题目

Problem

The share price of a company is monitored.

Exactly 3 years after monitoring began, the share price was £1.05

Exactly 5 years after monitoring began, the share price was £1.65

The share price, £ $V$ , of the company is modelled by the equation

\begin{align*} V=pt+q \end{align*}

where $t$ is the number of years after monitoring began and $p$ and $q$ are constants.

(a) Find the value of $p$ and the value of $q$ .

(3)

Exactly $T$ years after monitoring began, the share price was £2.50

(b) Find the value of $T$ , according to the model, giving your answer to one decimal place.

(2)

解答

展开

模型是一次函数。两个已知数据点分别是 $(3,1.05)$ 和 $(5,1.65)$ ，可以先求斜率 $p$ ，再代回求截距 $q$ 。

展开

Using $V=pt+q$ :

\begin{align*} 1.05=&\,3p+q,\\ 1.65=&\,5p+q. \end{align*}

Subtract the first equation from the second:

\begin{align*} 1.65-1.05=&\,5p-3p\\ 0.60=&\,2p\\ p=&\,0.30. \end{align*}

Substitute into $1.05=3p+q$ :

\begin{align*} 1.05=&\,3(0.30)+q\\ 1.05=&\,0.90+q\\ q=&\,0.15. \end{align*}

Therefore

\begin{align*} p=0.3, \qquad q=0.15. \end{align*}

展开

把 $V=2.50$ 代入模型 $V=0.3t+0.15$ ，解出对应的时间。

展开

When $V=2.50$ ,

\begin{align*} 2.50=&\,0.30T+0.15\\ 2.35=&\,0.30T\\ T=&\,\frac{2.35}{0.30}\\ =&\,7.833\ldots \end{align*}

\begin{align*} T=7.8 \end{align*}

to one decimal place.