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CIE 9231 2024 November Paper 21 Q1

A Level / CIE / FP2

CIE 9231 2024 Nov Paper 21 Paper · Question 1

题目

Problem

Find the set of values of kk for which the system of equations

x+5y+6z= 1,kx+2y+2z= 2,−3x+4y+8z= 3\begin{align*} x + 5y + 6z =&\, 1,\\ kx + 2y + 2z =&\, 2,\\ -3x + 4y + 8z =&\, 3 \end{align*}

has a unique solution and interpret this situation geometrically.

[4]
题目中文翻译

求使下列方程组有唯一解的 kk 值集合,并从几何角度解释这一情形:

x+5y+6z= 1,kx+2y+2z= 2,−3x+4y+8z= 3\begin{align*} x + 5y + 6z =&\, 1,\\ kx + 2y + 2z =&\, 2,\\ -3x + 4y + 8z =&\, 3 \end{align*}
(4)
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

解法一

思路

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求系数矩阵的行列式,并排除使它为零的参数。方程组的唯一解就是三个平面的唯一公共点。

答题过程

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The coefficient determinant is

 det⁡(156k22−348)= 8−5(8k+6)+6(4k+6)= 14−16k.\begin{align*} &\,\det\begin{pmatrix}1&5&6\\k&2&2\\-3&4&8\end{pmatrix}\\[4mm] =&\,8-5(8k+6)+6(4k+6)\\[4mm] =&\,14-16k. \end{align*}

There is a unique solution when k∈R∖{78}\boxed{k\in\mathbb R\setminus\{\frac78\}}. Geometrically, the three planes meet at exactly one point.