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CIE 9231 2023 November Paper 21 Q8

A Level / CIE / FP2

CIE 9231 2023 Nov Paper 21 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

1+cosθ+cos2θ++cos(n1)θ=12(1cosnθ+sinnθsinθ1cosθ).1+\cos\theta+\cos2\theta+\cdots+\cos(n-1)\theta=\frac{1}{2}\left(1-\cos n\theta+\frac{\sin n\theta\sin\theta}{1-\cos\theta}\right).
[7]

The diagram shows the curve with equation y=cosxy=\cos x for 0<x10<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+sin1nsin11cos1n).\int_0^1\cos x\,\mathrm{d}x<\frac{1}{2n}\left(1-\cos1+\frac{\sin\frac{1}{n}\sin1}{1-\cos\frac{1}{n}}\right).
[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

1+cosθ+cos2θ++cos(n1)θ=12(1cosnθ+sinnθsinθ1cosθ)1+\cos\theta+\cos2\theta+\cdots+\cos(n-1)\theta=\frac{1}{2}\left(1-\cos n\theta+\frac{\sin n\theta\sin\theta}{1-\cos\theta}\right)

图中显示曲线 y=cosxy=\cos x,其中 0<x10<x\le1,以及一组宽度均为 1n\frac{1}{n}nn 个矩形。

(c) 通过考虑这些矩形的面积之和,证明

01cosxdx<12n(1cos1+sin1nsin11cos1n)\int_0^1\cos x\,\mathrm{d}x<\frac{1}{2n}\left(1-\cos1+\frac{\sin\frac{1}{n}\sin1}{1-\cos\frac{1}{n}}\right)

(d) 使用类似方法,求关于 nn 的积分

01cosxdx\int_0^1\cos x\,\mathrm{d}x

的下界。

解答