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CIE 9709 2025 Nov Paper 12 Q5

A Level / CIE / P1

CIE 9709 2025 Nov Paper 12 Paper · Question 5

题目

Problem

The equation of a curve is y=4x12−xy=4x^{\frac{1}{2}}-x. The curve has a maximum point when x=ax=a and crosses the xx-axis at the point with coordinates (b,0)(b,0), where b>0b>0. The shaded region is bounded by the curve, the line x=ax=a and the xx-axis (see diagram).

(a) Find the value of aa.

[3]

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

[5]
题目中文翻译

曲线的方程为 y=4x12−xy=4x^{\frac{1}{2}}-x。当 x=ax=a 时,曲线有一个最大点;曲线在坐标为 (b,0)(b,0) 的点处与 xx 轴相交,其中 b>0b>0。阴影区域由曲线、直线 x=ax=a 和 xx 轴围成(见图)。

(a) 求 aa 的值。

(b) 求阴影区域的精确面积。

解答

(a)

解法一

思路

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对 4x−x4\sqrt x-x 求导。正的驻点左右导数符号相反,便可确认是最大点。

答题过程

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For x>0x>0, y′=2/x−1y'=2/\sqrt x-1. It is zero when x=2\sqrt x=2, so x=4x=4. The derivative is positive on (0,4)(0,4) and negative for x>4x>4. The curve is continuous at 00, so its maximum occurs at a=4a=4.

解法二

思路

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把 x\sqrt x 当作新变量。原函数就变成一个可以直接配方的二次式。

答题过程

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Let u=xu=\sqrt x, so u≥0u\ge0. Then

y=4u−u2=4−(u−2)2.y=4u-u^2=4-(u-2)^2.

The maximum occurs at u=2u=2, which is in the permitted range. Thus a=22=4a=2^2=4.

(b)

解法一

思路

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先找曲线的正横截距。图中的阴影从最大点所在的竖线延伸到这个截距,因此积分下限是 aa。

答题过程

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Setting 4x−x=04\sqrt x-x=0 gives x(4−x)=0\sqrt x(4-\sqrt x)=0. The positive intercept is b=16b=16. Since the curve lies above the axis on [4,16][4,16], the shaded area is

 ∫416(4x1/2−x) dx= [83x3/2−x22]416= (5123−128)−(643−8)=883\begin{align*} &\,\int_4^{16}(4x^{1/2}-x)\,\mathrm{d}x\\ =&\,\Big[\frac83x^{3/2}-\frac{x^2}2\Big]_4^{16}\\ =&\,\Big(\frac{512}3-128\Big)-\Big(\frac{64}3-8\Big) =\frac{88}3 \end{align*}

square units.