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P1 2025 Jan Q3

A Level / Edexcel / P1

IAL 2025 Jan Paper · Question 3

题目

Problem

The population of a town was monitored.

Exactly 55 years after monitoring began, the population was 58 00058\,000.

Exactly 1010 years after monitoring began, the population was 65 00065\,000.

Given that the population of the town, PP thousand, tt years after monitoring began can be modelled by the equation

P2=a+bt3\begin{align*} P^2=a+bt^3 \end{align*}

where aa and bb are constants,

(a) find the value of aa and the value of bb.

(3)

According to the model, exactly TT years after monitoring began, the population was 85 00085\,000.

Making your method clear,

(b) find the value of TT, giving your answer to one decimal place.

(Solutions relying entirely on calculator technology are not acceptable.)

(2)
题目中文翻译

对某城镇的人口进行监测。

监测开始整整 55 年后,人口为 58 00058\,000;监测开始整整 1010 年后,人口为 65 00065\,000。

已知监测开始 tt 年后,该城镇的人口为 PP 千人,并可用方程

P2=a+bt3\begin{align*} P^2=a+bt^3 \end{align*}

建模,其中 aa、bb 为常数。

(a) 求 aa 和 bb 的值。

(3)

根据该模型,监测开始整整 TT 年后,人口为 85 00085\,000。

(b) 清楚写出你的方法,求 TT 的值,并将答案精确到小数点后 11 位。

(完全依赖计算器技术的解答不予接受。)

(2)

解答

(a)

解法一

思路

展开

注意 PP 的单位是 thousand,所以 58 00058\,000 对应 P=58P=58,65 00065\,000 对应 P=65P=65。代入模型得到两个关于 a,ba,b 的方程。

答题过程

展开

When t=5t=5, P=58P=58, so

582= a+b(5)33364= a+125b.\begin{align*} 58^2=&\,a+b(5)^3\\ 3364=&\,a+125b. \end{align*}

When t=10t=10, P=65P=65, so

652= a+b(10)34225= a+1000b.\begin{align*} 65^2=&\,a+b(10)^3\\ 4225=&\,a+1000b. \end{align*}

Subtract:

4225−3364= (a+1000b)−(a+125b)861= 875bb= 861875= 0.984.\begin{align*} 4225-3364=&\,(a+1000b)-(a+125b)\\ 861=&\,875b\\ b=&\,\frac{861}{875}\\ =&\,0.984. \end{align*}

Then

3364= a+125(0.984)3364= a+123a= 3241.\begin{align*} 3364=&\,a+125(0.984)\\ 3364=&\,a+123\\ a=&\,3241. \end{align*}

Therefore

a=3241,b=0.984.\begin{align*} a=3241,\qquad b=0.984. \end{align*}

(b)

解法一

思路

展开

85 00085\,000 对应 P=85P=85。把 P=85P=85、(a) 中的 a,ba,b 代入模型,解出 T3T^3,再开三次方。

答题过程

展开

When the population is 85 00085\,000,

P=85.\begin{align*} P=85. \end{align*}

Use the model:

852= 3241+0.984T37225= 3241+0.984T33984= 0.984T3T3= 39840.984= 4048.780…\begin{align*} 85^2=&\,3241+0.984T^3\\ 7225=&\,3241+0.984T^3\\ 3984=&\,0.984T^3\\ T^3=&\,\frac{3984}{0.984}\\ =&\,4048.780\ldots \end{align*}

Therefore

T= 4048.780…3= 15.9…\begin{align*} T=&\,\sqrt[3]{4048.780\ldots}\\ =&\,15.9\ldots \end{align*}

So

T=15.9.\begin{align*} T=15.9. \end{align*}