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CIE 9231 2023 November Paper 11 Q3

A Level / CIE / FP1

CIE 9231 2023 Nov Paper 11 Paper · Question 3

题目

Problem

The quartic equation x4+bx3+cx2+dx−2=0x^4+bx^3+cx^2+dx-2=0 has roots α\alpha, β\beta, γ\gamma, δ\delta. It is given that

α+β+γ+δ= 3,α2+β2+γ2+δ2= 5,α−1+β−1+γ−1+δ−1= 6.\begin{align*} \alpha+\beta+\gamma+\delta=&\,3,\\ \alpha^2+\beta^2+\gamma^2+\delta^2=&\,5,\\ \alpha^{-1}+\beta^{-1}+\gamma^{-1}+\delta^{-1}=&\,6. \end{align*}

(a) Find the values of bb, cc and dd.

(6)

(b) Given also that α3+β3+γ3+δ3=−27\alpha^3+\beta^3+\gamma^3+\delta^3=-27, find the value of α4+β4+γ4+δ4\alpha^4+\beta^4+\gamma^4+\delta^4.

(2)
题目中文翻译

四次方程 x4+bx3+cx2+dx−2=0x^4+bx^3+cx^2+dx-2=0 的根为 α\alpha、β\beta、γ\gamma、δ\delta。已知

α+β+γ+δ= 3,α2+β2+γ2+δ2= 5,α−1+β−1+γ−1+δ−1= 6.\begin{align*} \alpha+\beta+\gamma+\delta=&\,3,\\ \alpha^2+\beta^2+\gamma^2+\delta^2=&\,5,\\ \alpha^{-1}+\beta^{-1}+\gamma^{-1}+\delta^{-1}=&\,6. \end{align*}

(a) 求 bb、cc 和 dd 的值。

(6)

(b) 又已知 α3+β3+γ3+δ3=−27\alpha^3+\beta^3+\gamma^3+\delta^3=-27，求 α4+β4+γ4+δ4\alpha^4+\beta^4+\gamma^4+\delta^4 的值。

(2)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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一次和给b，平方和给二次对称和c。倒数和等于三根乘积之和除四根积，注意韦达定理中d前面的负号。

答题过程

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Vieta’s formula gives

b=−(α+β+γ+δ)=−3.b=-(\alpha+\beta+\gamma+\delta)=\boxed{-3}.

The pairwise product sum is cc, so

5=(α+β+γ+δ)2−2c=9−2c,5=(\alpha+\beta+\gamma+\delta)^2-2c=9-2c,

giving c=2\boxed{c=2}.

Also αβγδ=−2≠0\alpha\beta\gamma\delta=-2\ne0, so all reciprocal roots are defined. The sum of the four triple products is −d-d. Hence

6= αβγ+αβδ+αγδ+βγδαβγδ= −d−2.\begin{align*} 6=&\,\frac{\alpha\beta\gamma+\alpha\beta\delta+ \alpha\gamma\delta+\beta\gamma\delta}{\alpha\beta\gamma\delta}\\[4mm] =&\,\frac{-d}{-2}. \end{align*}

Thus d=12\boxed{d=12}.

(b)

解法一

思路

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把四根分别代入已求出的四次式并相加，常数−2要累计四次。

答题过程

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The quartic is x4−3x3+2x2+12x−2=0x^4-3x^3+2x^2+12x-2=0. Summing this equation over its roots gives

S4−3S3+2S2+12S1−8=0,S_4-3S_3+2S_2+12S_1-8=0,

where Sj=αj+βj+γj+δjS_j=\alpha^j+\beta^j+\gamma^j+\delta^j. Therefore

S4= 3(−27)−2(5)−12(3)+8= −81−10−36+8= −119.\begin{align*} S_4=&\,3(-27)-2(5)-12(3)+8\\[4mm] =&\,-81-10-36+8\\[4mm] =&\,\boxed{-119}. \end{align*}