| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - at special curve features |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on logarithmic differentiation and numerical methods. Part (a) requires finding where y=0 (x=1) then applying the quotient rule; part (b) involves showing dy/dx changes sign in an interval; part (c) is routine trapezium rule application. All techniques are standard P2/C3 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differentiate using quotient rule (or product rule) | M1 | |
| Obtain \(\dfrac{(3x+1)\frac{2}{x} - 6\ln x}{(3x+1)^2}\) | A1 | OE |
| Substitute \(x = 1\) to obtain \(\frac{1}{2}\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate numerator of first derivative to zero | M1 | May be implied |
| Consider sign of \(\frac{2}{x}(3x+1) - 6\ln x\) for \(3.0\) and \(3.1\) | M1 | OE |
| Obtain \(0.074\ldots\) and \(-0.14\ldots\) or equivalents and justify conclusion | A1 | AG – necessary detail needed. \(0.00075\) and \(-0.001275\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use \(y\)-values \([0],\ \frac{2}{7}\ln 2\) or \(0.1980\) and \(\frac{2}{10}\ln 3\) or \(0.2197\) | B1 | |
| Use correct formula, or equivalent, with \(h = 1\) | M1 | |
| Obtain \(0.31\) | A1 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate using quotient rule (or product rule) | M1 | |
| Obtain $\dfrac{(3x+1)\frac{2}{x} - 6\ln x}{(3x+1)^2}$ | A1 | OE |
| Substitute $x = 1$ to obtain $\frac{1}{2}$ | A1 | OE |
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate numerator of first derivative to zero | M1 | May be implied |
| Consider sign of $\frac{2}{x}(3x+1) - 6\ln x$ for $3.0$ and $3.1$ | M1 | OE |
| Obtain $0.074\ldots$ and $-0.14\ldots$ or equivalents and justify conclusion | A1 | AG – necessary detail needed. $0.00075$ and $-0.001275$ |
---
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use $y$-values $[0],\ \frac{2}{7}\ln 2$ or $0.1980$ and $\frac{2}{10}\ln 3$ or $0.2197$ | B1 | |
| Use correct formula, or equivalent, with $h = 1$ | M1 | |
| Obtain $0.31$ | A1 | |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{389df578-e7a7-4d19-9416-5e580d107717-10_456_598_269_762}
The diagram shows the curve with equation $y = \frac { 2 \ln x } { 3 x + 1 }$. The curve crosses the $x$-axis at the point $A$ and has a maximum point $B$. The shaded region is bounded by the curve and the lines $x = 3$ and $y = 0$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the curve at $A$.
\item Show by calculation that the $x$-coordinate of $B$ lies between 3.0 and 3.1.
\item Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2022 Q7 [9]}}