| 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 differentiation of a product, integration by parts technique, and recursive application. Part (a) is straightforward (arcsin evaluation), part (b) requires guided algebraic manipulation following the hint, and part (c) involves applying the recurrence relation twice. While systematic, it demands careful algebraic handling and is more challenging than standard A-level integration, placing it well above average difficulty. |
| 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]{1de67949-6262-4ade-b986-02b6563ae404-11_78_1576_336_321}
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q6 [10]}}