OCR Further Additional Pure AS 2021 November — Question 8 7 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2021
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeNon-linear or complex iterative formula convergence
DifficultyChallenging +1.8 This is a Further Maths question requiring analysis of a non-standard recurrence relation's convergence behavior for all parameter values. Students must find fixed points, analyze stability conditions, and consider multiple cases (convergence, divergence, oscillation) - significantly beyond routine A-level work but structured enough to be accessible with systematic approach.
Spec8.01a Recurrence relations: general sequences, closed form and recurrence8.01c Sequence behaviour: periodic, convergent, divergent, oscillating, monotonic8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states

8 A sequence \(\left\{ \mathrm { u } _ { \mathrm { n } } \right\}\) is defined by the recurrence system \(u _ { 1 } = 1\) and \(\mathrm { u } _ { \mathrm { n } + 1 } = \mathrm { a } - \frac { \mathrm { a } ^ { 2 } } { 2 \mathrm { u } _ { \mathrm { n } } }\) for \(n \geqslant 1\), where \(a\) is a positive constant.
Determine with justification the behaviour of the sequence for all possible values of \(a\). \section*{END OF QUESTION PAPER}

Question 8:
AnswerMarks
8Sequence is periodic, with period 4
u =a− 1a2
2 2
1 a2 a−a2
u = a− 2 =
3 a− 1 a2 2−a
2
1a2(2−a) −a2
and u = a− 2 =
4 a−a2 2(1−a)
1 a2.2(1−a)
u = a− 2 = a+(1−a)=1
5 −a2
For a = 1 and a = 2 the sequence becomes undefined
Smaller period impossible (either or both of u , u )
2 3
since this requires a2 −2a+2=0, which has a
AnswerMarks
negative discriminantB1
B1
M1
A1
A1
B1
B1
AnswerMarks
[7]1.2
1.1
3.1a
1.1
1.1
2.3
AnswerMarks
2.4BC (e.g.)
Attempt at u , u and u using the given r.r.
3 4 5
u correct (allow unsimplified)
4
u correctly shown equal to 1
5
Stating clearly the “illegal” values of a
Justifying that all cases have been considered.
(NB Case for u is not required but may appear.)
4
PMT
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Question 8:
8 | Sequence is periodic, with period 4
u =a− 1a2
2 2
1 a2 a−a2
u = a− 2 =
3 a− 1 a2 2−a
2
1a2(2−a) −a2
and u = a− 2 =
4 a−a2 2(1−a)
1 a2.2(1−a)
u = a− 2 = a+(1−a)=1
5 −a2
For a = 1 and a = 2 the sequence becomes undefined
Smaller period impossible (either or both of u , u )
2 3
since this requires a2 −2a+2=0, which has a
negative discriminant | B1
B1
M1
A1
A1
B1
B1
[7] | 1.2
1.1
3.1a
1.1
1.1
2.3
2.4 | BC (e.g.)
Attempt at u , u and u using the given r.r.
3 4 5
u correct (allow unsimplified)
4
u correctly shown equal to 1
5
Stating clearly the “illegal” values of a
Justifying that all cases have been considered.
(NB Case for u is not required but may appear.)
4
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
8 A sequence $\left\{ \mathrm { u } _ { \mathrm { n } } \right\}$ is defined by the recurrence system\\
$u _ { 1 } = 1$ and $\mathrm { u } _ { \mathrm { n } + 1 } = \mathrm { a } - \frac { \mathrm { a } ^ { 2 } } { 2 \mathrm { u } _ { \mathrm { n } } }$ for $n \geqslant 1$, where $a$ is a positive constant.\\
Determine with justification the behaviour of the sequence for all possible values of $a$.

\section*{END OF QUESTION PAPER}

\hfill \mbox{\textit{OCR Further Additional Pure AS 2021 Q8 [7]}}