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P4 2024 Oct Q9

A Level / Edexcel / P4

IAL 2024 Oct Paper · Question 9

题目

Problem

(a) Express

1x(2x−1)\frac{1}{x(2x-1)}

in partial fractions.

(2)

The height above ground, hh metres, of a carriage on a fairground ride is modelled by the differential equation

dhdt=150h(2h−1)cos⁡(t10)\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{1}{50}h(2h-1)\cos\big(\frac{t}{10}\big)

where tt seconds is the time after the start of the ride.

Given that, at the start of the ride, the carriage is 2.52.5 m above ground,

(b) solve the differential equation to show that, according to the model,

h=510−8eksin⁡(t10)h=\frac{5}{10-8e^{k\sin\big(\frac{t}{10}\big)}}

where kk is a constant to be found.

(6)

(c) Hence find, according to the model, the time taken for the carriage to reach its maximum height above ground for the 3rd time.

Give your answer to the nearest second.

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

(2)
题目中文翻译

(a) 将

1x(2x−1)\frac{1}{x(2x-1)}

分解为部分分式。

(2)

游乐场设施中某个吊舱离地高度为 hh 米，其变化由微分方程

dhdt=150h(2h−1)cos⁡(t10)\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{1}{50}h(2h-1)\cos\big(\frac{t}{10}\big)

建模，其中 tt 秒为开始乘坐后的时间。

已知开始时吊舱离地 2.52.5 m，

(b) 解该微分方程并证明，根据模型，

h=510−8eksin⁡(t10)h=\frac{5}{10-8e^{k\sin\big(\frac{t}{10}\big)}}

其中 kk 是待求常数。

(6)

(c) 进而根据该模型，求吊舱第 3 次达到最大离地高度所用的时间。

答案取最接近的整秒。

（不接受完全依赖计算器技术的解法。）

(2)

解答

(a)

解法一

思路

展开

设两项部分分式，通分得到恒等式；代入使其中一项消失的值求常数，最后写回分式。

答题过程

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Write the partial fractions in the form

 1x(2x−1) =Ax+B2x−1.\begin{align*} &\,\frac1{x(2x-1)}\\ &\,=\frac{A}{x}+\frac{B}{2x-1}. \end{align*}

Multiplying by x(2x−1)x(2x-1) gives the identity

1=A(2x−1)+Bx.1=A(2x-1)+Bx.

Set x=0x=0 to obtain 1=−A1=-A, so A=−1A=-1. Set x=1/2x=1/2 to obtain 1=B/21=B/2, so B=2B=2.

Therefore

 1x(2x−1) =−1x+22x−1.\boxed{ \begin{align*} &\,\frac1{x(2x-1)}\\ &\,=-\frac1x+\frac2{2x-1} \end{align*} }.

(b)

解法一

思路

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分离变量后利用 (a) 积分。合并对数并指数化，代初始高度求常数，再收集所有含 hh 的项，化为指定形式。

答题过程

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Separate the variables:

dhh(2h−1)=150cos⁡(t10) dt.\frac{\mathrm{d}h}{h(2h-1)} =\frac1{50}\cos\bigg(\frac{t}{10}\bigg)\,\mathrm{d}t.

Using the partial fractions from part (a), the integral on the left is

 ∫(−1h+22h−1) dh =−ln⁡h+ln⁡(2h−1).\begin{align*} &\,\int\bigg(-\frac1h+\frac2{2h-1}\bigg)\,\mathrm{d}h\\ &\,=-\ln h+\ln(2h-1). \end{align*}

The integral on the right is

 150∫cos⁡(t10) dt =15sin⁡(t10).\begin{align*} &\,\frac1{50}\int\cos\bigg(\frac{t}{10}\bigg)\,\mathrm{d}t\\ &\,=\frac15\sin\bigg(\frac{t}{10}\bigg). \end{align*}

Combining the logarithms and including an integration constant gives

ln⁡(2h−1h)=15sin⁡(t10)+C.\ln\bigg(\frac{2h-1}{h}\bigg) =\frac15\sin\bigg(\frac{t}{10}\bigg)+C.

Exponentiating,

2h−1h=Aes,s=15sin⁡(t10).\frac{2h-1}{h}=Ae^{s}, \qquad s=\frac15\sin\bigg(\frac{t}{10}\bigg).

At t=0t=0, s=0s=0 and h=5/2h=5/2. Here 2h−1=42h-1=4, so

A=45/2=85.A=\frac4{5/2}=\frac85.

Consequently

2h−1=85hes.2h-1=\frac85he^s.

Multiply by 55 and collect the terms containing hh:

10h−5=8hes,10h-5=8he^s, h(10−8es)=5.h(10-8e^s)=5.

Substitute back for ss to obtain the required form:

h=510−8e15sin⁡(t/10).\boxed{h=\frac5{10-8e^{\frac15\sin(t/10)}}}.

Therefore

k=15.\boxed{k=\frac15}.

(c)

解法一

思路

展开

承接 (b) 的显式高度公式，先判断何时高度最大，再用正弦的周期列出各次最高点时间。第三次对应 n=2n=2。

答题过程

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By part (b),

h=510−8e15sin⁡(t/10).h=\frac5{10-8e^{\frac15\sin(t/10)}}.

To maximise hh, minimise the denominator. The exponential term is largest when

sin⁡(t10)=1.\sin\bigg(\frac{t}{10}\bigg)=1.

For t≥0t\ge0, the solutions are

t10=π2+2nπ,n=0,1,2,…\frac{t}{10}=\frac{\pi}{2}+2n\pi, \qquad n=0,1,2,\ldots

The third maximum corresponds to n=2n=2, so

t10=9π2.\frac{t}{10}=\frac{9\pi}{2}.

Hence

t=45π=141.37…t=45\pi=141.37\ldots

and the required time, to the nearest second, is

141 seconds.\boxed{141\text{ seconds}}.