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FP1 2021 June Q1

A Level / Edexcel / FP1

IAL 2021 June Paper · Question 1

题目

Problem

  1. (i)
f(x)=x3+4x−6f(x)=x^3+4x-6

(a) Show that the equation f(x)=0f(x) = 0 has a root α\alpha in the interval [1,1.5][1, 1.5]

(2)

(b) Taking 1.51.5 as a first approximation, apply the Newton Raphson process twice to f(x)f(x) to obtain an approximate value of α\alpha. Give your answer to 33 decimal places. Show your working clearly.

(4)

(ii)

 g(x)= 4x2+x−tan⁡x\begin{align*} &\,g(x)\\ =&\,4x^2+x-\tan x \end{align*}

where xx is measured in radians.

The equation g(x)=0g(x) = 0 has a single root β\beta in the interval [1.4,1.5][1.4, 1.5]

Use linear interpolation on the values at the end points of this interval to obtain an approximation to β\beta. Give your answer to 33 decimal places.

(4)
题目中文翻译
  1. (i)
f(x)=x3+4x−6f(x)=x^3+4x-6

(a) 证明方程 f(x)=0f(x) = 0 在区间 [1,1.5][1, 1.5] 内有一个根 α\alpha。

(2)

(b) 取 1.51.5 作为第一个近似值，对 f(x)f(x) 应用两次 Newton Raphson 法，求 α\alpha 的近似值。答案保留 33 位小数。清晰展示解题过程。

(4)

(ii)

 g(x)= 4x2+x−tan⁡x\begin{align*} &\,g(x)\\ =&\,4x^2+x-\tan x \end{align*}

其中 xx 以弧度为单位。

方程 g(x)=0g(x) = 0 在区间 [1.4,1.5][1.4, 1.5] 内有唯一根 β\beta。

使用该区间端点处的值进行线性插值，求 β\beta 的近似值。答案保留 33 位小数。

(4)

解答

(i)(a)

解法一

思路

展开

计算区间两端的函数值。由于多项式函数连续，且两个端点处的函数值异号，根据介值定理，区间内至少有一个根。

答题过程

展开  f(1)= 13+4(1)−6= −1, f(1.5)= (1.5)3+4(1.5)−6= 3.375.\begin{align*} &\,f(1)\\ =&\,1^3+4(1)-6\\ =&\,-1,\\ &\,f(1.5)\\ =&\,(1.5)^3+4(1.5)-6\\ =&\,3.375. \end{align*}

The function ff is continuous, and f(1)<0f(1)<0 while f(1.5)>0f(1.5)>0. Therefore, by the intermediate value theorem, f(x)=0f(x)=0 has a root α\alpha in

[1,1.5].\boxed{[1,1.5]}.

(i)(b)

解法一

思路

展开

先求 f′(x)f'(x)，再使用 Newton-Raphson 迭代式

xn+1=xn−f(xn)f′(xn).x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.

以 x1=1.5x_1=1.5 起步，连续计算两次，并只在最终答案处取三位小数。

答题过程

展开

Since

f(x)=x3+4x−6,f(x)=x^3+4x-6,

we have

f′(x)=3x2+4.f'(x)=3x^2+4.

The Newton-Raphson formula is

xn+1=xn−f(xn)f′(xn).x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.

Starting with x1=1.5x_1=1.5, the first application gives

x2= 1.5−f(1.5)f′(1.5)= 1.5−3.37510.75= 5143= 1.1860465…\begin{align*} x_2 =&\,1.5-\frac{f(1.5)}{f'(1.5)}\\ =&\,1.5-\frac{3.375}{10.75}\\ =&\,\frac{51}{43}\\ =&\,1.1860465\ldots \end{align*}

Using the unrounded value x2=51/43x_2=51/43,

 f(x2)= (5143)3+4(5143)−6= 0.412…,\begin{align*} &\,f(x_2)\\ =&\,\bigg(\frac{51}{43}\bigg)^3 +4\bigg(\frac{51}{43}\bigg)-6\\ =&\,0.412\ldots, \end{align*}

and

f′(x2)= 3(5143)2+4= 8.220….\begin{align*} f'(x_2)=&\,3\bigg(\frac{51}{43}\bigg)^2+4\\ =&\,8.220\ldots. \end{align*}

The second application gives

 x3= x2−x23+4x2−63x22+4= 5143−0.412…8.220…= 1.1358…\begin{align*} &\,x_3\\ =&\,x_2-\frac{x_2^3+4x_2-6}{3x_2^2+4}\\ =&\,\frac{51}{43} -\frac{0.412\ldots}{8.220\ldots}\\ =&\,1.1358\ldots \end{align*}

Therefore, to three decimal places,

α≈1.136.\boxed{\alpha\approx1.136}.

(ii)

解法一

思路

展开

先以弧度模式计算两个端点的函数值。把端点 (1.4,g(1.4))(1.4,g(1.4)) 与 (1.5,g(1.5))(1.5,g(1.5)) 用直线连接，再求这条弦与 xx 轴的交点，即为线性插值得到的根近似值。

答题过程

展开

Using radians,

 g(1.4)= 4(1.4)2+1.4−tan⁡(1.4)= 3.442116…, g(1.5)= 4(1.5)2+1.5−tan⁡(1.5)= −3.601419…\begin{align*} &\,g(1.4)\\ =&\,4(1.4)^2+1.4-\tan(1.4)\\ =&\,3.442116\ldots,\\ &\,g(1.5)\\ =&\,4(1.5)^2+1.5-\tan(1.5)\\ =&\,-3.601419\ldots \end{align*}

Linear interpolation gives

 β−1.41.5−1.4= 0−g(1.4)g(1.5)−g(1.4).\begin{align*} &\,\frac{\beta-1.4}{1.5-1.4}\\ =&\,\frac{0-g(1.4)}{g(1.5)-g(1.4)}. \end{align*}

The difference between the endpoint values is

 g(1.5)−g(1.4)= −3.601419… −3.442116…= −7.043536….\begin{align*} &\,g(1.5)-g(1.4)\\ =&\,-3.601419\ldots\\ &\,\hspace{2pt}-3.442116\ldots\\ =&\,-7.043536\ldots. \end{align*}

Hence

 β= 1.4 +0.1−3.442116…−7.043536…= 1.448869…\begin{align*} &\,\beta\\ =&\,1.4\\ &\,\hspace{2pt}+0.1 \frac{-3.442116\ldots} {-7.043536\ldots}\\ =&\,1.448869\ldots \end{align*}

Therefore, to three decimal places,

β≈1.449.\boxed{\beta\approx1.449}.