| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable variables |
| Difficulty | Challenging +1.2 This is a structured differential equation problem with clear guidance. Part (a) is routine differentiation using chain rule. Part (b) requires recognizing the substitution y=sec³θ from part (a), separating variables, and integrating—standard techniques for A-level. The substitution is heavily signposted, making this above-average difficulty but not requiring significant insight. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07l Derivative of ln(x): and related functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use of correct chain rule (and correct quotient rule) and \(\cos^{-3}\theta\) | M1 | Obtain \(k \times (\cos\theta)^{-4} \times \sin\theta\) or equivalent |
| \(\dfrac{\text{d}y}{\text{d}\theta} = -3 \times -\sin\theta(\cos\theta)^{-4} = 3\sin\theta\sec^4\theta\). Must be expressed in the given form | A1 | Obtain given answer from full and correct working (signs must be shown), but condone \(\frac{\text{d}}{\text{d}\theta}\left(\sec^3\theta\right) = \ldots\) and \(y'(\theta)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separate variables: \(\displaystyle\int \frac{\sin\theta}{\cos^4\theta}\,\text{d}\theta = \int \frac{(x+3)}{(x^2+9)}\,\text{d}x\) | B1 | Or \(\displaystyle\int\frac{3\sin\theta}{\cos^4\theta}\,\text{d}\theta = \int\frac{3x+9}{x^2+9}\,\text{d}x\). Condone missing integral signs or missing \(\text{d}x\) or \(\text{d}\theta\), but not both |
| Obtain \(p\sec^3\theta\,(+A)\) | B1 | Correct form, \(p\) any constant but not 0 |
| Use \(\displaystyle\int\frac{3x+9}{x^2+9}\,\text{d}x = \int\!\left(\frac{3x}{x^2+9}+\frac{9}{x^2+9}\right)\text{d}x\) and obtain \(q\ln(x^2+9)\) or \(r\tan^{-1}\frac{x}{3}\,(+C)\) | *M1 | Might have one third of both sides. Alt: substitute \(x = 3\tan\phi\) to obtain \(q\int 1+\tan\phi\,\text{d}\phi\); condone if have \(\theta\) in place of \(\phi\) |
| Obtain \(q\ln(x^2+9)\) and \(r\tan^{-1}\frac{x}{3}\,(+C)\) | DM1 | Obtain \(q\left(\phi \mp \ln(\cos\phi)\right)\) OE |
| Obtain \(\sec^3\theta = \frac{3}{2}\ln(x^2+9) + 3\tan^{-1}\frac{x}{3}\,(+C)\) or equivalent | A1 | Or might see a third of both sides. Must have 2 different variables |
| Use \(\theta = \frac{1}{3}\pi,\, x = 3\) in an equation including \(p\sec^3\theta\), \(q\ln(x^2+9)\) and \(r\tan^{-1}\frac{x}{3}\) to evaluate the constant of integration | M1 | Or as limits in a definite integral. Limits for \(\phi\) are 0 and \(\frac{1}{4}\pi\) |
| Obtain constant \(= 8 - \frac{3}{2}\ln 18 - \frac{3}{4}\pi\) | A1 | OE, e.g. \(1.308\ldots\) to at least 3sf |
| Obtain \(\cos\theta = 0.601\) | A1 | Accept AWRT \(0.601\) |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use of correct chain rule (and correct quotient rule) and $\cos^{-3}\theta$ | M1 | Obtain $k \times (\cos\theta)^{-4} \times \sin\theta$ or equivalent |
| $\dfrac{\text{d}y}{\text{d}\theta} = -3 \times -\sin\theta(\cos\theta)^{-4} = 3\sin\theta\sec^4\theta$. Must be expressed in the given form | A1 | Obtain given answer from full and correct working (signs must be shown), but condone $\frac{\text{d}}{\text{d}\theta}\left(\sec^3\theta\right) = \ldots$ and $y'(\theta)$ |
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## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables: $\displaystyle\int \frac{\sin\theta}{\cos^4\theta}\,\text{d}\theta = \int \frac{(x+3)}{(x^2+9)}\,\text{d}x$ | B1 | Or $\displaystyle\int\frac{3\sin\theta}{\cos^4\theta}\,\text{d}\theta = \int\frac{3x+9}{x^2+9}\,\text{d}x$. Condone missing integral signs or missing $\text{d}x$ or $\text{d}\theta$, but not both |
| Obtain $p\sec^3\theta\,(+A)$ | B1 | Correct form, $p$ any constant but not 0 |
| Use $\displaystyle\int\frac{3x+9}{x^2+9}\,\text{d}x = \int\!\left(\frac{3x}{x^2+9}+\frac{9}{x^2+9}\right)\text{d}x$ and obtain $q\ln(x^2+9)$ or $r\tan^{-1}\frac{x}{3}\,(+C)$ | *M1 | Might have one third of both sides. Alt: substitute $x = 3\tan\phi$ to obtain $q\int 1+\tan\phi\,\text{d}\phi$; condone if have $\theta$ in place of $\phi$ |
| Obtain $q\ln(x^2+9)$ and $r\tan^{-1}\frac{x}{3}\,(+C)$ | DM1 | Obtain $q\left(\phi \mp \ln(\cos\phi)\right)$ OE |
| Obtain $\sec^3\theta = \frac{3}{2}\ln(x^2+9) + 3\tan^{-1}\frac{x}{3}\,(+C)$ or equivalent | A1 | Or might see a third of both sides. Must have 2 different variables |
| Use $\theta = \frac{1}{3}\pi,\, x = 3$ in an equation including $p\sec^3\theta$, $q\ln(x^2+9)$ and $r\tan^{-1}\frac{x}{3}$ to evaluate the constant of integration | M1 | Or as limits in a definite integral. Limits for $\phi$ are 0 and $\frac{1}{4}\pi$ |
| Obtain constant $= 8 - \frac{3}{2}\ln 18 - \frac{3}{4}\pi$ | A1 | OE, e.g. $1.308\ldots$ to at least 3sf |
| Obtain $\cos\theta = 0.601$ | A1 | Accept AWRT $0.601$ |10
\begin{enumerate}[label=(\alph*)]
\item By writing $y = \sec ^ { 3 } \theta$ as $\frac { 1 } { \cos ^ { 3 } \theta }$, show that $\frac { \mathrm { d } y } { \mathrm {~d} \theta } = 3 \sin \theta \sec ^ { 4 } \theta$.
\item The variables $x$ and $\theta$ satisfy the differential equation
$$\left( x ^ { 2 } + 9 \right) \sin \theta \frac { d \theta } { d x } = ( x + 3 ) \cos ^ { 4 } \theta$$
It is given that $x = 3$ when $\theta = \frac { 1 } { 3 } \pi$.\\
Solve the differential equation to find the value of $\cos \theta$ when $x = 0$. Give your answer correct to 3 significant figures.\\
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 Q10 [10]}}