| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| 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 integration by parts via differentiation of a product, followed by algebraic manipulation and recursive application. While the technique is standard for FM students and the algebra is manageable, it requires careful execution across multiple steps with fractional powers and involves non-trivial exact value calculations with surds. The guided structure (suggesting the differentiation) reduces difficulty slightly from a completely unguided derivation. |
| Spec | 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_1 = \int_0^{\frac{1}{2}}\!(1-x^2)^{-\frac{1}{2}}\,dx = \left[\sin^{-1}x\right]_0^{\frac{1}{2}} = \frac{1}{6}\pi\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dx}\!\left(x(1-x^2)^{-\frac{1}{2}n}\right) = nx^2(1-x^2)^{-\frac{1}{2}n-1} + (1-x^2)^{-\frac{1}{2}n}\) | M1 A1 | |
| \(= n\!\left(1-(1-x^2)\right)(1-x^2)^{-\frac{1}{2}n-1} + (1-x^2)^{-\frac{1}{2}n}\) | M1 | |
| \(\left[x(1-x^2)^{-\frac{1}{2}n}\right]_0^{\frac{1}{2}} = nI_{n+2} - nI_n + I_n\) | M1 | |
| \(\frac{1}{2}\!\left(\frac{3}{4}\right)^{-\frac{1}{2}n} = nI_{n+2} - (n-1)I_n \Rightarrow nI_{n+2} = 2^{n-1}3^{-\frac{1}{2}n} + (n-1)I_n\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_3 = 3^{-\frac{1}{2}}\) | B1 | |
| \(3I_5 = 2^2 3^{-\frac{3}{2}} + 2I_3 \Rightarrow I_5 = \frac{10}{27}\sqrt{3}\) | M1 A1 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_1 = \int_0^{\frac{1}{2}}\!(1-x^2)^{-\frac{1}{2}}\,dx = \left[\sin^{-1}x\right]_0^{\frac{1}{2}} = \frac{1}{6}\pi$ | M1 A1 | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}\!\left(x(1-x^2)^{-\frac{1}{2}n}\right) = nx^2(1-x^2)^{-\frac{1}{2}n-1} + (1-x^2)^{-\frac{1}{2}n}$ | M1 A1 | |
| $= n\!\left(1-(1-x^2)\right)(1-x^2)^{-\frac{1}{2}n-1} + (1-x^2)^{-\frac{1}{2}n}$ | M1 | |
| $\left[x(1-x^2)^{-\frac{1}{2}n}\right]_0^{\frac{1}{2}} = nI_{n+2} - nI_n + I_n$ | M1 | |
| $\frac{1}{2}\!\left(\frac{3}{4}\right)^{-\frac{1}{2}n} = nI_{n+2} - (n-1)I_n \Rightarrow nI_{n+2} = 2^{n-1}3^{-\frac{1}{2}n} + (n-1)I_n$ | A1 | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_3 = 3^{-\frac{1}{2}}$ | B1 | |
| $3I_5 = 2^2 3^{-\frac{3}{2}} + 2I_3 \Rightarrow I_5 = \frac{10}{27}\sqrt{3}$ | M1 A1 | |
---6 The integral $\mathrm { I } _ { \mathrm { n } }$, where $n$ is an integer, is defined by $\mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { \frac { 1 } { 2 } } \left( 1 - \mathrm { x } ^ { 2 } \right) ^ { - \frac { 1 } { 2 } \mathrm { n } } \mathrm { dx }$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $I _ { 1 }$.
\item By considering $\frac { \mathrm { d } } { \mathrm { dx } } \left( \mathrm { x } \left( 1 - \mathrm { x } ^ { 2 } \right) ^ { - \frac { 1 } { 2 } \mathrm { n } } \right)$, or otherwise, show that
$$\mathrm { nl } _ { \mathrm { n } + 2 } = 2 ^ { \mathrm { n } - 1 } 3 ^ { - \frac { 1 } { 2 } \mathrm { n } } + ( \mathrm { n } - 1 ) \mathrm { I } _ { \mathrm { n } } .$$
\item Find the exact value of $I _ { 5 }$ giving the answer in the form $k \sqrt { 3 }$, where $k$ is a rational number to be determined.\\
\includegraphics[max width=\textwidth, alt={}, center]{20e14db3-0eb0-4954-91cf-027e16f8bf14-11_78_1576_336_321}
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q6 [10]}}