| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Rational function powers |
| Difficulty | Challenging +1.3 This is a standard Further Maths reduction formula question requiring integration by parts with a specific choice of u and dv, algebraic manipulation to derive the recurrence relation, and application to find a specific value. While it requires careful technique and multiple steps, the structure is formulaic and follows well-established patterns for FP2 reduction formulae questions. The integration by parts setup is guided, and the algebraic manipulation, though requiring attention to detail, is straightforward once the correct approach is identified. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Get \(x(1+x^2)^n - \int x.c(n+x^2)^{n-1}.2x \, dx\). Accurate use of parts. Clearly get A.G. | M1, A1, B1 | Reasonable attempt at parts. Include use of limits seen |
| (ii) Express \(x^2\) as \((1+x^2) - 1\). Get \(\frac{x^2}{(1+x^2)^{n+1}} = \frac{1}{(1+x^2)^n} - \frac{1}{(1+x^2)^{n+1}}\). Show \(I_n = 2^n +2n(I_n - I_{n+1})\). Tidy to A.G. | B1, M1, A1 | Justified |
| (iii) See \(2I_2 = 2^1 + I_1\). Work out \(I_1 = \frac{1}{4}\pi\). Get \(I_2 = \frac{1}{4} + \frac{1}{8}\pi\) | B1, M1, A1 | Quote/derive \(\tan^{-1}x\) |
**(i)** Get $x(1+x^2)^n - \int x.c(n+x^2)^{n-1}.2x \, dx$. Accurate use of parts. Clearly get A.G. | M1, A1, B1 | Reasonable attempt at parts. Include use of limits seen
**(ii)** Express $x^2$ as $(1+x^2) - 1$. Get $\frac{x^2}{(1+x^2)^{n+1}} = \frac{1}{(1+x^2)^n} - \frac{1}{(1+x^2)^{n+1}}$. Show $I_n = 2^n +2n(I_n - I_{n+1})$. Tidy to A.G. | B1, M1, A1 | Justified
**(iii)** See $2I_2 = 2^1 + I_1$. Work out $I_1 = \frac{1}{4}\pi$. Get $I_2 = \frac{1}{4} + \frac{1}{8}\pi$ | B1, M1, A1 | Quote/derive $\tan^{-1}x$7 It is given that, for integers $n \geqslant 1$,
$$I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { n } } \mathrm {~d} x$$
(i) Use integration by parts to show that $I _ { n } = 2 ^ { - n } + 2 n \int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) ^ { n + 1 } } \mathrm {~d} x$.\\
(ii) Show that $2 n I _ { n + 1 } = 2 ^ { - n } + ( 2 n - 1 ) I _ { n }$.\\
(iii) Find $I _ { 2 }$ in terms of $\pi$.
\hfill \mbox{\textit{OCR FP2 2008 Q7 [9]}}