| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find gradient at given parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = (dy/dθ)/(dx/dθ)) and substitution of given θ values. Part (i) involves routine differentiation of sin and tan functions with evaluation at a standard angle. Part (ii) requires solving a trigonometric equation, which is slightly more involved but still standard A-level technique. The question is slightly easier than average as it follows a predictable template with no conceptual surprises. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(\dfrac{dx}{d\theta} = 6\cos 2\theta\) | B1 | |
| Obtain \(\dfrac{dy}{d\theta} = 4\sec^2 2\theta\) | B1 | |
| Divide \(\dfrac{dy}{d\theta}\) by \(\dfrac{dx}{d\theta}\) with \(\theta\) equated to \(\frac{1}{6}\pi\) | M1 | |
| Obtain \(\frac{16}{3}\) or exact equivalent | A1 | Allow FT on A1 if \(\dfrac{dx}{d\theta} = 3\cos 2\theta\) and \(\dfrac{dy}{d\theta} = 2\sec^2 2\theta\) |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate expression for \(\dfrac{dy}{dx}\) to 2 with only one trigonometry ratio used | \*M1 | Either \(\cos 2\theta\) or \(\sec 2\theta\) |
| Obtain \(\cos^3 2\theta = \frac{1}{3}\) or \(\sec^3 2\theta = 3\) | A1 | |
| Attempt correct steps to find a value of \(\theta\) from \(\cos^3 2\theta = m\), \(0 < m < 1\) | DM1 | |
| Obtain \(\theta = 0.402\) and no others within the range | A1 | AWRT. SC: Allow FT if \(\dfrac{dx}{d\theta} = 3\cos 2\theta\) and \(\dfrac{dy}{d\theta} = 2\sec^2 2\theta\) |
| Total: 4 |
## Question 7:
### Part 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $\dfrac{dx}{d\theta} = 6\cos 2\theta$ | B1 | |
| Obtain $\dfrac{dy}{d\theta} = 4\sec^2 2\theta$ | B1 | |
| Divide $\dfrac{dy}{d\theta}$ by $\dfrac{dx}{d\theta}$ with $\theta$ equated to $\frac{1}{6}\pi$ | M1 | |
| Obtain $\frac{16}{3}$ or exact equivalent | A1 | Allow FT on A1 if $\dfrac{dx}{d\theta} = 3\cos 2\theta$ and $\dfrac{dy}{d\theta} = 2\sec^2 2\theta$ |
| **Total: 4** | | |
### Part 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate expression for $\dfrac{dy}{dx}$ to 2 with only one trigonometry ratio used | \*M1 | Either $\cos 2\theta$ or $\sec 2\theta$ |
| Obtain $\cos^3 2\theta = \frac{1}{3}$ or $\sec^3 2\theta = 3$ | A1 | |
| Attempt correct steps to find a value of $\theta$ from $\cos^3 2\theta = m$, $0 < m < 1$ | DM1 | |
| Obtain $\theta = 0.402$ and no others within the range | A1 | AWRT. SC: Allow FT if $\dfrac{dx}{d\theta} = 3\cos 2\theta$ and $\dfrac{dy}{d\theta} = 2\sec^2 2\theta$ |
| **Total: 4** | | |
---7 The parametric equations of a curve are
$$x = 3 \sin 2 \theta , \quad y = 1 + 2 \tan 2 \theta$$
for $0 \leqslant \theta < \frac { 1 } { 4 } \pi$.\\
(i) Find the exact gradient of the curve at the point for which $\theta = \frac { 1 } { 6 } \pi$.\\
(ii) Find the value of $\theta$ at the point where the gradient of the curve is 2 , giving the value correct to 3 significant figures.\\
\hfill \mbox{\textit{CAIE P2 2019 Q7 [8]}}