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CIE 9709 2024 June Paper 13 Q9

A Level / CIE / P1

CIE 9709 2024 June Paper 13 Paper · Question 9

题目

Problem

The diagram shows the curve with equation y=2x3+10y = \sqrt{2x^3 + 10}.

(a) Find the equation of the tangent to the curve at the point where x=3x = 3. Give your answer in the form ax+by+c=0ax + by + c = 0 where aa, bb and cc are integers.

[5]

(b) The region shaded in the diagram is enclosed by the curve and the straight lines x=1x = 1, x=3x = 3 and y=0y = 0.

Find the volume of the solid obtained when the shaded region is rotated through 360∘360^\circ about the xx-axis.

[3]
题目中文翻译

图中显示曲线,其方程为 y=2x3+10y = \sqrt{2x^3 + 10}。

(a) 求曲线在 x=3x = 3 处的切线方程。答案写成 ax+by+c=0ax + by + c = 0 的形式,其中 aa、bb 和 cc 是整数。

[5]

(b) 图中阴影区域由曲线以及直线 x=1x = 1、x=3x = 3 和 y=0y = 0 围成。

求该阴影区域绕 xx 轴旋转 360∘360^\circ 所得立体的体积。

[3]

解答

(a)

解法一

思路

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用链式法则求导,并在 x=3x=3 处同时计算曲线高度与斜率。

答题过程

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At x=3x=3, y=64=8y=\sqrt{64}=8. The derivative is

y′=3x2(2x3+10)−1/2,y'=3x^2(2x^3+10)^{-1/2},

so the tangent slope is 27/827/8. Its equation is y−8=278(x−3)y-8=\frac{27}{8}(x-3), or 27x−8y−17=027x-8y-17=0.

(b)

解法一

思路

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绕 xx 轴旋转时,半径是 yy,截面积用 πy2\pi y^2。平方后根号消失。

答题过程

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The volume is

V= π∫13(2x3+10) dx= π[x42+10x]13= 60π.\begin{align*} V=&\,\pi\int_1^3(2x^3+10)\,\mathrm{d}x\\ =&\,\pi\Big[\frac{x^4}2+10x\Big]_1^3\\ =&\,60\pi. \end{align*}