| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Tangent to exponential curve |
| Difficulty | Moderate -0.5 This is a straightforward C3 exponential question requiring standard techniques: solving an exponential equation using logarithms, then finding a tangent using differentiation and point-slope form. Both parts are routine applications with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06f Laws of logarithms: addition, subtraction, power rules1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(e^{2x+1} = 2\) leading to \(2x + 1 = \ln 2\) | M1 | |
| \(x = \frac{1}{2}(\ln 2 - 1)\) | A1 | (2) |
| (b) \(\frac{dy}{dx} = 8e^{2x+1}\) | B1 | |
| At \(x = \frac{1}{2}(\ln 2 - 1)\): \(\frac{dy}{dx} = 16\) | B1 | |
| \(y - 8 = 16\left(x - \frac{1}{2}(\ln 2 - 1)\right)\) | M1 | |
| \(y = 16x + 16 - 8\ln 2\) | A1 | (4) [6] |
**(a)** $e^{2x+1} = 2$ leading to $2x + 1 = \ln 2$ | M1 |
$x = \frac{1}{2}(\ln 2 - 1)$ | A1 | (2)
**(b)** $\frac{dy}{dx} = 8e^{2x+1}$ | B1 |
At $x = \frac{1}{2}(\ln 2 - 1)$: $\frac{dy}{dx} = 16$ | B1 |
$y - 8 = 16\left(x - \frac{1}{2}(\ln 2 - 1)\right)$ | M1 |
$y = 16x + 16 - 8\ln 2$ | A1 | (4) [6]
---\begin{enumerate}
\item The point $P$ lies on the curve with equation
\end{enumerate}
$$y = 4 \mathrm { e } ^ { 2 x + 1 }$$
The $y$-coordinate of $P$ is 8 .\\
(a) Find, in terms of $\ln 2$, the $x$-coordinate of $P$.\\
(b) Find the equation of the tangent to the curve at the point $P$ in the form $y = a x + b$, where $a$ and $b$ are exact constants to be found.\\
\hfill \mbox{\textit{Edexcel C3 2008 Q1 [6]}}