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CIE 9709 2025 November Paper 35 Q6

A Level / CIE / P3

CIE 9709 2025 Nov Paper 35 Paper · Question 6

题目

Problem

A curve has equation y=tan⁡x5+2sin⁡xy=\frac{\tan x}{5+2\sin x}.

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

(4)

(b) Show that the curve has no stationary points.

(2)
题目中文翻译

一条曲线的方程为 y=tan⁡x5+2sin⁡xy=\frac{\tan x}{5+2\sin x}。

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

(4)

(b) 证明该曲线没有驻点。

(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

展开

先用商法则,再把正切和正割写成正弦、余弦,通分后使用平方关系。

答题过程

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By the quotient rule,

dydx=(5+2sin⁡x)sec⁡2x−2cos⁡xtan⁡x(5+2sin⁡x)2.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{(5+2\sin x)\sec^2x-2\cos x\tan x} {(5+2\sin x)^2}.

Using tan⁡x=sin⁡x/cos⁡x\tan x=\sin x/\cos x and sec⁡2x=1/cos⁡2x\sec^2x=1/\cos^2x,

dydx= 5+2sin⁡x−2sin⁡xcos⁡2xcos⁡2x(5+2sin⁡x)2= 5+2sin⁡x(1−cos⁡2x)cos⁡2x(5+2sin⁡x)2= 5+2sin⁡3x(1−sin⁡2x)(5+2sin⁡x)2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{5+2\sin x-2\sin x\cos^2x} {\cos^2x(5+2\sin x)^2}\\ =&\,\frac{5+2\sin x(1-\cos^2x)} {\cos^2x(5+2\sin x)^2}\\ =&\,\boxed{\frac{5+2\sin^3x} {(1-\sin^2x)(5+2\sin x)^2}}. \end{align*}

(b)

解法一

思路

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用正弦的范围给分子估计下界,同时检查曲线定义域内分母为正。

答题过程

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Since −1⩽sin⁡x⩽1-1\leqslant\sin x\leqslant1,

3⩽5+2sin⁡3x⩽7.3\leqslant5+2\sin^3x\leqslant7.

Thus the numerator in (a) is never zero. On the domain of the curve, cos⁡x≠0\cos x\ne0, so 1−sin⁡2x>01-\sin^2x>0; also 5+2sin⁡x⩾3>05+2\sin x\geqslant3>0.

Therefore dy/dx>0\mathrm{d}y/\mathrm{d}x>0 at every point of the curve, and there are no stationary points\boxed{\text{there are no stationary points}}.