题目

Problem

Figure 1 shows a sketch of part of the curve with equation $y=f(x)$ where

f(x)=\frac{2x^2+3x-4}{e^x}-\frac{1}{x^2}\qquad x\in\mathbb{R},\ x\ne 0

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(a) Show that $f(x)=0$ has a root $\alpha$ in the interval $[1,2]$ .

(2)

(b) Show that the equation $f(x)=0$ can be written in the form

x=\sqrt[3]{\frac{e^x+4x^2}{2x+3}}

(2)

Using the iteration formula

x_{n+1}=\sqrt[3]{\frac{e^{x_n}+4x_n^2}{2x_n+3}}\qquad x_1=1

find, to 4 decimal places,

(ii) the value of $\alpha$

(3)

题目中文翻译

图 1 给出了曲线 $y=f(x)$ 的部分草图，其中

f(x)=\frac{2x^2+3x-4}{e^x}-\frac{1}{x^2}\qquad x\in\mathbb{R},\ x\ne 0

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

(a) 证明方程 $f(x)=0$ 在区间 $[1,2]$ 内有一根 $\alpha$ 。

(b) 证明方程 $f(x)=0$ 可写成

x=\sqrt[3]{\frac{e^x+4x^2}{2x+3}}

的形式。

利用迭代公式

x_{n+1}=\sqrt[3]{\frac{e^{x_n}+4x_n^2}{2x_n+3}}\qquad x_1=1

求下列各值，答案均精确到小数点后 4 位：

(ii) $\alpha$

解答

(a)

We evaluate $f(1)$ and $f(2)$ .

\begin{aligned} f(1) &=\frac{2(1)^2+3(1)-4}{e^1}-\frac{1}{1^2} \\ &=\frac{1}{e}-1 \\ &=-0.632\ldots \end{aligned}

f(1)<0

Also,

\begin{aligned} f(2) &=\frac{2(2)^2+3(2)-4}{e^2}-\frac{1}{2^2} \\ &=\frac{10}{e^2}-\frac14 \\ &=1.103\ldots \end{aligned}

f(2)>0

Since $f(x)$ is continuous on $[1,2]$ and changes sign between $x=1$ and $x=2$ , there is a root $\alpha$ in the interval $[1,2]$ .

(b)

Start with

f(x)=0

\frac{2x^2+3x-4}{e^x}-\frac{1}{x^2}=0

Move the second term to the other side:

\frac{2x^2+3x-4}{e^x}=\frac{1}{x^2}

Multiply by $e^x x^2$ :

x^2(2x^2+3x-4)=e^x

Expand and rearrange:

2x^4+3x^3-4x^2=e^x

2x^4+3x^3=e^x+4x^2

Factorise the left-hand side:

x^3(2x+3)=e^x+4x^2

Hence

x^3=\frac{e^x+4x^2}{2x+3}

Taking cube roots gives

x=\sqrt[3]{\frac{e^x+4x^2}{2x+3}}

as required.

(c)

The iteration formula is

x_{n+1}=\sqrt[3]{\frac{e^{x_n}+4x_n^2}{2x_n+3}},\qquad x_1=1

First,

\begin{aligned} x_2 &=\sqrt[3]{\frac{e^1+4(1)^2}{2(1)+3}} \\ &=1.1035\ldots \end{aligned}

Then

\begin{aligned} x_3 &=\sqrt[3]{\frac{e^{x_2}+4x_2^2}{2x_2+3}} \\ &=1.1484\ldots \end{aligned}

Therefore

\boxed{x_3=1.1484}

Continuing the iteration gives

x_4=1.1674\ldots,\qquad x_5=1.1755\ldots,\qquad x_6=1.1789\ldots

and the values settle to

\boxed{\alpha=1.1813}

to 4 decimal places.

IAL 2025 Oct Q2

题目

解答

(a)

(b)

(c)