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P2 2024 May Q8

A Level / Edexcel / P2

IAL 2024 May Paper · Question 8

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

(i) Solve, for 0<x≤π0<x\leq\pi, the equation

5sin⁡xtan⁡x+13=cos⁡x5\sin x\tan x+13=\cos x

giving your answer in radians to 33 significant figures.

(5)

(ii) The temperature inside a greenhouse is monitored on one particular day.

The temperature, H∘CH^\circ\text{C}, inside the greenhouse, tt hours after midnight, is modelled by the equation

H=10+12sin⁡(kt+18)∘0≤t<24H=10+12\sin(kt+18)^\circ \qquad 0\leq t<24

where kk is a constant.

Use the equation of the model to answer parts (a) to (c).

Given that

  • the temperature inside the greenhouse was 20∘C20^\circ\text{C} at 66 am
  • 0<k<200<k<20

(a) find all possible values for kk, giving each answer to 22 decimal places.

(4)

Given further that 0<k<100<k<10

(b) find the maximum temperature inside the greenhouse,

(1)

(c) find the time of day at which this maximum temperature occurs.

Give your answer to the nearest minute.

(2)
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。

(i) 在 0<x≤π0<x\leq\pi 内解方程 5sin⁡xtan⁡x+13=cos⁡x5\sin x\tan x+13=\cos x，以弧度给出答案，保留三位有效数字。（5分）

(ii) 某一天对温室内部温度进行监测。午夜后 tt 小时，温室内的温度 H ∘CH\,^\circ\mathrm{C} 用以下方程建模：

H= 10+12sin⁡(kt+18)∘, 0≤t<24,\begin{align*} H=&\,10+12\sin(kt+18)^\circ,\\ &\,0\leq t<24, \end{align*}

其中 kk 是常数。

使用该模型方程回答第(a)至(c)问。

已知早上6时温室内的温度为 20 ∘C20\,^\circ\mathrm{C}，且 0<k<200<k<20。

(a) 求 kk 的所有可能值，每个答案保留两位小数。（4分）

进一步已知 0<k<100<k<10。

(b) 求温室内的最高温度。（1分）

(c) 求一天中达到该最高温度的时刻，答案精确到最近的一分钟。（2分）

解答

(i)

解法一

思路

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把 tan⁡x\tan x 写成 sin⁡xcos⁡x\frac{\sin x}{\cos x}，再乘以 cos⁡x\cos x。之后用 sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x 化成关于 cos⁡x\cos x 的二次方程。

答题过程

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Since

tan⁡x=sin⁡xcos⁡x,\begin{align*} \tan x=\frac{\sin x}{\cos x}, \end{align*}

we have

5sin⁡x⋅sin⁡xcos⁡x+13=cos⁡x.\begin{align*} 5\sin x\cdot\frac{\sin x}{\cos x}+13=\cos x. \end{align*}

The original equation contains tan⁡x\tan x, so cos⁡x≠0\cos x\ne0. Multiply by cos⁡x\cos x:

5sin⁡2x+13cos⁡x=cos⁡2x.\begin{align*} 5\sin^2x+13\cos x=\cos^2x. \end{align*}

Using sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x,

5(1−cos⁡2x)+13cos⁡x= cos⁡2x5−5cos⁡2x+13cos⁡x= cos⁡2x6cos⁡2x−13cos⁡x−5= 0.\begin{align*} 5(1-\cos^2x)+13\cos x=&\,\cos^2x\\ 5-5\cos^2x+13\cos x=&\,\cos^2x\\ 6\cos^2x-13\cos x-5=&\,0. \end{align*}

Factorise:

(3cos⁡x+1)(2cos⁡x−5)=0.\begin{align*} (3\cos x+1)(2\cos x-5)=0. \end{align*}

So

cos⁡x=−13orcos⁡x=52.\cos x=-\frac13 \quad\text{or}\quad \cos x=\frac52.

Since cos⁡x=52\cos x=\frac52 is impossible,

cos⁡x=−13.\begin{align*} \cos x=-\frac13. \end{align*}

For 0<x≤π0<x\leq\pi,

x=1.9106….\begin{align*} x=1.9106\ldots. \end{align*}

Therefore,

x=1.91.\begin{align*} x=1.91. \end{align*}

(ii)(a)

解法一

思路

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6 am 表示 t=6t=6，代入 H=20H=20。因为角度单位是 degrees，所以解 sin⁡(6k+18)∘=56\sin(6k+18)^\circ=\frac56 时要用角度。

答题过程

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At 66 am, t=6t=6 and H=20H=20.

So

20= 10+12sin⁡(6k+18)∘10= 12sin⁡(6k+18)∘sin⁡(6k+18)∘= 56.\begin{align*} 20=&\,10+12\sin(6k+18)^\circ\\ 10=&\,12\sin(6k+18)^\circ\\ \sin(6k+18)^\circ=&\,\frac56. \end{align*}

Now

sin⁡−1(56)=56.442…∘.\begin{align*} \sin^{-1}\bigg(\frac56\bigg)=56.442\ldots^\circ. \end{align*}

Since 0<k<200<k<20,

18<6k+18<138.\begin{align*} 18<6k+18<138. \end{align*}

Thus

6k+18= 56.442…,or6k+18= 123.557….\begin{align*} 6k+18=&\,56.442\ldots,\\ \text{or}\quad6k+18=&\,123.557\ldots. \end{align*}

So

k= 56.442…−186= 6.407115…,ork= 123.557…−186= 17.592884….\begin{align*} k=&\,\frac{56.442\ldots-18}{6}\\ =&\,6.407115\ldots,\\ \text{or}\quad k=&\,\frac{123.557\ldots-18}{6}\\ =&\,17.592884\ldots. \end{align*}

Therefore,

k=6.41ork=17.59.\begin{align*} k=6.41\quad\text{or}\quad k=17.59. \end{align*}

(ii)(b)

解法一

思路

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sin⁡\sin 的最大值是 11，所以最高温度是 10+1210+12。

答题过程

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The maximum value of sin⁡(kt+18)∘\sin(kt+18)^\circ is 11.

So the maximum temperature is

10+12(1)=22.\begin{align*} 10+12(1)=22. \end{align*}

Therefore, the maximum temperature is

22∘C.\begin{align*} 22^\circ\text{C}. \end{align*}

(ii)(c)

解法一

思路

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现在额外给出 0<k<100<k<10，所以取未舍入的 k=6.407115…k=6.407115\ldots。最高温出现时，角度为 90∘90^\circ。

答题过程

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Since 0<k<100<k<10,

k=6.407….\begin{align*} k=6.407\ldots. \end{align*}

The maximum occurs when

kt+18=90.\begin{align*} kt+18=90. \end{align*}

So

6.407…t+18= 906.407…t= 72t= 11.237….\begin{align*} 6.407\ldots t+18=&\,90\\ 6.407\ldots t=&\,72\\ t=&\,11.237\ldots. \end{align*}

The fractional hour is

(11.237506…−11)×60=14.2504…(11.237506\ldots-11)\times60=14.2504\ldots

minutes. This rounds to 1414 minutes, so the time is 1111 hours and 1414 minutes after midnight.

Therefore, the time is

11:14 am.\begin{align*} 11{:}14\text{ am}. \end{align*}