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CIE 9709 2023 Nov Paper 31 Q9

A Level / CIE / P3

CIE 9709 2023 Nov Paper 31 Paper · Question 9

题目

Problem

The diagram shows the curve y=xe−14x2y=x\mathrm{e}^{-\frac14x^2}, for x≥0x\ge0, and its maximum point MM.

(a) Find the exact coordinates of MM.

(4)

(b) Using the substitution x=ux=\sqrt u, or otherwise, find by integration the exact area of the shaded region bounded by the curve, the xx-axis and the line x=3x=3.

(5)
题目中文翻译

图中显示了曲线 y=xe−14x2y=x\mathrm{e}^{-\frac14x^2}(x≥0x\ge0)及其极大值点 MM。

(a) 求 MM 的精确坐标。

(4)

(b) 使用代换 x=ux=\sqrt u 或其他方法,通过积分求由该曲线、xx 轴和直线 x=3x=3 围成的阴影区域的精确面积。

(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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用乘积法则与链式法则求导。指数因子恒正,因此导数符号由 1−x2/21-x^2/2 决定。

答题过程

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Differentiating,

dydx= e−x2/4+xe−x2/4(−x2)= e−x2/4(1−x22).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,e^{-x^2/4} +x e^{-x^2/4}\bigg(-\frac x2\bigg)\\[4mm] =&\,e^{-x^2/4}\bigg(1-\frac{x^2}{2}\bigg). \end{align*}

At a stationary point, x2=2x^2=2. Since x≥0x\ge0, x=2x=\sqrt2.

The derivative is positive for 0≤x<20\le x<\sqrt2 and negative for x>2x>\sqrt2, so this point is a maximum. Its ordinate is

y=2 e−2/4=2 e−1/2.y=\sqrt2\,e^{-2/4}=\sqrt2\,e^{-1/2}.

Thus M=(2,2 e−1/2)\boxed{M=(\sqrt2,\sqrt2\,e^{-1/2})}.

(b)

解法一

思路

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按题设代换 x=ux=\sqrt u,用 x dx=12 dux\,\mathrm{d}x=\tfrac12\,\mathrm{d}u 消去根式。上下限也要换成 uu。

答题过程

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Since y≥0y\ge0 for x≥0x\ge0, the area is

A=∫03xe−x2/4 dx.A=\int_0^3 xe^{-x^2/4}\,\mathrm{d}x.

Put u=x2u=x^2, so x=ux=\sqrt u and dx=12u−1/2 du\mathrm{d}x=\tfrac12u^{-1/2}\,\mathrm{d}u. The limits x=0,3x=0,3 become u=0,9u=0,9. Hence

A= 12∫09e−u/4 du= [−2e−u/4]09= 2−2e−9/4.\begin{align*} A=&\,\frac12\int_0^9 e^{-u/4}\,\mathrm{d}u\\[4mm] =&\,\big[-2e^{-u/4}\big]_0^9\\[4mm] =&\,\boxed{2-2e^{-9/4}}. \end{align*}

解法二

思路

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观察到指数的导数为 −x/2-x/2,因此被积函数与 e−x2/4e^{-x^2/4} 的导数只差常数,可直接逆用链式法则。

答题过程

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The area is

A=∫03xe−x2/4 dx.A=\int_0^3 xe^{-x^2/4}\,\mathrm{d}x.

Since

ddxe−x2/4=−x2e−x2/4,\frac{\mathrm{d}}{\mathrm{d}x}e^{-x^2/4} =-\frac x2e^{-x^2/4},

an antiderivative of xe−x2/4xe^{-x^2/4} is −2e−x2/4-2e^{-x^2/4}. Therefore

A=[−2e−x2/4]03=−2e−9/4+2.A=\big[-2e^{-x^2/4}\big]_0^3 =-2e^{-9/4}+2.

Thus A=2−2e−9/4\boxed{A=2-2e^{-9/4}}.