| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - given gradient condition |
| Difficulty | Standard +0.3 This is a straightforward chain rule differentiation of a logarithmic function followed by algebraic simplification using trigonometric identities, then solving a simple equation. The chain rule application is standard, the simplification to tan x uses basic trig identities (sin²x + cos²x = 1, tan x = sin x/cos x), and part (ii) is just solving a linear equation in tan x then using arctan. Slightly above routine due to the multi-step chain rule and trig manipulation, but still a standard textbook exercise with no novel insight required. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the chain rule | M1 | |
| Obtain correct derivative in any form | A1 | |
| Use correct trigonometry to express derivative in terms of \(\tan x\) | M1 | |
| Obtain \(\dfrac{dy}{dx} = -\dfrac{4\tan x}{4 + \tan^2 x}\), or equivalent | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate derivative to \(-1\) and solve a 3-term quadratic for \(\tan x\) | M1 | |
| Obtain answer \(x = 1.11\) and no other in the given interval | A1 | |
| Total: 2 |
## Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the chain rule | M1 | |
| Obtain correct derivative in any form | A1 | |
| Use correct trigonometry to express derivative in terms of $\tan x$ | M1 | |
| Obtain $\dfrac{dy}{dx} = -\dfrac{4\tan x}{4 + \tan^2 x}$, or equivalent | A1 | |
| **Total: 4** | | |
## Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate derivative to $-1$ and solve a 3-term quadratic for $\tan x$ | M1 | |
| Obtain answer $x = 1.11$ and no other in the given interval | A1 | |
| **Total: 2** | | |
---5 A curve has equation $y = \frac { 2 } { 3 } \ln \left( 1 + 3 \cos ^ { 2 } x \right)$ for $0 \leqslant x \leqslant \frac { 1 } { 2 } \pi$.\\
(i) Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\tan x$.\\
(ii) Hence find the $x$-coordinate of the point on the curve where the gradient is - 1 . Give your answer correct to 3 significant figures.\\
\hfill \mbox{\textit{CAIE P3 2017 Q5 [6]}}