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IAL 2021 June Q5

A Level / Edexcel / P1

IAL 2021 June Paper · Question 5

题目

Problem

The share value of two companies, company A and company B, has been monitored over a 15-year period.

The share value PAP_A of company A, in millions of pounds, is modelled by the equation

PA=53−0.4(t−8)2,t≥0\begin{align*} P_A=53-0.4(t-8)^2,\qquad t\ge0 \end{align*}

where tt is the number of years after monitoring began.

The share value PBP_B of company B, in millions of pounds, is modelled by the equation

PB=−1.6t+44.2,t≥0.\begin{align*} P_B=-1.6t+44.2,\qquad t\ge0. \end{align*}

Figure 2 shows a graph of both models.

Figure 2

Use the equations of one or both models to answer parts (a) to (d).

(a) Find the difference between the share value of company A and the share value of company B at the point monitoring began.

(2)

(b) State the maximum share value of company A during the 15-year period.

(1)

(c) Find, using algebra and showing your working, the times during this 15-year period when the share value of company A was greater than the share value of company B.

(4)

(d) Explain why the model for company A should not be used to predict its share value when t=20t=20.

(1)

解答

(a)

解法一

思路

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监测刚开始时 t=0t=0。分别代入两个模型，再相减。

答题过程

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When t=0t=0,

PA= 53−0.4(0−8)2= 53−0.4(64)= 27.4,\begin{align*} P_A=&\,53-0.4(0-8)^2\\ =&\,53-0.4(64)\\ =&\,27.4, \end{align*}

and

PB= −1.6(0)+44.2= 44.2.\begin{align*} P_B=&\,-1.6(0)+44.2\\ =&\,44.2. \end{align*}

The difference is

44.2−27.4=16.8.\begin{align*} 44.2-27.4=16.8. \end{align*}

So the difference is

16.8 million pounds.\begin{align*} 16.8\text{ million pounds}. \end{align*}

(b)

解法一

思路

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PA=53−0.4(t−8)2P_A=53-0.4(t-8)^2 是开口向下的二次模型，最大值在 (t−8)2=0(t-8)^2=0 时取得。

答题过程

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The maximum value of PAP_A is

53.\begin{align*} 53. \end{align*}

So the maximum share value of company A is

53 million pounds.\begin{align*} 53\text{ million pounds}. \end{align*}

(c)

解法一

思路

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先求两公司价值相等的时间，再用图像或不等式判断哪一侧是 PA>PBP_A>P_B。题目只考虑 15 年期间，所以还要限制 0≤t≤150\le t\le15。

答题过程

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Set PA=PBP_A=P_B:

53−0.4(t−8)2= −1.6t+44.28.8+1.6t−0.4(t2−16t+64)= 08.8+1.6t−0.4t2+6.4t−25.6= 0−0.4t2+8t−16.8= 0.\begin{align*} 53-0.4(t-8)^2=&\,-1.6t+44.2\\ 8.8+1.6t-0.4(t^2-16t+64)=&\,0\\ 8.8+1.6t-0.4t^2+6.4t-25.6=&\,0\\ -0.4t^2+8t-16.8=&\,0. \end{align*}

Multiply by −10-10:

4t2−80t+168= 0t2−20t+42= 0.\begin{align*} 4t^2-80t+168=&\,0\\ t^2-20t+42=&\,0. \end{align*}

Using the quadratic formula,

t= 20±202−4(1)(42)2= 20±2322= 10±58.\begin{align*} t =&\,\frac{20\pm\sqrt{20^2-4(1)(42)}}{2}\\ =&\,\frac{20\pm\sqrt{232}}{2}\\ =&\,10\pm\sqrt{58}. \end{align*}

Now

10−58=2.38…,10+58=17.61….\begin{align*} 10-\sqrt{58}=2.38\ldots,\qquad 10+\sqrt{58}=17.61\ldots. \end{align*}

Only the first crossing is within the 15-year period. Since PA<PBP_A<P_B at t=0t=0 and PA>PBP_A>P_B after this crossing,

10−58<t≤15.\begin{align*} 10-\sqrt{58}<t\le15. \end{align*}

Equivalently,

2.38…<t≤15.\begin{align*} 2.38\ldots<t\le15. \end{align*}

(d)

解法一

思路

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模型是根据 15 年监测期建立的，t=20t=20 超出给定时间范围，因此不应外推使用。

答题过程

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The model was based on a 15-year period, but t=20t=20 is outside this period. Therefore it should not be used to predict the share value at t=20t=20.