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CIE 9709 2023 November Paper 13 Q11

A Level / CIE / P1

CIE 9709 2023 Nov Paper 13 Paper · Question 11

题目

Problem

The diagram shows part of the curve with equation y=x+2(2x−1)2y=x+\dfrac{2}{(2x-1)^2}. The lines x=1x=1 and x=2x=2 intersect the curve at PP and QQ respectively and RR is the stationary point on the curve.

(a) Verify that the xx-coordinate of RR is 32\dfrac{3}{2} and find the yy-coordinate of RR.

(4)

(b) Find the exact value of the area of the shaded region.

(6)
题目中文翻译

图中显示方程为 y=x+2(2x−1)2y=x+\dfrac{2}{(2x-1)^2} 的曲线的一部分。直线 x=1x=1 和 x=2x=2 分别与曲线相交于 PP 和 QQ,RR 为曲线上的驻点。

(a) 验证 RR 的 xx 坐标为 32\dfrac{3}{2},并求 RR 的 yy 坐标。

(4)

(b) 求阴影区域面积的精确值。

(6)
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

展开

用链式法则求导,代入指定横坐标验证导数为零,再代回曲线求纵坐标。

答题过程

展开

The derivative is 1−8(2x−1)−31-8(2x-1)^{-3}. At x=3/2x=3/2, it equals 1−8/23=01-8/2^3=0. The height is 3/2+2/22=23/2+2/2^2=2, so R=(3/2,2)R=(3/2,2).

(b)

解法一

思路

展开

用两个端点的纵坐标求直线下方的梯形面积,再减去曲线下方的定积分。

答题过程

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At x=1x=1 and x=2x=2, the respective heights are 33 and 20/920/9. The area under PQPQ is

12(3+209)(2−1)=4718.\frac12\Big(3+\frac{20}{9}\Big)(2-1)=\frac{47}{18}.

The area under the curve is

[x22−12x−1]12=136.\Big[\frac{x^2}{2}-\frac1{2x-1}\Big]_1^2=\frac{13}{6}.

The shaded area is 47/18−13/6=4/947/18-13/6=4/9.

解法二

思路

展开

也可以先求弦 PQPQ 的方程,然后直接积分弦与曲线的高度差。

答题过程

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The line joining (1,3)(1,3) to (2,20/9)(2,20/9) is y=−7x/9+34/9y=-7x/9+34/9. The exact shaded area is

 ∫12(−16x9+349−2(2x−1)2) dx= [−8x29+34x9+12x−1]12= 49.\begin{align*} &\,\int_1^2\Big(-\frac{16x}{9}+\frac{34}{9}-\frac2{(2x-1)^2}\Big)\,\mathrm{d}x\\ =&\,\Big[-\frac{8x^2}{9}+\frac{34x}{9}+\frac1{2x-1}\Big]_1^2\\ =&\,\frac49. \end{align*}