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CIE 9709 2025 June Paper 31 Q10

A Level / CIE / P3

CIE 9709 2025 June Paper 31 Paper · Question 10

题目

Problem

(a) Find the quotient and remainder when x3+5x2−2x−15x^3+5x^2-2x-15 is divided by x2−3x^2-3.

(3)

(b) The variables xx and yy satisfy the differential equation

dydx=x3+5x2−2x−156y(x2−3).\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^3+5x^2-2x-15}{6y(x^2-3)}.

It is given that y=2y=2 when x=2x=2.

Solve the differential equation to obtain an expression for y2y^2 in terms of xx.

(5)
题目中文翻译

(a) 求 x3+5x2−2x−15x^3+5x^2-2x-15 除以 x2−3x^2-3 所得的商和余数。

(3)

(b) 变量 xx 和 yy 满足微分方程

dydx=x3+5x2−2x−156y(x2−3).\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^3+5x^2-2x-15}{6y(x^2-3)}.

已知当 x=2x=2 时,y=2y=2。

解此微分方程,求出 y2y^2 关于 xx 的表达式。

(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

先除去最高次项,再除去二次项,余式次数必须小于除式次数。

答题过程

展开

Subtracting x(x2−3)x(x^2-3) from the numerator leaves

5x2+x−15.5x^2+x-15.

Subtracting 5(x2−3)5(x^2-3) then leaves xx. Therefore

x3+5x2−2x−15= (x+5)(x2−3) +x.\begin{align*} x^3+5x^2-2x-15=&\,(x+5)(x^2-3)\\[4mm] &\,\hspace{2pt}+x. \end{align*}

The quotient is x+5\boxed{x+5} and the remainder is x\boxed{x}.

(b)

解法一

思路

展开

分离变量后利用(a)的商与余式拆分。余下的分子是分母导数的一半,积分产生对数。

答题过程

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Separating variables and using (a),

6y dy=(x+5+xx2−3) dx.6y\,\mathrm{d}y =\bigg(x+5+\frac{x}{x^2-3}\bigg)\,\mathrm{d}x.

For the last term, let u=x2−3u=x^2-3, so du=2x dx\mathrm{d}u=2x\,\mathrm{d}x and ∫x/(x2−3) dx=12ln⁡∣x2−3∣+C1\int x/(x^2-3)\,\mathrm{d}x=\tfrac12\ln|x^2-3|+C_1. Integrating both sides gives

3y2=12x2+5x+12ln⁡∣x2−3∣+C.3y^2=\frac12x^2+5x+\frac12\ln|x^2-3|+C.

The initial condition x=2x=2, y=2y=2 gives

12=2+10+12ln⁡1+C⟹C=0.12=2+10+\frac12\ln1+C \quad\Longrightarrow\quad C=0.

On the branch containing x=2x=2, x>3x>\sqrt3, so x2−3>0x^2-3>0. Hence

y2=x26+53x+16ln⁡(x2−3).\boxed{y^2=\frac{x^2}{6}+\frac53x+\frac16\ln(x^2-3)}.

The initial condition selects y>0y>0 on the interval around x=2x=2 where the right-hand side remains positive.