| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2016 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find equation of tangent |
| Difficulty | Standard +0.3 This is a straightforward application of the product rule to find dy/dx for an exponential-trigonometric product, then evaluating at x=0 for the tangent (which simplifies nicely), and solving dy/dx=0 for the stationary point using standard techniques. While it requires multiple steps, each is routine for A-level—slightly easier than average due to the clean algebra at the origin. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use product rule to obtain expression of form \(k_1e^{-x}\sin 2x + k_2e^{-x}\cos 2x\) | M1 | |
| Obtain correct \(-3e^{-x}\sin 2x + 6e^{-x}\cos 2x\) | A1 | |
| Substitute \(x=0\) in first derivative to obtain equation of form \(y = mx\) | M1 | |
| Obtain \(y = 6x\) or equivalent with no errors in solution | A1 | [4] |
| (ii) Equate first derivative to zero and obtain \(\tan 2x = k\) | M1* | |
| Carry out correct process to find value of \(x\) | dep M1* | |
| Obtain \(x = 0.554\) | A1 | |
| Obtain \(y = 1.543\) | A1 | [4] |
(i) Use product rule to obtain expression of form $k_1e^{-x}\sin 2x + k_2e^{-x}\cos 2x$ | M1 |
Obtain correct $-3e^{-x}\sin 2x + 6e^{-x}\cos 2x$ | A1 |
Substitute $x=0$ in first derivative to obtain equation of form $y = mx$ | M1 |
Obtain $y = 6x$ or equivalent with no errors in solution | A1 | [4]
(ii) Equate first derivative to zero and obtain $\tan 2x = k$ | M1* |
Carry out correct process to find value of $x$ | dep M1* |
Obtain $x = 0.554$ | A1 |
Obtain $y = 1.543$ | A1 | [4]6\\
\includegraphics[max width=\textwidth, alt={}, center]{d53a2d6b-4c5e-4bc6-8aa1-587e97c87920-2_371_839_1409_651}
The diagram shows the part of the curve $y = 3 \mathrm { e } ^ { - x } \sin 2 x$ for $0 \leqslant x \leqslant \frac { 1 } { 2 } \pi$, and the stationary point $M$.\\
(i) Find the equation of the tangent to the curve at the origin.\\
(ii) Find the coordinates of $M$, giving each coordinate correct to 3 decimal places.
\hfill \mbox{\textit{CAIE P2 2016 Q6 [8]}}