| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Derive reduction formula by differentiation |
| Difficulty | Challenging +1.8 This is a substantial Further Maths reduction formula question requiring integration with surds (part a), deriving a reduction formula via differentiation and integration by parts (part b), and applying it recursively (part c). While the technique is standard for FM students, the algebraic manipulation and multi-step nature elevate it above average difficulty, though it remains a recognizable exam pattern rather than requiring novel insight. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.08h Integration by substitution8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_1 = \int_0^3 (4+x^2)^{-\frac{1}{2}} = \left[\sinh^{-1}\left(\frac{1}{2}x\right)\right]_0^3\) | M1 A1 | Recognises integral or uses appropriate substitution |
| \(\ln 2\) | A1 | Insert limits, must simplify |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dx}\left(x(4+x^2)^{-\frac{1}{2}n}\right) = -nx^2(4+x^2)^{-\frac{1}{2}n-1} + (4+x^2)^{-\frac{1}{2}n}\) | M1 A1 | Uses the product rule to differentiate |
| \(-n(4+x^2-4)(4+x^2)^{-\frac{1}{2}n-1} + (4+x^2)^{-\frac{1}{2}n}\) | M1 | Applies \(x^2 = 4+x^2-4\) |
| \(\left[x(4+x^2)^{-\frac{1}{2}n}\right]_0^3 = -nI_n + 4nI_{n+2} + I_n\) | M1 | Integrates both sides using the limits given |
| \(\frac{3}{2}\left(\frac{2}{5}\right)^n = (1-n)I_n + 4nI_{n+2} \Rightarrow 4nI_{n+2} = \frac{3}{2}\left(\frac{2}{5}\right)^n + (n-1)I_n\) | A1 | Substitutes limits and rearranges, AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_n = \left[x(4+x^2)^{-\frac{1}{2}n}\right]_0^3 + n\int_0^3 x^2(4+x^2)^{-\frac{1}{2}n-1}\,dx\) | M1 A1 | Integrates by parts |
| \(I_n = \left[x(4+x^2)^{-\frac{1}{2}n}\right]_0^3 + n\int_0^3 (4+x^2-4)(4+x^2)^{-\frac{1}{2}n-1}\,dx\) | M1 | Applies \(x^2 = 4+x^2-4\) |
| \(I_n = \frac{3}{2}\left(\frac{2}{5}\right)^n + nI_n - 4I_{n+2} \Rightarrow 4nI_{n+2} = \frac{3}{2}\left(\frac{2}{5}\right)^n + (n-1)I_n\) | M1 A1 | Substitutes limits and rearranges, AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_3 = \frac{3}{20}\) | B1 | Applies reduction formula with \(n=1\) |
| \(12I_5 = \frac{3}{2}\left(\frac{8}{125}\right) + 2I_3 \Rightarrow I_5 = \frac{33}{1000} = 0.033\) | M1 A1 | Applies reduction formula with \(n=3\) |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_1 = \int_0^3 (4+x^2)^{-\frac{1}{2}} = \left[\sinh^{-1}\left(\frac{1}{2}x\right)\right]_0^3$ | M1 A1 | Recognises integral or uses appropriate substitution |
| $\ln 2$ | A1 | Insert limits, must simplify |
**Total: 3 marks**
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}\left(x(4+x^2)^{-\frac{1}{2}n}\right) = -nx^2(4+x^2)^{-\frac{1}{2}n-1} + (4+x^2)^{-\frac{1}{2}n}$ | M1 A1 | Uses the product rule to differentiate |
| $-n(4+x^2-4)(4+x^2)^{-\frac{1}{2}n-1} + (4+x^2)^{-\frac{1}{2}n}$ | M1 | Applies $x^2 = 4+x^2-4$ |
| $\left[x(4+x^2)^{-\frac{1}{2}n}\right]_0^3 = -nI_n + 4nI_{n+2} + I_n$ | M1 | Integrates both sides using the limits given |
| $\frac{3}{2}\left(\frac{2}{5}\right)^n = (1-n)I_n + 4nI_{n+2} \Rightarrow 4nI_{n+2} = \frac{3}{2}\left(\frac{2}{5}\right)^n + (n-1)I_n$ | A1 | Substitutes limits and rearranges, AG |
**Alternative method for 7(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_n = \left[x(4+x^2)^{-\frac{1}{2}n}\right]_0^3 + n\int_0^3 x^2(4+x^2)^{-\frac{1}{2}n-1}\,dx$ | M1 A1 | Integrates by parts |
| $I_n = \left[x(4+x^2)^{-\frac{1}{2}n}\right]_0^3 + n\int_0^3 (4+x^2-4)(4+x^2)^{-\frac{1}{2}n-1}\,dx$ | M1 | Applies $x^2 = 4+x^2-4$ |
| $I_n = \frac{3}{2}\left(\frac{2}{5}\right)^n + nI_n - 4I_{n+2} \Rightarrow 4nI_{n+2} = \frac{3}{2}\left(\frac{2}{5}\right)^n + (n-1)I_n$ | M1 A1 | Substitutes limits and rearranges, AG |
**Total: 5 marks**
---
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_3 = \frac{3}{20}$ | B1 | Applies reduction formula with $n=1$ |
| $12I_5 = \frac{3}{2}\left(\frac{8}{125}\right) + 2I_3 \Rightarrow I_5 = \frac{33}{1000} = 0.033$ | M1 A1 | Applies reduction formula with $n=3$ |
**Total: 3 marks**
---7 The integral $\mathrm { I } _ { \mathrm { n } }$, where n is an integer, is defined by $\mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { \frac { 3 } { 2 } } \left( 4 + \mathrm { x } ^ { 2 } \right) ^ { - \frac { 1 } { 2 } \mathrm { n } } \mathrm { dx }$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $I _ { 1 }$, expressing your answer in logarithmic form.
\item By considering $\frac { d } { d x } \left( x \left( 4 + x ^ { 2 } \right) ^ { - \frac { 1 } { 2 } n } \right)$, or otherwise, show that
$$4 n l _ { n + 2 } = \frac { 3 } { 2 } \left( \frac { 2 } { 5 } \right) ^ { n } + ( n - 1 ) l _ { n } .$$
\item Find the value of $I _ { 5 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q7 [11]}}