| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - direct evaluation |
| Difficulty | Moderate -0.8 Part (a) requires direct differentiation of an exponential function using the chain rule and substitution of x=2 - a routine procedural task. Part (b) involves standard integration of an exponential function with limits. Both parts are straightforward applications of basic calculus techniques with no problem-solving or insight required, making this easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate to obtain form \(ke^{-\frac{1}{2}x}\) | M1 | For any non-zero \(k\) except 6 |
| Substitute \(x=2\) to obtain \(-3e^{-1}\) or \(-\frac{3}{e}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Integrate to obtain \(-12e^{-\frac{1}{2}x}\) | B1 | OE |
| Use limits 0 and 2 correctly to an integral of the form \(ke^{-\frac{x}{2}}\), retaining exactness | M1 | For any non-zero \(k\) except 6, or equivalent perhaps involving integration of \(6-6e^{-\frac{1}{2}x}\) |
| Subtract from 12 to obtain final answer \(12e^{-1}\) or \(\frac{12}{e}\) | A1 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate to obtain form $ke^{-\frac{1}{2}x}$ | M1 | For any non-zero $k$ except 6 |
| Substitute $x=2$ to obtain $-3e^{-1}$ or $-\frac{3}{e}$ | A1 | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to obtain $-12e^{-\frac{1}{2}x}$ | B1 | OE |
| Use limits 0 and 2 correctly to an integral of the form $ke^{-\frac{x}{2}}$, retaining exactness | M1 | For any non-zero $k$ except 6, or equivalent perhaps involving integration of $6-6e^{-\frac{1}{2}x}$ |
| Subtract from 12 to obtain final answer $12e^{-1}$ or $\frac{12}{e}$ | A1 | |3\\
\includegraphics[max width=\textwidth, alt={}, center]{b104e2a7-06c8-4e2e-a4f9-5095ad56897a-04_652_392_274_872}
The diagram shows the curve with equation $y = 6 \mathrm { e } ^ { - \frac { 1 } { 2 } x }$. The points on the curve with $x$-coordinates 0 and 2 are denoted by $A$ and $B$ respectively. The shaded region is enclosed by the curve, the line through $A$ parallel to the $x$-axis and the line through $B$ parallel to the $y$-axis.
\begin{enumerate}[label=(\alph*)]
\item Find the exact gradient of the curve at $B$.
\item Find the exact area of the shaded region.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q3 [5]}}