| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2018 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Show gradient expression then find coordinates |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding dy/dx using the chain rule, determining the parameter value at the origin, finding the tangent equation, and locating stationary points where dy/dx = 0. While it involves multiple steps and requires careful algebraic manipulation (especially with the product rule for y), all techniques are routine for P2 level with no novel problem-solving insight needed. Slightly above average difficulty due to the multi-step nature and need for numerical solving, but well within standard parametric questions. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks |
|---|---|
| 5(i) | Use product rule to differentiate y obtaining ke2t +k te2t |
| 1 2 | M1 |
| Obtain correct 3e2t +6te2t | A1 |
| Answer | Marks |
|---|---|
| t+1 | B1 |
| Answer | Marks |
|---|---|
| dx dt dt | M1 |
| Answer | Marks |
|---|---|
| 2 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 5(ii) | dy dy |
| Answer | Marks | Guidance |
|---|---|---|
| dx dt | M1 | dx |
| Answer | Marks |
|---|---|
| 2 | A1 |
| Obtain x=−1.19 | A1 |
| Obtain y=−0.55 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | Use product rule to differentiate y obtaining ke2t +k te2t
1 2 | M1
Obtain correct 3e2t +6te2t | A1
1
State derivative of x is 1+
t+1 | B1
dy dy dx
Use = / with t =0 to find gradient
dx dt dt | M1
Obtain y= 3x or equivalent
2 | A1
5
Question | Answer | Marks | Guidance
--- 5(ii) ---
5(ii) | dy dy
Equate or to zero and solve for t
dx dt | M1 | dx
Allow full marks if correct solution is obtained but
dt
is incorrect
Obtain t =−1
2 | A1
Obtain x=−1.19 | A1
Obtain y=−0.55 | A1
4
Question | Answer | Marks | GuidanceA curve has parametric equations
$$x = t + \ln(t + 1), \quad y = 3te^{2t}.$$
\begin{enumerate}[label=(\roman*)]
\item Find the equation of the tangent to the curve at the origin. [5]
\item Find the coordinates of the stationary point, giving each coordinate correct to 2 decimal places. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2018 Q5 [9]}}