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CIE 9709 2024 March Paper 12 Q11

A Level / CIE / P1

CIE 9709 2024 March Paper 12 Paper · Question 11

题目

Problem

The diagram shows the curve with equation

y=2x−23−3x−13+1y = 2x^{-\frac{2}{3}} - 3x^{-\frac{1}{3}} + 1

for x>0x > 0. The curve crosses the xx-axis at points AA and BB and has a minimum point MM.

(a) Find the exact coordinates of MM.

[4]

(b) Find the area of the region bounded by the curve and the line segment ABAB.

[7]
题目中文翻译

图中显示了曲线

y=2x−23−3x−13+1y = 2x^{-\frac{2}{3}} - 3x^{-\frac{1}{3}} + 1

其中 x>0x > 0。该曲线与 xx 轴相交于点 AA 和 BB,并有一个极小点 MM。

(a) 求 MM 的精确坐标。

[4]

(b) 求由曲线和线段 ABAB 围成区域的面积。

[7]

解答

(a)

解法一

思路

展开

直接对两个负指数项求导。提出共同的幂后,只需解关于 x1/3x^{1/3} 的一次方程。

答题过程

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For x>0x>0,

y′= −43x−5/3+x−4/3= x−5/3(x1/3−43).\begin{align*} y'=&\,-\frac43x^{-5/3}+x^{-4/3}\\ =&\,x^{-5/3}\Big(x^{1/3}-\frac43\Big). \end{align*}

This changes from negative to positive at x=(4/3)3=64/27x=(4/3)^3=64/27, so that is the minimum. Here x−1/3=3/4x^{-1/3}=3/4, giving y=2(3/4)2−3(3/4)+1=−1/8y=2(3/4)^2-3(3/4)+1=-1/8. Thus M=(64/27,−1/8)M=(64/27,-1/8).

解法二

思路

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把 x−1/3x^{-1/3} 当作新变量,配方就能直接读出最小值。

答题过程

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Let u=x−1/3u=x^{-1/3}, which takes all positive values. Then

y=2u2−3u+1=2(u−34)2−18.y=2u^2-3u+1=2\Big(u-\frac34\Big)^2-\frac18.

The minimum occurs at the permitted value u=3/4u=3/4, giving x=u−3=64/27x=u^{-3}=64/27 and y=−1/8y=-1/8.

(b)

解法一

思路

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先用 u=x−1/3u=x^{-1/3} 求两个横截距。曲线在这两个交点之间位于 xx 轴下方,所以面积是积分的相反数。

答题过程

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With u=x−1/3u=x^{-1/3}, the intercept condition is (2u−1)(u−1)=0(2u-1)(u-1)=0. Thus u=1/2u=1/2 or u=1u=1, corresponding to x=8x=8 and x=1x=1. For 1<x<81<x<8, 1/2<u<11/2<u<1 and the product is negative. Hence the area is

 −∫18(2x−2/3−3x−1/3+1) dx= −[6x1/3−92x2/3+x]18= −(2−52)=12.\begin{align*} &\,-\int_1^8\Big(2x^{-2/3}-3x^{-1/3}+1\Big)\,\mathrm{d}x\\ =&\,-\Big[6x^{1/3}-\frac92 x^{2/3}+x\Big]_1^8\\ =&\,-\Big(2-\frac52\Big)=\frac12. \end{align*}