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IAL 2020 Oct Q7

A Level / Edexcel / P3

IAL 2020 Oct Paper · Question 7

题目

Problem

(a) Express cos⁡x+4sin⁡x\cos x+4\sin x in the form Rcos⁡(x−α)R\cos(x-\alpha) where R>0R>0 and 0<α<π20<\alpha<\dfrac{\pi}{2}.

Give the exact value of RR and give the value of α\alpha, in radians, to 3 decimal places.

(3)

A scientist is studying the behaviour of seabirds in a colony.

She models the height above sea level, HH metres, of one of the birds in the colony by the equation

H=243+cos⁡(12t)+4sin⁡(12t)0≤t≤6.5H=\frac{24}{3+\cos\left(\frac{1}{2}t\right)+4\sin\left(\frac{1}{2}t\right)}\qquad 0\le t\le 6.5

where tt seconds is the time after it leaves the nest.

Find, according to the model,

(b) the minimum height of the seabird above sea level, giving your answer to the nearest cm,

(2)

(c) the value of tt, to 2 decimal places, when H=10H=10.

(4)
题目中文翻译

(a) 将 cos⁡x+4sin⁡x\cos x+4\sin x 化为 Rcos⁡(x−α)R\cos(x-\alpha) 的形式，其中 R>0R>0 且 0<α<π20<\alpha<\dfrac{\pi}{2}。

写出 RR 的精确值，并将 α\alpha 的弧度值精确到小数点后 3 位。

一位科学家正在研究一个海鸟群落中海鸟的行为。

她用下式来建立其中一只海鸟离巢后相对于海平面的高度 HH（单位：米）模型：

H=243+cos⁡(12t)+4sin⁡(12t)0≤t≤6.5H=\frac{24}{3+\cos\left(\frac{1}{2}t\right)+4\sin\left(\frac{1}{2}t\right)}\qquad 0\le t\le 6.5

其中 tt（秒）表示海鸟离开巢穴后的时间。

根据该模型，求

(b) 海鸟高于海平面的最小高度，答案精确到最接近的厘米；

(c) 当 H=10H=10 时的 tt 值，答案精确到小数点后 2 位。

解答

(a)

We want

cos⁡x+4sin⁡x=Rcos⁡(x−α)\cos x+4\sin x=R\cos(x-\alpha)

Expand the right hand side:

Rcos⁡(x−α)=Rcos⁡xcos⁡α+Rsin⁡xsin⁡αR\cos(x-\alpha)=R\cos x\cos\alpha+R\sin x\sin\alpha

Compare coefficients:

Rcos⁡α=1,Rsin⁡α=4R\cos\alpha=1,\qquad R\sin\alpha=4

Therefore

R2=12+42=17R^2=1^2+4^2=17

so

R=17R=\sqrt{17}

Also,

tan⁡α=41=4\tan\alpha=\frac{4}{1}=4

Since

0<α<π20<\alpha<\frac{\pi}{2}

we get

α=arctan⁡4=1.326\alpha=\arctan4=1.326

to 3 decimal places.

Thus

cos⁡x+4sin⁡x=17cos⁡(x−1.326)\boxed{\cos x+4\sin x=\sqrt{17}\cos(x-1.326)}

where

R=17\boxed{R=\sqrt{17}}

(b)

Using part (a),

cos⁡(12t)+4sin⁡(12t)=17cos⁡(12t−α)\cos\left(\frac12t\right)+4\sin\left(\frac12t\right) =\sqrt{17}\cos\left(\frac12t-\alpha\right)

So

H=243+17cos⁡(12t−α)H=\frac{24}{3+\sqrt{17}\cos\left(\frac12t-\alpha\right)}

To make HH as small as possible, the denominator must be as large as possible.

The largest possible value of

cos⁡(12t−α)\cos\left(\frac12t-\alpha\right)

is 11. This value is possible in the given interval.

Therefore the minimum height is

Hmin⁡=243+17H_{\min}=\frac{24}{3+\sqrt{17}}

Using a calculator,

Hmin⁡=3.369316…H_{\min}=3.369316\ldots

So, to the nearest cm,

Hmin⁡=3.37 m\boxed{H_{\min}=3.37\text{ m}}

(c)

We need

H=10H=10

So

10=243+17cos⁡(12t−α)10=\frac{24}{3+\sqrt{17}\cos\left(\frac12t-\alpha\right)}

Rearrange:

3+17cos⁡(12t−α)=24103+\sqrt{17}\cos\left(\frac12t-\alpha\right)=\frac{24}{10}

Hence

3+17cos⁡(12t−α)=2.43+\sqrt{17}\cos\left(\frac12t-\alpha\right)=2.4

So

17cos⁡(12t−α)=−0.6\sqrt{17}\cos\left(\frac12t-\alpha\right)=-0.6

Therefore

cos⁡(12t−α)=−0.617\cos\left(\frac12t-\alpha\right)=-\frac{0.6}{\sqrt{17}}

Using

α=1.326…\alpha=1.326\ldots

we solve

cos⁡(12t−α)=−0.617\cos\left(\frac12t-\alpha\right)=-\frac{0.6}{\sqrt{17}}

In the given interval

0≤t≤6.50\leq t\leq 6.5

the solution is

12t−α=1.716836…\frac12t-\alpha=1.716836\ldots

Therefore

12t=1.716836…+α\frac12t=1.716836\ldots+\alpha

So

t=6.085307…t=6.085307\ldots

Hence

t=6.09\boxed{t=6.09}

to 2 decimal places.