题目

Problem

The diagram shows the curve $y=x\mathrm{e}^{2x}-5x$ and its minimum point $M$ , where $x=\alpha$ .

(a) Show that $\alpha$ satisfies the equation $\alpha=\dfrac12\ln\bigg(\dfrac5{1+2\alpha}\bigg)$ .

(3)

(b) Verify by calculation that $\alpha$ lies between $0.4$ and $0.5$ .

(2)

(c) Use an iterative formula based on the equation in part (a) to determine $\alpha$ correct to $2$ decimal places. Give the result of each iteration to $4$ decimal places.

(3)

题目中文翻译

图中显示了曲线 $y=x\mathrm{e}^{2x}-5x$ 及其极小值点 $M$ ，该点的 $x$ 坐标为 $\alpha$ 。

(a) 证明 $\alpha$ 满足方程 $\alpha=\dfrac12\ln\bigg(\dfrac5{1+2\alpha}\bigg)$ 。

(3)

(b) 通过计算验证 $\alpha$ 位于 $0.4$ 与 $0.5$ 之间。

(2)

(3)

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

在极小点导数为零。用乘积法则求导后，先把指数函数单独移到一边，再取对数得到指定形式。

答题过程

展开

For $y=xe^{2x}-5x$ ,

\frac{\mathrm{d}y}{\mathrm{d}x} =e^{2x}+2xe^{2x}-5.

At $x=\alpha$ , the derivative is zero, so

\begin{align*} (1+2\alpha)e^{2\alpha}=&\,5,\\[4mm] e^{2\alpha}=&\,\frac5{1+2\alpha}. \end{align*}

Taking logarithms,

2\alpha=\ln\bigg(\frac5{1+2\alpha}\bigg),

hence

\boxed{\alpha=\frac12\ln\bigg(\frac5{1+2\alpha}\bigg)}.

(b)

解法一

思路

展开

把驻点条件写成连续函数的零点条件，计算区间两端的符号变化。

答题过程

展开

Let $g(x)=(1+2x)e^{2x}-5$ , which is continuous. Then

\begin{align*} g(0.4)=&\,1.8e^{0.8}-5=-0.994026\ldots<0,\\[4mm] g(0.5)=&\,2e-5=0.436563\ldots>0. \end{align*}

Therefore there is a zero between $0.4$ and $0.5$ . Also

g'(x)=4(1+x)e^{2x}>0\qquad(0.4\le x\le0.5),

so this zero is unique there. Hence $\boxed{0.4<\alpha<0.5}$ .

(c)

解法一

思路

展开

采用(a)给出的迭代式。计算时保留内部精度，表中列到四位小数；继续到数值稳定于同一个两位小数结果。

答题过程

展开

Use

\alpha_{n+1}=\frac12\ln\bigg(\frac5{1+2\alpha_n}\bigg), \qquad \alpha_0=0.4.

Keeping unrounded values in the calculation gives:

$n$	$\alpha_n$ to 4 decimal places
0	0.4000
1	0.5108
2	0.4528
3	0.4823
4	0.4671
5	0.4749
6	0.4709
7	0.4729
8	0.4719
9	0.4724
10	0.4721

The values converge to a number which rounds to $0.47$ . Therefore

\boxed{\alpha=0.47\text{ to 2 decimal places}}.

CIE 9709 2024 March Paper 32 Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程