题目

Problem

Find the exact coordinates of the points on the curve $y=\dfrac{x^2}{1-3x}$ at which the gradient of the tangent is equal to $8$ .

(5)

题目中文翻译

求曲线 $y=\dfrac{x^2}{1-3x}$ 上切线斜率等于 $8$ 的各点的精确坐标。

(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

解法一

思路

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用商法则求导，再令导数为 $8$ 。解得横坐标后，逐个代回原曲线，注意分母不能为零。

答题过程

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By the quotient rule,

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

Setting this equal to $8$ ,

\begin{align*} 2x-3x^2=&\,8(1-6x+9x^2),\\[4mm] 75x^2-50x+8=&\,0,\\[4mm] (15x-4)(5x-2)=&\,0. \end{align*}

Thus $x=4/15$ or $x=2/5$ , both different from $1/3$ .

For $x=4/15$ ,

y=\frac{16/225}{1-4/5}=\frac{16}{45}.

For $x=2/5$ ,

y=\frac{4/25}{1-6/5}=-\frac45.

The required points are

\boxed{\bigg(\frac4{15},\frac{16}{45}\bigg) \quad\text{and}\quad \bigg(\frac25,-\frac45\bigg)}.

CIE 9709 2023 Nov Paper 31 Q1

题目

解答

解法一

思路

答题过程