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CIE 9709 2023 November Paper 12 Q5

A Level / CIE / P1

CIE 9709 2023 Nov Paper 12 Paper · Question 5

题目

Problem

The first, second and third terms of a geometric progression are sin⁡θ\sin\theta, cos⁡θ\cos\theta and 2−sin⁡θ2-\sin\theta respectively, where θ\theta radians is an acute angle.

(a) Find the value of θ\theta.

(3)

(b) Using this value of θ\theta, find the sum of the first 10 terms of the progression. Give the answer in the form bc−1\dfrac{b}{\sqrt{c}-1}, where bb and cc are integers to be found.

(3)
题目中文翻译

一个等比数列的第一、第二和第三项分别为 sin⁡θ\sin\theta、cos⁡θ\cos\theta 和 2−sin⁡θ2-\sin\theta,其中 θ\theta 弧度是一个锐角。

(a) 求 θ\theta 的值。

(3)

(b) 利用这个 θ\theta 值,求该数列前 10 项的和。答案写成 bc−1\dfrac{b}{\sqrt{c}-1} 的形式,其中 bb 和 cc 是需要求出的整数。

(3)
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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等比中项的平方等于首项与第三项的乘积。用平方关系把方程全部写成正弦。

答题过程

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The geometric progression gives

cos⁡2θ=sin⁡θ(2−sin⁡θ).\cos^2\theta=\sin\theta(2-\sin\theta).

Replacing cos⁡2θ\cos^2\theta by 1−sin⁡2θ1-\sin^2\theta gives 1=2sin⁡θ1=2\sin\theta. As θ\theta is acute, θ=π/6\theta=\pi/6 radians.

(b)

解法一

思路

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用 (a) 求首项和公比。计算公比的十次方后,直接用有限等比和公式。

答题过程

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The first term is 1/21/2 and the ratio is (3/2)/(1/2)=3(\sqrt3/2)/(1/2)=\sqrt3. Hence

S10= 12(3)10−13−1= 1213−1.\begin{align*} S_{10}=&\,\frac12\frac{(\sqrt3)^{10}-1}{\sqrt3-1}\\ =&\,\frac{121}{\sqrt3-1}. \end{align*}

Thus b=121b=121 and c=3c=3.