CAIE Further Paper 2 2020 November — Question 3 9 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2020
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimultaneous equations
TypeSystem of three linear equations
DifficultyStandard +0.8 This is a structured Further Maths question requiring understanding of consistency/uniqueness conditions for linear systems and geometric interpretation of three planes. While systematic, it demands matrix reasoning (recognizing dependent equations), case analysis across different k values, and translating algebraic conditions into geometric configurations (line of intersection vs parallel planes). More conceptually demanding than routine simultaneous equations but follows a guided structure.
Spec4.03s Consistent/inconsistent: systems of equations4.03t Plane intersection: geometric interpretation

3
  1. Show that the system of equations $$\begin{array} { r } x - 2 y - 4 z = 1 \\ x - 2 y + k z = 1 \\ - x + 2 y + 2 z = 1 \end{array}$$ where \(k\) is a constant, does not have a unique solution.
  2. Given that \(k = - 4\), show that the system of equations in part (a) is consistent. Interpret this situation geometrically.
  3. Given instead that \(k = - 2\), show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically.
  4. For the case where \(k \neq - 2\) and \(k \neq - 4\), show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. \includegraphics[max width=\textwidth, alt={}, center]{7da7fa35-1b97-4708-a1a2-cba9e35c8bf0-06_894_841_260_612} The diagram shows the curve with equation \(\mathrm { y } = 1 - \mathrm { x } ^ { 3 }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(n\) rectangles of width \(\frac { 1 } { n }\).

Question 3(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(\begin{vmatrix} 1 & -2 & -4 \\ 1 & -2 & k \\ -1 & 2 & 2 \end{vmatrix} = -4 - 2k + 2(2+k) - 4(0) = 0\)M1 A1 Shows that determinant is zero
Total: 2
Question 3(b):
AnswerMarks Guidance
AnswerMarks Guidance
\(x - 2y - 4z = 1,\quad x - 2y - 4z = 1 \Rightarrow z = -1,\ x - 2y = -3\)B1 Derives one equation with two unknowns or states that third plane is not parallel to the repeated one
\(-x + 2y + 2z = 1\) (or \(-x+2y+2z=1\) is not parallel to \(x-2y-4z=1\))
Two of the planes are identical (or coincident).B1
There is a line of intersection with the other plane.B1
Total: 3
Question 3(c):
AnswerMarks Guidance
AnswerMarks Guidance
\(x - 2y - 4z = 1,\quad x - 2y - 2z = 1 \Rightarrow -1 = 1\)B1 Derives contradiction
Two parallel planes, not identical.B1
Total: 2
Question 3(d):
AnswerMarks Guidance
AnswerMarks Guidance
\(x - 2y - 4z = 1,\quad x - 2y + kz = 1 \Rightarrow (-4-k)z = 0,\ -2z = 2 \Rightarrow 0 = 2\)B1 Derives contradiction
The three planes form a triangular prism.B1
Total: 2
## Question 3(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} 1 & -2 & -4 \\ 1 & -2 & k \\ -1 & 2 & 2 \end{vmatrix} = -4 - 2k + 2(2+k) - 4(0) = 0$ | M1 A1 | Shows that determinant is zero |
| **Total: 2** | | |

## Question 3(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x - 2y - 4z = 1,\quad x - 2y - 4z = 1 \Rightarrow z = -1,\ x - 2y = -3$ | B1 | Derives one equation with two unknowns or states that third plane is not parallel to the repeated one |
| $-x + 2y + 2z = 1$ (or $-x+2y+2z=1$ is not parallel to $x-2y-4z=1$) | | |
| Two of the planes are identical (or coincident). | B1 | |
| There is a line of intersection with the other plane. | B1 | |
| **Total: 3** | | |

## Question 3(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x - 2y - 4z = 1,\quad x - 2y - 2z = 1 \Rightarrow -1 = 1$ | B1 | Derives contradiction |
| Two parallel planes, not identical. | B1 | |
| **Total: 2** | | |

## Question 3(d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x - 2y - 4z = 1,\quad x - 2y + kz = 1 \Rightarrow (-4-k)z = 0,\ -2z = 2 \Rightarrow 0 = 2$ | B1 | Derives contradiction |
| The three planes form a triangular prism. | B1 | |
| **Total: 2** | | |
3
\begin{enumerate}[label=(\alph*)]
\item Show that the system of equations

$$\begin{array} { r } 
x - 2 y - 4 z = 1 \\
x - 2 y + k z = 1 \\
- x + 2 y + 2 z = 1
\end{array}$$

where $k$ is a constant, does not have a unique solution.
\item Given that $k = - 4$, show that the system of equations in part (a) is consistent. Interpret this situation geometrically.
\item Given instead that $k = - 2$, show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically.
\item For the case where $k \neq - 2$ and $k \neq - 4$, show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically.\\

\includegraphics[max width=\textwidth, alt={}, center]{7da7fa35-1b97-4708-a1a2-cba9e35c8bf0-06_894_841_260_612}

The diagram shows the curve with equation $\mathrm { y } = 1 - \mathrm { x } ^ { 3 }$ for $0 \leqslant x \leqslant 1$, together with a set of $n$ rectangles of width $\frac { 1 } { n }$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q3 [9]}}