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CIE 9231 2024 June Paper 13 Q2

A Level / CIE / FP1

CIE 9231 2024 June Paper 13 Paper · Question 2

题目

Problem

The cubic equation x3+2x2+3x+1=0x^3+2x^2+3x+1=0 has roots α\alpha, β\beta, γ\gamma.

(a) Find a cubic equation whose roots are α2+1\alpha^2+1, β2+1\beta^2+1, γ2+1\gamma^2+1.

(3)

(b) Find the value of (α2+1)2+(β2+1)2+(γ2+1)2(\alpha^2+1)^2+(\beta^2+1)^2+(\gamma^2+1)^2.

(2)

(c) Find the value of (α2+1)3+(β2+1)3+(γ2+1)3(\alpha^2+1)^3+(\beta^2+1)^3+(\gamma^2+1)^3.

(2)
题目中文翻译

三次方程 x3+2x2+3x+1=0x^3+2x^2+3x+1=0 的根为 α\alpha、β\beta、γ\gamma。

(a) 求一个以 α2+1\alpha^2+1、β2+1\beta^2+1、γ2+1\gamma^2+1 为根的三次方程。

(3)

(b) 求 (α2+1)2+(β2+1)2+(γ2+1)2(\alpha^2+1)^2+(\beta^2+1)^2+(\gamma^2+1)^2 的值。

(2)

(c) 求 (α2+1)3+(β2+1)3+(γ2+1)3(\alpha^2+1)^3+(\beta^2+1)^3+(\gamma^2+1)^3 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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令y=x平方加1,原方程中奇次项可提取x。把含x的两边平方后用x平方等于y−1消去x,避免对复数根随意选择平方根符号。

答题过程

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For a root xx of the original cubic, put y=x2+1y=x^2+1, so x2=y−1x^2=y-1. The cubic gives

x(x2+3)= −2x2−1,x(y+2)= 1−2y.\begin{align*} x(x^2+3)=&\,-2x^2-1,\\[4mm] x(y+2)=&\,1-2y. \end{align*}

Squaring and substituting x2=y−1x^2=y-1,

(y−1)(y+2)2= (1−2y)2,y3+3y2−4= 1−4y+4y2.\begin{align*} (y-1)(y+2)^2=&\,(1-2y)^2,\\[4mm] y^3+3y^2-4=&\,1-4y+4y^2. \end{align*}

Hence a required cubic is

y3−y2+4y−5=0.\boxed{y^3-y^2+4y-5=0}.

Its roots are α2+1\alpha^2+1, β2+1\beta^2+1 and γ2+1\gamma^2+1.

(b)

解法一

思路

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把三个新根作为整体,直接用新方程系数给出的一次和与二次对称和。

答题过程

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Let the new roots be u,v,wu,v,w. From (a),

u+v+w=1,uv+vw+wu=4.u+v+w=1,\qquad uv+vw+wu=4.

Thus the required sum is

u2+v2+w2= (u+v+w)2 −2(uv+vw+wu)= 1−8=−7.\begin{align*} u^2+v^2+w^2=&\,(u+v+w)^2\\[4mm] &\,-2(uv+vw+wu)\\[4mm] =&\,1-8=\boxed{-7}. \end{align*}

(c)

解法一

思路

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每个新根满足新三次方程,分别代入并相加,把立方和降为平方和与一次和。

答题过程

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Each of u,v,wu,v,w satisfies y3=y2−4y+5y^3=y^2-4y+5. Summing and using (b),

u3+v3+w3= (u2+v2+w2)−4(u+v+w)+15= −7−4+15= 4.\begin{align*} u^3+v^3+w^3 =&\,(u^2+v^2+w^2)-4(u+v+w)+15\\[4mm] =&\,-7-4+15\\[4mm] =&\,\boxed{4}. \end{align*}

解法二

思路

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也可用三变量立方和恒等式,直接读取新根的三个基本对称量。

答题过程

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From (a), u+v+w=1u+v+w=1, uv+vw+wu=4uv+vw+wu=4 and uvw=5uvw=5. The identity for the sum of cubes gives

u3+v3+w3= (u+v+w)3 −3(u+v+w)(uv+vw+wu)+3uvw= 13−3(1)(4)+3(5)= 4.\begin{align*} u^3+v^3+w^3 =&\,(u+v+w)^3\\[4mm] &\,\hspace{2pt}-3(u+v+w)(uv+vw+wu)+3uvw\\[4mm] =&\,1^3-3(1)(4)+3(5)\\[4mm] =&\,\boxed{4}. \end{align*}