| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Compound expressions with binomial expansion |
| Difficulty | Challenging +1.2 This is a standard reduction formula question requiring integration by parts to establish the recurrence relation, then applying it twice with the base case I₀. While it involves Further Maths content (reduction formulae), the technique is routine and well-practiced, with clear structure and no novel insight required. The algebraic manipulation is straightforward, placing it moderately above average difficulty. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(I_n = \left[-\frac{2}{3}x^n(1-x)^2\right]_0^1 + \frac{2}{3}\int_0^1 x^{n-1}(1-x)^2 \, dx\) | M1 | For using integration by parts |
| \(-\) | A1 | For correct first stage result |
| \(= \frac{2}{3}n\int_0^1 x^{n-1}(1-x)y(1-x) \, dx\) | M1 | For use of limits in integrated term |
| \(= \frac{2}{3}n(I_{n-1} - I_n)\) | M1 | For splitting the remaining integral up |
| Hence \((2n+3)I_n = 2nI_{n-1}\), as required | A1 | For correct relation between \(I_n\) and \(I_{n-1}\) |
| \(-\) | A1 | For showing given answer correctly |
| \(-\) | 6 | |
| (ii) \(I_2 = \frac{8}{7} = \frac{8}{7} \times \frac{2}{5}I_0\) | M1 | For two uses of the recurrence relation |
| \(-\) | A1 | For correct expression in terms of \(I_0\) |
| Hence \(I_2 = \frac{8}{33}\left[−\frac{2}{3}(1−x)^3\right]_0^1 = \frac{16}{105}\) | M1 | For evaluation of \(I_0\) |
| \(-\) | A1 | For correct answer |
| \(-\) | 4 |
(i) $I_n = \left[-\frac{2}{3}x^n(1-x)^2\right]_0^1 + \frac{2}{3}\int_0^1 x^{n-1}(1-x)^2 \, dx$ | M1 | For using integration by parts |
$-$ | A1 | For correct first stage result |
$= \frac{2}{3}n\int_0^1 x^{n-1}(1-x)y(1-x) \, dx$ | M1 | For use of limits in integrated term |
$= \frac{2}{3}n(I_{n-1} - I_n)$ | M1 | For splitting the remaining integral up |
Hence $(2n+3)I_n = 2nI_{n-1}$, as required | A1 | For correct relation between $I_n$ and $I_{n-1}$ |
$-$ | A1 | For showing given answer correctly |
$-$ | **6** | |
(ii) $I_2 = \frac{8}{7} = \frac{8}{7} \times \frac{2}{5}I_0$ | M1 | For two uses of the recurrence relation |
$-$ | A1 | For correct expression in terms of $I_0$ |
Hence $I_2 = \frac{8}{33}\left[−\frac{2}{3}(1−x)^3\right]_0^1 = \frac{16}{105}$ | M1 | For evaluation of $I_0$ |
$-$ | A1 | For correct answer |
$-$ | **4** | |
---6 (i) Given that $I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \sqrt { } ( 1 - x ) \mathrm { d } x$, prove that, for $n \geqslant 1$,
$$( 2 n + 3 ) I _ { n } = 2 n I _ { n - 1 } .$$
(ii) Hence find the exact value of $I _ { 2 }$.
\hfill \mbox{\textit{OCR FP2 Q6 [10]}}