| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Standard +0.8 This question requires applying the product rule to differentiate e^(-2x)tan(x), then algebraically manipulating the result into the specific form (a + b tan x)^2, which demands recognizing a perfect square pattern and working with trigonometric identities (sec^2 x = 1 + tan^2 x). Parts (ii) and (iii) add conceptual depth by requiring interpretation of the derivative form and optimization. While the individual techniques are standard A-level, the algebraic manipulation to show the required form and the multi-part reasoning elevate this above routine exercises. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks |
|---|---|
| (i) State or imply that the derivative of \(e^{-2x}\) is \(-2e^{-2x}\) | B1 |
| Use product or quotient rule | M1 |
| Obtain correct derivative in any form | A1 |
| Use Pythagoras | M1 |
| Justify the given form | A1 |
| [5] | |
| (ii) Fully justify the given statement | B1 |
| [1] | |
| (iii) State answer \(x = \frac{1}{4}\pi\) | B1 |
| [1] |
**(i)** State or imply that the derivative of $e^{-2x}$ is $-2e^{-2x}$ | B1 |
Use product or quotient rule | M1 |
Obtain correct derivative in any form | A1 |
Use Pythagoras | M1 |
Justify the given form | A1 |
| [5] |
**(ii)** Fully justify the given statement | B1 |
| [1] |
**(iii)** State answer $x = \frac{1}{4}\pi$ | B1 |
| [1] |5 The equation of a curve is $y = \mathrm { e } ^ { - 2 x } \tan x$, for $0 \leqslant x < \frac { 1 } { 2 } \pi$.\\
(i) Obtain an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and show that it can be written in the form $\mathrm { e } ^ { - 2 x } ( a + b \tan x ) ^ { 2 }$, where $a$ and $b$ are constants.\\
(ii) Explain why the gradient of the curve is never negative.\\
(iii) Find the value of $x$ for which the gradient is least.
\hfill \mbox{\textit{CAIE P3 2015 Q5 [7]}}