OCR Further Additional Pure 2020 November — Question 3 9 marks

Exam BoardOCR
ModuleFurther Additional Pure (Further Additional Pure)
Year2020
SessionNovember
Marks9
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Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeMethod of differences
DifficultyChallenging +1.8 This is a Further Maths reduction formula question requiring (a) algebraic manipulation to establish the recurrence relation, (b)(i) telescoping application of the formula with careful arithmetic, and (b)(ii) a non-trivial insight connecting the integral to π via arctan(1). The method of differences and the π inequality deduction elevate this beyond routine A-level work, though the techniques are standard for Further Maths students.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)8.06a Reduction formulae: establish, use, and evaluate recursively
3 For integers \(n \geqslant 0 , \mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { 1 } \frac { \mathrm { x } ^ { \mathrm { n } } } { 1 + \mathrm { x } ^ { 2 } } \mathrm { dx }\).
  1. For integers \(n \geqslant 2\), show that \(I _ { n } + I _ { n - 2 } = \frac { 1 } { n - 1 }\).
    1. Determine the exact value of \(I _ { 10 }\).
    2. Deduce that \(\pi < 3 \frac { 107 } { 315 }\).
Question 3:
AnswerMarks Guidance
3(a) ( )
1 xn − 2 1+x2 1 xn − 11 1
I n + I n – 2 = ∫ dx = ∫xn − 2 dx =   =
1+x2 n−1 n−1
AnswerMarks
0 0 0M1
M1
A1
AnswerMarks
[3]1.1 1.1
1.1Combining as a single integral (added numerators)
Factorising and cancelling
AG legitimately obtained
AnswerMarks Guidance
(b)(i) I = 1 π
0
4
I = 1 – 1 π , I = 1 – I = 1 π – 2 , I = 1 – I = 13– 1 π
2 4 2 6 4
4 3 4 3 5 15 4
I = 1 – I = 1 π – 76 , I = 1 – I = …
8 6 10 8
7 4 105 9
or I = 1−1+ 1 − 1 + 1– 1 π =263– 1 π
10
AnswerMarks
3 5 7 9 4 315 4B1
M1
M1
A1
AnswerMarks
[4]1.1
1.1
1.1
AnswerMarks
1.1BC
Use of the Reduction Formula repeatedly (≥ ×3)
All the way to I
10
Correct final answer
AnswerMarks
(ii)I > 0 ⇒ 1 π < 263
10
4 315
⇒ π < 263×4 = 1052 = 3107
AnswerMarks
315 315 315M1
A1
AnswerMarks
[2]3.1a
2.2aClear statement and consequence noted
AG from clear demonstration
Question 3:
3 | (a) | ( )
1 xn − 2 1+x2 1 xn − 11 1
I n + I n – 2 = ∫ dx = ∫xn − 2 dx =   =
1+x2 n−1 n−1
0 0 0 | M1
M1
A1
[3] | 1.1 1.1
1.1 | Combining as a single integral (added numerators)
Factorising and cancelling
AG legitimately obtained
(b) | (i) | I = 1 π
0
4
I = 1 – 1 π , I = 1 – I = 1 π – 2 , I = 1 – I = 13– 1 π
2 4 2 6 4
4 3 4 3 5 15 4
I = 1 – I = 1 π – 76 , I = 1 – I = …
8 6 10 8
7 4 105 9
or I = 1−1+ 1 − 1 + 1– 1 π =263– 1 π
10
3 5 7 9 4 315 4 | B1
M1
M1
A1
[4] | 1.1
1.1
1.1
1.1 | BC
Use of the Reduction Formula repeatedly (≥ ×3)
All the way to I
10
Correct final answer
(ii) | I > 0 ⇒ 1 π < 263
10
4 315
⇒ π < 263×4 = 1052 = 3107
315 315 315 | M1
A1
[2] | 3.1a
2.2a | Clear statement and consequence noted
AG from clear demonstration
3 For integers $n \geqslant 0 , \mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { 1 } \frac { \mathrm { x } ^ { \mathrm { n } } } { 1 + \mathrm { x } ^ { 2 } } \mathrm { dx }$.
\begin{enumerate}[label=(\alph*)]
\item For integers $n \geqslant 2$, show that $I _ { n } + I _ { n - 2 } = \frac { 1 } { n - 1 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Determine the exact value of $I _ { 10 }$.
\item Deduce that $\pi < 3 \frac { 107 } { 315 }$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure 2020 Q3 [9]}}
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Q1 5 Q2 9 Q3 9 Q4 7 Q5 13 Q6 10 Q7 10 Q8 12