题目

Problem

The share price, \pounds $V$ , of a company is being monitored.

A graph is drawn of $\log_{10}V$ against $t$ , where $t$ is the number of years after monitoring began.

The graph is a straight line passing through the points $(0,2)$ and $(5,2.25)$ .

Using this information,

(a) find an equation for the line in the form

\log_{10}V=mt+c

where $m$ and $c$ are constants.

(2)

(b) Write the answer to part (a) in the form

V=ab^t

where $a$ and $b$ are constants to be found.

Give the exact value of $a$ and the value of $b$ to 3 significant figures.

(3)

When $t=T$ , the rate of increase in the share price of the company was \pounds $50$ per year.

(Solutions relying entirely on calculator technology are not acceptable.)

(4)

题目中文翻译

某公司的股价为 \pounds $V$ ，现对其进行监测。

以 $\log_{10}V$ 对 $t$ 作图，其中 $t$ 表示开始监测后经过的年数。

该图像是一条经过点 $(0,2)$ 和 $(5,2.25)$ 的直线。

根据这些信息，

(a) 求这条直线的方程，并写成

\log_{10}V=mt+c

的形式，其中 $m,c$ 为常数。

(b) 将(a)中的结果写成

V=ab^t

的形式，其中 $a,b$ 为待求常数。

写出 $a$ 的精确值，并将 $b$ 保留 3 位有效数字。

当 $t=T$ 时，该公司股价的增长率为每年 \pounds $50$ 。

（不接受完全依赖计算器技术的解法。）

解答

(a)

The line passes through $(0,2)$ and $(5,2.25)$ .

Its gradient is

m=\frac{2.25-2}{5-0}=\frac{0.25}{5}=0.05

When $t=0$ ,

\log_{10}V=2

c=2

Therefore

\boxed{\log_{10}V=0.05t+2}

(b)

From part (a),

\log_{10}V=0.05t+2

V=10^{0.05t+2}

Rewrite:

\begin{aligned} V&=10^2\cdot 10^{0.05t} \\ &=100\left(10^{0.05}\right)^t \end{aligned}

Thus

a=100

and

b=10^{0.05}=1.122\ldots

So, to 3 significant figures,

b=1.12

Therefore

\boxed{V=100(1.12)^t}

with

\boxed{a=100,\quad b=1.12\text{ to 3 s.f.}}

(c)

Using

V=100(1.12)^t

we get

\frac{\mathrm{d}V}{\mathrm{d}t} =100\ln(1.12)(1.12)^t

When $t=T$ , the rate of increase is $50$ pounds per year, so

100\ln(1.12)(1.12)^T=50

Hence

(1.12)^T=\frac{50}{100\ln(1.12)}

Take logs:

T=\frac{\ln\left(\frac{50}{100\ln(1.12)}\right)}{\ln(1.12)}

Therefore

T=13.0\ldots

So, to the nearest integer,

\boxed{T=13}

IAL 2025 Oct Q3

题目

解答

(a)

(b)

(c)