| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a straightforward C3 question requiring standard differentiation of exponential and logarithmic functions, finding a tangent equation, then calculating intercepts and triangle area. All steps are routine applications of learned techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations |
| Answer | Marks |
|---|---|
| (a) \(\frac{dy}{dx} = 2e^x - \frac{6}{x}\) | M1 |
| \(x = 1, y = 2e\), grad \(= 2e - 6\) | A1 |
| \(\therefore y - 2e = (2e-6)(x-1)\) [\(y = (2e-6)x + 6\)] | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 0 \Rightarrow (2e-6)x + 6 = 0\) | M1 A1 | |
| \(x = \frac{-6}{2e-6} = \frac{3}{3-e}\) | M1 A1 | |
| area \(= \frac{1}{2} \times 6 \times \frac{3}{3-e} = \frac{9}{3-e}\) | M1 A1 | (8 marks) |
**(a)** $\frac{dy}{dx} = 2e^x - \frac{6}{x}$ | M1 |
$x = 1, y = 2e$, grad $= 2e - 6$ | A1 |
$\therefore y - 2e = (2e-6)(x-1)$ [$y = (2e-6)x + 6$] | M1 A1 |
**(b)** $x = 0 \Rightarrow y = 6$
$y = 0 \Rightarrow (2e-6)x + 6 = 0$ | M1 A1 |
$x = \frac{-6}{2e-6} = \frac{3}{3-e}$ | M1 A1 |
area $= \frac{1}{2} \times 6 \times \frac{3}{3-e} = \frac{9}{3-e}$ | M1 A1 | (8 marks)The curve $C$ has the equation $y = 2e^x - 6 \ln x$ and passes through the point $P$ with $x$-coordinate $1$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the tangent to $C$ at $P$. [4]
\end{enumerate}
The tangent to $C$ at $P$ meets the coordinate axes at the points $Q$ and $R$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the area of triangle $OQR$, where $O$ is the origin, is $\frac{9}{3-e}$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q3 [8]}}