| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2012 |
| Session | Specimen |
| Marks | 7 |
| Topic | Stationary points and optimisation |
| Type | Second derivative test justification |
| Difficulty | Moderate -0.8 This is a straightforward application of standard calculus techniques: differentiate a polynomial, solve a quadratic equation for stationary points, then apply the second derivative test. All steps are routine with no problem-solving insight required, making it easier than average but not trivial since it requires correct execution of multiple standard procedures. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
**(i)** Attempt to differentiate **M1**
Obtain $6x^2 - 10x - 4$ **A1**
**(ii)** Setting *their* $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ **M1**
Solving quadratic to obtain $x = 2$, $x = -\dfrac{1}{3}$ **A1**
**(iii)** Looks at sign of $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$, derived correctly from *their* $\dfrac{\mathrm{d}y}{\mathrm{d}x}$, or other correct method **M1**
When $x = 2$, $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$ therefore minimum **A1**
When $x = -\dfrac{1}{3}$, $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0$ therefore maximum **A1**
**Total: 7 marks**8 A curve has equation $y = 2 x ^ { 3 } - 5 x ^ { 2 } - 4 x + 1$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(ii) Hence find the $x$-coordinates of the stationary points of the curve.\\
(iii) By using the second derivative, determine whether each of the stationary points is a maximum or a minimum.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2012 Q8 [7]}}