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IAL 2022 Oct Q5

A Level / Edexcel / P3

IAL 2022 Oct Paper · Question 5

题目

Problem

The profit made by a company, £ PP million, tt years after the company started trading, is modelled by the equation

P=4t−110+34ln⁡[t+1(2t+1)2]P=\frac{4t-1}{10}+\frac34\ln\left[\frac{t+1}{(2t+1)^2}\right]

The graph of PP against tt is shown in Figure 2.

According to the model,

(a) show that exactly one year after it started trading, the company had made a loss of approximately £ 830 000830\,000

(2)

A manager of the company wants to know the value of tt for which P=0P=0

(b) Show that this value of tt occurs in the interval [6,7][6,7]

(2)

(c) Show that the equation P=0P=0 can be expressed in the form

t=14+158ln⁡[(2t+1)2t+1]t=\frac14+\frac{15}{8}\ln\left[\frac{(2t+1)^2}{t+1}\right]
(2)

(d) Using the iteration formula

tn+1=14+158ln⁡[(2tn+1)2tn+1] with t1=6t_{n+1}=\frac14+\frac{15}{8}\ln\left[\frac{(2t_n+1)^2}{t_n+1}\right]\text{ with }t_1=6

find the value of t2t_2 and the value of t6t_6, giving your answers to 3 decimal places.

(3)

(e) Hence find, according to the model, how many months it takes in total, from when the company started trading, for it to make a profit.

(2)
题目中文翻译

某公司在开始营业后 tt 年时所获得的利润为 £ PP million，其模型方程为

P=4t−110+34ln⁡[t+1(2t+1)2]P=\frac{4t-1}{10}+\frac34\ln\left[\frac{t+1}{(2t+1)^2}\right]

图 2 给出了 PP 关于 tt 的图像。

根据该模型，

(a) 证明公司在开业恰好一年后亏损约 £ 830 000830\,000。

一位经理想知道 P=0P=0 时的 tt 值。

(b) 证明该 tt 值位于区间 [6,7][6,7] 内。

(c) 证明方程 P=0P=0 可写成

t=14+158ln⁡[(2t+1)2t+1]t=\frac14+\frac{15}{8}\ln\left[\frac{(2t+1)^2}{t+1}\right]

的形式。

(d) 使用迭代公式

tn+1=14+158ln⁡[(2tn+1)2tn+1]，其中 t1=6t_{n+1}=\frac14+\frac{15}{8}\ln\left[\frac{(2t_n+1)^2}{t_n+1}\right]\text{，其中 }t_1=6

求 t2t_2 和 t6t_6 的值，答案精确到小数点后 3 位。

(e) 由此根据模型求公司从开始营业起到开始盈利总共用了多少个月。

解答

(a)

Substitute t=1t=1 into the model:

P=4(1)−110+34ln⁡[1+1(2(1)+1)2]P=\frac{4(1)-1}{10} +\frac34\ln\left[\frac{1+1}{(2(1)+1)^2}\right]

So

P=310+34ln⁡(29)P=\frac{3}{10}+\frac34\ln\left(\frac{2}{9}\right)

Hence

P=−0.828…P=-0.828\ldots

Since PP is measured in millions of pounds, this means a loss of

0.828… million pounds0.828\ldots\text{ million pounds}

which is approximately

£ 830 000\boxed{\text{£ }830\,000}

Therefore, exactly one year after it started trading, the company had made a loss of approximately £ 830 000830\,000.

(b)

Evaluate PP at the endpoints.

At t=6t=6,

P(6)=−0.0879…P(6)=-0.0879\ldots

At t=7t=7,

P(7)=0.1975…P(7)=0.1975\ldots

So

P(6)<0andP(7)>0P(6)<0 \qquad\text{and}\qquad P(7)>0

Since PP is continuous on [6,7][6,7], there is a root of P=0P=0 in the interval

[6,7]\boxed{[6,7]}

(c)

Start with

P=0P=0

So

4t−110+34ln⁡[t+1(2t+1)2]=0\frac{4t-1}{10} +\frac34\ln\left[\frac{t+1}{(2t+1)^2}\right]=0

Move the logarithm term to the other side:

4t−110=−34ln⁡[t+1(2t+1)2]\frac{4t-1}{10} =-\frac34\ln\left[\frac{t+1}{(2t+1)^2}\right]

Multiply by 1010:

4t−1=−152ln⁡[t+1(2t+1)2]4t-1=-\frac{15}{2}\ln\left[\frac{t+1}{(2t+1)^2}\right]

So

4t=1−152ln⁡[t+1(2t+1)2]4t=1-\frac{15}{2}\ln\left[\frac{t+1}{(2t+1)^2}\right]

Divide by 44:

t=14−158ln⁡[t+1(2t+1)2]t=\frac14-\frac{15}{8}\ln\left[\frac{t+1}{(2t+1)^2}\right]

Using

−ln⁡A=ln⁡(1A)-\ln A=\ln\left(\frac1A\right)

we get

t=14+158ln⁡[(2t+1)2t+1]t=\frac14+\frac{15}{8}\ln\left[\frac{(2t+1)^2}{t+1}\right]

as required.

(d)

Using

tn+1=14+158ln⁡[(2tn+1)2tn+1]t_{n+1}=\frac14+\frac{15}{8}\ln\left[\frac{(2t_n+1)^2}{t_n+1}\right]

with t1=6t_1=6,

t2=14+158ln⁡[1327]t_2=\frac14+\frac{15}{8}\ln\left[\frac{13^2}{7}\right]

So

t2=6.219978…t_2=6.219978\ldots

Therefore

t2=6.220\boxed{t_2=6.220}

Continuing the iteration gives

t3=6.287,t4=6.307,t5=6.312,t6=6.314\begin{aligned} t_3&=6.287,\\ t_4&=6.307,\\ t_5&=6.312,\\ t_6&=6.314 \end{aligned}

Hence

t6=6.314\boxed{t_6=6.314}

(e)

From part (d), the company starts to make a profit after approximately

6.3146.314

years.

Convert this to months:

6.314×12=75.7686.314\times12=75.768

So the total time is approximately

76 months\boxed{76\text{ months}}