| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - direct evaluation |
| Difficulty | Easy -1.2 This is a straightforward differentiation exercise requiring direct application of standard rules (chain rule for tan 2x and the quotient/chain rule for the exponential function) followed by simple substitution of x=0. Both parts are routine calculations with no problem-solving element, making this easier than average but not trivial since it involves transcendental functions rather than just polynomials. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Differentiate to obtain form \(k_1 \cos x + k_2 \sec^2 2x\) | M1 | |
| Obtain correct second term \(2\sec^2 2x\) | A1 | |
| Obtain \(3\cos x + 2\sec^2 2x\) and hence answer \(5\) | A1 | [3] |
| (ii) Differentiate to obtain form \(e^{2x}\left(1 + e^{2x}\right)^2\) | M1 | |
| Obtain correct \(-12e^{2x}\left(1 + e^{2x}\right)^2\) or equivalent (may be implied) | A1 | |
| Obtain \(-3\) | A1 | [3] |
**(i)** Differentiate to obtain form $k_1 \cos x + k_2 \sec^2 2x$ | M1 |
Obtain correct second term $2\sec^2 2x$ | A1 |
Obtain $3\cos x + 2\sec^2 2x$ and hence answer $5$ | A1 | [3]
**(ii)** Differentiate to obtain form $e^{2x}\left(1 + e^{2x}\right)^2$ | M1 |
Obtain correct $-12e^{2x}\left(1 + e^{2x}\right)^2$ or equivalent (may be implied) | A1 |
Obtain $-3$ | A1 | [3]2 Find the gradient of each of the following curves at the point for which $x = 0$.\\
(i) $y = 3 \sin x + \tan 2 x$\\
(ii) $y = \frac { 6 } { 1 + \mathrm { e } ^ { 2 x } }$
\hfill \mbox{\textit{CAIE P2 2014 Q2 [6]}}