| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Solve equation involving derivatives |
| Difficulty | Standard +0.8 This question requires quotient rule differentiation with exponentials, algebraic manipulation to reach a specific factored form, and solving an exponential equation. While the techniques are standard P2 content, the algebraic manipulation to achieve the given factorization and the multi-step nature (differentiate, set to zero, manipulate, solve) makes it moderately challenging—above average but not requiring novel insight. |
| 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 |
| Use quotient rule (or equivalent) to find first derivative | \*M1 | Condone poor use of brackets if recovered later |
| Obtain \(\frac{18e^{2x}(e^x-1) - e^x(9e^{2x}+16)}{(e^x-1)^2}\) | A1 | OE |
| Equate first derivative to zero and attempt factorisation | DM1 | Need to be working with \(9e^{3x} - 18e^{2x} \pm 16e^x = 0\) |
| Obtain \(e^x(3e^x - 8)(3e^x + 2) = 0\) | A1 | AG – necessary detail needed; SC B3 if numerator in incorrect order and result obtained; SC B3 if denominator not squared and result obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Observe \(e^x \neq 0\) and \(3e^x + 2 \neq 0\), hence one stationary point | B1 | Allow if discounted by crossing through OE |
| Attempt exact solution for \(x\) and for \(y\) | M1 | |
| Obtain \(x = \ln\!\left(\frac{8}{3}\right)\) or exact equivalent | A1 | |
| Obtain \(y = 48\) | A1 | Not from a rounded decimal |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use quotient rule (or equivalent) to find first derivative | \*M1 | Condone poor use of brackets if recovered later |
| Obtain $\frac{18e^{2x}(e^x-1) - e^x(9e^{2x}+16)}{(e^x-1)^2}$ | A1 | OE |
| Equate first derivative to zero and attempt factorisation | DM1 | Need to be working with $9e^{3x} - 18e^{2x} \pm 16e^x = 0$ |
| Obtain $e^x(3e^x - 8)(3e^x + 2) = 0$ | A1 | AG – necessary detail needed; **SC B3** if numerator in incorrect order and result obtained; **SC B3** if denominator not squared and result obtained |
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## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Observe $e^x \neq 0$ and $3e^x + 2 \neq 0$, hence one stationary point | B1 | Allow if discounted by crossing through OE |
| Attempt exact solution for $x$ and for $y$ | M1 | |
| Obtain $x = \ln\!\left(\frac{8}{3}\right)$ or exact equivalent | A1 | |
| Obtain $y = 48$ | A1 | Not from a rounded decimal |
---6 A curve has equation $y = \frac { 9 \mathrm { e } ^ { 2 x } + 16 } { \mathrm { e } ^ { x } - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of any stationary point on the curve satisfies the equation
$$\mathrm { e } ^ { x } \left( 3 \mathrm { e } ^ { x } - 8 \right) \left( 3 \mathrm { e } ^ { x } + 2 \right) = 0$$
\item Hence show that the curve has only one stationary point and find its exact coordinates.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2022 Q6 [8]}}