| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Topic | Stationary points and optimisation |
| Type | Find stationary points coordinates |
| Difficulty | Standard +0.3 This is a straightforward stationary point question requiring differentiation of a simple function (x² - ln x), setting the derivative to zero, and solving a basic equation. The only mild challenge is handling the natural logarithm and expressing the answer in exact form, but this is standard A-level technique with no novel problem-solving required. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives |
$\dfrac{dy}{dx} = 2x - \dfrac{1}{x}$
M1 Attempt integration – one correct term
A1 Fully correct
solve to obtain $x = (\pm)\dfrac{1}{\sqrt{2}}$
M1 Equate to 0 and attempt to solve
A1 Obtain at least the positive root
A1 Obtain correct stationary point having selected the positive root only from $\pm\dfrac{1}{\sqrt{2}}$. Allow $x = ..., y = ...$. Exact final answer only, else A0
only one stationary point at $\left(\dfrac{1}{\sqrt{2}}, \dfrac{1}{2} - \ln\dfrac{1}{\sqrt{2}}\right)$ AEF
as $x \leq 0$ cannot be valid due to $\ln$
A1 [6] Explanation of why there is only one root, referring to $\ln x$. Only considering $+$ve solution will get max 4/68 Show that the graph of $y = x ^ { 2 } - \ln x$ has only one stationary point and give the coordinates of that point in exact form.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2014 Q8 [6]}}