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FP1 2022 Jan Q8

A Level / Edexcel / FP1

IAL 2022 Jan Paper · Question 8

题目

Problem

 f(x)= 2x−23+12x −13x−5−52\begin{align*} &\,f(x)\\ =&\,2x^{-\frac23}+\frac12x\\ &\,\hspace{2pt}-\frac1{3x-5}-\frac52 \end{align*} x≠53.x\neq\frac53.

The table below shows values of f(x)f(x) for some values of xx, with values of f(x)f(x) given to 44 decimal places where appropriate.

xx1122334455
f(x)f(x)0.50.5−0.2885-0.28850.58400.5840

(a) Complete the table giving the values to 44 decimal places.

(2)

The equation f(x)=0f(x) = 0 has exactly one positive root, α\alpha.

Using the values in the completed table and explaining your reasoning,

(b) determine an interval of width one that contains α\alpha.

(2)

(c) Hence use interval bisection twice to obtain an interval of width 0.250.25 that contains α\alpha.

(3)

Given also that the equation f(x)=0f(x) = 0 has a negative root, β\beta, in the interval [−1,−0.5][-1, -0.5]

(d) use linear interpolation once on this interval to find an approximation for β\beta.

Give your answer to 33 significant figures.

(3)
题目中文翻译  f(x)= 2x−23+12x −13x−5−52\begin{align*} &\,f(x)\\ =&\,2x^{-\frac23}+\frac12x\\ &\,\hspace{2pt}-\frac1{3x-5}-\frac52 \end{align*} x≠53.x\neq\frac53.

下表显示了 f(x)f(x) 在某些 xx 值处的值，f(x)f(x) 的值在适当情况下保留 44 位小数。

xx1122334455
f(x)f(x)0.50.5−0.2885-0.28850.58400.5840

(a) 完成表格，值保留 44 位小数。

(2)

方程 f(x)=0f(x) = 0 恰好有一个正根 α\alpha。

使用完成表格中的值并解释推理过程，

(b) 确定宽度为 11 且包含 α\alpha 的区间。

(2)

(c) 由此使用区间二分法两次，获得宽度为 0.250.25 且包含 α\alpha 的区间。

(3)

已知方程 f(x)=0f(x) = 0 在区间 [−1,−0.5][-1, -0.5] 内有一个负根 β\beta，

(d) 在此区间上使用一次线性插值法，求 β\beta 的近似值。 答案保留 33 位有效数字。

(3)
题目勘误

官方原题表格将 f(5)f(5) 误印为 0.58340.5834，实际上应为 0.58400.5840。本页已采用更正值。

解答

(a)

解法一

思路

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把 x=2x=2 与 x=4x=4 分别代入函数式，并按题意保留四位小数。完整表格也会在后续小题用于比较函数值的正负。

答题过程

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Substituting x=2x=2 and x=4x=4 gives

 f(2)= 2(2)−2/3+12(2) −13(2)−5−52= 222/3+1−1−52= −1.240078…≈ −1.2401.\begin{align*} &\,f(2)\\ =&\,2(2)^{-2/3}+\frac12(2)\\ &\,\hspace{2pt}-\frac1{3(2)-5}-\frac52\\ =&\,\frac2{2^{2/3}}+1-1-\frac52\\ =&\,-1.240078\ldots\\ \approx&\,-1.2401. \end{align*}

and

 f(4)= 2(4)−2/3+12(4) −13(4)−5−52= 242/3+2−17−52= 0.150843…≈ 0.1508.\begin{align*} &\,f(4)\\ =&\,2(4)^{-2/3}+\frac12(4)\\ &\,\hspace{2pt}-\frac1{3(4)-5}-\frac52\\ =&\,\frac2{4^{2/3}}+2-\frac17-\frac52\\ =&\,0.150843\ldots\\ \approx&\,0.1508. \end{align*}

Hence the completed table is

xx1122334455
f(x)f(x)0.50.5−1.2401-1.2401−0.2885-0.28850.15080.15080.58400.5840

(b)

解法一

思路

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表中在 [1,2][1,2] 和 [3,4][3,4] 都出现异号端点，但 x=5/3x=5/3 是函数在 [1,2][1,2] 内的不连续点，因此该处的符号变化不能证明存在零点。函数在 [3,4][3,4] 连续，故应选择这一段。

答题过程

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Since

f(3)≈−0.2885<0f(3)\approx-0.2885<0

and

f(4)≈0.1508>0,f(4)\approx0.1508>0,

ff changes sign on [3,4][3,4]. Also, ff is continuous on this interval, so

α∈[3,4].\boxed{\alpha\in[3,4]}.

