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CIE 9709 2024 June Paper 32 Q6

A Level / CIE / P3

CIE 9709 2024 June Paper 32 Paper · Question 6

题目

Problem

The diagram shows the curve y=xe−axy=x\mathrm{e}^{-ax}, where aa is a positive constant, and its maximum point MM.

(a) Find the exact coordinates of MM.

(4)

(b) Find the exact value of ∫02axe−ax dx\displaystyle\int_0^{\frac2a}x\mathrm{e}^{-ax}\,\mathrm{d}x.

(5)
题目中文翻译

图中显示了曲线 y=xe−axy=x\mathrm{e}^{-ax} 及其极大值点 MM,其中 aa 为正常数。

(a) 求 MM 的精确坐标。

(4)

(b) 求 ∫02axe−ax dx\displaystyle\int_0^{\frac2a}x\mathrm{e}^{-ax}\,\mathrm{d}x 的精确值。

(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

(a)

解法一

思路

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乘积求导后提出指数因子,求驻点并代回。指数为正,可用导数变号确认最大值。

答题过程

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Differentiating,

dydx=e−ax−axe−ax=e−ax(1−ax).\frac{\mathrm{d}y}{\mathrm{d}x} =e^{-ax}-axe^{-ax}=e^{-ax}(1-ax).

Since e−ax>0e^{-ax}>0, the stationary point requires 1−ax=01-ax=0, giving x=1/ax=1/a.

Then

y=1ae−1=1ae.y=\frac1a e^{-1}=\frac1{ae}.

The derivative changes from positive to negative at x=1/ax=1/a because a>0a>0. Thus

M=(1a,1ae).\boxed{M=\bigg(\frac1a,\frac1{ae}\bigg)}.

(b)

解法一

思路

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分部积分时,积分指数会带来 −1/a-1/a。代入上限 2/a2/a 后指数变成 e−2e^{-2}。

答题过程

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Let I=∫02/axe−ax dxI=\int_0^{2/a}xe^{-ax}\,\mathrm{d}x. For integration by parts, take u=xu=x and dv=e−ax dx\mathrm{d}v=e^{-ax}\,\mathrm{d}x, so du=dx\mathrm{d}u=\mathrm{d}x and v=−e−ax/av=-e^{-ax}/a. Then

I= [−xae−ax]02/a+1a∫02/ae−ax dx= [−xae−ax−1a2e−ax]02/a= −2a2e−2−1a2e−2+1a2= 1−3e−2a2.\begin{align*} I=&\,\bigg[-\frac xa e^{-ax}\bigg]_0^{2/a} +\frac1a\int_0^{2/a}e^{-ax}\,\mathrm{d}x\\[4mm] =&\,\bigg[-\frac xa e^{-ax}-\frac1{a^2}e^{-ax}\bigg]_0^{2/a}\\[4mm] =&\,-\frac2{a^2}e^{-2}-\frac1{a^2}e^{-2} +\frac1{a^2}\\[4mm] =&\,\boxed{\frac{1-3e^{-2}}{a^2}}. \end{align*}