| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find range where function increasing/decreasing |
| Difficulty | Moderate -0.8 This is a straightforward C1 stationary points question requiring standard differentiation, solving a quadratic, second derivative test, and interpreting dy/dx > 0. All steps are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the three-part structure. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = x^3 - 3x^2 + 4\); \(\frac{dy}{dx} = 3x^2 - 6x\); \(3x^2 - 6x = 0\); \(3x(x-2) = 0\); \(x = 0, x = 2\); \(y = 4, y = 0\) | B1, B1, M1, M1, A1, A1√, 6 | \(3x^2 - 6x\) (1 term correct; Completely correct); \(\frac{dy}{dx} = 0\); Correct method to solve quadratic; \(x = 0, 2\); \(y = 4, 0\); SR: one correct (x,y) pair _www_ B1 |
| (ii) \(\frac{d^2y}{dx^2} = 6x - 6\); \(x = 0\) \(y'' = -6\) \(-\) ve max; \(x = 2\) \(y'' = 6\) \(+\) ve min | M1, B1, B1, 3 | Correct method to find nature of stationary points (can be a sketch); \(x = 0\) max; \(x = 2\) min (N.B. If no method shown but both min and max correctly stated, award all 3 marks) |
| (iii) Increasing: \(x < 0\) \(x > 2\) | M1, A1, 2 | Any inequality (or inequalities) involving both their x values from part (i); Allow \(x \le 0\) \(x \ge 2\) |
**(i)** $y = x^3 - 3x^2 + 4$; $\frac{dy}{dx} = 3x^2 - 6x$; $3x^2 - 6x = 0$; $3x(x-2) = 0$; $x = 0, x = 2$; $y = 4, y = 0$ | B1, B1, M1, M1, A1, A1√, 6 | $3x^2 - 6x$ (1 term correct; Completely correct); $\frac{dy}{dx} = 0$; Correct method to solve quadratic; $x = 0, 2$; $y = 4, 0$; **SR:** one correct (x,y) pair _www_ B1
**(ii)** $\frac{d^2y}{dx^2} = 6x - 6$; $x = 0$ $y'' = -6$ $-$ ve max; $x = 2$ $y'' = 6$ $+$ ve min | M1, B1, B1, 3 | Correct method to find nature of stationary points (can be a sketch); $x = 0$ max; $x = 2$ min (**N.B. If no method shown but both min and max correctly stated, award all 3 marks**)
**(iii)** Increasing: $x < 0$ $x > 2$ | M1, A1, 2 | Any inequality (or inequalities) involving both their x values from part (i); Allow $x \le 0$ $x \ge 2$6 (i) Find the coordinates of the stationary points on the curve $y = x ^ { 3 } - 3 x ^ { 2 } + 4$.\\
(ii) Determine whether each stationary point is a maximum point or a minimum point.\\
(iii) For what values of $x$ does $x ^ { 3 } - 3 x ^ { 2 } + 4$ increase as $x$ increases?
\hfill \mbox{\textit{OCR C1 2006 Q6 [11]}}