| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| 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 product rule differentiation question with exponential and polynomial functions. Part (a) requires finding where y=0, then differentiating and substituting. Part (b) requires setting dy/dx=0 and solving. While it involves multiple steps, the techniques are standard A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt use of product rule to find first derivative | *M1 | |
| Obtain \(-\frac{1}{2}e^{-\frac{1}{2}x}(x^2-5x+4) + e^{-\frac{1}{2}x}(2x-5)\) | A1 | OE |
| Obtain \(x = 4\) for point \(B\) | B1 | |
| Substitute \(x = 4\) to find the value of the derivative | DM1 | |
| Obtain \(3e^{-2}\) | A1 | Or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equate *their* first derivative to zero and simplify as far as quadratic equation | *M1 | Allow if it appears in part (a) |
| Obtain at least \(x^2 - 9x + 14 = 0\) | A1 | OE |
| Solve to find relevant \(x\) value and substitute to find the value of \(y\) | DM1 | |
| Obtain \(x = 7\) and \(y = 18e^{-\frac{7}{2}}\) | A1 | Or exact equivalent |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt use of product rule to find first derivative | *M1 | |
| Obtain $-\frac{1}{2}e^{-\frac{1}{2}x}(x^2-5x+4) + e^{-\frac{1}{2}x}(2x-5)$ | A1 | OE |
| Obtain $x = 4$ for point $B$ | B1 | |
| Substitute $x = 4$ to find the value of the derivative | DM1 | |
| Obtain $3e^{-2}$ | A1 | Or exact equivalent |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equate *their* first derivative to zero and simplify as far as quadratic equation | *M1 | Allow if it appears in part (a) |
| Obtain at least $x^2 - 9x + 14 = 0$ | A1 | OE |
| Solve to find relevant $x$ value and substitute to find the value of $y$ | DM1 | |
| Obtain $x = 7$ and $y = 18e^{-\frac{7}{2}}$ | A1 | Or exact equivalent |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{4ce3208e-8ceb-4848-a9c7-fcda166319f4-06_526_947_276_591}
The diagram shows the curve with equation $y = \mathrm { e } ^ { - \frac { 1 } { 2 } x } \left( x ^ { 2 } - 5 x + 4 \right)$. The curve crosses the $x$-axis at the points $A$ and $B$, and has a maximum at the point $C$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact gradient of the curve at $B$.
\item Find the exact coordinates of $C$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q5 [9]}}