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CIE 9709 2023 March Paper 32 Q8

A Level / CIE / P3

CIE 9709 2023 March Paper 32 Paper · Question 8

题目

Problem

The diagram shows the curve y=x3ln⁡xy=x^3\ln x, for x>0x>0, and its minimum point MM.

(a) Find the exact coordinates of MM.

(4)

(b) Find the exact area of the shaded region bounded by the curve, the xx-axis and the line x=12x=\frac12.

(5)
题目中文翻译

图中显示了曲线 y=x3ln⁡xy=x^3\ln x(x>0x>0)及其最小值点 MM。

(a) 求 MM 的精确坐标。

(4)

(b) 求曲线、xx 轴及直线 x=12x=\frac12 所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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用乘积法则求导,注意 x>0x>0,所以 x2x^2 不能为零。解出对数条件后代回求纵坐标。

答题过程

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Differentiating,

dydx= 3x2ln⁡x+x31x= x2(3ln⁡x+1).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,3x^2\ln x+x^3\frac1x\\[4mm] =&\,x^2(3\ln x+1). \end{align*}

Since x>0x>0, the stationary point satisfies

3ln⁡x+1=0,x=e−1/3.3\ln x+1=0,\qquad x=e^{-1/3}.

The derivative changes from negative to positive at this value, so it is a minimum. Substituting,

y=(e−1/3)3ln⁡(e−1/3)=e−1(−13)=−13e.y=(e^{-1/3})^3\ln(e^{-1/3}) =e^{-1}\bigg(-\frac13\bigg)=-\frac1{3e}.

Thus

M=(e−1/3,−13e).\boxed{M=\bigg(e^{-1/3},-\frac1{3e}\bigg)}.

(b)

解法一

思路

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曲线在 1/2≤x≤11/2\le x\le1 位于横轴下方,面积是负积分。用分部积分处理 x3ln⁡xx^3\ln x,再取相反数。

答题过程

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The curve meets the xx-axis at x=1x=1. On 1/2≤x≤11/2\le x\le1, x3ln⁡x≤0x^3\ln x\le0, so

A=−∫1/21x3ln⁡x dx.A=-\int_{1/2}^{1}x^3\ln x\,\mathrm{d}x.

Integrating by parts with u=ln⁡xu=\ln x, dv=x3 dx\mathrm{d}v=x^3\,\mathrm{d}x,

∫x3ln⁡x dx= x44ln⁡x−14∫x3 dx= x44ln⁡x−x416+C.\begin{align*} \int x^3\ln x\,\mathrm{d}x =&\,\frac{x^4}{4}\ln x-\frac14\int x^3\,\mathrm{d}x\\[4mm] =&\,\frac{x^4}{4}\ln x-\frac{x^4}{16}+C. \end{align*}

Thus

A= −[x44ln⁡x−x416]1/21= −[−116−(−ln⁡264−1256)]= 15256−ln⁡264.\begin{align*} A =&\,-\bigg[\frac{x^4}{4}\ln x-\frac{x^4}{16}\bigg]_{1/2}^{1}\\[4mm] =&\,-\bigg[-\frac1{16} -\bigg(-\frac{\ln2}{64}-\frac1{256}\bigg)\bigg]\\[4mm] =&\,\boxed{\frac{15}{256}-\frac{\ln2}{64}}. \end{align*}