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IAL 2025 May FP1 Q3

A Level / Edexcel / FP1

IAL 2025 May Paper · Question 3

Question

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+585xf(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)=3sinx3cosxg(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)

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+585xf(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)=3sinx3cosxg(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+585x,f(x)=x^2+5-8^{5x},

we have

f(0)=4,f(0.5)=(0.5)2+582.5=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+581.25=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[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)=3sin43cos4=2.396>0,\begin{align*} g(4) =&\,3^{\sin4}-3\cos4\\ =&\,2.396\ldots>0, \end{align*}

whereas

g(5)=3sin53cos5=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))(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,

β4+0g(4)g(5)g(4)(54)=4+2.3960.5022.396=4.826\begin{align*} \beta \approx&\,4+ \frac{0-g(4)}{g(5)-g(4)}(5-4)\\ =&\,4+ \frac{-2.396\ldots} {-0.502\ldots-2.396\ldots}\\ =&\,4.826\ldots \end{align*}

Therefore, to 44 significant figures,

β4.827.\boxed{\beta\approx4.827}.