| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a straightforward application of the quotient rule to find dy/dx, setting it equal to zero to find the stationary point, then using the second derivative test. While it involves exponential functions and requires careful algebraic manipulation, it follows a standard procedure with no conceptual surprises, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use correct quotient or product rule | M1 | |
| Obtain correct derivative in any form, e.g. \(\frac{2e^{2x}}{x^3} - \frac{3e^{2x}}{x^4}\) | A1 | |
| Equate derivative to zero and solve a 2-term equation for non-zero \(x\) | M1 | |
| Obtain \(x = \frac{3}{2}\) correctly | A1 | [4] |
| (ii) Carry out a method for determining the nature of a stationary point, e.g. test derivative either side | M1 | |
| Show point is a minimum with no errors seen | A1 | [2] |
**(i)** Use correct quotient or product rule | M1 |
Obtain correct derivative in any form, e.g. $\frac{2e^{2x}}{x^3} - \frac{3e^{2x}}{x^4}$ | A1 |
Equate derivative to zero and solve a 2-term equation for non-zero $x$ | M1 |
Obtain $x = \frac{3}{2}$ correctly | A1 | [4]
**(ii)** Carry out a method for determining the nature of a stationary point, e.g. test derivative either side | M1 |
Show point is a minimum with no errors seen | A1 | [2]
---4 The curve with equation $y = \frac { \mathrm { e } ^ { 2 x } } { x ^ { 3 } }$ has one stationary point.\\
(i) Find the $x$-coordinate of this point.\\
(ii) Determine whether this point is a maximum or a minimum point.
\hfill \mbox{\textit{CAIE P3 2012 Q4 [6]}}