The diagram shows the curve with equation y=6x+5. The shaded region is bounded by the curve, the x-axis and the lines x=a and x=2a, where a is a positive constant.
The shaded region is rotated through 360∘ about the x-axis to form a solid. The solid has volume, V, such that V≥46π.
(a) Show that 9a2+5a−46≥0.
[4]
(b) Find the range of possible values of a.
[3]
题目中文翻译
图中显示的是方程为 y=6x+5 的曲线。阴影区域由曲线、x 轴以及直线 x=a 和 x=2a 围成,其中 a 是正的常数。
阴影区域绕 x 轴旋转 360∘ 形成一个立体。该立体的体积为 V,且 V≥46π。
(a) 证明 9a2+5a−46≥0。
(b) 求 a 的可能取值范围。
解答
(a)
解法一
思路
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垂直于 x 轴的截面是圆盘,半径是 y。将截面积 πy2 在 [a,2a] 上积分,再代入体积下限。
The zeros are 2 and −923. Because the leading coefficient is positive, the inequality holds for a≤−923 or a≥2. As a is positive, the required range is a≥2.