Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9231 2024 November Paper 22 Q2

A Level / CIE / FP2

CIE 9231 2024 Nov Paper 22 Paper · Question 2

题目

Problem

The curve CC has equation

4y2+4ln⁡(xy)=1.4y^2 + 4\ln(xy) = 1.

(a) Show that, at the point (2,12)\big(2, \frac{1}{2}\big) on CC,

dydx=−16.\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{6}.
[3]

(b) Find the value of d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} at the point (2,12)\big(2, \frac{1}{2}\big).

[4]
题目中文翻译

曲线 CC 的方程为

4y2+4ln⁡(xy)=14y^2 + 4\ln(xy) = 1

(a) 证明在 CC 上的点 (2,12)\big(2, \frac{1}{2}\big) 处,

dydx=−16\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{6}
(3)

(b) 求点 (2,12)\big(2, \frac{1}{2}\big) 处 d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} 的值。

(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

展开

对 ln⁡(xy)\ln(xy) 用链式法则。把导数写成 1x+y′y\frac1x+\frac{y'}y,再代入给定点。

答题过程

展开

Since xy>0xy>0, the logarithm is defined, and

y+xy′xy=1x+y′y.\frac{y+xy'}{xy}=\frac1x+\frac{y'}y.

Differentiating gives

8yy′+4y+xy′xy=(8y+4y)y′+4x=0.8yy'+4\frac{y+xy'}{xy} =\bigg(8y+\frac4y\bigg)y'+\frac4x=0.

At (2,12)(2,\frac12), this becomes 12y′+2=012y'+2=0, so y′=−16y'=-\frac16.

(b)

解法一

思路

展开

对 (a) 的隐式导数等式再求导。把 y′y' 前的括号作为一个整体,用乘积法则。

答题过程

展开

Differentiating the equation from (a) yields

(8y+4y)y′′+(8−4y2)(y′)2−4x2=0.\bigg(8y+\frac4y\bigg)y''+ \bigg(8-\frac4{y^2}\bigg)(y')^2-\frac4{x^2}=0.

Substituting x=2x=2, y=12y=\frac12 and y′=−16y'=-\frac16 gives

12y′′= 1+836=119,y′′= 11108.\begin{align*} 12y''=&\,1+\frac8{36}=\frac{11}{9},\\[4mm] y''=&\,\boxed{\frac{11}{108}}. \end{align*}

解法二

思路

展开

也可先把 y′y' 单独表示成 x,yx,y 的函数,再用乘积法则和链式法则求导。

答题过程

展开

From (a), y′=−x−1(2y+y−1)−1y'=-x^{-1}(2y+y^{-1})^{-1}. Let H=2y+y−1H=2y+y^{-1}. Then

y′′=x−2H−1+x−1H−2(2−y−2)y′.y''=x^{-2}H^{-1}+x^{-1}H^{-2}(2-y^{-2})y'.

At the point, H=3H=3, x=2x=2, 2−y−2=−22-y^{-2}=-2 and y′=−16y'=-\frac16. Therefore

y′′=112+154=11108.y''=\frac1{12}+\frac1{54}=\frac{11}{108}.