| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - at special curve features |
| Difficulty | Moderate -0.3 This is a straightforward differentiation question requiring chain rule for (ln x)² and standard derivative of ln x, followed by substitution. Finding where y=0 involves simple factorization of ln x. The minimum requires setting dy/dx=0. All steps are routine A-level techniques with no novel problem-solving, making it slightly easier than average. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(x = 1\) | B1 | Allow \(e^0\); must come from correct work, e.g. \(\ln x = 0\) |
| Obtain \(x = e^2\) | B1 | |
| Differentiate to obtain at least one correct term | *M1 | |
| Obtain correct first derivative \(\frac{2\ln x}{x} - \frac{2}{x}\) | A1 | Allow \(\frac{\ln x}{x} + \frac{\ln x}{x} - \frac{2}{x}\); Allow \(-\frac{2x}{x^2}\) |
| Substitute at least one of *their* \(x\)-values corresponding to \(y = 0\) to find gradient | DM1 | Allow unsimplified |
| Obtain gradient \(-2\) [at \(A\)] and gradient \(2e^{-2}\) at [\(B\)] | A1 | Must be simplified |
| Total | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate first derivative to zero | M1 | *Their* derivative must have at least 2 terms |
| Obtain \(x = e\) | A1 | Allow \(e^1\) |
| Total | 2 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $x = 1$ | B1 | Allow $e^0$; must come from correct work, e.g. $\ln x = 0$ |
| Obtain $x = e^2$ | B1 | |
| Differentiate to obtain at least one correct term | *M1 | |
| Obtain correct first derivative $\frac{2\ln x}{x} - \frac{2}{x}$ | A1 | Allow $\frac{\ln x}{x} + \frac{\ln x}{x} - \frac{2}{x}$; Allow $-\frac{2x}{x^2}$ |
| Substitute at least one of *their* $x$-values corresponding to $y = 0$ to find gradient | DM1 | Allow unsimplified |
| Obtain gradient $-2$ [at $A$] and gradient $2e^{-2}$ at [$B$] | A1 | Must be simplified |
| **Total** | **6** | |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate first derivative to zero | M1 | *Their* derivative must have at least 2 terms |
| Obtain $x = e$ | A1 | Allow $e^1$ |
| **Total** | **2** | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{61df367d-741f-4906-8ab9-2f32e8711aa6-08_451_1086_260_525}
The diagram shows the curve with equation
$$y = ( \ln x ) ^ { 2 } - 2 \ln x$$
The curve crosses the $x$-axis at the points $A$ and $B$, and has a minimum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of the gradient of the curve at each of the points $A$ and $B$.
\item Find the exact $x$-coordinate of $M$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2021 Q6 [8]}}