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CIE 9709 2026 June Paper 32 Q5

A Level / CIE / P3

CIE 9709 2026 June Paper 32 Paper · Question 5

题目

Problem

The parametric equations of a curve are

x=2θ−tan⁡2θ,y=6θ+2ln⁡(cos⁡2θ),x=2\theta-\tan2\theta,\qquad y=6\theta+2\ln(\cos2\theta),

for 0≤θ≤14π0\le\theta\le\frac14\pi.

(a) Find an expression for dydx\frac{\mathrm{d}y}{\mathrm{d}x} in terms of tan⁡2θ\tan2\theta.

(6)

(b) Find the exact coordinates of the point on the curve at which the gradient of the tangent is equal to −1-1.

(4)
题目中文翻译

一条曲线的参数方程为

x=2θ−tan⁡2θ,y=6θ+2ln⁡(cos⁡2θ),x=2\theta-\tan2\theta,\qquad y=6\theta+2\ln(\cos2\theta),

其中 0≤θ≤14π0\le\theta\le\frac14\pi。

(a) 求用 tan⁡2θ\tan2\theta 表示的 dydx\frac{\mathrm{d}y}{\mathrm{d}x} 的表达式。

(6)

(b) 求曲线上切线斜率等于 −1-1 的点的精确坐标。

(4)
原卷通用作答要求

Answer all questions. Write your answers on the answer booklet. Follow the instructions on the front cover of the answer booklet. If you need additional answer paper, ask the invigilator for a continuation booklet.

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。在答题册上作答，并遵守答题册封面的指示。如需额外答题纸，请向监考人员索取续答册。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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先分别对参数求导。用 dy/dx=(dy/dθ)/(dx/dθ)\mathrm{d}y/\mathrm{d}x=(\mathrm{d}y/\mathrm{d}\theta)/(\mathrm{d}x/\mathrm{d}\theta)，再将正割平方换成正切平方。

答题过程

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For parameters where the curve and gradient are defined,

dxdθ= 2−2sec⁡22θ= −2tan⁡22θ,dydθ= 6+2−2sin⁡2θcos⁡2θ= 6−4tan⁡2θ.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}\theta} =&\,2-2\sec^22\theta\\[4mm] =&\,-2\tan^22\theta,\\[4mm] \frac{\mathrm{d}y}{\mathrm{d}\theta} =&\,6+2\frac{-2\sin2\theta}{\cos2\theta}\\[4mm] =&\,6-4\tan2\theta. \end{align*}

Therefore

dydx= 6−4tan⁡2θ−2tan⁡22θ= 2tan⁡2θ−3tan⁡22θ.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{6-4\tan2\theta}{-2\tan^22\theta}\\[4mm] =&\,\frac{2\tan2\theta-3}{\tan^22\theta}. \end{align*}

This finite-gradient expression applies for 0<θ<π/40<\theta<\pi/4. At θ=0\theta=0 the denominator vanishes, and at θ=π/4\theta=\pi/4 the parametric expressions are undefined.

(b)

解法一

思路

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将上一问的梯度等于 −1-1，对 tan⁡2θ\tan2\theta 解二次方程。根据参数范围选根，再代入原来的参数式。

答题过程

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Put z=tan⁡2θz=\tan2\theta. Setting the gradient to −1-1 gives

2z−3= −z2,z2+2z−3= 0,(z−1)(z+3)= 0.\begin{align*} 2z-3=&\,-z^2,\\[4mm] z^2+2z-3=&\,0,\\[4mm] (z-1)(z+3)=&\,0. \end{align*}

For 0<θ<π/40<\theta<\pi/4, z>0z>0, so z=1z=1 and θ=π/8\theta=\pi/8. Hence the coordinates are

x= 2π8−tan⁡π4= π4−1,y= 6π8+2ln⁡(cos⁡π4)= 3π4−ln⁡2.\begin{align*} x=&\,2\frac\pi8-\tan\frac\pi4\\[4mm] =&\,\frac\pi4-1,\\[4mm] y=&\,6\frac\pi8+2\ln\bigg(\cos\frac\pi4\bigg)\\[4mm] =&\,\frac{3\pi}{4}-\ln2. \end{align*}