| 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 | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| 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 |
| Total | 5 |
| 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 |
| Total | 4 |
## 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 |
| **Total** | **5** | |
## 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 |
| **Total** | **4** | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{a1ea242a-c7f4-46b0-b4b8-bd13b3880557-06_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]}}