OCR Further Additional Pure 2019 June — Question 4 10 marks

Exam BoardOCR
ModuleFurther Additional Pure (Further Additional Pure)
Year2019
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and Series
TypeSecond-Order Homogeneous Recurrence Relations
DifficultyChallenging +1.2 This is a non-homogeneous second-order recurrence relation from Further Maths, requiring finding the auxiliary equation, complementary function, and particular integral. While the topic is advanced, the execution is relatively straightforward: the auxiliary equation m² + 2 = 0 gives complex roots, the particular integral is a constant easily found, and applying initial conditions is routine. The calculation for T₂₀ is mechanical once the general solution is established.
Spec8.01g Second-order recurrence: solve with distinct, repeated, or complex roots
4
  1. Solve the second-order recurrence relation \(T _ { n + 2 } + 2 T _ { n } = - 87\) given that \(T _ { 0 } = - 27\) and \(T _ { 1 } = 27\).
  2. Determine the value of \(T _ { 20 }\).
Question 4:
AnswerMarks Guidance
4(a) PS is T =−29
n
( ) ( )
n n
CS from λ 2 + 2 = 0, λ = ± i 2 , is T = Ai 2 + B −i 2
n
( ) ( )
n n
General Solution is T = Ai 2 + B −i 2 −29
n
Use of T = −27 and T = 27 to create simultaneous equations in A, B
0 1
Two correct equations: A + B = 2 and A – B = −28i 2
Solving attempt at two equations: A = 1−14 i 2 and B = 1+14 i 2
( )( )n ( )( )n
i.e. T = 1−14 i 2 i 2 + 1+14 i 2 −i 2 −29
AnswerMarks
n1.1
1.1a
1.1
1.2
1.1a
1.1
1.1a
AnswerMarks
1.1B1
M1
A1
B1
M1
A1
M1
AnswerMarks
A1Particular Solution
Complementary Solution attempted from auxiliary equation;
correct
FT GS = CS (with 2 arbitrary constants) + PS (with none)
Or unsimplified equivalents
cao
AnswerMarks Guidance
Alt. versionB1 Particular Solution
PS is T = −29
n
2)n( )
CS from λ 2 + 2 = 0, λ = ± i 2 , is T = ( Ccosnπ +Dsinnπ
AnswerMarks Guidance
n 2 2M1 Complementary Solution attempted from auxiliary equation;
A1correct
2)n( )
General Solution is ( Ccosnπ +Dsinnπ – 29
AnswerMarks Guidance
2 2B1 FT GS = CS (with 2 arbitrary constants) + PS (with none)
Use of T = −27 and T = 27 to create simultaneous equations in C, D
AnswerMarks Guidance
0 1M1
Two correct equations: C – 29 = –27 and2 D – 29 = 27 A1
Solving attempt at two equations: C = 2 and D = 282 A1
( )
i.e. T = ( 2)n 2cosnπ +28 2sinnπ – 29
AnswerMarks Guidance
n 2 2A1 cao
[8]
AnswerMarks
(b)Substituting n = 20 ⇒
( ) ( )
T = 1−14 i 2 ×1024 + 1+14 i 2 × 1024 −29
20
AnswerMarks
= 2048 – 29 = 20192.1
1.1M1
A1
AnswerMarks
[2]Or by applying the given recurrence relation (e.g. BC)
B1
Particular Solution
Question 4:
4 | (a) | PS is T =−29
n
( ) ( )
n n
CS from λ 2 + 2 = 0, λ = ± i 2 , is T = Ai 2 + B −i 2
n
( ) ( )
n n
General Solution is T = Ai 2 + B −i 2 −29
n
Use of T = −27 and T = 27 to create simultaneous equations in A, B
0 1
Two correct equations: A + B = 2 and A – B = −28i 2
Solving attempt at two equations: A = 1−14 i 2 and B = 1+14 i 2
( )( )n ( )( )n
i.e. T = 1−14 i 2 i 2 + 1+14 i 2 −i 2 −29
n | 1.1
1.1a
1.1
1.2
1.1a
1.1
1.1a
1.1 | B1
M1
A1
B1
M1
A1
M1
A1 | Particular Solution
Complementary Solution attempted from auxiliary equation;
correct
FT GS = CS (with 2 arbitrary constants) + PS (with none)
Or unsimplified equivalents
cao
Alt. version | B1 | Particular Solution
PS is T = −29
n
2)n( )
CS from λ 2 + 2 = 0, λ = ± i 2 , is T = ( Ccosnπ +Dsinnπ
n 2 2 | M1 | Complementary Solution attempted from auxiliary equation;
A1 | correct
2)n( )
General Solution is ( Ccosnπ +Dsinnπ – 29
2 2 | B1 | FT GS = CS (with 2 arbitrary constants) + PS (with none)
Use of T = −27 and T = 27 to create simultaneous equations in C, D
0 1 | M1
Two correct equations: C – 29 = –27 and | 2 D – 29 = 27 | A1 | Unsimplified
Solving attempt at two equations: C = 2 and D = 28 | 2 | A1
( )
i.e. T = ( 2)n 2cosnπ +28 2sinnπ – 29
n 2 2 | A1 | cao
[8]
(b) | Substituting n = 20 ⇒
( ) ( )
T = 1−14 i 2 ×1024 + 1+14 i 2 × 1024 −29
20
= 2048 – 29 = 2019 | 2.1
1.1 | M1
A1
[2] | Or by applying the given recurrence relation (e.g. BC)
B1
Particular Solution
4
\begin{enumerate}[label=(\alph*)]
\item Solve the second-order recurrence relation $T _ { n + 2 } + 2 T _ { n } = - 87$ given that $T _ { 0 } = - 27$ and $T _ { 1 } = 27$.
\item Determine the value of $T _ { 20 }$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure 2019 Q4 [10]}}
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