B\begin{align*} &\, (156.25A + 200 + 5B) - (20.25A + 36 + 3B) \\ =&\, (156.25 - 20.25)A + (200 - 36) + (5B - 3B) \\ =&\, 136A + 164 + 2B \end{align*}

Equate the result to the given value of 176176:

136A+164+2B=176  ⟹  136A+2B=12(2)136A + 164 + 2B = 176 \implies 136A + 2B = 12 \tag{2}

Divide Equation 2 by 22 to simplify:

68A+B=6(3)68A + B = 6 \tag{3}

Step 3: Solve the simultaneous equations

Substitute Equation 1 (B=8A−32B = 8A - 32) into Equation 3:

68A+(8A−32)=668A + (8A - 32) = 6 76A−32=676A - 32 = 6 76A=3876A = 38 A=3876=0.5A = \frac{38}{76} = 0.5

Now substitute A=0.5A = 0.5 back into Equation 1 to find BB:

B=8(0.5)−32=4−32=−28B = 8(0.5) - 32 = 4 - 32 = -28

Therefore, the constants are:

A=0.5andB=−28A = 0.5 \quad \text{and} \quad B = -28

解法二:多项式除法

思路

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把 f(x)f(x) 除以 x+2x+2,余式必须为零。所得 A,BA,B 关系与因式定理相同,再承接定积分方程联立求解。

答题过程

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Dividing by x+2x+2 gives quotient

Ax2+(6−2A)x+(4A−16).Ax^2+(6-2A)x+(4A-16).

Indeed,

f(x)= (x+2)[Ax2+(6−2A)x +(4A−16)]+(B−8A+32).\begin{align*} f(x)=&\,(x+2)\big[Ax^2+(6-2A)x\\ &\,\hspace{2pt}+(4A-16)\big]+(B-8A+32). \end{align*}

Since x+2x+2 is a factor, the remainder is zero:

B−8A+32=0,B=8A−32.B-8A+32=0,\qquad B=8A-32.

From the same definite integral calculation as in 解法一,

136A+164+2B=176.136A+164+2B=176.

Substituting B=8A−32B=8A-32 gives

136A+164+16A−64= 176,152A= 76,A=12,B= 8(12)−32=−28.\begin{align*} 136A+164+16A-64=&\,176,\\ 152A=&\,76,\qquad A=\frac12,\\ B=&\,8\Big(\frac12\Big)-32=-28. \end{align*}