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CIE 9709 2022 June Paper 31 Q8

A Level / CIE / P3

CIE 9709 2022 June Paper 31 Paper · Question 8

题目

Problem

The equation of a curve is x3+y3+2xy+8=0x^3+y^3+2xy+8=0.

(a) Express dydx\dfrac{\mathrm{d}y}{\mathrm{d}x} in terms of xx and yy.

(4)

The tangent to the curve at the point where x=0x=0 and the tangent at the point where y=0y=0 intersect at the acute angle α\alpha.

(b) Find the exact value of tan⁡α\tan\alpha.

(5)
题目中文翻译

一条曲线的方程为 x3+y3+2xy+8=0x^3+y^3+2xy+8=0。

(a) 用 xx 和 yy 表示 dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}。

(4)

曲线在 x=0x=0 处的切线与在 y=0y=0 处的切线相交,所成的锐角为 α\alpha。

(b) 求 tan⁡α\tan\alpha 的精确值。

(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

对含 yy 的项用链式法则,对 2xy2xy 用乘积法则,最后收集导数项。

答题过程

展开

Differentiating implicitly,

 3x2+3y2dydx +2y+2xdydx=0.\begin{align*} &\,3x^2+3y^2\frac{\mathrm{d}y}{\mathrm{d}x}\\[4mm] &\,\hspace{2pt}+2y+2x\frac{\mathrm{d}y}{\mathrm{d}x}=0. \end{align*}

Hence

(3y2+2x)dydx=−(3x2+2y).(3y^2+2x)\frac{\mathrm{d}y}{\mathrm{d}x}=-(3x^2+2y).

Therefore

dydx=−3x2+2y3y2+2x.\boxed{\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac{3x^2+2y}{3y^2+2x}}.

(b)

解法一

思路

展开

先分别求两个坐标轴交点及切线斜率。两条切线的倾角作差,使用正切差角公式得到锐角的正切。

答题过程

展开

At x=0x=0, the curve gives y3+8=0y^3+8=0, so the point is (0,−2)(0,-2). From part (a), its tangent gradient is

m1=−−412=13.m_1=-\frac{-4}{12}=\frac13.

At y=0y=0, x3+8=0x^3+8=0, so the point is (−2,0)(-2,0). Its tangent gradient is

m2=−12−4=3.m_2=-\frac{12}{-4}=3.

If their inclination angles are β1,β2\beta_1,\beta_2, then tan⁡β1=1/3\tan\beta_1=1/3 and tan⁡β2=3\tan\beta_2=3. The acute angle is α=β2−β1\alpha=\beta_2-\beta_1. Thus

tan⁡α=3−1/31+3(1/3)=43.\boxed{\tan\alpha =\frac{3-1/3}{1+3(1/3)} =\frac43}.