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CIE 9709 2023 Nov Paper 33 Q5

A Level / CIE / P3

CIE 9709 2023 Nov Paper 33 Paper · Question 5

题目

Problem

Find the exact coordinates of the stationary points of the curve y=e3x2−11−x2y=\dfrac{\mathrm{e}^{3x^2-1}}{1-x^2}.

(6)
题目中文翻译

求曲线 y=e3x2−11−x2y=\dfrac{\mathrm{e}^{3x^2-1}}{1-x^2} 的各驻点的精确坐标。

(6)
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

解法一

思路

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用商法则求导,提出恒正的指数因子,再解分子为零。代回函数前确认各根不等于 ±1\pm1。

答题过程

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For x≠±1x\ne\pm1, the quotient rule gives

dydx= 6xe3x2−1(1−x2)+2xe3x2−1(1−x2)2= 2xe3x2−1(4−3x2)(1−x2)2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{6xe^{3x^2-1}(1-x^2) +2xe^{3x^2-1}}{(1-x^2)^2}\\[4mm] =&\,\frac{2xe^{3x^2-1}(4-3x^2)}{(1-x^2)^2}. \end{align*}

At a stationary point the derivative is zero. Since the exponential is positive,

2x(4−3x2)=0.2x(4-3x^2)=0.

Hence

x=0orx=±233.x=0\quad\text{or}\quad x=\pm\frac{2\sqrt3}{3}.

All three values are in the domain. At x=0x=0, y=e−1y=e^{-1}. At either of the other values, x2=4/3x^2=4/3, so

y=e4−11−4/3=−3e3.y=\frac{e^{4-1}}{1-4/3}=-3e^3.

Thus the stationary points are

 (0,e−1), (233,−3e3), (−233,−3e3).\begin{align*} &\,\boxed{(0,e^{-1})},\\[4mm] &\,\boxed{\bigg(\frac{2\sqrt3}{3},-3e^3\bigg)},\\[4mm] &\,\boxed{\bigg(-\frac{2\sqrt3}{3},-3e^3\bigg)}. \end{align*}