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CIE 9709 2025 June Paper 32 Q8

A Level / CIE / P3

CIE 9709 2025 June Paper 32 Paper · Question 8

题目

Problem

The variables xx and θ\theta satisfy the differential equation

sin⁡2θdxdθ=(4x+3)cos⁡2θ,\sin2\theta\frac{\mathrm{d}x}{\mathrm{d}\theta}=(4x+3)\cos2\theta,

and x=0x=0 when θ=112π\theta=\frac1{12}\pi.

Solve the differential equation and obtain an expression for xx in terms of θ\theta.

(7)
题目中文翻译

变量 xx 和 θ\theta 满足微分方程

sin⁡2θdxdθ=(4x+3)cos⁡2θ,\sin2\theta\frac{\mathrm{d}x}{\mathrm{d}\theta}=(4x+3)\cos2\theta,

且当 θ=112π\theta=\frac1{12}\pi 时,x=0x=0。

解此微分方程,并求出 xx 关于 θ\theta 的表达式。

(7)
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

解法一

思路

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分离变量后,左边是一次函数倒数,右边是对数正弦的导数。初值确定比例常数,再把 xx 显式解出。

答题过程

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Separating variables,

14x+3 dx=cos⁡2θsin⁡2θ dθ.\frac1{4x+3}\,\mathrm{d}x =\frac{\cos2\theta}{\sin2\theta}\,\mathrm{d}\theta.

The derivatives of 4x+34x+3 and sin⁡2θ\sin2\theta are 44 and 2cos⁡2θ2\cos2\theta, respectively. Thus the logarithmic integrals give

14ln⁡∣4x+3∣=12ln⁡∣sin⁡2θ∣+C.\frac14\ln|4x+3| =\frac12\ln|\sin2\theta|+C.

Using x=0x=0 at θ=π/12\theta=\pi/12,

14ln⁡3=12ln⁡12+C,\frac14\ln3=\frac12\ln\frac12+C,

so C=14ln⁡12C=\frac14\ln12.

On the interval 0<θ<π/20<\theta<\pi/2 containing the initial point, sin⁡2θ>0\sin2\theta>0 and the initial branch has 4x+3>04x+3>0. Therefore

ln⁡(4x+3)=2ln⁡(sin⁡2θ)+ln⁡12.\ln(4x+3)=2\ln(\sin2\theta)+\ln12.

Exponentiating,

4x+3=12sin⁡22θ.4x+3=12\sin^22\theta.

Hence

x=3sin⁡22θ−34.\boxed{x=3\sin^22\theta-\frac34}.