Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2026 June Paper 35 Q3

A Level / CIE / P3

CIE 9709 2026 June Paper 35 Paper · Question 3

题目

Problem

The variables xx and yy satisfy the equation yxa=kyx^a=k, where aa and kk are constants.

(a) Show that the graph of ln⁡y\ln y against ln⁡x\ln x is a straight line.

(2)

(b) When x=5x=5, y=0.72y=0.72 and when x=2.3x=2.3, y=5y=5.

Find the values of aa and kk. Give your answers correct to 22 significant figures.

(3)
题目中文翻译

变量 xx 和 yy 满足方程 yxa=kyx^a=k,其中 aa 和 kk 为常数。

(a) 证明以 ln⁡x\ln x 为横坐标、ln⁡y\ln y 为纵坐标的图像是一条直线。

(2)

(b) 当 x=5x=5 时,y=0.72y=0.72;当 x=2.3x=2.3 时,y=5y=5。

求 aa 和 kk 的值。答案保留 22 位有效数字。

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

对两边取对数,把横坐标和纵坐标明确写成 ln⁡x\ln x 与 ln⁡y\ln y。

答题过程

展开

For positive x,yx,y, taking logarithms gives

ln⁡y+aln⁡x= ln⁡k,ln⁡y= −aln⁡x+ln⁡k.\begin{align*} \ln y+a\ln x=&\,\ln k,\\[4mm] \ln y=&\,-a\ln x+\ln k. \end{align*}

This is a straight line in coordinates (ln⁡x,ln⁡y)(\ln x,\ln y), with gradient −a-a and intercept ln⁡k\ln k.

(b)

解法一

思路

展开

将两组数据代入同一常数关系,相除消去 kk,再取对数求 aa。

答题过程

展开

The data give 0.72(5a)=k=5(2.3a)0.72(5^a)=k=5(2.3^a). Therefore

(52.3)a= 50.72,a= ln⁡(5/0.72)ln⁡(5/2.3)= 2.495647….\begin{align*} \bigg(\frac5{2.3}\bigg)^a=&\,\frac5{0.72},\\[4mm] a=&\,\frac{\ln(5/0.72)}{\ln(5/2.3)}\\[4mm] =&\,2.495647\ldots. \end{align*}

Using this unrounded value, k=0.72(5a)=39.968247…k=0.72(5^a)=39.968247\ldots. To two significant figures, a=2.5a=2.5 and k=40k=40.