| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Volume and surface area of revolution |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring integration by parts to derive a reduction formula, then applying it to find a volume of revolution. While the techniques are standard for FP1 (reduction formulae and volumes of revolution), the multi-step nature, algebraic manipulation required in part (ii), and the need to connect the reduction formula to the volume calculation make this significantly harder than average A-level questions but still within expected FP1 scope. |
| Spec | 4.08d Volumes of revolution: about x and y axes8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_2 = \int_0^{\frac{1}{4}\pi} \sec^2 x\, dx = [\tan x]_0^{\frac{1}{4}\pi} = 1\) | M1A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_n = \int_0^{\frac{1}{4}\pi} \sec^{n-2}x \cdot \sec^2 x\, dx\) | M1 | |
| \(= \left[\sec^{n-2}x\tan x\right]_0^{\frac{1}{4}\pi} - \int_0^{\frac{1}{4}\pi}(n-2)\sec^{n-3}x(\sec x\tan x)\tan x\, dx\) | M1A1 | |
| \(= \left[\sec^{n-2}x\tan x\right]_0^{\frac{1}{4}\pi} - (n-2)\int_0^{\frac{1}{4}\pi}\sec^{n-2}x(\sec^2 x - 1)\, dx\) | M1A1 | |
| \(\Rightarrow (n-1)I_n = 2^{\frac{n}{2}-1} + (n-2)I_{n-2}\) | AG | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Volume of revolution \(= \pi\int y^2\, dx = \pi\int_0^{\frac{1}{4}\pi}\sec^6 x\, dx\) | M1 | |
| \(3I_4 = 2 + 2\times 1 \Rightarrow I_4 = \frac{4}{3}\) | M1 | |
| \(5I_6 = 4 + 4\times\frac{4}{3} \Rightarrow I_6 = \frac{28}{15}\) | M1 | |
| Volume of revolution \(= \frac{28\pi}{15}\) | A1 | |
| 4 |
## Question 8:
**8(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_2 = \int_0^{\frac{1}{4}\pi} \sec^2 x\, dx = [\tan x]_0^{\frac{1}{4}\pi} = 1$ | M1A1 | |
| | **2** | |
**8(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_n = \int_0^{\frac{1}{4}\pi} \sec^{n-2}x \cdot \sec^2 x\, dx$ | M1 | |
| $= \left[\sec^{n-2}x\tan x\right]_0^{\frac{1}{4}\pi} - \int_0^{\frac{1}{4}\pi}(n-2)\sec^{n-3}x(\sec x\tan x)\tan x\, dx$ | M1A1 | |
| $= \left[\sec^{n-2}x\tan x\right]_0^{\frac{1}{4}\pi} - (n-2)\int_0^{\frac{1}{4}\pi}\sec^{n-2}x(\sec^2 x - 1)\, dx$ | M1A1 | |
| $\Rightarrow (n-1)I_n = 2^{\frac{n}{2}-1} + (n-2)I_{n-2}$ | | AG |
| | **5** | |
**8(iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Volume of revolution $= \pi\int y^2\, dx = \pi\int_0^{\frac{1}{4}\pi}\sec^6 x\, dx$ | M1 | |
| $3I_4 = 2 + 2\times 1 \Rightarrow I_4 = \frac{4}{3}$ | M1 | |
| $5I_6 = 4 + 4\times\frac{4}{3} \Rightarrow I_6 = \frac{28}{15}$ | M1 | |
| Volume of revolution $= \frac{28\pi}{15}$ | A1 | |
| | **4** | |(i) Find the value of $I _ { 2 }$.\\
(ii) Show that, for $n > 2$,
$$( n - 1 ) I _ { n } = 2 ^ { \frac { 1 } { 2 } n - 1 } + ( n - 2 ) I _ { n - 2 }$$
(iii) The curve $C$ has equation $y = \sec ^ { 3 } x$ for $0 \leqslant x \leqslant \frac { 1 } { 4 } \pi$. The region $R$ is bounded by $C$, the $x$-axis, the $y$-axis and the line $x = \frac { 1 } { 4 } \pi$. Find the volume of revolution generated when $R$ is rotated through $2 \pi$ radians about the $x$-axis.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q8 [11]}}