| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a straightforward calculus question requiring standard techniques: differentiation to find a maximum point (setting derivative to zero and solving a simple exponential equation), then integration of exponential functions with substitution of limits. Both parts follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State derivative \(-e^{-x} - (-2)e^{-2x}\), or equivalent | B1 + B1 | — |
| Equate derivative to zero and solve for \(x\) | M1 | — |
| Obtain \(p = \ln 2\), or exact equivalent | A1 | [4] |
| (ii) State indefinite integral \(-e^{-x} - (-\frac{1}{2})e^{-2x}\), or equivalent | B1 + B1 | — |
| Substitute limits \(x = 0\) and \(x = p\) correctly | M1 | — |
| Obtain given answer following full and correct working | A1 | [4] |
**(i)** State derivative $-e^{-x} - (-2)e^{-2x}$, or equivalent | B1 + B1 | —
Equate derivative to zero and solve for $x$ | M1 | —
Obtain $p = \ln 2$, or exact equivalent | A1 | [4]
**(ii)** State indefinite integral $-e^{-x} - (-\frac{1}{2})e^{-2x}$, or equivalent | B1 + B1 | —
Substitute limits $x = 0$ and $x = p$ correctly | M1 | —
Obtain given answer following full and correct working | A1 | [4]
---\includegraphics{figure_5}
The diagram shows the curve $y = e^{-x} - e^{-2x}$ and its maximum point $M$. The $x$-coordinate of $M$ is denoted by $p$.
\begin{enumerate}[label=(\roman*)]
\item Find the exact value of $p$. [4]
\item Show that the area of the shaded region bounded by the curve, the $x$-axis and the line $x = p$ is equal to $\frac{1}{4}$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2010 Q5 [8]}}