| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Derive reduction formula by differentiation |
| Difficulty | Challenging +1.8 This is a Further Maths reduction formula question requiring differentiation of a product, integration by parts reasoning, and recursive calculation. The hint guides students to the key step, but they must manipulate the resulting equation algebraically to isolate the recurrence relation, then apply it twice with careful arithmetic. More demanding than standard A-level but the structure is relatively standard for Further Maths reduction formulae. |
| Spec | 1.08h Integration by substitution8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}\left(x(1+x^5)^n\right) = 5nx^5(1+x^5)^{n-1} + (1+x^5)^n\) | M1 A1 | Uses the product rule to differentiate |
| \(5n(1+x^5-1)(1+x^5)^{n-1} + (1+x^5)^n\) | M1* | Uses \(x^5 = x^5 + 1 - 1\) |
| \(\left[x(1+x^5)^n\right]_0^1 = 5nI_n - 5nI_{n-1} + I_n\) | DM1 | Integrates both sides using the limits given. Requires previous method mark |
| \(2^n = (5n+1)I_n - 5nI_{n-1} \Rightarrow (5n+1)I_n = 2^n + 5nI_{n-1}\) | A1 | Substitutes limits and rearranges. AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_n = \int_0^1(1+x^5)^n\,dx = \left[x(1+x^5)^n\right]_0^1 - 5n\int_0^1 x^5(1+x^5)^{n-1}\,dx\) | M1 A1 | Integrates by parts |
| \(I_n = \left[x(1+x^5)^n\right]_0^1 - 5n\int_0^1(1+x^5)^n\,dx + 5n\int_0^1(1+x^5)^{n-1}\,dx\) | M1* | Uses \(x^5 = x^5+1-1\) |
| \(I_n = \left[x(1+x^5)^n\right]_0^1 - 5nI_n + 5nI_{n-1}\) | DM1 | Forms recursive formula. Requires previous method mark |
| \((5n+1)I_n = 2^n + 5nI_{n-1}\) | A1 | Substitutes limits and rearranges. AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_1 = \left[x + \frac{1}{6}x^6\right]_0^1 = \frac{7}{6}\) or \(I_0 = 1\) | B1 | |
| \(11I_2 = 2^2 + 10I_1 \Rightarrow I_2 = \frac{47}{33}\) | M1 A1 | Applies reduction formula |
| \(16I_3 = 2^3 + 15I_2 \Rightarrow I_3 = \frac{323}{176}\) | A1 |
## Question 4(a):
| $\frac{d}{dx}\left(x(1+x^5)^n\right) = 5nx^5(1+x^5)^{n-1} + (1+x^5)^n$ | M1 A1 | Uses the product rule to differentiate |
|---|---|---|
| $5n(1+x^5-1)(1+x^5)^{n-1} + (1+x^5)^n$ | M1* | Uses $x^5 = x^5 + 1 - 1$ |
| $\left[x(1+x^5)^n\right]_0^1 = 5nI_n - 5nI_{n-1} + I_n$ | DM1 | Integrates both sides using the limits given. Requires previous method mark |
| $2^n = (5n+1)I_n - 5nI_{n-1} \Rightarrow (5n+1)I_n = 2^n + 5nI_{n-1}$ | A1 | Substitutes limits and rearranges. AG |
**Alternative method for 4(a):**
| $I_n = \int_0^1(1+x^5)^n\,dx = \left[x(1+x^5)^n\right]_0^1 - 5n\int_0^1 x^5(1+x^5)^{n-1}\,dx$ | M1 A1 | Integrates by parts |
|---|---|---|
| $I_n = \left[x(1+x^5)^n\right]_0^1 - 5n\int_0^1(1+x^5)^n\,dx + 5n\int_0^1(1+x^5)^{n-1}\,dx$ | M1* | Uses $x^5 = x^5+1-1$ |
| $I_n = \left[x(1+x^5)^n\right]_0^1 - 5nI_n + 5nI_{n-1}$ | DM1 | Forms recursive formula. Requires previous method mark |
| $(5n+1)I_n = 2^n + 5nI_{n-1}$ | A1 | Substitutes limits and rearranges. AG |
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## Question 4(b):
| $I_1 = \left[x + \frac{1}{6}x^6\right]_0^1 = \frac{7}{6}$ or $I_0 = 1$ | B1 | |
|---|---|---|
| $11I_2 = 2^2 + 10I_1 \Rightarrow I_2 = \frac{47}{33}$ | M1 A1 | Applies reduction formula |
| $16I_3 = 2^3 + 15I_2 \Rightarrow I_3 = \frac{323}{176}$ | A1 | |
---4 The integral $\mathrm { I } _ { \mathrm { n } }$ is defined by $\mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { 1 } \left( 1 + \mathrm { x } ^ { 5 } \right) ^ { \mathrm { n } } \mathrm { dx }$.
\begin{enumerate}[label=(\alph*)]
\item By considering $\frac { d } { d x } \left( x \left( 1 + x ^ { 5 } \right) ^ { n } \right)$, or otherwise, show that
$$( 5 n + 1 ) l _ { n } = 2 ^ { n } + 5 n l _ { n - 1 }$$
\item Find the exact value of $I _ { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q4 [9]}}