| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Classify nature of stationary points |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring standard techniques: basic polynomial differentiation, solving a quadratic for stationary points, using the second derivative test, and finding a point given tangent gradient. All parts are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = x^2 - 9\) | B1 | 1 term correct |
| B1 2 | Both terms correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 - 9 = 0\) | *M1 | Uses \(\frac{dy}{dx} = 0\) |
| \(x = 3, -3\) | A1 | |
| \(y = -18, 18\) | A1 3 | (1 correct pair A1 A0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = 2x\) | DM1 | Looks at sign of \(\frac{d^2y}{dx^2}\) or other correct method |
| \(x = 3\): \(\frac{d^2y}{dx^2} = 6\) | A1 | \(x = 3\) minimum |
| \(x = -3\): \(\frac{d^2y}{dx^2} = -6\) | A1 3 | \(x = -3\) maximum. (NB: If no method shown but min and max correctly stated, award all 3 marks unless earlier incorrect working) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of \(24x + 3y + 2 = 0\) is \(-8\) | B1 | Gradient \(= -8\) |
| \(x^2 - 9 = -8\) | M1 | \(x^2 - 9 = -8\) |
| \(x = \pm 1\) | M1 | One of their \(x\) values substituted in both line and curve |
| \(x = 1\), \(y = -8\frac{2}{3}\) | M1 | Second \(x\) value substituted in both line and curve or justification that first point is the correct one |
| \(x = -1\), \(y = 7\frac{1}{3}\) | ||
| \(\therefore p = 1\), \(q = -8\frac{2}{3}\) | A1 5 | \(p = 1\), \(q = -8\frac{2}{3}\) seen |
## Question 10(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = x^2 - 9$ | B1 | 1 term correct |
| | B1 **2** | Both terms correct |
## Question 10(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 - 9 = 0$ | *M1 | Uses $\frac{dy}{dx} = 0$ |
| $x = 3, -3$ | A1 | |
| $y = -18, 18$ | A1 **3** | (1 correct pair A1 A0) |
## Question 10(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = 2x$ | DM1 | Looks at sign of $\frac{d^2y}{dx^2}$ or other correct method |
| $x = 3$: $\frac{d^2y}{dx^2} = 6$ | A1 | $x = 3$ minimum |
| $x = -3$: $\frac{d^2y}{dx^2} = -6$ | A1 **3** | $x = -3$ maximum. (NB: If no method shown but min and max correctly stated, award all 3 marks unless earlier incorrect working) |
## Question 10(iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of $24x + 3y + 2 = 0$ is $-8$ | B1 | Gradient $= -8$ |
| $x^2 - 9 = -8$ | M1 | $x^2 - 9 = -8$ |
| $x = \pm 1$ | M1 | One of their $x$ values substituted in both line and curve |
| $x = 1$, $y = -8\frac{2}{3}$ | M1 | Second $x$ value substituted in both line and curve **or** justification that first point is the correct one |
| $x = -1$, $y = 7\frac{1}{3}$ | | |
| $\therefore p = 1$, $q = -8\frac{2}{3}$ | A1 **5** | $p = 1$, $q = -8\frac{2}{3}$ seen |10 (i) Given that $y = \frac { 1 } { 3 } x ^ { 3 } - 9 x$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(ii) Find the coordinates of the stationary points on the curve $y = \frac { 1 } { 3 } x ^ { 3 } - 9 x$.\\
(iii) Determine whether each stationary point is a maximum point or a minimum point.\\
(iv) Given that $24 x + 3 y + 2 = 0$ is the equation of the tangent to the curve at the point ( $p , q$ ), find $p$ and $q$.
\hfill \mbox{\textit{OCR C1 2005 Q10 [13]}}