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IAL 2025 May Q7

A Level / Edexcel / P3

IAL 2025 May Paper · Question 7

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

A continuous curve has equation

y=e−x2sin⁡3x0≤x≤π3y=e^{-x^2}\sin 3x\qquad 0\le x\le \frac{\pi}{3}

The curve has a stationary point at the point PP.

(a) Show, using calculus, that the xx coordinate of PP is a solution of the equation

x=13arctan⁡(32x)x=\frac{1}{3}\arctan\left(\frac{3}{2x}\right)
(4)

Using the iteration formula

xn+1=13arctan⁡(32xn)x1=0.4x_{n+1}=\frac{1}{3}\arctan\left(\frac{3}{2x_n}\right)\qquad x_1=0.4

(b) find the value of

(i) x2x_2

(ii) x4x_4

giving your answers to 4 decimal places.

(3)

(c) Using a suitable interval and a suitable function which should be stated, show that the xx coordinate of PP is 0.4300.430 correct to 3 decimal places.

(2)
题目中文翻译

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。

一条连续曲线的方程为

y=e−x2sin⁡3x0≤x≤π3y=e^{-x^2}\sin 3x\qquad 0\le x\le \frac{\pi}{3}

该曲线在点 PP 处有一个驻点。

(a) 用求导证明，点 PP 的 xx 坐标满足方程

x=13arctan⁡(32x)x=\frac{1}{3}\arctan\left(\frac{3}{2x}\right)

(b) 使用迭代公式

xn+1=13arctan⁡(32xn)x1=0.4x_{n+1}=\frac{1}{3}\arctan\left(\frac{3}{2x_n}\right)\qquad x_1=0.4

求

(i) x2x_2

(ii) x4x_4

答案均精确到小数点后 4 位。

(c) 选取合适区间并写出应使用的函数，证明点 PP 的 xx 坐标为 0.4300.430（精确到小数点后 3 位）。

解答

(a)

Given

y=e−x2sin⁡3xy=e^{-x^2}\sin3x

Differentiate using the product rule:

dydx=e−x2(3cos⁡3x)+sin⁡3x(−2xe−x2)=3e−x2cos⁡3x−2xe−x2sin⁡3x\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} &=e^{-x^2}(3\cos3x)+\sin3x(-2xe^{-x^2}) \\ &=3e^{-x^2}\cos3x-2xe^{-x^2}\sin3x \end{aligned}

At the stationary point PP,

dydx=0\frac{\mathrm{d}y}{\mathrm{d}x}=0

So

3e−x2cos⁡3x−2xe−x2sin⁡3x=03e^{-x^2}\cos3x-2xe^{-x^2}\sin3x=0

Since

e−x2>0e^{-x^2}>0

we can divide by e−x2e^{-x^2}:

3cos⁡3x−2xsin⁡3x=03\cos3x-2x\sin3x=0

Thus

3cos⁡3x=2xsin⁡3x3\cos3x=2x\sin3x

Divide by 2xcos⁡3x2x\cos3x:

tan⁡3x=32x\tan3x=\frac{3}{2x}

Therefore

3x=arctan⁡(32x)3x=\arctan\left(\frac{3}{2x}\right)

and hence

x=13arctan⁡(32x)\boxed{x=\frac13\arctan\left(\frac{3}{2x}\right)}

as required.

(b)

The iteration formula is

xn+1=13arctan⁡(32xn),x1=0.4x_{n+1}=\frac13\arctan\left(\frac{3}{2x_n}\right),\qquad x_1=0.4

First,

x2=13arctan⁡(32(0.4))=0.436731…\begin{aligned} x_2 &=\frac13\arctan\left(\frac{3}{2(0.4)}\right) \\ &=0.436731\ldots \end{aligned}

So

x2=0.4367\boxed{x_2=0.4367}

Next,

x3=0.429158…x_3=0.429158\ldots

and

x4=0.430711…x_4=0.430711\ldots

Therefore

x4=0.4307\boxed{x_4=0.4307}

(c)

Define

F(x)=x−13arctan⁡(32x)F(x)=x-\frac13\arctan\left(\frac{3}{2x}\right)

To show that the root is 0.4300.430 correct to 3 decimal places, use the interval

[0.4295,0.4305][0.4295,0.4305]

Now

F(0.4295)=0.4295−13arctan⁡(32(0.4295))=−0.001141…F(0.4295) =0.4295-\frac13\arctan\left(\frac{3}{2(0.4295)}\right) =-0.001141\ldots

and

F(0.4305)=0.4305−13arctan⁡(32(0.4305))=0.0000639…F(0.4305) =0.4305-\frac13\arctan\left(\frac{3}{2(0.4305)}\right) =0.0000639\ldots

So

F(0.4295)<0F(0.4295)<0

and

F(0.4305)>0F(0.4305)>0

There is a sign change in the interval [0.4295,0.4305][0.4295,0.4305], so the root lies in this interval.

Therefore the xx coordinate of PP is

0.430\boxed{0.430}

correct to 3 decimal places.