CalcGospel 國際數學圖譜
返回

CIE 9709 2024 November Paper 32 Q6

A Level / CIE / P3

CIE 9709 2024 Nov Paper 32 Paper · Question 6

题目

Problem

The variables xx and yy satisfy the equation ay=bxay=b^x, where aa and bb are constants. The graph of ln⁡y\ln y against xx is a straight line passing through the points (0.50,2.24)(0.50,2.24) and (3.40,8.27)(3.40,8.27), as shown in the diagram.

Find the values of aa and bb. Give each value correct to 11 significant figure.

(4)
题目中文翻译

变量 xx 和 yy 满足方程 ay=bxay=b^x，其中 aa 和 bb 为常数。如图所示，ln⁡y\ln y 对 xx 的图像是一条经过点 (0.50,2.24)(0.50,2.24) 和 (3.40,8.27)(3.40,8.27) 的直线。

求 aa 和 bb 的值。每个值保留 11 位有效数字。

(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

对数图的斜率为 ln⁡b\ln b、纵截距为 −ln⁡a-\ln a。保留中间精度，最后按题意只保留一位有效数字。

答题过程

展开

Taking logarithms of ay=bxay=b^x gives

ln⁡y=xln⁡b−ln⁡a.\ln y=x\ln b-\ln a.

Thus the gradient of the plotted straight line is ln⁡b\ln b:

ln⁡b=8.27−2.243.40−0.50=6.032.90.\ln b=\frac{8.27-2.24}{3.40-0.50} =\frac{6.03}{2.90}.

Hence

b=exp⁡(6.03/2.90)=7.998950….b=\exp(6.03/2.90)=7.998950\ldots.

Using the first point,

ln⁡a=0.50ln⁡b−2.24,\ln a=0.50\ln b-2.24,

so

a=exp⁡(0.50(6.03/2.90)−2.24)=0.301090….a=\exp\big(0.50(6.03/2.90)-2.24\big) =0.301090\ldots.

To 1 significant figure,

a=0.3,b=8.\boxed{a=0.3,\qquad b=8}.