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CIE 9231 2023 November Paper 23 Q8

A Level / CIE / FP2

CIE 9231 2023 Nov Paper 23 Paper · Question 8

题目

Problem

(a) State the sum of the series 1+z+z2++zn11+z+z^2+\cdots+z^{n-1}, for z1z\ne1.

[1]

(b) By letting z=cosθ+isinθz=\cos\theta+\mathrm{i}\sin\theta, where cosθ1\cos\theta\ne1, show that

\begin{aligned} 1+\cos\theta+\cos 2\theta+\cdots+\cos(n-1)\theta =&\,\frac{1}{2}\left(1-\cos n\theta\\ &\,+\frac{\sin n\theta\sin\theta}{1-\cos\theta}\right). \end{aligned}
[7]

The diagram shows the curve with equation y=cosxy=\cos x for 0x10\le x\le1, together with a set of nn rectangles of width 1n\frac{1}{n}.

(c) By considering the sum of the areas of these rectangles, show that

01cosxdx<12n(1cos1+sin1sin(1n)1cos(1n)).\begin{aligned} \int_0^1\cos x\,\mathrm{d}x<&\,\frac{1}{2n}\Bigg(1-\cos 1\\ &\,+\frac{\sin 1\sin\left(\frac{1}{n}\right)}{1-\cos\left(\frac{1}{n}\right)}\Bigg). \end{aligned}
[4]

(d) Use a similar method to find, in terms of nn, a lower bound for 01cosxdx\int_0^1\cos x\,\mathrm{d}x.

[3]
题目中文翻译

(a) 写出级数 1+z+z2++zn11+z+z^2+\cdots+z^{n-1} 的和,其中 z1z\ne1

(b) 令 z=cosθ+isinθz=\cos\theta+\mathrm{i}\sin\theta,其中 cosθ1\cos\theta\ne1,证明

\begin{aligned} 1+\cos\theta+\cos 2\theta+\cdots+\cos(n-1)\theta ={}&\,\frac{1}{2}\left(1-\cos n\theta\\ &\,+\frac{\sin n\theta\sin\theta}{1-\cos\theta}\right) \end{aligned}

图中显示曲线 y=cosxy=\cos x,其中 0x10\le x\le1,并显示了一组宽度为 1n\frac{1}{n}nn 个矩形。

(c) 考虑这些矩形面积之和,证明

01cosxdx<12n(1cos1+sin1sin(1n)1cos(1n))\begin{aligned} \int_0^1\cos x\,\mathrm{d}x<&\,\frac{1}{2n}\Bigg(1-\cos 1\\ &\,+\frac{\sin 1\sin\left(\frac{1}{n}\right)}{1-\cos\left(\frac{1}{n}\right)}\Bigg) \end{aligned}

(d) 使用类似方法,求出 01cosxdx\int_0^1\cos x\,\mathrm{d}x 的一个关于 nn 的下界。

解答