| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points coordinates |
| Difficulty | Moderate -0.3 This is a straightforward application of product and quotient rules to find stationary points. Both parts require standard differentiation techniques (product rule for (i), quotient rule for (ii)) followed by solving dy/dx = 0. The algebra is routine and the question structure is typical textbook fare, making it slightly easier than average but not trivial since it requires correct application of two different differentiation rules. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use product rule to obtain form \(k_1 e^{3x} + k_2 xe^{-3x}\) | M1 | |
| Obtain correct \(4e^{-3x} - 12xe^{-3x}\) | A1 | |
| Obtain \(x = \frac{1}{3}\) or \(0.333\) or better and no other | A1 | [3] |
| (ii) Use quotient rule or equivalent | M1* | |
| Obtain correct numerator \(8x(x+1) - 4x^2\) or equivalent | A1 | |
| Equate numerator to zero and solve to find at least one value | M1 dep | |
| Obtain \(x = -2\) | A1 | |
| Obtain \(x = 0\) | A1 | [5] |
(i) Use product rule to obtain form $k_1 e^{3x} + k_2 xe^{-3x}$ | M1 |
Obtain correct $4e^{-3x} - 12xe^{-3x}$ | A1 |
Obtain $x = \frac{1}{3}$ or $0.333$ or better and no other | A1 | [3]
(ii) Use quotient rule or equivalent | M1* |
Obtain correct numerator $8x(x+1) - 4x^2$ or equivalent | A1 |
Equate numerator to zero and solve to find at least one value | M1 dep |
Obtain $x = -2$ | A1 |
Obtain $x = 0$ | A1 | [5]
---5 Find the $x$-coordinates of the stationary points of the following curves:\\
(i) $y = 4 x \mathrm { e } ^ { - 3 x }$;\\
(ii) $y = \frac { 4 x ^ { 2 } } { x + 1 }$.
\hfill \mbox{\textit{CAIE P2 2015 Q5 [8]}}