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CIE 9709 2025 June Paper 12 Q6

A Level / CIE / P1

CIE 9709 2025 June Paper 12 Paper · Question 6

题目

Problem

The diagram shows the curve with equation y=9(5x+4)12y = \frac{9}{(5x + 4)^{\frac{1}{2}}} and the line y=6−3xy = 6 - 3x. The line and the curve intersect at the point PP which has yy-coordinate 3.

Find the area of the shaded region.

[6]
题目中文翻译

图中显示了方程为 y=9(5x+4)12y = \frac{9}{(5x + 4)^{\frac{1}{2}}} 的曲线和直线 y=6−3xy = 6 - 3x。直线与曲线交于点 PP,点 PP 的 yy 坐标为 3。

求阴影区域的面积。

[6]

解答

解法一

思路

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左边的阴影上界是曲线,右边是直线。用交点 PP 将阴影切成积分区域和直角三角形。

答题过程

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At PP, 6−3x=36-3x=3, so x=1x=1. The area below the curve from x=0x=0 to x=1x=1 is

 ∫019(5x+4)−1/2 dx= [185(5x+4)1/2]01= 185(3−2)=185.\begin{align*} &\,\int_0^1 9(5x+4)^{-1/2}\,\mathrm{d}x\\ =&\,\Big[\frac{18}{5}(5x+4)^{1/2}\Big]_0^1\\ =&\,\frac{18}{5}(3-2)=\frac{18}{5}. \end{align*}

The line meets the xx-axis at x=2x=2. The remaining region is a triangle of width 2−1=12-1=1 and height 33, so the total area is 185+32=5110\frac{18}{5}+\frac32=\frac{51}{10} square units.