题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

The number of dormice and the number of voles on an island are being monitored.

Initially there are $2000$ dormice on the island.

A model predicts that the number of dormice will increase by $3\%$ each year, so that the numbers of dormice on the island at the end of each year form a geometric sequence.

(a) Find, according to the model, the number of dormice on the island $6$ years after monitoring began. Give your answer to $3$ significant figures.

(2)

The number of voles on the island is being monitored over the same period of time.

Given that

$4$ years after monitoring began there were $3690$ voles on the island
$7$ years after monitoring began there were $3470$ voles on the island
the number of voles on the island at the end of each year is modelled as a geometric sequence

(b) find the equation of this model in the form

N=ab^t

where $N$ is the number of voles, $t$ years after monitoring began and $a$ and $b$ are constants.

Give the value of $a$ and the value of $b$ to $2$ significant figures.

(3)

When $t=T$ , the number of dormice on the island is equal to the number of voles on the island.

(3)

题目中文翻译

本题必须写出所有计算步骤。完全依赖计算器技术的解答不予接受。

正在监测一个岛上的睡鼠和田鼠数量。起初岛上有 $2000$ 只睡鼠。

一个模型预测睡鼠数量每年增长 $3\%$ ，因此每年年末岛上的睡鼠数量构成等比数列。

(a) 根据该模型，求监测开始六年后岛上的睡鼠数量，答案保留三位有效数字。（2分）

在同一时期内，也监测岛上的田鼠数量。已知监测开始四年后有 $3690$ 只田鼠，七年后有 $3470$ 只田鼠；每年年末的田鼠数量用等比数列建模。

(b) 求该模型的方程，写成 $N=ab^t$ 的形式，其中 $N$ 是监测开始 $t$ 年后的田鼠数量， $a$ 、 $b$ 是常数。 $a$ 、 $b$ 的值均保留两位有效数字。（3分）

当 $t=T$ 时，岛上的睡鼠数量等于田鼠数量。

解答

(a)

解法一

思路

展开

每年增加 $3\%$ ，所以公比是 $1.03$ 。6 年后是 $2000(1.03)^6$ 。

答题过程

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The model for the dormice is

\begin{align*} D=2000(1.03)^t. \end{align*}

After $6$ years,

\begin{align*} D=&\,2000(1.03)^6\\ =&\,2388.10\ldots. \end{align*}

To $3$ significant figures, the number of dormice is

\begin{align*} 2390. \end{align*}

(b)

解法一

思路

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对田鼠模型 $N=ab^t$ ，把 $t=4$ 和 $t=7$ 代入。两式相除可以先求出 $b^3$ ，再求 $b$ 和 $a$ 。

答题过程

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For the voles,

\begin{align*} N=ab^t. \end{align*}

Using the data,

\begin{align*} 3690=&\,ab^4,\\ 3470=&\,ab^7. \end{align*}

Divide the second equation by the first:

\begin{align*} \frac{3470}{3690}=&\,\frac{ab^7}{ab^4}\\ b^3=&\,\frac{3470}{3690}\\ b=&\,\bigg(\frac{3470}{3690}\bigg)^{\frac13}\\ =&\,0.9797\ldots. \end{align*}

Now

\begin{align*} a=&\,\frac{3690}{b^4}\\ =&\,\frac{3690}{(0.9797\ldots)^4}\\ =&\,4005.181682\ldots. \end{align*}

To $2$ significant figures,

\begin{align*} a=4000,\qquad b=0.98. \end{align*}

Therefore, the model is

\begin{align*} N=4000(0.98)^t. \end{align*}

(c)

解法一

思路

展开

令睡鼠模型和田鼠模型相等。因为未知数在指数上，需要取对数。

答题过程

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When the two populations are equal,

\begin{align*} 2000(1.03)^T=4000(0.98)^T. \end{align*}

Divide by $2000(0.98)^T$ :

\begin{align*} \bigg(\frac{1.03}{0.98}\bigg)^T=&\,2. \end{align*}

Take logarithms:

\begin{align*} T\log\bigg(\frac{1.03}{0.98}\bigg)=&\,\log 2\\ L=&\,\log\bigg(\frac{1.03}{0.98}\bigg),\\ T=&\,\frac{\log2}{L}\\ =&\,13.929384\ldots. \end{align*}

Therefore,

\begin{align*} T=13.9. \end{align*}

解法二

思路

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不先将两个指数式相除，而是对原等式两边取对数，再用对数乘积法则和幂法则把含 $T$ 的项移到同一边。沿用第(b)问按要求舍入的模型。

答题过程

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Using the models from parts (a) and (b),

2000(1.03)^T=4000(0.98)^T.

Taking logarithms of both sides gives

\begin{align*} &\,\log2000+T\log1.03\\ =&\,\log4000+T\log0.98. \end{align*}

Rearranging,

\begin{align*} T(\log1.03-\log0.98) =&\,\log4000-\log2000\\ =&\,\log2. \end{align*}

Hence

\begin{align*} T=&\,\frac{\log2}{\log1.03-\log0.98}\\ =&\,13.929384\ldots. \end{align*}

Therefore, to one decimal place, $T=13.9$ .

P2 2024 May Q10

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

解法二

思路

答题过程