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CIE 9709 2025 Nov Paper 15 Q8

A Level / CIE / P1

CIE 9709 2025 Nov Paper 15 Paper · Question 8

题目

Problem

The diagram shows the curve with equation y=6x+5y=\sqrt{6x+5}. The shaded region is bounded by the curve, the xx-axis and the lines x=ax=a and x=2ax=2a, where aa is a positive constant.

The shaded region is rotated through 360∘360^\circ about the xx-axis to form a solid. The solid has volume, VV, such that V≥46πV\ge 46\pi.

(a) Show that 9a2+5a−46≥09a^2+5a-46\ge 0.

[4]

(b) Find the range of possible values of aa.

[3]
题目中文翻译

图中显示的是方程为 y=6x+5y=\sqrt{6x+5} 的曲线。阴影区域由曲线、xx 轴以及直线 x=ax=a 和 x=2ax=2a 围成,其中 aa 是正的常数。

阴影区域绕 xx 轴旋转 360∘360^\circ 形成一个立体。该立体的体积为 VV,且 V≥46πV\ge 46\pi。

(a) 证明 9a2+5a−46≥09a^2+5a-46\ge 0。

(b) 求 aa 的可能取值范围。

解答

(a)

解法一

思路

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垂直于 xx 轴的截面是圆盘,半径是 yy。将截面积 πy2\pi y^2 在 [a,2a][a,2a] 上积分,再代入体积下限。

答题过程

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The volume is

V= π∫a2a(6x+5) dx=π[3x2+5x]a2a= π(12a2+10a−3a2−5a)=π(9a2+5a).\begin{align*} V=&\,\pi\int_a^{2a}(6x+5)\,\mathrm{d}x =\pi\Big[3x^2+5x\Big]_a^{2a}\\ =&\,\pi\bigl(12a^2+10a-3a^2-5a\bigr) =\pi(9a^2+5a). \end{align*}

Since V≥46πV\ge46\pi and π>0\pi>0, we obtain 9a2+5a−46≥09a^2+5a-46\ge0.

(b)

解法一

思路

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求出二次式的两个零点。开口向上的二次式在两根外侧非负,最后还要取与 a>0a>0 的交集。

答题过程

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The inequality factors as

(a−2)(9a+23)≥0.(a-2)(9a+23)\ge0.

The zeros are 22 and −239-\frac{23}{9}. Because the leading coefficient is positive, the inequality holds for a≤−239a\le-\frac{23}{9} or a≥2a\ge2. As aa is positive, the required range is a≥2a\ge2.