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CIE 9709 2022 November Paper 12 Q11

A Level / CIE / P1

CIE 9709 2022 Nov Paper 12 Paper · Question 11

题目

Problem

(a) Find the coordinates of the minimum point of the curve y=94x2−12x+18y=\dfrac{9}{4}x^2-12x+18.

(3)

The diagram shows the curves with equations y=94x2−12x+18y=\dfrac{9}{4}x^2-12x+18 and y=18−38x52y=18-\dfrac{3}{8}x^{\frac{5}{2}}. The curves intersect at the points (0,18)(0,18) and (4,6)(4,6).

(b) Find the area of the shaded region.

(5)

(c) A point PP is moving along the curve y=18−38x52y=18-\dfrac{3}{8}x^{\frac{5}{2}} in such a way that the xx-coordinate of PP is increasing at a constant rate of 2 units per second.

Find the rate at which the yy-coordinate of PP is changing when x=4x=4.

(3)
题目中文翻译

(a) 求曲线 y=94x2−12x+18y=\dfrac{9}{4}x^2-12x+18 最低点的坐标。

(3)

图中两条曲线的方程分别为 y=94x2−12x+18y=\dfrac{9}{4}x^2-12x+18 和 y=18−38x52y=18-\dfrac{3}{8}x^{\frac{5}{2}}。两条曲线相交于点 (0,18)(0,18) 和 (4,6)(4,6)。

(b) 求阴影区域的面积。

(5)

(c) 点 PP 沿曲线 y=18−38x52y=18-\dfrac{3}{8}x^{\frac{5}{2}} 运动,其 xx 坐标以每秒 2 个单位的恒定速率增加。

求当 x=4x=4 时,PP 的 yy 坐标的变化率。

(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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把二次式配方。平方项最小为零,直接读出顶点。

答题过程

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The equation can be written as y=94(x−8/3)2+2y=\frac94(x-8/3)^2+2. Its minimum point is (8/3,2)(8/3,2).

解法二

思路

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也可令一阶导数为零。由于二阶导数为正,所得驻点是极小点。

答题过程

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Differentiation gives y′=92x−12y'=\frac92x-12, so y′=0y'=0 at x=8/3x=8/3. The second derivative is 9/2>09/2>0, and substitution gives y=2y=2. Hence the minimum point is (8/3,2)(8/3,2).

(b)

解法一

思路

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用两个交点作积分端点。图中分数次幂曲线在上,面积是两式之差的积分。

答题过程

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The shaded area is

 ∫04(12x−94x2−38x5/2) dx= [6x2−34x3−328x7/2]04= 96−48−967=2407.\begin{align*} &\,\int_0^4\Big(12x-\frac94x^2-\frac38x^{5/2}\Big)\,\mathrm{d}x\\ =&\,\Big[6x^2-\frac34x^3-\frac3{28}x^{7/2}\Big]_0^4\\ =&\,96-48-\frac{96}{7}=\frac{240}{7}. \end{align*}

(c)

解法一

思路

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先求纵坐标对横坐标的导数,再乘以横坐标对时间的变化率。负号表示纵坐标减小。

答题过程

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For the moving point, dydx=−15x3/2/16\frac{\mathrm{d}y}{\mathrm{d}x}=-15x^{3/2}/16. Therefore at x=4x=4,

 dydt=dydxdxdt= −1516(8)(2)=−15.\begin{align*} &\,\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}\\ =&\,-\frac{15}{16}(8)(2)=-15. \end{align*}

The height is decreasing at 1515 units per second.