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IAL 2026 Jan Q4

A Level / Edexcel / P3

IAL 2026 Jan Paper · Question 4

题目

Problem

The resting heart rate, hh beats per minute, of a particular group of mammals, can be modelled by the equation

log10h=2.40.25log10m\log_{10}h=2.4-0.25\log_{10}m

where mm kg is the mass of the mammal.

The mass of one of these mammals is 33 kg.

According to the model,

(a) find the resting heart rate of this mammal. Give your answer to 3 significant figures.

(2)

(b) Show that the equation of the model can be written in the form

h=pmqh=\frac{p}{m^q}

giving each of the values of the constants pp and qq to 3 significant figures where appropriate.

(3)

(c) With reference to the model, interpret the value of the constant pp.

(1)
题目中文翻译

某一类哺乳动物的静息心率 hh(单位:次/分钟)可由下式建模:

log10h=2.40.25log10m\log_{10}h=2.4-0.25\log_{10}m

其中 mm kg 表示该哺乳动物的质量。

现有一只该类哺乳动物,其质量为 33 kg。

根据模型,

(a) 求这只哺乳动物的静息心率,答案保留 3 位有效数字;

(b) 证明模型方程可写成

h=pmqh=\frac{p}{m^q}

的形式,并求常数 p,qp,q 的值,在适当处保留 3 位有效数字;

(c) 结合模型实际意义,解释常数 pp 的含义。

解答

(a)

解法一

思路

展开

m=3m=3 代入模型:

log10h=2.40.25log103\log_{10}h=2.4-0.25\log_{10}3

然后用 10()10^{(\cdot)} 消去 log10\log_{10}

答题过程

展开

Substitute m=3m=3 into the model.

log10h=2.40.25log103\begin{align*} \log_{10}h =&\,2.4-0.25\log_{10}3 \end{align*}

Therefore

h=102.40.25log103=190.862\begin{align*} h =&\,10^{2.4-0.25\log_{10}3} \\[2mm] =&\,190.862\ldots \end{align*}

So the resting heart rate is

191\boxed{191}

beats per minute, to 3 significant figures.

(b)

解法一

思路

展开

用对数法则:

klog10m=log10(mk)k\log_{10}m=\log_{10}(m^k)

并且

2.4=log10(102.4)2.4=\log_{10}(10^{2.4})

把右边合并成一个对数后,就可以去掉 log10\log_{10}

答题过程

展开

Starting from

log10h=2.40.25log10m\log_{10}h=2.4-0.25\log_{10}m

write

2.4=log10(102.4)2.4=\log_{10}(10^{2.4})

and

0.25log10m=log10(m0.25)-0.25\log_{10}m=\log_{10}(m^{-0.25})

So

log10h=log10(102.4)+log10(m0.25)=log10(102.4m0.25)\begin{align*} \log_{10}h =&\,\log_{10}(10^{2.4})+\log_{10}(m^{-0.25}) \\[2mm] =&\,\log_{10}(10^{2.4}m^{-0.25}) \end{align*}

Hence

h=102.4m0.25h=10^{2.4}m^{-0.25}

Since

102.4=251.18810^{2.4}=251.188\ldots

we get

h=251m0.25h=\frac{251}{m^{0.25}}

to 3 significant figures where appropriate.

Therefore

p=251,q=0.25\boxed{p=251,\qquad q=0.25}

(c)

解法一

思路

展开

h=pmqh=\frac{p}{m^q}

中,若 m=1m=1,则 mq=1m^q=1,所以 h=ph=p。因此 pp 表示质量为 11 kg 时模型预测的静息心率。

答题过程

展开

When m=1m=1,

h=p1q=ph=\frac{p}{1^q}=p

Therefore pp represents the resting heart rate, in beats per minute, of a mammal of mass 11 kg according to the model.