| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| 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 -0.3 This is a standard C1 differentiation question requiring routine application of differentiation rules, second derivative test, and interpretation of results. While it has multiple parts, each step follows a predictable algorithm with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 9 - 6x - 3x^2\) | *M1, A1 | Attempt to differentiate \(y\) or \(-y\) (at least one correct term); 3 correct terms |
| At stationary points, \(9 - 6x - 3x^2 = 0\) | M1 | Use of \(\frac{dy}{dx}=0\) (for \(y\) or \(-y\)) |
| \(3(3+x)(1-x) = 0\) | DM1 | Correct method to solve 3 term quadratic |
| \(x = -3\) or \(x = 1\) | A1 | \(x = -3,\ 1\) |
| \(y = 0,\ 32\) | A1 ft [6] | \(y=0,\ 32\) (1 correct pair www A1 A0) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = -6x - 6\) | M1 | Looks at sign of \(\frac{d^2y}{dx^2}\), derived correctly from \(k\frac{dy}{dx}\), or other correct method |
| When \(x=-3\), \(\frac{d^2y}{dx^2} > 0\) | A1 | \(x = -3\) minimum |
| When \(x=1\), \(\frac{d^2y}{dx^2} < 0\) | A1 [3] | \(x = 1\) maximum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-3 < x < 1\) | M1, A1 [2+6+3=11] | Uses the \(x\) values of both turning points in inequality/inequalities; Correct inequality or inequalities. Allow \(\leq\) |
## Question 8:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 9 - 6x - 3x^2$ | *M1, A1 | Attempt to differentiate $y$ or $-y$ (at least one correct term); 3 correct terms |
| At stationary points, $9 - 6x - 3x^2 = 0$ | M1 | Use of $\frac{dy}{dx}=0$ (for $y$ or $-y$) |
| $3(3+x)(1-x) = 0$ | DM1 | Correct method to solve 3 term quadratic |
| $x = -3$ or $x = 1$ | A1 | $x = -3,\ 1$ |
| $y = 0,\ 32$ | A1 ft [6] | $y=0,\ 32$ (1 correct pair www A1 A0) |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = -6x - 6$ | M1 | Looks at sign of $\frac{d^2y}{dx^2}$, derived correctly from $k\frac{dy}{dx}$, or other correct method |
| When $x=-3$, $\frac{d^2y}{dx^2} > 0$ | A1 | $x = -3$ minimum |
| When $x=1$, $\frac{d^2y}{dx^2} < 0$ | A1 [3] | $x = 1$ maximum |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-3 < x < 1$ | M1, A1 [2+6+3=11] | Uses the $x$ values of both turning points in inequality/inequalities; Correct inequality or inequalities. Allow $\leq$ |
---8 (i) Find the coordinates of the stationary points of the curve $y = 27 + 9 x - 3 x ^ { 2 } - x ^ { 3 }$.\\
(ii) Determine, in each case, whether the stationary point is a maximum or minimum point.\\
(iii) Hence state the set of values of $x$ for which $27 + 9 x - 3 x ^ { 2 } - x ^ { 3 }$ is an increasing function.\\
$9 \quad A$ is the point $( 2,7 )$ and $B$ is the point $( - 1 , - 2 )$.\\
\hfill \mbox{\textit{OCR C1 2007 Q8 [11]}}