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IAL 2025 Oct A Q2

A Level / Edexcel / P3

IAL 2025 Oct A Paper · Question 2

题目

Problem

Given that

y=(2x2−3)tan⁡(12x)0<x<πy=(2x^2-3)\tan\bigg(\frac12x\bigg) \qquad 0<x<\pi

dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x}=0 when x=αx=\alpha.

(a) show that

2α2−3+4αsin⁡α=02\alpha^2-3+4\alpha\sin\alpha=0
(6)

The iterative formula

xn+1=32xn+4sin⁡xnx_{n+1}=\frac{3}{2x_n+4\sin x_n}

can be used to find an approximation for α\alpha.

(b) Taking x1=0.7x_1 = 0.7, find the values of x2x_2 and x3x_3, giving each answer to 4 decimal places.

(2)

(c) By choosing a suitable interval and a suitable function that should be stated, show that α=0.7283\alpha = 0.7283 to 4 decimal places.

(2)
题目中文翻译

已知

y=(2x2−3)tan⁡(12x)0<x<πy=(2x^2-3)\tan\bigg(\frac12x\bigg) \qquad 0<x<\pi

并且当 x=αx=\alpha 时，dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x}=0。

(a) 证明

2α2−3+4αsin⁡α=02\alpha^2-3+4\alpha\sin\alpha=0

可用迭代公式

xn+1=32xn+4sin⁡xnx_{n+1}=\frac{3}{2x_n+4\sin x_n}

求 α\alpha 的近似值。

(b) 取 x1=0.7x_1 = 0.7，求 x2x_2 和 x3x_3，每个答案保留 4 位小数。

(c) 选取适当区间和适当函数并写明，证明 α=0.7283\alpha = 0.7283，精确到 4 位小数。

解答

(a)

解法一

思路

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先对两个因子的乘积求导，并对 tan⁡(12x)\tan\big(\frac12x\big) 使用链式法则。令 x=αx=\alpha 且导数为 00 后，把正切和正割写成正弦、余弦，再用二倍角公式自然推出题目要求的等式。

答题过程

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Using the product rule and the chain rule,

dydx= 4xtan⁡(x2) +12(2x2−3)sec⁡2(x2).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,4x\tan\bigg(\frac{x}{2}\bigg)\\ &\,+\frac12(2x^2-3) \sec^2\bigg(\frac{x}{2}\bigg). \end{align*}

When x=αx=\alpha, dydx=0\dfrac{\mathrm{d}y}{\mathrm{d}x}=0. Hence

 4αtan⁡(α2) +12(2α2−3)sec⁡2(α2)=0.\begin{align*} &\,4\alpha\tan\bigg(\frac{\alpha}{2}\bigg)\\ &\,+\frac12(2\alpha^2-3) \sec^2\bigg(\frac{\alpha}{2}\bigg)=0. \end{align*}

Multiplying by 2cos⁡2(α2)2\cos^2\big(\frac{\alpha}{2}\big) gives

 8αsin⁡(α2)cos⁡(α2) +(2α2−3)=0.\begin{align*} &\,8\alpha \sin\bigg(\frac{\alpha}{2}\bigg) \cos\bigg(\frac{\alpha}{2}\bigg)\\ &\,+(2\alpha^2-3)=0. \end{align*}

Using

2sin⁡(α2)cos⁡(α2)=sin⁡α,2\sin\bigg(\frac{\alpha}{2}\bigg) \cos\bigg(\frac{\alpha}{2}\bigg) =\sin\alpha,

we obtain

4αsin⁡α+2α2−3=0.4\alpha\sin\alpha+2\alpha^2-3=0.

Therefore,

2α2−3+4αsin⁡α=0.\boxed{2\alpha^2-3+4\alpha\sin\alpha=0}.

(b)

解法一

思路

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严格按照给定公式，把 x1=0.7x_1=0.7 代入求 x2x_2，再把未过早取整的 x2x_2 代入求 x3x_3。最终才把两个答案分别保留四位小数。

答题过程

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Using x1=0.7x_1=0.7,

x2= 32(0.7)+4sin⁡(0.7)= 0.754361956…\begin{align*} x_2 =&\,\frac{3}{2(0.7)+4\sin(0.7)}\\ =&\,0.754361956\ldots \end{align*}

Therefore,

x3= 32x2+4sin⁡x2= 0.706211472…\begin{align*} x_3 =&\,\frac{3}{2x_2+4\sin x_2}\\ =&\,0.706211472\ldots \end{align*}

Hence, to 4 decimal places,

x2=0.7544,x3=0.7062.\boxed{x_2=0.7544,\qquad x_3=0.7062}.

(c)

解法一

思路

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定义 (a) 左边对应的连续函数，并选取四位小数舍入边界 0.728250.72825 与 0.728350.72835。若端点函数值异号，便能确定根严格位于两者之间，从而四舍五入为 0.72830.7283。本小题不能只继续做重复迭代。

答题过程

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Let

f(x)=2x2−3+4xsin⁡x.f(x)=2x^2-3+4x\sin x.

This function is continuous. At the endpoints of the interval,

f(0.72825)= 2(0.72825)2−3 +4(0.72825)sin⁡(0.72825)= −0.00051431…,\begin{align*} f(0.72825) =&\,2(0.72825)^2-3\\ &\,+4(0.72825)\sin(0.72825)\\ =&\,-0.00051431\ldots, \end{align*}

whereas

f(0.72835)= 2(0.72835)2−3 +4(0.72835)sin⁡(0.72835)= 0.00026066….\begin{align*} f(0.72835) =&\,2(0.72835)^2-3\\ &\,+4(0.72835)\sin(0.72835)\\ =&\,0.00026066\ldots. \end{align*}

Since there is a change of sign,

0.72825<α<0.72835.0.72825<\alpha<0.72835.

Therefore, to 4 decimal places,

α=0.7283.\boxed{\alpha=0.7283}.