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IAL 2024 Jan Q3

A Level / Edexcel / P3

IAL 2024 Jan Paper · Question 3

题目

Problem

The amount of money raised for a charity is being monitored.

The total amount raised in the tt months after monitoring began, £DD, is modelled by the equation

log10D=1.04+0.38t\log_{10}D=1.04+0.38t

(a) Write this equation in the form

D=abtD=ab^t

where aa and bb are constants to be found. Give each value to 4 significant figures.

(3)

When t=Tt=T, the total amount of money raised is £45000

According to the model,

(b) find the value of TT, giving your answer to 3 significant figures.

(2)

The charity aims to raise a total of £350000 within the first 12 months of monitoring.

According to the model,

(c) determine whether or not the charity will achieve its aim.

(2)
题目中文翻译

某慈善机构筹得的款项正在被监测。

监测开始后 tt 个月,累计筹得的总金额 £DD 满足模型方程

log10D=1.04+0.38t\log_{10}D=1.04+0.38t

(a) 将此方程写成

D=abtD=ab^t

的形式,其中 aabb 是需要求出的常数。两个值都精确到 4 位有效数字。

t=Tt=T 时,筹得的总金额为 £45000。

根据该模型,

(b) 求 TT 的值,答案精确到 3 位有效数字。

该慈善机构的目标是在监测开始后的前 12 个月内筹得总计 £350000。

根据该模型,

(c) 判断该慈善机构能否实现其目标。

解答

(a)

解法一

思路

展开

把常用对数方程两边以 1010 为底取指数,再利用指数律把含 tt 的部分写成某个常数的 tt 次方。

答题过程

展开

From

log10D=1.04+0.38t,\log_{10}D=1.04+0.38t,

we obtain

D=101.04+0.38t=101.04(100.38)t.\begin{align*} D=&\,10^{1.04+0.38t} \\[2mm] =&\,10^{1.04}\big(10^{0.38}\big)^t. \end{align*}

Now

101.04=10.964710^{1.04}=10.9647\ldots

and

100.38=2.39883.10^{0.38}=2.39883\ldots.

Therefore, to 44 significant figures,

D=10.96(2.399)t,\boxed{D=10.96(2.399)^t},

so a=10.96a=10.96 and b=2.399b=2.399.

(b)

解法一

思路

展开

直接把 D=45000D=45000 代回题目给出的对数模型,可以避免受到 (a) 中四舍五入数值的影响。

答题过程

展开

Substituting D=45000D=45000 and t=Tt=T into the model gives

log1045000=1.04+0.38T.\log_{10}45000=1.04+0.38T.

Hence

T=log10450001.040.38=9.50845.\begin{align*} T=&\,\frac{\log_{10}45000-1.04}{0.38} \\[2mm] =&\,9.50845\ldots. \end{align*}

Therefore, to 33 significant figures,

T=9.51 months.\boxed{T=9.51\text{ months}}.

(c)

解法一

思路

展开

计算 t=12t=12 时模型预测的累计金额,并与目标 £350000 比较。因为模型中的金额随 tt 增大,所以第 12 个月时达到或超过目标,就表示能在前 12 个月内完成目标。

答题过程

展开

At t=12t=12,

log10D=1.04+0.38(12)=5.60.\begin{align*} \log_{10}D=&\,1.04+0.38(12) \\[2mm] =&\,5.60. \end{align*}

Therefore,

D=105.60=398107.17.\begin{align*} D=&\,10^{5.60} \\[2mm] =&\,398107.17\ldots. \end{align*}

Since

398107.17>350000,398107.17\ldots>350000,

the charity is predicted to raise more than £350000 by the end of the first 1212 months. Hence,

the charity will achieve its aim.\boxed{\text{the charity will achieve its aim}.}