| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find gradient at point |
| Difficulty | Moderate -0.8 This is a straightforward application of product rule (part i) and quotient rule (part ii) followed by direct substitution. Both are standard textbook exercises with no problem-solving required—students simply apply memorized differentiation rules and evaluate at the given point. Easier than average A-level questions. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Differentiate \(\ln(x-3)\) to obtain \(\frac{1}{x-3}\) | B1 | |
| Attempt to use product rule | M1 | |
| Obtain \(\ln(x-3) + \frac{x}{x-3}\) or equivalent | A1 | |
| Substitute \(4\) to obtain \(4\) | A1 | [4] |
| (ii) Use correct quotient or product rule | M1 | |
| Obtain correct derivative in any form, e.g. \(\frac{(x+1)-(x-1)}{(x+1)^2}\) | A1 | |
| Substitute \(4\) to obtain \(\frac{2}{25}\) | A1 | [3] |
**(i)** Differentiate $\ln(x-3)$ to obtain $\frac{1}{x-3}$ | B1 |
Attempt to use product rule | M1 |
Obtain $\ln(x-3) + \frac{x}{x-3}$ or equivalent | A1 |
Substitute $4$ to obtain $4$ | A1 | [4]
**(ii)** Use correct quotient or product rule | M1 |
Obtain correct derivative in any form, e.g. $\frac{(x+1)-(x-1)}{(x+1)^2}$ | A1 |
Substitute $4$ to obtain $\frac{2}{25}$ | A1 | [3]5 Find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when $x = 4$ in each of the following cases:\\
(i) $y = x \ln ( x - 3 )$,\\
(ii) $y = \frac { x - 1 } { x + 1 }$.
\hfill \mbox{\textit{CAIE P2 2011 Q5 [7]}}