| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Trigonometric power reduction |
| Difficulty | Challenging +1.2 This is a standard reduction formula question requiring integration by parts to derive the recurrence relation, then repeated application to find Iāā. While it involves multiple steps and careful algebraic manipulation, the technique is well-practiced in FP2 and follows a predictable pattern. The derivation is routine for students at this level, though the repeated application to n=11 requires accuracy. |
| Spec | 8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u = \cos^{n-1}x\), \(dv = \cos x\, dx\) | M1* | By parts the right way round |
| \(du = -(n-1)\cos^{n-2}x\sin x\), \(v = \sin x\) | A1 | |
| \(\Rightarrow I_n = \left[\cos^{n-1}x\sin x\right]_0^{\frac{\pi}{2}} + (n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x\sin^2 x\, dx\) | A1 | Integral so far |
| \(= 0 + (n-1)(I_{n-2} - I_n)\) | *M1 | Correct use of \(\sin^2 x = 1-\cos^2 x\); Dependent on 1st M |
| \(\Rightarrow nI_n = (n-1)I_{n-2} \Rightarrow I_n = \frac{n-1}{n}I_{n-2}\) | A1 | www AG |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_1 = 1\) | B1 | For \(I_1\) soi |
| \(I_{11} = \frac{10}{11}I_9 = \ldots = \frac{10}{11}\cdot\frac{8}{9}\cdot\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}I_1\) | M1 | Use of (i) to give product of 5 fractions |
| \(\Rightarrow I_{11} = \frac{3840}{10395} = \frac{256}{693}\) | A1 | Correct fraction |
| [3] |
# Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = \cos^{n-1}x$, $dv = \cos x\, dx$ | M1* | By parts the right way round |
| $du = -(n-1)\cos^{n-2}x\sin x$, $v = \sin x$ | A1 | |
| $\Rightarrow I_n = \left[\cos^{n-1}x\sin x\right]_0^{\frac{\pi}{2}} + (n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x\sin^2 x\, dx$ | A1 | Integral so far |
| $= 0 + (n-1)(I_{n-2} - I_n)$ | *M1 | Correct use of $\sin^2 x = 1-\cos^2 x$; Dependent on 1st M |
| $\Rightarrow nI_n = (n-1)I_{n-2} \Rightarrow I_n = \frac{n-1}{n}I_{n-2}$ | A1 | **www** AG |
| **[5]** | | |
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# Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_1 = 1$ | B1 | For $I_1$ soi |
| $I_{11} = \frac{10}{11}I_9 = \ldots = \frac{10}{11}\cdot\frac{8}{9}\cdot\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}I_1$ | M1 | Use of **(i)** to give product of 5 fractions |
| $\Rightarrow I_{11} = \frac{3840}{10395} = \frac{256}{693}$ | A1 | Correct fraction |
| **[3]** | | |
---4 It is given that $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { n } x \mathrm {~d} x$ for $n \geqslant 0$.\\
(i) Show that $I _ { n } = \frac { n - 1 } { n } I _ { n - 2 }$ for $n \geqslant 2$.\\
(ii) Hence find $I _ { 11 }$ as an exact fraction.
\hfill \mbox{\textit{OCR FP2 2013 Q4 [8]}}