| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Integration Using Polynomial Division |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring polynomial long division (a standard A-level technique) followed by integration of the resulting quotient plus a simple rational function. Part (b) involves recognizing that the remainder term integrates to an arctangent, which is routine for P3/C4 level. The calculation is mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Divide to obtain quotient \(x^2 + k\) | M1 | \(k\) is a constant. |
| Obtain quotient \(x^2 - 4\) | A1 | If quotient stated separately, mark at this stage. |
| Obtain remainder \(32\) | A1 | If remainder stated separately, mark at this stage. Need not state which is quotient and remainder, but if stated wrongly, max 2/3. After a correct division, still allow the marks if then written as \(x^2 - 4 + \frac{32}{x^2+4}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Expands brackets to get \(B = 0\) | M1 | \((x^2+4)(x^2+Bx+C)+D = x^4 + Bx^3 + (C+4)x^2 + 4Bx + 4C + D\) |
| \(C = -4\) | A1 | |
| \(D = 32\) | A1 | Need not state which is quotient and remainder, but if stated wrongly, max 2/3. |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{3}x^3 - 4x\) | B1 FT | Follow *their* quotient of form \(Ax^2 + B\). |
| Obtain \(p\tan^{-1} qx\) where \(q = 2\) or \(q = \frac{1}{2}\) | M1 | |
| Obtain \(16\tan^{-1}\frac{1}{2}x\) | A1 FT | Follow *their* constant remainder, i.e. \(\left(\frac{\textit{their constant remainder}}{2}\right)\tan^{-1}\frac{1}{2}x\). |
| Use limits correctly in an expression containing \(p\tan^{-1} qx\) where \(q = 2\) or \(q = \frac{1}{2}\) and \(rx^3 + sx\) | M1 | Terms need not be evaluated, e.g. \(\left[8\sqrt{3} - 8\sqrt{3}\right] + 16\tan^{-1}\sqrt{3} - \left(\frac{8}{3} - 8 + 16\tan^{-1}1\right)\) or \(\frac{8}{3} - 8\) can be \(-\frac{16}{3}\), \(16\tan^{-1}\sqrt{3}\) can be \(\frac{16\pi}{3}\), \(16\tan^{-1}1\) can be \(4\pi\). |
| Obtain \(\frac{4}{3}(\pi + 4)\) from full and correct working | A1 | AG |
| Total | 5 |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Divide to obtain quotient $x^2 + k$ | M1 | $k$ is a constant. |
| Obtain quotient $x^2 - 4$ | A1 | If quotient stated separately, mark at this stage. |
| Obtain remainder $32$ | A1 | If remainder stated separately, mark at this stage. Need not state which is quotient and remainder, but if stated wrongly, max 2/3. After a correct division, still allow the marks if then written as $x^2 - 4 + \frac{32}{x^2+4}$. |
**Alternative Method for Question 9(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expands brackets to get $B = 0$ | M1 | $(x^2+4)(x^2+Bx+C)+D = x^4 + Bx^3 + (C+4)x^2 + 4Bx + 4C + D$ |
| $C = -4$ | A1 | |
| $D = 32$ | A1 | Need not state which is quotient and remainder, but if stated wrongly, max 2/3. |
| **Total** | **3** | |
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{3}x^3 - 4x$ | B1 FT | Follow *their* quotient of form $Ax^2 + B$. |
| Obtain $p\tan^{-1} qx$ where $q = 2$ or $q = \frac{1}{2}$ | M1 | |
| Obtain $16\tan^{-1}\frac{1}{2}x$ | A1 FT | Follow *their* constant remainder, i.e. $\left(\frac{\textit{their constant remainder}}{2}\right)\tan^{-1}\frac{1}{2}x$. |
| Use limits correctly in an expression containing $p\tan^{-1} qx$ where $q = 2$ or $q = \frac{1}{2}$ and $rx^3 + sx$ | M1 | Terms need not be evaluated, e.g. $\left[8\sqrt{3} - 8\sqrt{3}\right] + 16\tan^{-1}\sqrt{3} - \left(\frac{8}{3} - 8 + 16\tan^{-1}1\right)$ or $\frac{8}{3} - 8$ can be $-\frac{16}{3}$, $16\tan^{-1}\sqrt{3}$ can be $\frac{16\pi}{3}$, $16\tan^{-1}1$ can be $4\pi$. |
| Obtain $\frac{4}{3}(\pi + 4)$ from full and correct working | A1 | AG |
| **Total** | **5** | |9
\begin{enumerate}[label=(\alph*)]
\item Find the quotient and remainder when $x ^ { 4 } + 16$ is divided by $x ^ { 2 } + 4$.
\item Hence show that $\int _ { 2 } ^ { 2 \sqrt { 3 } } \frac { x ^ { 4 } + 16 } { x ^ { 2 } + 4 } \mathrm {~d} x = \frac { 4 } { 3 } ( \pi + 4 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q9 [8]}}