题目

Problem

A fungal disease is affecting some of the trees in a forest. The fraction of the trees affected after $t$ years is denoted by $x$ . The rate of increase of $x$ is proportional to the product of the fraction of the trees affected and the fraction of the trees not affected.

(a) Explain why, after $t$ years, $\frac{\mathrm{d}x}{\mathrm{d}t}=kx(1-x)$ , where $k$ is a constant.

(1)

(b) When the disease is first detected, one quarter of the trees are affected. Two years later, one third of the trees are affected.

Solve the differential equation to find the number of years from the time when the disease is first detected until the time when three quarters of the trees are affected. Give your answer correct to the nearest year.

(8)

题目中文翻译

一种真菌病害正在感染森林中的部分树木。 $t$ 年后受感染树木所占的比例记为 $x$ 。 $x$ 的增长速率与受感染树木所占的比例和未受感染树木所占的比例的乘积成正比。

(a) 解释为什么在 $t$ 年后，有 $\frac{\mathrm{d}x}{\mathrm{d}t}=kx(1-x)$ ，其中 $k$ 为常数。

(1)

(b) 首次发现该病害时，四分之一的树木已受感染。两年后，三分之一的树木受感染。

解此微分方程，求从首次发现该病害到四分之三的树木受感染所经过的年数。答案取最接近的整数年。

(8)

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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受感染比例为 $x$ ，未感染比例便是 $1-x$ 。把文字中的正比关系直接写成微分方程。

答题过程

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The affected fraction is $x$ , so the unaffected fraction is $1-x$ .

The rate of increase is proportional to their product. Introducing a positive constant of proportionality $k$ gives

\boxed{\frac{\mathrm{d}x}{\mathrm{d}t}=kx(1-x)}.

(b)

解法一

思路

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分离变量后做部分分式，得到感染与未感染比例之比的对数。两个时间条件确定常数，最后代入四分之三。

答题过程

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Take $t=0$ when the disease is first detected. Separating variables,

\frac1{x(1-x)}\,\mathrm{d}x=k\,\mathrm{d}t.

For partial fractions, write

\frac1{x(1-x)}=\frac A x+\frac B{1-x}.

Multiplying by $x(1-x)$ gives

1=A(1-x)+Bx.

Comparing coefficients gives $A=1$ and $B-A=0$ , so $B=1$ . Hence

\frac1{x(1-x)}=\frac1x+\frac1{1-x}.

Integration therefore gives

\ln x-\ln(1-x)=kt+C,

where $0<x<1$ .

Using $x=1/4$ at $t=0$ ,

C=\ln\frac{1/4}{3/4}=-\ln3.

Using $x=1/3$ at $t=2$ ,

\ln\frac12=2k-\ln3,

k=\frac12\ln\frac32.

The particular solution is therefore

\ln\frac{x}{1-x} =\frac t2\ln\frac32-\ln3.

Exponentiating,

\frac{x}{1-x}=\frac13\bigg(\frac32\bigg)^{t/2}.

Thus $3x=(1-x)(3/2)^{t/2}$ . Collecting the terms in $x$ and dividing by the positive factor gives

x=\frac1{1+3(2/3)^{t/2}}.

When $x=3/4$ ,

\ln3=\frac t2\ln\frac32-\ln3,

giving

\begin{align*} t=&\,\frac{4\ln3}{\ln(3/2)}\\[4mm] =&\,10.838\ldots. \end{align*}

The time from first detection is $\boxed{11\text{ years}}$ to the nearest year.

CIE 9709 2025 November Paper 35 Q11

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程