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CIE 9231 2022 November Paper 12 Q2

A Level / CIE / FP1

CIE 9231 2022 Nov Paper 12 Paper · Question 2

题目

Problem

The equation x4+3x2+2x+6=0x^4+3x^2+2x+6=0 has roots α\alpha, β\beta, γ\gamma, δ\delta.

(a) Find a quartic equation whose roots are 1α2\dfrac1{\alpha^2}, 1β2\dfrac1{\beta^2}, 1γ2\dfrac1{\gamma^2}, 1δ2\dfrac1{\delta^2} and state the value of 1α2+1β2+1γ2+1δ2\dfrac1{\alpha^2}+\dfrac1{\beta^2}+\dfrac1{\gamma^2}+\dfrac1{\delta^2}.

(4)

(b) Find the value of β2γ2δ2+α2γ2δ2+α2β2δ2+α2β2γ2\beta^2\gamma^2\delta^2+\alpha^2\gamma^2\delta^2+\alpha^2\beta^2\delta^2+\alpha^2\beta^2\gamma^2.

(3)

(c) Find the value of 1α4+1β4+1γ4+1δ4\dfrac1{\alpha^4}+\dfrac1{\beta^4}+\dfrac1{\gamma^4}+\dfrac1{\delta^4}.

(2)
题目中文翻译

方程 x4+3x2+2x+6=0x^4+3x^2+2x+6=0 的根为 α\alpha、β\beta、γ\gamma、δ\delta。

(a) 求一个以 1α2\dfrac1{\alpha^2}、1β2\dfrac1{\beta^2}、1γ2\dfrac1{\gamma^2}、1δ2\dfrac1{\delta^2} 为根的四次方程，并写出 1α2+1β2+1γ2+1δ2\dfrac1{\alpha^2}+\dfrac1{\beta^2}+\dfrac1{\gamma^2}+\dfrac1{\delta^2} 的值。

(4)

(b) 求 β2γ2δ2+α2γ2δ2+α2β2δ2+α2β2γ2\beta^2\gamma^2\delta^2+\alpha^2\gamma^2\delta^2+\alpha^2\beta^2\delta^2+\alpha^2\beta^2\gamma^2 的值。

(3)

(c) 求 1α4+1β4+1γ4+1δ4\dfrac1{\alpha^4}+\dfrac1{\beta^4}+\dfrac1{\gamma^4}+\dfrac1{\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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原根均非零，可令y为x的负二次幂。先把原式中唯一的奇次项隔离并乘适当的y幂，再平方消去x。

答题过程

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The original quartic has nonzero constant term, so its roots are nonzero. Put y=x−2y=x^{-2}, giving x2=1/yx^2=1/y. Multiplying the original equation by y2y^2 gives

1+3y+2xy2+6y2=0.1+3y+2xy^2+6y^2=0.

Hence 2xy2=−(6y2+3y+1)2xy^2=-(6y^2+3y+1). Squaring and using x2=1/yx^2=1/y,

4y3= (6y2+3y+1)2= 36y4+36y3+21y2+6y+1.\begin{align*} 4y^3=&\,(6y^2+3y+1)^2\\[4mm] =&\,36y^4+36y^3+21y^2+6y+1. \end{align*}

Thus a required quartic is

36y4+32y3+21y2+6y+1=0.\boxed{36y^4+32y^3+21y^2+6y+1=0}.

Its roots are 1/α2,1/β2,1/γ2,1/δ21/\alpha^2,1/\beta^2,1/\gamma^2,1/\delta^2, so their sum is

 1α2+1β2+1γ2+1δ2= −3236= −89.\begin{align*} &\,\frac1{\alpha^2}+\frac1{\beta^2} +\frac1{\gamma^2}+\frac1{\delta^2}\\[4mm] =&\,-\frac{32}{36}\\[4mm] =&\,\boxed{-\frac89}. \end{align*}

(b)

解法一

思路

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倒数平方和的分母是四根积的平方，乘回去便是所求的四个三根平方乘积之和。

答题过程

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From the original quartic, αβγδ=6\alpha\beta\gamma\delta=6. Therefore

 β2γ2δ2+α2γ2δ2 +α2β2δ2+α2β2γ2= (αβγδ)2 ×(1α2+1β2+1γ2+1δ2)= 36(−89)= −32.\begin{align*} &\,\beta^2\gamma^2\delta^2+\alpha^2\gamma^2\delta^2\\[4mm] &\,\hspace{2pt}+\alpha^2\beta^2\delta^2+\alpha^2\beta^2\gamma^2\\[4mm] =&\,(\alpha\beta\gamma\delta)^2\\[4mm] &\,\times\bigg(\frac1{\alpha^2}+\frac1{\beta^2} +\frac1{\gamma^2}+\frac1{\delta^2}\bigg)\\[4mm] =&\,36\big(-\tfrac89\big)\\[4mm] =&\,\boxed{-32}. \end{align*}

(c)

解法一

思路

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这是新根的平方和，用新根的一次和及二次对称和即可。

答题过程

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From the quartic in (a), the sum of the new roots is −8/9-8/9 and their pairwise product sum is 21/36=7/1221/36=7/12. Hence

 1α4+1β4+1γ4+1δ4= (−89)2−2(712)= 6481−76= −61162.\begin{align*} &\,\frac1{\alpha^4}+\frac1{\beta^4} +\frac1{\gamma^4}+\frac1{\delta^4}\\[4mm] =&\,\big(-\tfrac89\big)^2-2\big(\tfrac7{12}\big)\\[4mm] =&\,\frac{64}{81}-\frac76\\[4mm] =&\,\boxed{-\frac{61}{162}}. \end{align*}