OCR MEI Further Pure Core AS 2018 June — Question 6 4 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2018
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind line of invariant points
DifficultyStandard +0.3 This is a straightforward Further Maths question requiring students to find an invariant line by solving (A - λI)v = 0 for eigenvalues and eigenvectors, or by using the direct method y = mx where the line maps to itself. While it's a Further Maths topic, the matrix is simple (triangular with obvious eigenvalues 2 and -1), making it easier than average even for Further Pure content.
Spec4.03g Invariant points and lines

Find the invariant line of the transformation of the \(x\)-\(y\) plane represented by the matrix \(\begin{pmatrix} 2 & 0 \\ 4 & -1 \end{pmatrix}\). [4]

Question 6:
AnswerMarks
62 0x x'
 
4 1 y y'
 x’ = 2x, y’ = 4x  y
y = mx + c and y’ = mx’ + c
 4x  mx  c = m.2x + c
 m 4 , c = 0
3
so invariant line is y 4x
AnswerMarks
3M1
M1
A1
A1
AnswerMarks
[4]2.1
1.1b
2.4
AnswerMarks
1.1bsoi [condone x = 2x, y=4x  y]
condition for invariance
forming identity in x
AnswerMarks
(may assume c = 0)assuming c = 0:
y = mx and y’ = mx’
4x  mx = m.2x
m4
3
Question 6:
6 | 2 0x x'
 
4 1 y y'
 x’ = 2x, y’ = 4x  y
y = mx + c and y’ = mx’ + c
 4x  mx  c = m.2x + c
 m 4 , c = 0
3
so invariant line is y 4x
3 | M1
M1
A1
A1
[4] | 2.1
1.1b
2.4
1.1b | soi [condone x = 2x, y=4x  y]
condition for invariance
forming identity in x
(may assume c = 0) | assuming c = 0:
y = mx and y’ = mx’
4x  mx = m.2x
m4
3
Find the invariant line of the transformation of the $x$-$y$ plane represented by the matrix $\begin{pmatrix} 2 & 0 \\ 4 & -1 \end{pmatrix}$. [4]

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2018 Q6 [4]}}