Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2024 June Paper 31 Q3

A Level / CIE / P3

CIE 9709 2024 June Paper 31 Paper · Question 3

题目

Problem

The variables xx and yy satisfy the equation ay=bxa^y=bx, where aa and bb are constants. The graph of yy against ln⁡x\ln x is a straight line passing through the points (0.336,1.00)(0.336,1.00) and (1.31,1.50)(1.31,1.50), as shown in the diagram.

Find the values of aa and bb. Give each value correct to the nearest integer.

(4)
题目中文翻译

变量 xx 和 yy 满足方程 ay=bxa^y=bx,其中 aa 和 bb 为常数。如图所示,yy 对 ln⁡x\ln x 的图像是一条经过点 (0.336,1.00)(0.336,1.00) 和 (1.31,1.50)(1.31,1.50) 的直线。

求 aa 和 bb 的值。每个值四舍五入至最接近的整数。

(4)
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

解法一

思路

展开

取对数后,注意横轴是 ln⁡x\ln x,斜率是 1/ln⁡a1/\ln a。也可将两个点直接代入同一个线性关系求解。

答题过程

展开

Taking logarithms,

yln⁡a=ln⁡b+ln⁡x.y\ln a=\ln b+\ln x.

The two plotted points give

ln⁡a=ln⁡b+0.336,\ln a=\ln b+0.336,

and

1.50ln⁡a=ln⁡b+1.31.1.50\ln a=\ln b+1.31.

Subtracting,

0.50ln⁡a=0.974⟹ln⁡a=1.948.0.50\ln a=0.974 \quad\Longrightarrow\quad \ln a=1.948.

Thus

a=e1.948=7.01464….a=e^{1.948}=7.01464\ldots.

Also,

ln⁡b=1.948−0.336=1.612,\ln b=1.948-0.336=1.612,

so b=e1.612=5.01282…b=e^{1.612}=5.01282\ldots.

To the nearest integer,

a=7,b=5.\boxed{a=7,\qquad b=5}.