CAIE FP1 2019 November — Question 3 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
Marks7
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Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSimple recurrence evaluation
DifficultyChallenging +1.8 This is a challenging Further Maths integration problem requiring integration by parts twice to establish a recurrence relation, then applying it iteratively. The algebraic manipulation is non-trivial and the multi-step nature (5+2 marks) with the need to work backwards from the recurrence makes it significantly harder than standard A-level questions, though the techniques themselves are within FP1 scope.
Spec4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n
The integral \(I_n\), where \(n\) is a positive integer, is defined by $$I_n = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{-n} \sin \pi x \, dx.$$
  1. Show that $$n(n+1)I_{n+2} = 2^{n+1} n + \pi - \pi^2 I_n.$$ [5]
  2. Find \(I_5\) in terms of \(\pi\) and \(I_1\). [2]
Question 3:
AnswerMarks
3(i) x−n−1  1 1 x−n−1
I n+2 =  sinπx −π∫ cosπx dx
−n−1 −n−1
 1
1
2
AnswerMarks Guidance
2M1 A1 Integrates by parts.
 
2n+1 π x−n  1 1 x−n 
= +  cosπx +π∫ sinπx dx
n+1 n+1 −n −n
 1 1 
 2 
AnswerMarks Guidance
2M1 Integrates by parts again.
2n+1
π 1 π 
= + − I
 n
n+1 n+1 n n 
⇒ ( n+1 ) I =2n+1+π  1 − π I 
AnswerMarks Guidance
n+2   n n n  M1 Uses I .
n
⇒n ( n+1 ) I =2n+1n+π−π2I
AnswerMarks Guidance
n+2 nA1 AG
5
AnswerMarks
3(ii)2I =4+π−π2I
3 1
π2( )
12I =48+π− 4+π−π2I
AnswerMarks Guidance
5 2 1M1 Substitutes I into reduction formula.
3
1 ( )
⇒I =4+ 2π−4π2 −π3 +π4I
AnswerMarks Guidance
5 24 1A1 AEF, must be exact with fractions simplified.
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(i) ---
3(i) |  x−n−1  1 1 x−n−1
I n+2 =  sinπx −π∫ cosπx dx
−n−1 −n−1
 1
1
2
2 | M1 A1 | Integrates by parts.
 
2n+1 π x−n  1 1 x−n 
= +  cosπx +π∫ sinπx dx
n+1 n+1 −n −n
 1 1 
 2 
2 | M1 | Integrates by parts again.
2n+1
π 1 π 
= + − I
 n
n+1 n+1 n n 
⇒ ( n+1 ) I =2n+1+π  1 − π I 
n+2   n n n   | M1 | Uses I .
n
⇒n ( n+1 ) I =2n+1n+π−π2I
n+2 n | A1 | AG
5
--- 3(ii) ---
3(ii) | 2I =4+π−π2I
3 1
π2( )
12I =48+π− 4+π−π2I
5 2 1 | M1 | Substitutes I into reduction formula.
3
1 ( )
⇒I =4+ 2π−4π2 −π3 +π4I
5 24 1 | A1 | AEF, must be exact with fractions simplified.
2
Question | Answer | Marks | Guidance
The integral $I_n$, where $n$ is a positive integer, is defined by
$$I_n = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{-n} \sin \pi x \, dx.$$

\begin{enumerate}[label=(\roman*)]
\item Show that
$$n(n+1)I_{n+2} = 2^{n+1} n + \pi - \pi^2 I_n.$$ [5]

\item Find $I_5$ in terms of $\pi$ and $I_1$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2019 Q3 [7]}}
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