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CIE 9231 2023 November Paper 12 Q4

A Level / CIE / FP1

CIE 9231 2023 Nov Paper 12 Paper · Question 4

题目

Problem

The cubic equation 27x3+18x2+6x−1=027x^3+18x^2+6x-1=0 has roots α\alpha, β\beta, γ\gamma.

(a) Show that a cubic equation with roots 3α+13\alpha+1, 3β+13\beta+1, 3γ+13\gamma+1 is

y3−y2+y−2=0.y^3-y^2+y-2=0.
(3)

The sum (3α+1)n+(3β+1)n+(3γ+1)n(3\alpha+1)^n+(3\beta+1)^n+(3\gamma+1)^n is denoted by SnS_n.

(b) Find the values of S2S_2 and S3S_3.

(4)

(c) Find the values of S−1S_{-1} and S−2S_{-2}.

(3)
题目中文翻译

三次方程 27x3+18x2+6x−1=027x^3+18x^2+6x-1=0 的根为 α\alpha、β\beta、γ\gamma。

(a) 证明，以 3α+13\alpha+1、3β+13\beta+1、3γ+13\gamma+1 为根的一个三次方程是

y3−y2+y−2=0.y^3-y^2+y-2=0.
(3)

和式 (3α+1)n+(3β+1)n+(3γ+1)n(3\alpha+1)^n+(3\beta+1)^n+(3\gamma+1)^n 记为 SnS_n。

(b) 求 S2S_2 和 S3S_3 的值。

(4)

(c) 求 S−1S_{-1} 和 S−2S_{-2} 的值。

(3)
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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新根与原根是一次关系，直接反解x再代回三次方程，完整展开得到目标式。

答题过程

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Put y=3x+1y=3x+1, so x=(y−1)/3x=(y-1)/3. Substituting into the original cubic gives

0= 27(y−13)3 +18(y−13)2 +6(y−13)−1= (y−1)3+2(y−1)2+2(y−1)−1= y3−3y2+3y−1 +2y2−4y+2+2y−2−1= y3−y2+y−2.\begin{align*} 0=&\,27\bigg(\frac{y-1}{3}\bigg)^3\\[4mm] &\,\hspace{2pt}+18\bigg(\frac{y-1}{3}\bigg)^2\\[4mm] &\,\hspace{4pt}+6\bigg(\frac{y-1}{3}\bigg)-1\\[4mm] =&\,(y-1)^3+2(y-1)^2+2(y-1)-1\\[4mm] =&\,y^3-3y^2+3y-1\\[4mm] &\,+2y^2-4y+2+2y-2-1\\[4mm] =&\,y^3-y^2+y-2. \end{align*}

Therefore the required cubic is y3−y2+y−2=0\boxed{y^3-y^2+y-2=0}.

(b)

解法一

思路

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对三个新根用韦达定理。平方和用恒等式，立方和通过新方程降幂。

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=1,uvw=2.u+v+w=1,\qquad uv+vw+wu=1,\qquad uvw=2.

Hence

S2= (u+v+w)2−2(uv+vw+wu)= 1−2=−1.\begin{align*} S_2=&\,(u+v+w)^2-2(uv+vw+wu)\\[4mm] =&\,1-2=\boxed{-1}. \end{align*}

Each new root satisfies y3=y2−y+2y^3=y^2-y+2, so summing gives

S3=S2−S1+6=−1−1+6=4.\boxed{S_3=S_2-S_1+6=-1-1+6=4}.

(c)

解法一

思路

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新根积非零，可用倒数和公式。将新三次式除以y平方后求和，直接得倒数平方和。

答题过程

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Since uvw=2≠0uvw=2\ne0, the reciprocal roots are defined and

S−1=uv+vw+wuuvw=12.\boxed{S_{-1}=\frac{uv+vw+wu}{uvw}=\frac12}.

Dividing y3−y2+y−2=0y^3-y^2+y-2=0 by y2y^2 gives

2y−2=y−1+y−1.2y^{-2}=y-1+y^{-1}.

Summing over the three roots,

2S−2=S1−3+S−1=1−3+12=−32,2S_{-2}=S_1-3+S_{-1}=1-3+\frac12=-\frac32,

so S−2=−34\boxed{S_{-2}=-\frac34}.

解法二

思路

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也可把三个倒数看成新的一组数，分别求其一次和与二次对称和，再用平方和恒等式。

答题过程

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Using u+v+w=1u+v+w=1, uv+vw+wu=1uv+vw+wu=1 and uvw=2uvw=2,

S−1= 1u+1v+1w= uv+vw+wuuvw=12.\begin{align*} S_{-1}=&\,\frac1u+\frac1v+\frac1w\\[4mm] =&\,\frac{uv+vw+wu}{uvw}=\boxed{\frac12}. \end{align*}

The pairwise product sum of the reciprocals is

1uv+1vw+1wu=u+v+wuvw=12.\frac1{uv}+\frac1{vw}+\frac1{wu} =\frac{u+v+w}{uvw}=\frac12.

Therefore

S−2= S−12 −2(1uv+1vw+1wu)= 14−1=−34.\begin{align*} S_{-2}=&\,S_{-1}^2\\[4mm] &\,-2\bigg(\frac1{uv}+\frac1{vw}+\frac1{wu}\bigg)\\[4mm] =&\,\frac14-1=\boxed{-\frac34}. \end{align*}