Although f(1)f(1) and f(2)f(2) have opposite signs, ff is not continuous on [1,2][1,2] because x=5/3x=5/3 lies in this interval. Therefore [1,2][1,2] cannot be used to locate the root.

(c)

解法一

思路

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从 (b) 的区间 [3,4][3,4] 开始连续取两次中点。第一次根据 f(3.5)f(3.5) 的符号保留右半段；第二次根据 f(3.75)f(3.75) 的符号保留左半段。

答题过程

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The first midpoint is

3+42=3.5.\frac{3+4}{2}=3.5.

Evaluating the function gives

 f(3.5)= 2(3.5)−2/3+3.52 −13(3.5)−5−52= −0.064221…<0.\begin{align*} &\,f(3.5)\\ =&\,2(3.5)^{-2/3}+\frac{3.5}{2}\\ &\,\hspace{2pt}-\frac1{3(3.5)-5}-\frac52\\ =&\,-0.064221\ldots<0. \end{align*}

Since f(4)>0f(4)>0, the root lies in [3.5,4][3.5,4].

The second midpoint is

3.5+42=3.75.\frac{3.5+4}{2}=3.75.

Evaluating again gives

 f(3.75)= 2(3.75)−2/3+3.752 −13(3.75)−5−52= 0.043595…>0.\begin{align*} &\,f(3.75)\\ =&\,2(3.75)^{-2/3}+\frac{3.75}{2}\\ &\,\hspace{2pt}-\frac1{3(3.75)-5}-\frac52\\ =&\,0.043595\ldots>0. \end{align*}

Since f(3.5)<0f(3.5)<0, the required interval is

α∈[3.5,3.75].\boxed{\alpha\in[3.5,3.75]}.

Its width is 3.75−3.5=0.253.75-3.5=0.25.

(d)

解法一

思路

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计算区间两端的函数值，再用连接两端点的直线近似函数图像。按函数值从负值上升到零所占的比例，估计横坐标从 −1-1 向 −0.5-0.5 移动的距离。

答题过程

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For negative xx, evaluate the power using the real cube root:

x−2/3=1(x3)2.x^{-2/3}=\frac1{(\sqrt[3]{x})^2}.

At the left endpoint,

 f(−1)= 2(−1)−2/3+−12 −13(−1)−5−52= 2−12+18−52= 16−4+1−208= −78= −0.875.\begin{align*} &\,f(-1)\\ =&\,2(-1)^{-2/3}+\frac{-1}{2}\\ &\,\hspace{2pt}-\frac1{3(-1)-5}-\frac52\\ =&\,2-\frac12+\frac18-\frac52\\ =&\,\frac{16-4+1-20}{8}\\ =&\,-\frac78\\ =&\,-0.875. \end{align*}

At the right endpoint,

 f(−0.5)= 2(−0.5)−2/3+−0.52 −13(−0.5)−5−52= 2(−0.53)2−14 +213−52= 0.578648…\begin{align*} &\,f(-0.5)\\ =&\,2(-0.5)^{-2/3}+\frac{-0.5}{2}\\ &\,\hspace{2pt}-\frac1{3(-0.5)-5}-\frac52\\ =&\,\frac2{(\sqrt[3]{-0.5})^2}-\frac14\\ &\,\hspace{2pt}+\frac2{13}-\frac52\\ =&\,0.578648\ldots \end{align*}

The interval width is

−0.5−(−1)=0.5.-0.5-(-1)=0.5.

Write DD for the difference between the two function values:

 D=f(−0.5)−f(−1)= 0.578648…−(−0.875)= 1.453648…\begin{align*} &\,D=f(-0.5)-f(-1)\\ =&\,0.578648\ldots-(-0.875)\\ =&\,1.453648\ldots \end{align*}

Using linear interpolation,

 β≈ −1+0−(−0.875)D ×(−0.5−(−1))= −1+0.875D×0.5= −0.699033…\begin{align*} &\,\beta\\ \approx&\,-1+\frac{0-(-0.875)}{D}\\ &\,\hspace{2pt}\times\big(-0.5-(-1)\big)\\ =&\,-1+\frac{0.875}{D}\times0.5\\ =&\,-0.699033\ldots \end{align*}

The calculation uses the unrounded function values. Therefore, to 33 significant figures,

β≈−0.699.\boxed{\beta\approx-0.699}.