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CIE 9709 2022 March Paper 32 Q8

A Level / CIE / P3

CIE 9709 2022 March Paper 32 Paper · Question 8

题目

Problem

(a) Find the quotient and remainder when 8x3+4x2+2x+78x^3+4x^2+2x+7 is divided by 4x2+14x^2+1.

(3)

(b) Hence find the exact value of

∫0128x3+4x2+2x+74x2+1 dx.\int_0^{\frac12}\frac{8x^3+4x^2+2x+7}{4x^2+1}\,\mathrm{d}x.
(5)
题目中文翻译

(a) 求 8x3+4x2+2x+78x^3+4x^2+2x+7 除以 4x2+14x^2+1 的商和余数。

(3)

(b) 由此求

∫0128x3+4x2+2x+74x2+1 dx\int_0^{\frac12}\frac{8x^3+4x^2+2x+7}{4x^2+1}\,\mathrm{d}x

的精确值。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

依次除最高次项,再减去对应倍数的除式,剩下常数余式。

答题过程

展开

Subtracting 2x(4x2+1)2x(4x^2+1),

(8x3+4x2+2x+7)−2x(4x2+1)=4x2+7.(8x^3+4x^2+2x+7)-2x(4x^2+1)=4x^2+7.

Subtracting one more copy of the divisor gives

(4x2+7)−(4x2+1)=6.(4x^2+7)-(4x^2+1)=6.

Therefore

8x3+4x2+2x+7=(4x2+1)(2x+1)+6.8x^3+4x^2+2x+7=(4x^2+1)(2x+1)+6.

The quotient is 2x+1\boxed{2x+1} and the remainder is 6\boxed{6}.

(b)

解法一

思路

展开

承接(a),将被积函数写成一次多项式加反正切型分式,逐项积分。

答题过程

展开

Using part (a),

8x3+4x2+2x+74x2+1=2x+1+64x2+1.\frac{8x^3+4x^2+2x+7}{4x^2+1} =2x+1+\frac6{4x^2+1}.

Since ddxtan⁡−1(2x)=2/(1+4x2)\frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}(2x)=2/(1+4x^2), the remaining integral is 3tan⁡−1(2x)3\tan^{-1}(2x). Thus

I= [x2+x+3tan⁡−1(2x)]01/2= 14+12+3π4= 34(1+π).\begin{align*} I =&\,\big[x^2+x+3\tan^{-1}(2x)\big]_0^{1/2}\\[4mm] =&\,\frac14+\frac12+\frac{3\pi}{4}\\[4mm] =&\,\boxed{\frac34(1+\pi)}. \end{align*}