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IAL 2024 May Q8

A Level / Edexcel / P2

IAL 2024 May Paper · Question 8

题目

Problem

(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.

(2)

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*}

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}\left(\frac56\right)=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….6k+18=56.442\ldots \quad\text{or}\quad 6k+18=123.557\ldots.

So

k= 56.442…−186=6.407…,k= 123.557…−186=17.592….\begin{align*} k=&\,\frac{56.442\ldots-18}{6}=6.407\ldots,\\ k=&\,\frac{123.557\ldots-18}{6}=17.592\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 k=6.41…k=6.41\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*}

This is about 1111 hours and 1414 minutes after midnight.

Therefore, the time is

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