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FP1 2025 May Q3

A Level / Edexcel / FP1

IAL 2025 May Paper · Question 3

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

(i)

f(x)=x2+5−85xf(x) = x^2 + 5 - 8^{5x}

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

use interval bisection to determine an interval of width 0.25 that contains α\alpha.

(3)

(ii)

g(x)=3sin⁡x−3cos⁡xg(x) = 3^{\sin x} - 3\cos x

where xx is in radians.

(a) Show that the equation g(x)=0g(x) = 0 has a root, β\beta, in the interval [4,5][4, 5]

(2)

(b) Use linear interpolation once on the interval [4,5][4, 5] to determine an approximation for β\beta.

Give your answer to 4 significant figures.

(2)
题目中文翻译

本题必须展示所有运算步骤。

完全依赖计算器技术的解答不可接受。

(i)

f(x)=x2+5−85xf(x) = x^2 + 5 - 8^{5x}

已知方程 f(x)=0f(x) = 0 在区间 [0,1][0, 1] 内有唯一根 α\alpha

使用区间二分法确定宽度为 0.25 且包含 α\alpha 的区间。

(3)

(ii)

g(x)=3sin⁡x−3cos⁡xg(x) = 3^{\sin x} - 3\cos x

其中 xx 以弧度为单位。

(a) 证明方程 g(x)=0g(x) = 0 在区间 [4,5][4, 5] 内有根 β\beta

(2)

(b) 在区间 [4,5][4, 5] 上使用一次线性插值法求 β\beta 的近似值。

答案精确到 4 位有效数字。

(2)

解答

(i)

解法一

思路

展开

从已知区间 [0,1][0,1] 开始二分。先计算中点 0.50.5 的函数值,根据端点异号把区间缩为 [0,0.5][0,0.5];再计算新中点 0.250.25 的函数值,得到宽度恰为 0.250.25 的含根区间。必须写出函数值,不能只写正负号。

答题过程

展开

For

f(x)=x2+5−85x,f(x)=x^2+5-8^{5x},

we have

f(0)= 4,f(0.5)= (0.5)2+5−82.5= −175.769…\begin{align*} f(0)=&\,4,\\ f(0.5)=&\,(0.5)^2+5-8^{2.5}\\ =&\,-175.769\ldots \end{align*}

Since f(0)>0f(0)>0 and f(0.5)<0f(0.5)<0, the root lies in [0,0.5][0,0.5]. Bisecting this interval again,

 f(0.25)= (0.25)2+5−81.25= −8.39184…\begin{align*} &\,f(0.25)\\ =&\,(0.25)^2+5-8^{1.25}\\ =&\,-8.39184\ldots \end{align*}

As f(0)>0f(0)>0 and f(0.25)<0f(0.25)<0, and ff is continuous, the required interval is

[0,0.25].\boxed{[0,0.25]}.

(ii)(a)

解法一

思路

展开

分别计算区间两端的函数值。因为指数函数、正弦和余弦都连续,所以 gg 在 [4,5][4,5] 上连续;端点函数值异号后,使用介值定理即可证明区间内有根。

答题过程

展开

Using radians,

g(4)= 3sin⁡4−3cos⁡4= 2.396…>0,\begin{align*} g(4) =&\,3^{\sin4}-3\cos4\\ =&\,2.396\ldots>0, \end{align*}

whereas

g(5)= 3sin⁡5−3cos⁡5= −0.502…<0.\begin{align*} g(5) =&\,3^{\sin5}-3\cos5\\ =&\,-0.502\ldots<0. \end{align*}

The function gg is continuous on [4,5][4,5]. Since g(4)g(4) and g(5)g(5) have opposite signs, the intermediate value theorem shows that g(x)=0g(x)=0 has a root β\beta in

[4,5].\boxed{[4,5]}.

(ii)(b)

解法一

思路

展开

承接 (a) 的两个端点函数值,用连接 (4,g(4))(4,g(4)) 与 (5,g(5))(5,g(5)) 的弦和 xx 轴的交点近似根。由于右端函数值为负数,分母必须保留减去负数的符号。

答题过程

展开

Using linear interpolation once,

 β≈ 4 +0−g(4)g(5)−g(4) ×(5−4).\begin{align*} &\,\beta\\ \approx&\,4\\ &\,\hspace{2pt}+\frac{0-g(4)}{g(5)-g(4)}\\ &\,\hspace{4pt}\times(5-4). \end{align*}

The denominator is

D= g(5)−g(4)= −0.502… −2.396…= −2.898622…\begin{align*} D=&\,g(5)-g(4)\\ =&\,-0.502\ldots\\ &\,\hspace{2pt}-2.396\ldots\\ =&\,-2.898622\ldots \end{align*}

Thus

β≈ 4+−2.396…D= 4+−2.396…−2.898622…= 4.826722…\begin{align*} \beta\approx&\,4+\frac{-2.396\ldots}{D}\\ =&\,4+\frac{-2.396\ldots}{-2.898622\ldots}\\ =&\,4.826722\ldots \end{align*}

Therefore, to 44 significant figures,

β≈4.827.\boxed{\beta\approx4.827}.