| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - direct evaluation |
| Difficulty | Moderate -0.8 This is a straightforward application of the product rule to differentiate 3xe^(-2x), followed by direct substitution of x = -1/2. The integration part requires integration by parts but is also routine. Both parts are standard textbook exercises requiring only technique application with no problem-solving or insight needed, making this easier than average. |
| Spec | 1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either: Use correct product rule | M1 | |
| Obtain \(3e^{-2x} - 6xe^{-2x}\) or equivalent | A1 | |
| Substitute \(-\frac{1}{2}\) and obtain \(6e\) | A1 | |
| Or: Take ln of both sides and use implicit differentiation correctly | M1 | |
| Obtain \(\frac{dy}{dx} = y\!\left(\frac{1}{x}-2\right)\) or equivalent | A1 | |
| Substitute \(-\frac{1}{2}\) and obtain \(6e\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use integration by parts to reach \(kxe^{-2x} \pm \int ke^{-2x}\,dx\) | M1 | |
| Obtain \(-\frac{3}{2}xe^{-2x} + \int\frac{3}{2}e^{-2x}\,dx\) or equivalent | A1 | |
| Obtain \(-\frac{3}{2}xe^{-2x} - \frac{3}{4}e^{-2x}\) or equivalent | A1 | |
| Substitute correct limits correctly | DM1 | |
| Obtain \(-\frac{3}{4}\) with no errors or inexact work seen | A1 | [5] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Either:** Use correct product rule | M1 | |
| Obtain $3e^{-2x} - 6xe^{-2x}$ or equivalent | A1 | |
| Substitute $-\frac{1}{2}$ and obtain $6e$ | A1 | |
| **Or:** Take ln of both sides and use implicit differentiation correctly | M1 | |
| Obtain $\frac{dy}{dx} = y\!\left(\frac{1}{x}-2\right)$ or equivalent | A1 | |
| Substitute $-\frac{1}{2}$ and obtain $6e$ | A1 | [3] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use integration by parts to reach $kxe^{-2x} \pm \int ke^{-2x}\,dx$ | M1 | |
| Obtain $-\frac{3}{2}xe^{-2x} + \int\frac{3}{2}e^{-2x}\,dx$ or equivalent | A1 | |
| Obtain $-\frac{3}{2}xe^{-2x} - \frac{3}{4}e^{-2x}$ or equivalent | A1 | |
| Substitute correct limits correctly | DM1 | |
| Obtain $-\frac{3}{4}$ with no errors or inexact work seen | A1 | [5] |
---5 The expression $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 3 x \mathrm { e } ^ { - 2 x }$.\\
(i) Find the exact value of $\mathrm { f } ^ { \prime } \left( - \frac { 1 } { 2 } \right)$.\\
(ii) Find the exact value of $\int _ { - \frac { 1 } { 2 } } ^ { 0 } \mathrm { f } ( x ) \mathrm { d } x$.
\hfill \mbox{\textit{CAIE P3 2012 Q5 [8]}}