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CIE 9709 2025 November Paper 35 Q3

A Level / CIE / P3

CIE 9709 2025 Nov Paper 35 Paper · Question 3

题目

Problem

The variables xx and yy satisfy the equation Ay=bxAy=b^x, where AA and bb are constants.

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

(2)

(b) When x=3.4x=3.4, ln⁡y=0.86\ln y=0.86 and when x=5.7x=5.7, ln⁡y=2.56\ln y=2.56.

Find the value of AA and the value of bb. Give your answers correct to 22 significant figures.

(3)
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。

(a) 证明 ln⁡y\ln y 对 xx 的图像是一条直线。

(2)

(b) 当 x=3.4x=3.4 时,ln⁡y=0.86\ln y=0.86;当 x=5.7x=5.7 时,ln⁡y=2.56\ln y=2.56。

求 AA 和 bb 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

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取自然对数后,纵变量 ln⁡y\ln y 与横变量 xx 满足一次函数关系。

答题过程

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Taking natural logarithms of Ay=bxAy=b^x,

ln⁡A+ln⁡y=xln⁡b.\ln A+\ln y=x\ln b.

Thus

ln⁡y=(ln⁡b)x−ln⁡A.\boxed{\ln y=(\ln b)x-\ln A}.

This has the form Y=mX+cY=mX+c with Y=ln⁡yY=\ln y and X=xX=x, so the graph is a straight line with gradient ln⁡b\ln b and intercept −ln⁡A-\ln A.

(b)

解法一

思路

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由两点先求直线斜率,即 ln⁡b\ln b。代回一点求 ln⁡A\ln A,最后指数化;中间不提前舍入。

答题过程

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The two given points satisfy

ln⁡A+0.86= 3.4ln⁡b,ln⁡A+2.56= 5.7ln⁡b.\begin{align*} \ln A+0.86=&\,3.4\ln b,\\[4mm] \ln A+2.56=&\,5.7\ln b. \end{align*}

Subtracting the first equation from the second, the gradient is

ln⁡b=2.56−0.865.7−3.4=1723.\ln b=\frac{2.56-0.86}{5.7-3.4} =\frac{17}{23}.

Hence

b=e17/23=2.094113….b=e^{17/23}=2.094113\ldots.

Using the first point,

ln⁡A=3.4ln⁡b−0.86=3.41723−0.86,\ln A=3.4\ln b-0.86 =3.4\frac{17}{23}-0.86,

so

A=e3.4(17/23)−0.86=5.222851….A=e^{3.4(17/23)-0.86}=5.222851\ldots.

To 2 significant figures,

A=5.2,b=2.1.\boxed{A=5.2,\qquad b=2.1}.