| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| 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 multiple techniques: evaluating a definite integral with (1+x²)^(-1/2) using standard substitution/recognition, deriving a reduction formula via differentiation of a product (non-trivial algebraic manipulation), and applying arc length with substitution. While methodical, it demands careful execution across three connected parts with no single step being routine A-level material. |
| Spec | 1.08i Integration by parts4.08c Improper integrals: infinite limits or discontinuous integrands4.08f Integrate using partial fractions8.06a Reduction formulae: establish, use, and evaluate recursively8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_{-1} = \int_0^{\sqrt{3}}(1+x^2)^{-\frac{1}{2}}\,dx = \left[\sinh^{-1}x\right]_0^{\sqrt{3}} = \ln 3\) | M1 A1 | Recognises integral. |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{d}{dx}\left(x\left(1+x^2\right)^{\frac{1}{2}n}\right) = nx^2\left(1+x^2\right)^{\frac{1}{2}n-1} + \left(1+x^2\right)^{\frac{1}{2}n}\) | M1 A1 | Uses the product rule to differentiate. |
| \(n\left(1+x^2-1\right)\left(1+x^2\right)^{\frac{1}{2}n-1} + \left(1+x^2\right)^{\frac{1}{2}n}\) | M1* | Uses \(x^2 = 1+x^2-1\). |
| \(\left[x\left(1+x^2\right)^{\frac{1}{2}n}\right]_0^{\sqrt{3}} = nI_n - nI_{n-2} + I_n\) | DM1 | Integrates both sides using limits given. Requires previous method mark. |
| \(\frac{4}{3}\left(\frac{5}{3}\right)^n = (n+1)I_n - nI_{n-2} \Rightarrow (n+1)I_n = nI_{n-2} + \frac{4}{3}\left(\frac{5}{3}\right)^n\) | A1 | Substitutes limits and rearranges. AG. |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = \int_0^{\frac{2}{3}}\sqrt{1+4x^2}\,dx\) | M1 | Forms correct integral with correct limits. |
| \(\frac{1}{2}\int_0^{\sqrt{3}}\sqrt{1+u^2}\,du = \frac{1}{2}I_1\) | A1 | AG. |
| \(2I_1 = I_{-1} + \frac{4}{3}\left(\frac{5}{3}\right)\) | M1 | Applies reduction formula with \(n=1\). |
| \(s = \frac{1}{4}\ln 3 + \frac{5}{9}\) | A1 | |
| 4 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_{-1} = \int_0^{\sqrt{3}}(1+x^2)^{-\frac{1}{2}}\,dx = \left[\sinh^{-1}x\right]_0^{\sqrt{3}} = \ln 3$ | M1 A1 | Recognises integral. |
| | **2** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{d}{dx}\left(x\left(1+x^2\right)^{\frac{1}{2}n}\right) = nx^2\left(1+x^2\right)^{\frac{1}{2}n-1} + \left(1+x^2\right)^{\frac{1}{2}n}$ | M1 A1 | Uses the product rule to differentiate. |
| $n\left(1+x^2-1\right)\left(1+x^2\right)^{\frac{1}{2}n-1} + \left(1+x^2\right)^{\frac{1}{2}n}$ | M1* | Uses $x^2 = 1+x^2-1$. |
| $\left[x\left(1+x^2\right)^{\frac{1}{2}n}\right]_0^{\sqrt{3}} = nI_n - nI_{n-2} + I_n$ | DM1 | Integrates both sides using limits given. Requires previous method mark. |
| $\frac{4}{3}\left(\frac{5}{3}\right)^n = (n+1)I_n - nI_{n-2} \Rightarrow (n+1)I_n = nI_{n-2} + \frac{4}{3}\left(\frac{5}{3}\right)^n$ | A1 | Substitutes limits and rearranges. AG. |
| | **5** | |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = \int_0^{\frac{2}{3}}\sqrt{1+4x^2}\,dx$ | M1 | Forms correct integral with correct limits. |
| $\frac{1}{2}\int_0^{\sqrt{3}}\sqrt{1+u^2}\,du = \frac{1}{2}I_1$ | A1 | AG. |
| $2I_1 = I_{-1} + \frac{4}{3}\left(\frac{5}{3}\right)$ | M1 | Applies reduction formula with $n=1$. |
| $s = \frac{1}{4}\ln 3 + \frac{5}{9}$ | A1 | |
| | **4** | |7 The integral $\mathrm { I } _ { \mathrm { n } }$, where n is an integer, is defined by $\mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { \frac { 4 } { 3 } } \left( 1 + \mathrm { x } ^ { 2 } \right) ^ { \frac { 1 } { 2 } \mathrm { n } } \mathrm { dx }$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $I _ { - 1 }$ giving your answer in the form $\ln a$, where $a$ is an integer to be determined.
\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
$$( \mathrm { n } + 1 ) \mathrm { I } _ { \mathrm { n } } = \mathrm { nl } _ { \mathrm { n } - 2 } + \frac { 4 } { 3 } \left( \frac { 5 } { 3 } \right) ^ { \mathrm { n } }$$
\item A curve has equation $y = x ^ { 2 }$, for $0 \leqslant x \leqslant \frac { 2 } { 3 }$. The arc length of the curve is denoted by $s$.
Use the substitution $\mathrm { u } = 2 \mathrm { x }$ to show that $\mathrm { s } = \frac { 1 } { 2 } \mathrm { l } _ { 1 }$ and find the exact value of $s$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q7 [11]}}