| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Establish bounds or inequalities |
| Difficulty | Challenging +1.8 This is a sophisticated Further Maths question requiring derivation of a reduction formula via integration by parts, inequality reasoning about monotonic sequences, and a creative application to bound π. While the integration by parts is standard for FP2, parts (ii) and (iii) require genuine mathematical insight about comparing integrals and manipulating the recurrence relation—significantly harder than routine reduction formula exercises but still within expected FP2 scope. |
| Spec | 8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow nI_n = I_{n-2} \Rightarrow I_n = \frac{n-1}{n} I_{n-2}\) | M1, M1, A1 | Correct start for reduction; Deal with \(\cos^2\) dep on 1st M; www |
| Answer | Marks | Guidance |
|---|---|---|
| Alternative: \(I_n = \frac{n-1}{n} I_{n-2}, \frac{n-1}{n} < 1 \Rightarrow I_n < I_{n-2} \Rightarrow I_{n+2} < I_n\) for all \(n \geq\) \(I_{2n+1} < I_{2n-1}\) | M1, A1 | Allow using \(n\) instead of \(2n+1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow 3.0022 < \pi < 3.3024\) | B1, B1, M1, M1, A1, A1 | For \(I_1\) soi; For \(I_0\) soi; Applying reduction formula for at least one of \(I_9\) and \(I_{11}\); Applying reduction formula for \(I_{10}\); Lhs fraction or decimal equivalent correct to 4dp; Likewise Rhs |
| [6] |
**(i)** $I_n = \int_0^{\pi/2} \sin^n x \, dx = I_n = \int_0^{\pi/2} \sin^{n-1} x \sin x \, dx$
$\Rightarrow I_n = \left[\sin^{n-1} xx(-\cos x)\right]_0^{\pi/2} + \int_0^{\pi/2} \cos x(n-1)\sin^{n-2} x \cos x \, dx$
$= 0 + (n-1) \int_0^{\pi/2} \sin^{n-2} x 1-\sin^2 x \, dx = (n-1) I_{n-2} - I_n$
$\Rightarrow nI_n = I_{n-2} \Rightarrow I_n = \frac{n-1}{n} I_{n-2}$ | M1, M1, A1 | Correct start for reduction; Deal with $\cos^2$ dep on 1st M; www
**(ii)** $I_n = \frac{n-1}{n} I_{n-2} \Rightarrow I_{2n+1} = \frac{2n}{2n+1} I_{2n-1} = \frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1} I_{2n-3}$
and $\frac{2n}{2n+1} < 1$
Alternative: $I_n = \frac{n-1}{n} I_{n-2}, \frac{n-1}{n} < 1 \Rightarrow I_n < I_{n-2} \Rightarrow I_{n+2} < I_n$ for all $n \geq$ $I_{2n+1} < I_{2n-1}$ | M1, A1 | Allow using $n$ instead of $2n+1$
**(iii)** $I_{11} = \frac{256}{693}$, $I_{10} = \frac{63}{512}\pi$, $I_9 = \frac{128}{315}$
$\Rightarrow \frac{256}{693} < \frac{63}{512}\pi < \frac{128}{315}$
$\Rightarrow \frac{131072}{43659} < \pi < \frac{65536}{19845}$
$\Rightarrow 3.0022 < \pi < 3.3024$ | B1, B1, M1, M1, A1, A1 | For $I_1$ soi; For $I_0$ soi; Applying reduction formula for at least one of $I_9$ and $I_{11}$; Applying reduction formula for $I_{10}$; Lhs fraction or decimal equivalent correct to 4dp; Likewise Rhs
| [6] |
---7 It is given that, for non-negative integers $n , I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { n } x \mathrm {~d} x$.\\
(i) Show that $I _ { n } = \frac { n - 1 } { n } I _ { n - 2 }$ for $n \geqslant 2$.\\
(ii) Explain why $I _ { 2 n + 1 } < I _ { 2 n - 1 }$.\\
(iii) It is given that $I _ { 2 n + 1 } < I _ { 2 n } < I _ { 2 n - 1 }$. Take $n = 5$ to find an interval within which the value of $\pi$ lies.
\hfill \mbox{\textit{OCR FP2 2014 Q7 [11]}}