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CIE 9709 2025 June Paper 61 Q7

A Level / CIE / S2

CIE 9709 2025 June Paper 61 Paper · Question 7

题目

Problem

XX is a random variable with probability density function given by

f(x)={(1+cos⁡πx)0≤x≤1,0otherwise.f(x)= \begin{cases} (1+\cos\pi x) & 0\le x\le 1,\\ 0 & \text{otherwise}. \end{cases}

(a) Show that P(X<12)=12+1π\mathrm{P}\big(X<\frac12\big)=\frac12+\frac1\pi.

(3)

(b) Show that E(X)=12−2π2\mathrm{E}(X)=\frac12-\frac{2}{\pi^2}.

(5)
题目中文翻译

随机变量 XX 的概率密度函数为

f(x)={(1+cos⁡πx)0≤x≤1,0其他情况.f(x)= \begin{cases} (1+\cos\pi x) & 0\le x\le 1,\\ 0 & \text{其他情况}. \end{cases}

(a) 证明 P(X<12)=12+1π\mathrm{P}\big(X<\frac12\big)=\frac12+\frac1\pi。

(3)

(b) 证明 E(X)=12−2π2\mathrm{E}(X)=\frac12-\frac{2}{\pi^2}。

(5)
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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直接积分密度到一半处，代入 sin⁡(π/2)=1\sin(\pi/2)=1，自然推出给定结果。

答题过程

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Using the density,

P(X<12)= ∫01/2(1+cos⁡πx) dx= [x+sin⁡πxπ]01/2= 12+sin⁡(π/2)π−0= 12+1π.\begin{align*} \mathrm P\bigg(X<\frac12\bigg) =&\,\int_0^{1/2}(1+\cos\pi x)\,\mathrm{d}x\\[4mm] =&\,\bigg[x+\frac{\sin\pi x}{\pi}\bigg]_0^{1/2}\\[4mm] =&\,\frac12+\frac{\sin(\pi/2)}\pi-0\\[4mm] =&\,\boxed{\frac12+\frac1\pi}. \end{align*}

(b)

解法一

思路

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期望积分中的 xcos⁡πxx\cos\pi x 使用分部积分，注意最后在两个端点都要代入余弦项。

答题过程

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The expectation is

E(X)=∫01x(1+cos⁡πx) dx.\mathrm E(X)=\int_0^1x(1+\cos\pi x)\,\mathrm{d}x.

For ∫xcos⁡πx dx\int x\cos\pi x\,\mathrm{d}x, integrate by parts with u=xu=x and dv=cos⁡πx dx\mathrm{d}v=\cos\pi x\,\mathrm{d}x. Then du=dx\mathrm{d}u=\mathrm{d}x and v=sin⁡πx/πv=\sin\pi x/\pi, so

 ∫xcos⁡πx dx= xsin⁡πxπ −1π∫sin⁡πx dx= xsin⁡πxπ+cos⁡πxπ2.\begin{align*} &\,\int x\cos\pi x\,\mathrm{d}x\\[4mm] =&\,\frac{x\sin\pi x}{\pi}\\[4mm] &\,-\frac1\pi\int\sin\pi x\,\mathrm{d}x\\[4mm] =&\,\frac{x\sin\pi x}{\pi}+\frac{\cos\pi x}{\pi^2}. \end{align*}

Therefore

E(X)= [x22]01 +[xsin⁡πxπ+cos⁡πxπ2]01= 12+(0−1π2) −(0+1π2)= 12−2π2.\begin{align*} \mathrm E(X) =&\,\bigg[\frac{x^2}{2}\bigg]_0^1\\[4mm] &\,+\bigg[\frac{x\sin\pi x}{\pi}+\frac{\cos\pi x}{\pi^2}\bigg]_0^1\\[4mm] =&\,\frac12+\bigg(0-\frac1{\pi^2}\bigg)\\[4mm] &\,-\bigg(0+\frac1{\pi^2}\bigg)\\[4mm] =&\,\boxed{\frac12-\frac2{\pi^2}}. \end{align*}