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CIE 9709 2023 November Paper 12 Q3

A Level / CIE / P1

CIE 9709 2023 Nov Paper 12 Paper · Question 3

题目

Problem

The equation of a curve is such that dydx=12x+72x4\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{2}x+\dfrac{72}{x^4}. The curve passes through the point P(2,8)P(2,8).

(a) Find the equation of the normal to the curve at PP.

(2)

(b) Find the equation of the curve.

(4)
题目中文翻译

一条曲线满足 dydx=12x+72x4\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{2}x+\dfrac{72}{x^4},且经过点 P(2,8)P(2,8)。

(a) 求曲线在 PP 点处的法线方程。

(2)

(b) 求曲线的方程。

(4)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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代入 x=2x=2 得到切线斜率。法线垂直于切线,因此取负倒数,再代入点 PP。

答题过程

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The tangent gradient at PP is 1+72/16=11/21+72/16=11/2, so the normal gradient is −2/11-2/11. The normal is

y−8=−211(x−2).y-8=-\frac2{11}(x-2).

(b)

解法一

思路

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把第二项写成负指数,逐项积分。最后用曲线上已知点求积分常数。

答题过程

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On the branch through x=2x=2, integration gives

y=x24−24x3+C.y=\frac{x^2}{4}-\frac{24}{x^3}+C.

At PP, we have 8=1−3+C8=1-3+C, so C=10C=10. The curve is y=x2/4−24/x3+10y=x^2/4-24/x^3+10 on x>0x>0.