CalcGospel 國際數學圖譜
返回

FP1 2023 May Q5

A Level / Edexcel / FP1

IAL 2023 May Paper · Question 5

题目

Problem

5. f(x)=x2−6x+3f(x) = x^2 - 6x + 3

The equation f(x)=0f(x) = 0 has roots α\alpha and β\beta

Without solving the equation,

(a) determine the value of

(α2+1)(β2+1)(\alpha^2 + 1)(\beta^2 + 1)
(4)

(b) find a quadratic equation which has roots

αα2+1\dfrac{\alpha}{\alpha^2 + 1}

and

ββ2+1\dfrac{\beta}{\beta^2 + 1}

giving your answer in the form px2+qx+r=0px^2 + qx + r = 0 where pp, qq and rr are integers to be determined.

(6)
题目中文翻译
  1. f(x)=x2−6x+3f(x) = x^2 - 6x + 3

方程 f(x)=0f(x) = 0 的根为 α\alpha 和 β\beta。

不解方程，

(a) 求 (α2+1)(β2+1)(\alpha^2 + 1)(\beta^2 + 1) 的值。

(4)

(b) 求一个二次方程，使其根为

αα2+1\dfrac{\alpha}{\alpha^2 + 1}

和

ββ2+1\dfrac{\beta}{\beta^2 + 1}

答案写成 px2+qx+r=0px^2 + qx + r = 0 的形式，其中 pp、qq 和 rr 是待定整数。

(6)

解答

(a)

解法一

思路

展开

先用根与系数关系得到 α+β\alpha+\beta 和 αβ\alpha\beta。展开目标式后，把 α2+β2\alpha^2+\beta^2 改写成 (α+β)2−2αβ(\alpha+\beta)^2-2\alpha\beta，即可完全用根的和与积计算。

答题过程

展开

By the sum and product of roots,

α+β=6,\alpha+\beta=6, αβ=3.\alpha\beta=3.

Therefore,

 (α2+1)(β2+1)= α2β2+α2+β2+1= (αβ)2+(α+β)2 −2αβ+1= 32+62−2(3)+1= 40.\begin{align*} &\,(\alpha^2+1)(\beta^2+1)\\ =&\,\alpha^2\beta^2 +\alpha^2+\beta^2+1\\ =&\,(\alpha\beta)^2 +(\alpha+\beta)^2\\ &\,-2\alpha\beta+1\\ =&\,3^2+6^2-2(3)+1\\ =&\,\boxed{40}. \end{align*}

(b)

解法一

思路

展开

设新根为 u,vu,v，分别求它们的和与积。(a) 的结果正好是共同分母；分子则用 α+β\alpha+\beta 与 αβ\alpha\beta 化简。最后使用“x2−x^2-根的和 x+\,x+根的积=0=0”构造方程，并化为整数系数。

答题过程

展开

Let

u=αα2+1,u=\frac{\alpha}{\alpha^2+1}, v=ββ2+1.v=\frac{\beta}{\beta^2+1}.

Using part (a), the common denominator is 4040. Write the numerator of the sum as NN:

 N= α(β2+1) +β(α2+1)= αβ2+α +βα2+β= αβ(α+β) +(α+β)= 3(6)+6= 24.\begin{align*} &\,N\\ =&\,\alpha(\beta^2+1)\\ &\,\hspace{2pt}+\beta(\alpha^2+1)\\ =&\,\alpha\beta^2+\alpha\\ &\,\hspace{2pt}+\beta\alpha^2+\beta\\ =&\,\alpha\beta(\alpha+\beta)\\ &\,\hspace{2pt}+(\alpha+\beta)\\ =&\,3(6)+6\\ =&\,24. \end{align*}

Therefore

 u+v= αα2+1+ββ2+1= N(α2+1)(β2+1)= 2440= 35.\begin{align*} &\,u+v\\ =&\,\frac{\alpha}{\alpha^2+1} +\frac{\beta}{\beta^2+1}\\ =&\,\frac{N}{(\alpha^2+1)(\beta^2+1)}\\ =&\,\frac{24}{40}\\ =&\,\frac35. \end{align*}

Also,

 uv= αβ(α2+1)(β2+1)= 340.\begin{align*} &\,uv\\ =&\,\frac{\alpha\beta} {(\alpha^2+1)(\beta^2+1)}\\ =&\,\frac3{40}. \end{align*}

Hence a quadratic equation with roots uu and vv is

x2−35x+340=0.x^2-\frac35x+\frac3{40}=0.

Multiplying by 4040 gives

40x2−24x+3=0.\boxed{40x^2-24x+3=0}.