| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Rational function powers |
| Difficulty | Challenging +1.8 This is a substantial reduction formula question requiring differentiation of a product, integration by parts manipulation, application of a standard formula, and recursive calculation. While the steps are guided, it demands careful algebraic manipulation across multiple stages and proper handling of definite integral bounds—significantly harder than routine A-level but follows a clear template typical of Further Maths reduction formula questions. |
| Spec | 1.08d Evaluate definite integrals: between limits4.02q De Moivre's theorem: multiple angle formulae8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}\left\{x(4+x^2)^{-n}\right\} = (4+x^2)^{-n} - 2nx^2(4+x^2)^{-n-1}\) | M1A1 | |
| Integrate w.r.t. \(x\): \(\left[x(4+x^2)^{-n}\right]_0^2 = I_n - 2n\int_0^2(x^2-4)(4+x^2)^{-n-1}dx\) | M1 | |
| M1 | ||
| \(\Rightarrow 2 \cdot 8^{-n} = I_n - 2nI_n + 8nI_{n+1} \Rightarrow 8nI_{n+1} = (2n-1)I_n + 2\cdot8^{-n}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_1 = \int_0^2 \frac{1}{4+x^2}dx = \left[\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)\right]_0^2 = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8}\) | M1A1 | |
| \(8I_2 = \frac{\pi}{8} + \frac{1}{4} \Rightarrow I_2 = \frac{\pi}{64} + \frac{1}{32}\) | M1A1FT | |
| \(16I_3 = \frac{3\pi}{64} + \frac{3}{32} + \frac{1}{32} \Rightarrow I_3 = \frac{3\pi}{1024} + \frac{1}{128}\) | A1 | AG |
## Question 6(i):
| $\frac{d}{dx}\left\{x(4+x^2)^{-n}\right\} = (4+x^2)^{-n} - 2nx^2(4+x^2)^{-n-1}$ | M1A1 | |
|---|---|---|
| Integrate w.r.t. $x$: $\left[x(4+x^2)^{-n}\right]_0^2 = I_n - 2n\int_0^2(x^2-4)(4+x^2)^{-n-1}dx$ | M1 | |
| | M1 | |
| $\Rightarrow 2 \cdot 8^{-n} = I_n - 2nI_n + 8nI_{n+1} \Rightarrow 8nI_{n+1} = (2n-1)I_n + 2\cdot8^{-n}$ | A1 | AG |
**Total: 5**
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## Question 6(ii):
| $I_1 = \int_0^2 \frac{1}{4+x^2}dx = \left[\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)\right]_0^2 = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8}$ | M1A1 | |
|---|---|---|
| $8I_2 = \frac{\pi}{8} + \frac{1}{4} \Rightarrow I_2 = \frac{\pi}{64} + \frac{1}{32}$ | M1A1FT | |
| $16I_3 = \frac{3\pi}{64} + \frac{3}{32} + \frac{1}{32} \Rightarrow I_3 = \frac{3\pi}{1024} + \frac{1}{128}$ | A1 | AG |
**Total: 5**
---6 Let $I _ { n }$ denote $\int _ { 0 } ^ { 2 } \left( 4 + x ^ { 2 } \right) ^ { - n } \mathrm {~d} x$.\\
(i) Find $\frac { \mathrm { d } } { \mathrm { d } x } \left( x \left( 4 + x ^ { 2 } \right) ^ { - n } \right)$ and hence show that
$$8 n I _ { n + 1 } = ( 2 n - 1 ) I _ { n } + 2 \times 8 ^ { - n } .$$
(ii) Use the result for integrating $\frac { 1 } { x ^ { 2 } + a ^ { 2 } }$ with respect to $x$, in the List of Formulae (MF10), to find the value of $I _ { 1 }$ and deduce that
$$I _ { 3 } = \frac { 3 } { 1024 } \pi + \frac { 1 } { 128 }$$
\hfill \mbox{\textit{CAIE FP1 2017 Q6 [10]}}