| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Geometric curve properties |
| Difficulty | Challenging +1.2 This is a multi-step differential equations problem requiring geometric interpretation, forming a DE from a geometric condition, and solving by separation of variables. Part (a) requires understanding normal gradients and triangle area, while part (b) involves standard integration techniques (substitution for tan x). The geometric setup is moderately sophisticated but the calculus is routine for P3 level, making it slightly above average difficulty. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Justify the given statement \(\frac{MN}{y} = \frac{dy}{dx}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Express the area of \(PMN\) in terms of \(y\) and \(\frac{dy}{dx}\) and equate to \(\tan x\) | M1 | |
| Obtain the given equation correctly | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Separate variables and integrate at least one side | M1 | |
| Obtain term \(\frac{1}{6}y^3\) | A1 | |
| Obtain term of the form \(\pm\ln\cos x\) | M1 | |
| Evaluate a constant or use \(x = 0\) and \(y = 1\) in a solution containing terms \(ay^3\) and \(\pm\ln\cos x\), or equivalent | M1 | |
| Obtain correct answer in any form, e.g. \(\frac{1}{6}y^3 = -\ln\cos x + \frac{1}{6}\) | A1 | |
| Obtain final answer \(y = \sqrt[3]{(1 - 6\ln\cos x)}\) | A1 | OE |
## Question 7(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Justify the given statement $\frac{MN}{y} = \frac{dy}{dx}$ | B1 | |
## Question 7(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Express the area of $PMN$ in terms of $y$ and $\frac{dy}{dx}$ and equate to $\tan x$ | M1 | |
| Obtain the given equation correctly | A1 | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Separate variables and integrate at least one side | M1 | |
| Obtain term $\frac{1}{6}y^3$ | A1 | |
| Obtain term of the form $\pm\ln\cos x$ | M1 | |
| Evaluate a constant or use $x = 0$ and $y = 1$ in a solution containing terms $ay^3$ and $\pm\ln\cos x$, or equivalent | M1 | |
| Obtain correct answer in any form, e.g. $\frac{1}{6}y^3 = -\ln\cos x + \frac{1}{6}$ | A1 | |
| Obtain final answer $y = \sqrt[3]{(1 - 6\ln\cos x)}$ | A1 | OE |7\\
\includegraphics[max width=\textwidth, alt={}, center]{1990cbac-d96f-4484-be4b-67dab35b3147-10_647_519_260_813}
For the curve shown in the diagram, the normal to the curve at the point $P$ with coordinates $( x , y )$ meets the $x$-axis at $N$. The point $M$ is the foot of the perpendicular from $P$ to the $x$-axis.
The curve is such that for all values of $x$ in the interval $0 \leqslant x < \frac { 1 } { 2 } \pi$, the area of triangle $P M N$ is equal to $\tan x$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\frac { M N } { y } = \frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence show that $x$ and $y$ satisfy the differential equation $\frac { 1 } { 2 } y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \tan x$.
\end{enumerate}\item Given that $y = 1$ when $x = 0$, solve this differential equation to find the equation of the curve, expressing $y$ in terms of $x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q7 [9]}}