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IAL 2026 Jan A FP1 Q7

A Level / Edexcel / FP1

IAL 2026 Jan A Paper · Question 7

Question

Problem

f(x)=30+7xx5f(x) = 30 + \dfrac{7}{\sqrt{x}}-x^5, x>0x > 0

The only real root, α\alpha, of the equation f(x)=0f(x) = 0 lies in the interval [2,2.1][2, 2.1].

(a) Starting with the interval [2,2.1][2, 2.1], use interval bisection twice to find an interval of width 0.025 that contains α\alpha.

(4)

(b) Find f(x)f'(x).

(2)

(c) Taking 2 as a first approximation to α\alpha, apply the Newton-Raphson process once to f(x)f(x) to find a second approximation to α\alpha, giving your answer to 2 decimal places.

(2)

中文翻译

f(x)=30+7xx5f(x) = 30 + \dfrac{7}{\sqrt{x}}-x^5x>0x > 0

方程 f(x)=0f(x) = 0 的唯一实根 α\alpha 在区间 [2,2.1][2, 2.1] 内。

(a) 从区间 [2,2.1][2, 2.1] 开始,使用区间二分法两次,找到宽度为 0.025 且包含 α\alpha 的区间。

(4)

(b) 求 f(x)f'(x)

(2)

(c) 取 2 作为 α\alpha 的初始近似值,对 f(x)f(x) 应用一次牛顿-拉夫森过程求 α\alpha 的第二个近似值,答案精确到 2 位小数。

(2)

解答

(a)

解法一

思路

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每次二分都计算当前区间中点的函数值,并与一个端点的函数值比较符号。连续二分两次后,区间宽度由 0.10.1 变为 0.050.05,再变为 0.0250.025

答题过程

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At the endpoints of the initial interval,

f(2)=2.9497>0,f(2.1)=6.0105<0.\begin{align*} f(2)=&\,2.9497\ldots>0,\\ f(2.1)=&\,-6.0105\ldots<0. \end{align*}

The first midpoint is 2.052.05, and

f(2.05)=1.3160<0.f(2.05)=-1.3160\ldots<0.

Since f(2)>0f(2)>0 and f(2.05)<0f(2.05)<0, the root lies in

[2,2.05].[2,2.05].

The midpoint of this new interval is 2.0252.025, and

f(2.025)=0.8684>0.f(2.025)=0.8684\ldots>0.

Since f(2.025)>0f(2.025)>0 and f(2.05)<0f(2.05)<0,

2.025<α<2.05.\boxed{2.025<\alpha<2.05}.

This interval has width 2.052.025=0.0252.05-2.025=0.025.

(b)

解法一

思路

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先把根式分母写成幂 x1/2x^{-1/2},再逐项求导,注意负指数求导后的系数与指数符号。

答题过程

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Writing

f(x)=30+7x1/2x5,f(x)=30+7x^{-1/2}-x^5,

we obtain

f(x)=72x3/25x4.\boxed{ f'(x)=-\frac{7}{2}x^{-3/2}-5x^4 }.

(c)

解法一

思路

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使用牛顿-拉夫森公式 xn+1=xnf(xn)f(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},只迭代一次。题目指定 x0=2x_0=2,所以代入 (a)、(b) 已算出的函数值和导数值即可。

答题过程

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The Newton-Raphson formula is

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

Using x0=2x_0=2,

f(2)=2.9497f(2)=2.9497\ldots

and

f(2)=81.2374.f'(2)=-81.2374\ldots.

Therefore,

x1=2f(2)f(2)=22.949781.2374=2.0363\begin{align*} x_1 =&\,2-\frac{f(2)}{f'(2)}\\ =&\,2-\frac{2.9497\ldots}{-81.2374\ldots}\\ =&\,2.0363\ldots \end{align*}

Hence the second approximation, to two decimal places, is

x1=2.04.\boxed{x_1=2.04}.