| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Multi-part questions combining substitution with curve/area analysis |
| Difficulty | Standard +0.8 This is a two-part question requiring (a) differentiation of a product involving trigonometric and composite functions to find a minimum, then (b) integration by substitution with trigonometric expressions. Part (b) requires careful handling of limits transformation and recognizing the integral structure after substitution. While the substitution is given, executing it correctly with proper bounds and simplification requires solid technique beyond routine exercises. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use of correct product rule and correct chain rule | M1 | \(\frac{dy}{dx} = A\cos x\sqrt{2+\cos x} + \frac{B\sin x\sin x}{\sqrt{2+\cos x}}\) |
| Obtain \(\frac{dy}{dx} = 2\cos x\sqrt{2+\cos x} - \frac{2\sin^2 x}{2\sqrt{2+\cos x}}\) | A1 | OE |
| Equate derivative to zero and obtain a horizontal 3-term quadratic equation in \(\cos a\) or 4-term quartic equation in \(\cos x\). The only error in the form of the differential allowed is for \((2+\cos x)^{-\frac{1}{2}}\) to be \((2+\cos x)^{\frac{1}{2}}\) or \((2+\cos x)^{-\frac{3}{2}}\) | *M1 | Accept in \(\cos x\). E.g. \(3\cos^2 x + 4\cos x - 1 = 0\). E.g. \(3\cos^4 x + 16\cos^3 x + 18\cos^2 x - 1 = 0\) |
| Solve for \(\cos a\) | DM1 | \(\cos a = \frac{-2+\sqrt{7}}{3}\) or \(0.215\). Allow presence of other solution(s). |
| Obtain \(a = 4.93\) | A1 | Allow more accurate e.g. \(4.929\ldots\) even though question states 2 d.p. If \(x=1.35\) leads to \(x=4.93\) award A1 BOD. If \(x=1.35\) and \(x=4.93\) award A0. |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(du = -\sin x\,dx\) | B1 | OE. If B0, max M1M1M1. |
| Substitute throughout for \(u\) and \(du\) | M1 | |
| Obtain \(-\int 2\sqrt{u}\,du\) | A1 | OE. Ignore limits if \(-\int 2\sqrt{u}\,du\), but if \(+\int 2\sqrt{u}\,du\), then must have correct limits \(\int_1^3 2\sqrt{u}\,du\). (See final M1) |
| Integrate to obtain \(ku^{\frac{3}{2}}\ (+C)\) | M1 | Constant of integration not required |
| Use correct limits correctly in an expression of the form \(ku^{\frac{3}{2}}\) or \(k(2+\cos x)^{\frac{3}{2}}\) | M1 | 1 and 3 for \(u\), or 0 and \(\pi\) for \(x\). |
| Obtain \(\frac{4}{3}(3\sqrt{3}-1)\) or \(4\sqrt{3} - \frac{4}{3}\) or \(\frac{4}{3}\sqrt{27} - \frac{4}{3}\) | A1 | OE. Allow e.g. \(\sqrt{3}^3\) for \(\sqrt{27}\). ISW but don't ignore e.g. multiplying throughout by 3. If answer changed from negative to positive value at end, then A0. Last M1A1 can use modulus, providing no errors seen. |
| Total | 6 |
## Question 11(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use of correct product rule and correct chain rule | M1 | $\frac{dy}{dx} = A\cos x\sqrt{2+\cos x} + \frac{B\sin x\sin x}{\sqrt{2+\cos x}}$ |
| Obtain $\frac{dy}{dx} = 2\cos x\sqrt{2+\cos x} - \frac{2\sin^2 x}{2\sqrt{2+\cos x}}$ | A1 | OE |
| Equate derivative to zero and obtain a horizontal 3-term quadratic equation in $\cos a$ or 4-term quartic equation in $\cos x$. The only error in the form of the differential allowed is for $(2+\cos x)^{-\frac{1}{2}}$ to be $(2+\cos x)^{\frac{1}{2}}$ or $(2+\cos x)^{-\frac{3}{2}}$ | *M1 | Accept in $\cos x$. E.g. $3\cos^2 x + 4\cos x - 1 = 0$. E.g. $3\cos^4 x + 16\cos^3 x + 18\cos^2 x - 1 = 0$ |
| Solve for $\cos a$ | DM1 | $\cos a = \frac{-2+\sqrt{7}}{3}$ or $0.215$. Allow presence of other solution(s). |
| Obtain $a = 4.93$ | A1 | Allow more accurate e.g. $4.929\ldots$ even though question states 2 d.p. If $x=1.35$ leads to $x=4.93$ award A1 BOD. If $x=1.35$ and $x=4.93$ award A0. |
| **Total** | **5** | |
---
## Question 11(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $du = -\sin x\,dx$ | B1 | OE. If B0, max M1M1M1. |
| Substitute throughout for $u$ and $du$ | M1 | |
| Obtain $-\int 2\sqrt{u}\,du$ | A1 | OE. Ignore limits if $-\int 2\sqrt{u}\,du$, but if $+\int 2\sqrt{u}\,du$, then must have correct limits $\int_1^3 2\sqrt{u}\,du$. (See final M1) |
| Integrate to obtain $ku^{\frac{3}{2}}\ (+C)$ | M1 | Constant of integration not required |
| Use correct limits correctly in an expression of the form $ku^{\frac{3}{2}}$ or $k(2+\cos x)^{\frac{3}{2}}$ | M1 | 1 and 3 for $u$, or 0 and $\pi$ for $x$. |
| Obtain $\frac{4}{3}(3\sqrt{3}-1)$ or $4\sqrt{3} - \frac{4}{3}$ or $\frac{4}{3}\sqrt{27} - \frac{4}{3}$ | A1 | OE. Allow e.g. $\sqrt{3}^3$ for $\sqrt{27}$. ISW but don't ignore e.g. multiplying throughout by 3. If answer changed from negative to positive value at end, then A0. Last M1A1 can use modulus, providing no errors seen. |
| **Total** | **6** | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-18_565_634_260_717}
The diagram shows the curve $y = 2 \sin x \sqrt { 2 + \cos x }$, for $0 \leqslant x \leqslant 2 \pi$, and its minimum point $M$, where $x = a$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ correct to 2 decimal places.\\
\includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-18_2716_38_109_2012}\\
\includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-19_2726_33_97_22}
\item Use the substitution $u = 2 + \cos x$ to find the exact area of the shaded region $R$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q11 [11]}}