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CIE 9709 2026 June Paper 31 Q2

A Level / CIE / P3

CIE 9709 2026 June Paper 31 Paper · Question 2

题目

Problem

The variables xx and yy satisfy the equation y3=Aekxy^3=A\mathrm{e}^{kx}, where AA and kk are constants. The graph of ln⁡y\ln y against xx is a straight line passing through the points (−1.32,5.09)(-1.32,5.09) and (0.25,1.16)(0.25,1.16).

Find the values of AA and kk.

(5)
题目中文翻译

变量 xx 和 yy 满足方程 y3=Aekxy^3=A\mathrm{e}^{kx},其中 AA 和 kk 为常数。以 xx 为横坐标、ln⁡y\ln y 为纵坐标的图像是一条经过点 (−1.32,5.09)(-1.32,5.09) 和 (0.25,1.16)(0.25,1.16) 的直线。

求 AA 和 kk 的值。

(5)
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

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纵坐标是 ln⁡y\ln y。对原方程取对数后,直线的斜率为 k/3k/3、截距为 ln⁡A/3\ln A/3。

答题过程

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Taking logarithms gives

3ln⁡y=ln⁡A+kx.3\ln y=\ln A+kx.

The slope of the given line is

k3=1.16−5.090.25−(−1.32)=−3.931.57.\frac{k}{3}=\frac{1.16-5.09}{0.25-(-1.32)} =-\frac{3.93}{1.57}.

Thus k=−7.509554…≈−7.51k=-7.509554\ldots\approx-7.51. Using (0.25,1.16)(0.25,1.16),

ln⁡A=3(1.16)−0.25k=5.357388….\ln A=3(1.16)-0.25k=5.357388\ldots.

Hence A=e5.357388…=212.170…≈212A=e^{5.357388\ldots}=212.170\ldots\approx212.

解法二

思路

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也可先由图中两点恢复 yy,直接代入指数方程。将两个方程相除,消去 AA。

答题过程

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The two values of yy are e5.09e^{5.09} and e1.16e^{1.16}. Substituting them into the original equation gives

e15.27= Ae−1.32k,e3.48= Ae0.25k.\begin{align*} e^{15.27}=&\,Ae^{-1.32k},\\[4mm] e^{3.48}=&\,Ae^{0.25k}. \end{align*}

Dividing the second by the first gives e−11.79=e1.57ke^{-11.79}=e^{1.57k}, so k=−11.79/1.57≈−7.51k=-11.79/1.57\approx-7.51. Then A=e3.48−0.25k≈212A=e^{3.48-0.25k}\approx212.