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CIE 9709 2026 June Paper 35 Q5

A Level / CIE / P3

CIE 9709 2026 June Paper 35 Paper · Question 5

题目

Problem

A curve has equation y=sin⁡2x3+cos⁡xy=\frac{\sin2x}{3+\cos x} for 0<x<π0<x<\pi.

(a) Express dydx\frac{\mathrm{d}y}{\mathrm{d}x} as a simplified fraction in terms of cos⁡x\cos x.

(4)

(b) Hence, find an equation of the normal to the curve at the point where x=12πx=\frac12\pi. Give your answer in an exact form.

(2)
题目中文翻译

一条曲线的方程为 y=sin⁡2x3+cos⁡xy=\frac{\sin2x}{3+\cos x},其中 0<x<π0<x<\pi。

(a) 将 dydx\frac{\mathrm{d}y}{\mathrm{d}x} 表示为用 cos⁡x\cos x 表示的最简分式。

(4)

(b) 由此求曲线上 x=12πx=\frac12\pi 对应的点处的法线方程。答案写成精确形式。

(2)
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

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使用商法则后,将二倍角的正弦、余弦都换成 sin⁡x\sin x、cos⁡x\cos x,再消去 sin⁡2x\sin^2x。

答题过程

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The quotient rule gives

dydx=2cos⁡2x(3+cos⁡x)+sin⁡2xsin⁡x(3+cos⁡x)2.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{2\cos2x(3+\cos x)+\sin2x\sin x}{(3+\cos x)^2}.

With c=cos⁡xc=\cos x, use cos⁡2x=2c2−1\cos2x=2c^2-1 and sin⁡2xsin⁡x=2(1−c2)c\sin2x\sin x=2(1-c^2)c. The numerator becomes

 2(2c2−1)(3+c)+2c(1−c2)= 12c2+4c3−6−2c+2c−2c3= 2c3+12c2−6.\begin{align*} &\,2(2c^2-1)(3+c)+2c(1-c^2)\\[4mm] =&\,12c^2+4c^3-6-2c+2c-2c^3\\[4mm] =&\,2c^3+12c^2-6. \end{align*}

Therefore

dydx=2cos⁡3x+12cos⁡2x−6(3+cos⁡x)2.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{2\cos^3x+12\cos^2x-6}{(3+\cos x)^2}.

(b)

解法一

思路

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代入 x=π/2x=\pi/2 求切线梯度,法线梯度是它的负倒数。原曲线此时经过 y=0y=0。

答题过程

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At x=π/2x=\pi/2, y=0y=0 and the tangent gradient is −6/9=−2/3-6/9=-2/3. The normal gradient is 3/23/2. Thus the normal is

y=32(x−π2).y=\frac32\bigg(x-\frac\pi2\bigg).