| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Differentiation of reciprocal functions |
| Difficulty | Standard +0.8 This question requires product rule differentiation with trigonometric functions, manipulation using double angle formulas and reciprocal trig identities to reach the given form, then solving a transcendental equation numerically. The algebraic manipulation to 'show that' the derivative takes the given form is non-trivial, requiring insight to express everything in terms of cos x, and the numerical solution adds another layer. This is above average difficulty but within reach of a well-prepared P2 student. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks |
|---|---|
| (i) Differentiate using product rule | M1 |
| Obtain \(\sec^2 x\cos 2x - 2\tan x\sin 2x\) | A1 |
| Use \(\cos 2x = 2\cos^2 x - 1\) or \(\sin 2x = 2\sin x\cos x\) or both | B1 |
| Express derivative in terms of \(\sec x\) and \(\cos x\) only | M1 |
| Obtain \(4\cos^3 x - \sec^2 x - 2\) with no errors seen (AG) | A1 [5] |
| (ii) State \(4\cos^4 x - 2\cos^2 x - 1 = 0\) | B1 |
| Apply quadratic formula to a 3 term quadratic equation in terms of \(\cos^2 x\) to find the least positive value of \(\cos^2 x\) | M1 |
| Obtain or imply \(\cos^2 x = \frac{1 + \sqrt{5}}{4}\) or \(0.809...\) | A1 |
| Obtain \(0.45\) | A1 [4] |
(i) Differentiate using product rule | M1 |
Obtain $\sec^2 x\cos 2x - 2\tan x\sin 2x$ | A1 |
Use $\cos 2x = 2\cos^2 x - 1$ or $\sin 2x = 2\sin x\cos x$ or both | B1 |
Express derivative in terms of $\sec x$ and $\cos x$ only | M1 |
Obtain $4\cos^3 x - \sec^2 x - 2$ with no errors seen (AG) | A1 [5] |
(ii) State $4\cos^4 x - 2\cos^2 x - 1 = 0$ | B1 |
Apply quadratic formula to a 3 term quadratic equation in terms of $\cos^2 x$ to find the least positive value of $\cos^2 x$ | M1 |
Obtain or imply $\cos^2 x = \frac{1 + \sqrt{5}}{4}$ or $0.809...$ | A1 |
Obtain $0.45$ | A1 [4] |8\\
\includegraphics[max width=\textwidth, alt={}, center]{de8af872-9f77-4787-8e66-ed199405ca25-3_581_650_1272_744}
The diagram shows the curve
$$y = \tan x \cos 2 x , \text { for } 0 \leqslant x < \frac { 1 } { 2 } \pi$$
and its maximum point $M$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 \cos ^ { 2 } x - \sec ^ { 2 } x - 2$.\\
(ii) Hence find the $x$-coordinate of $M$, giving your answer correct to 2 decimal places.
\hfill \mbox{\textit{CAIE P2 2014 Q8 [9]}}