| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a straightforward application of the quotient rule to find stationary points. Part (a) requires recognizing ln(x)=0 when x=1. Part (b) involves differentiating y=ln(x)/x using the quotient rule, setting equal to zero, and solving to get x=e. Part (c) requires evaluating the derivative at x=e³ and finding the negative reciprocal. All steps are standard C3 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation |
| Answer | Marks |
|---|---|
| \(A = (1, 0)\) | B1 |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = \frac{\frac{1}{x}\cdot x - \ln x}{x^2} = \frac{1 - \ln x}{x^2} = 0\) | M1 A1 |
| \(\ln x = 1 \Rightarrow x = e\) | M1 |
| \(y = \frac{1}{e}\); \(B = \left(e, \frac{1}{e}\right)\) | A1 A1 |
| Answer | Marks |
|---|---|
| At \(x = e^3\): \(\frac{dy}{dx} = \frac{1 - \ln e^3}{(e^3)^2} = \frac{1-3}{e^6} = \frac{-2}{e^6}\) | M1 A1 |
| Gradient of normal \(= \frac{e^6}{2}\) | A1 |
# Question 6:
## Part (a)
| $A = (1, 0)$ | B1 | |
## Part (b)
| $\frac{dy}{dx} = \frac{\frac{1}{x}\cdot x - \ln x}{x^2} = \frac{1 - \ln x}{x^2} = 0$ | M1 A1 | |
| $\ln x = 1 \Rightarrow x = e$ | M1 | |
| $y = \frac{1}{e}$; $B = \left(e, \frac{1}{e}\right)$ | A1 A1 | |
## Part (c)
| At $x = e^3$: $\frac{dy}{dx} = \frac{1 - \ln e^3}{(e^3)^2} = \frac{1-3}{e^6} = \frac{-2}{e^6}$ | M1 A1 | |
| Gradient of normal $= \frac{e^6}{2}$ | A1 | |
---6 The diagram shows the curve $y = \frac { \ln x } { x }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{33ca7e6d-b9eb-46be-b5b0-c5685212d7ff-4_586_1034_1612_513}
The curve crosses the $x$-axis at $A$ and has a stationary point at $B$.
\begin{enumerate}[label=(\alph*)]
\item State the coordinates of $A$.
\item Find the coordinates of the stationary point, $B$, of the curve, giving your answer in an exact form.
\item Find the exact value of the gradient of the normal to the curve at the point where $x = \mathrm { e } ^ { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2010 Q6 [9]}}