OCR Further Pure Core AS 2020 November — Question 4 6 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2020
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMatrices
TypeSolving linear systems using matrices
DifficultyStandard +0.3 This is a straightforward Further Maths matrix question requiring Cramer's rule or matrix inversion to solve a 2×2 system, followed by checking the determinant is non-zero. While it's Further Maths content (making it slightly above average), the technique is routine and the determinant a²b² + 2 is clearly always positive, requiring minimal insight.
Spec4.03r Solve simultaneous equations: using inverse matrix
You are given the system of equations $$a^2x - 2y = 1$$ $$x + b^2y = 3$$ where \(a\) and \(b\) are real numbers.
  1. Use a matrix method to find \(x\) and \(y\) in terms of \(a\) and \(b\). [4]
  2. Explain why the method used in part (a) works for all values of \(a\) and \(b\). [2]
Question 4:
AnswerMarks Guidance
4(a) a2 −2x
  
 1 b2 y
−1
a2 −2 1 b2 2 
  =  
 1 b2  a2b2 +2−1 a2 
x 1 b2 2 1
 =   
y a2b2 +2−1 a2 3
b2 +6 3a2 −1
x= , y=
AnswerMarks
a2b2 +2 a2b2 +2B1
M1
M1
A1
AnswerMarks
[4]3.1a
1.1
1.1
AnswerMarks
1.1Rewriting LHS in matrix form
correct process for finding the inverse
Multiplication by their inverse
Need to see x = …, y = …
AnswerMarks
(b)Since a2b2 = (ab)2 ≥ 0 then a2b2 + 2 > 0 for all values
of a and b the determinant of the matrix cannot be 0
(so the matrix is never singular)
so the inverse always exists and the method always
AnswerMarks
works.B1
B1
AnswerMarks
[2]2.4
2.4Argument must be complete and
correct. eg a2b2 + 2 ≥ 0 is B0.
4
or eg 2(d2 + 9) + (2(d – 3)2 + 3)i
AnswerMarks
(b)When d = 3, the PoI would be 36 + 3i and
 1 
C =z:arg(z−(36+3i))= π
2  4 
But 36 + 3i is not in C since arg0 is not defined
AnswerMarks
2M1
A13.1a
3.2bBoth
AG. Or arg0 is not π/4NB C ∩C =∅ without
1 2
justification is M0A0.
AnswerMarks
Alternative MethodM1
A1Set up inequality using x coordinates of
vertical line and starting point of half
line
Or “d = 3” is not valid etc.
For the intersection to exist we need
2d2 +18>12d
d2 −6d +9>0
(d −3)2 >0
[ d ≠3 ]
Hence d cannot be equal to 3
[2]
M1
A1
Set up inequality using x coordinates of
vertical line and starting point of half
line
Or “d = 3” is not valid etc.
PMT
OCR (Oxford Cambridge and RSA Examinations)
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CB2 8EA
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Question 4:
4 | (a) | a2 −2x
  
 1 b2 y
−1
a2 −2 1 b2 2 
  =  
 1 b2  a2b2 +2−1 a2 
x 1 b2 2 1
 =   
y a2b2 +2−1 a2 3
b2 +6 3a2 −1
x= , y=
a2b2 +2 a2b2 +2 | B1
M1
M1
A1
[4] | 3.1a
1.1
1.1
1.1 | Rewriting LHS in matrix form
correct process for finding the inverse
Multiplication by their inverse
Need to see x = …, y = …
(b) | Since a2b2 = (ab)2 ≥ 0 then a2b2 + 2 > 0 for all values
of a and b the determinant of the matrix cannot be 0
(so the matrix is never singular)
so the inverse always exists and the method always
works. | B1
B1
[2] | 2.4
2.4 | Argument must be complete and
correct. eg a2b2 + 2 ≥ 0 is B0.
4
or eg 2(d2 + 9) + (2(d – 3)2 + 3)i
(b) | When d = 3, the PoI would be 36 + 3i and
 1 
C =z:arg(z−(36+3i))= π
2  4 
But 36 + 3i is not in C since arg0 is not defined
2 | M1
A1 | 3.1a
3.2b | Both
AG. Or arg0 is not π/4 | NB C ∩C =∅ without
1 2
justification is M0A0.
Alternative Method | M1
A1 | Set up inequality using x coordinates of
vertical line and starting point of half
line
Or “d = 3” is not valid etc.
For the intersection to exist we need
2d2 +18>12d
d2 −6d +9>0
(d −3)2 >0
[ d ≠3 ]
Hence d cannot be equal to 3
[2]
M1
A1
Set up inequality using x coordinates of
vertical line and starting point of half
line
Or “d = 3” is not valid etc.
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
You are given the system of equations
$$a^2x - 2y = 1$$
$$x + b^2y = 3$$
where $a$ and $b$ are real numbers.

\begin{enumerate}[label=(\alph*)]
\item Use a matrix method to find $x$ and $y$ in terms of $a$ and $b$. [4]

\item Explain why the method used in part (a) works for all values of $a$ and $b$. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core AS 2020 Q4 [6]}}
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