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CIE 9231 2024 June Paper 23 Q3

A Level / CIE / FP2

CIE 9231 2024 June Paper 23 Paper · Question 3

题目

Problem

The curve CC has equation

x3+2xy+8y3=−12.x^3+2xy+8y^3=-12.

(a) Show that, at the point (−2,−1)(-2,-1) on CC,

dydx=−12.\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{2}.
[3]

(b) Find the value of d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} at the point (−2,−1)(-2,-1).

[5]
题目中文翻译

曲线 CC 的方程为

x3+2xy+8y3=−12x^3+2xy+8y^3=-12

(a) 证明在曲线 CC 上的点 (−2,−1)(-2,-1) 处,

dydx=−12\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{2}
(3)

(b) 求点 (−2,−1)(-2,-1) 处 d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} 的值。

(5)
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

展开

隐式求导,将含 y′y' 的项收集在一起,再代入 (−2,−1)(-2,-1)。

答题过程

展开

Differentiating with respect to xx gives

3x2+2y+(2x+24y2)y′=0.3x^2+2y+(2x+24y^2)y'=0.

At (−2,−1)(-2,-1) this is 10+20y′=010+20y'=0, so y′=−12y'=-\frac12.

(b)

解法一

思路

展开

对 (a) 的整个等式再求一次导数。用乘积法则展开后,把 x,y,y′x,y,y' 同时代入。

答题过程

展开

Differentiating the equation from (a), using the product rule on (2x+24y2)y′(2x+24y^2)y', gives

6x+2y′+(2+48yy′)y′+(2x+24y2)y′′=0.6x+2y'+(2+48yy')y'+(2x+24y^2)y''=0.

Collecting terms gives

6x+4y′+48y(y′)2+(2x+24y2)y′′=0.6x+4y'+48y(y')^2+(2x+24y^2)y''=0.

At (−2,−1)(-2,-1), with y′=−12y'=-\frac12, this becomes

−12−2−12+20y′′=0.-12-2-12+20y''=0.

Hence y′′=1310\boxed{y''=\frac{13}{10}}.