| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Determine constant from stationary point condition |
| Difficulty | Moderate -0.3 This is a straightforward C1 stationary points question requiring standard differentiation, solving dy/dx=0, and using the second derivative test. While it has multiple parts, each step follows routine procedures with no conceptual challenges beyond basic calculus mechanics, 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 | Marks | Guidance |
| \(\frac{dy}{dx}=6x^2-2ax+8\) | M1 | Attempt to differentiate, at least two non-zero terms correct |
| A1 | Fully correct | |
| When \(x=4\), \(\frac{dy}{dx}=104-8a\) | M1 | Substitutes \(x=4\) into their \(\frac{dy}{dx}\). These Ms may be awarded in either order |
| \(\frac{dy}{dx}=0\) gives \(a=13\) | M1 | Sets their \(\frac{dy}{dx}\) to 0. Must be seen |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2}=12x-26\) | M1 | Correct method to find nature of stationary point e.g. substituting \(x=4\) into second derivative (at least one term correct from their first derivative in (i)) and consider the sign. Alternate valid methods include: 1) Evaluating gradient at either side of \(4\left(x>\frac{1}{3}\right)\) e.g. at 3, \(-16\) at 5, 28; 2) Evaluating \(y=-46\) at 4 and either side of \(4\left(x>\frac{1}{3}\right)\) e.g. \((3,-37)\), \((5,-33)\) |
| When \(x=4\), \(\frac{d^2y}{dx^2}>0\) so minimum | A1 | www. If using alternatives, working must be fully correct to obtain the A mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6x^2-26x+8=0\) | M1 | Sets their derivative to zero |
| \((3x-1)(x-4)=0\) | M1 | Correct method to solve quadratic (appx 1). Could be \((6x-2)(x-4)=0\) or \((3x-1)(2x-8)=0\) |
| \(x=\frac{1}{3}\) | A1 | oe |
## Question 9:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx}=6x^2-2ax+8$ | M1 | Attempt to differentiate, at least two non-zero terms correct |
| | A1 | Fully correct |
| When $x=4$, $\frac{dy}{dx}=104-8a$ | M1 | Substitutes $x=4$ into their $\frac{dy}{dx}$. These Ms may be awarded in either order |
| $\frac{dy}{dx}=0$ gives $a=13$ | M1 | Sets their $\frac{dy}{dx}$ to 0. Must be seen |
| | A1 | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2}=12x-26$ | M1 | Correct method to find nature of stationary point e.g. substituting $x=4$ into second derivative (at least one term correct from their first derivative in (i)) and consider the sign. **Alternate valid methods include:** 1) Evaluating gradient at either side of $4\left(x>\frac{1}{3}\right)$ e.g. at 3, $-16$ at 5, 28; 2) Evaluating $y=-46$ at 4 and either side of $4\left(x>\frac{1}{3}\right)$ e.g. $(3,-37)$, $(5,-33)$ |
| When $x=4$, $\frac{d^2y}{dx^2}>0$ so minimum | A1 | **www**. If using alternatives, working must be fully correct to obtain the A mark |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6x^2-26x+8=0$ | M1 | Sets their derivative to zero |
| $(3x-1)(x-4)=0$ | M1 | Correct method to solve quadratic **(appx 1)**. Could be $(6x-2)(x-4)=0$ or $(3x-1)(2x-8)=0$ |
| $x=\frac{1}{3}$ | A1 | oe |9 The curve $y = 2 x ^ { 3 } - a x ^ { 2 } + 8 x + 2$ passes through the point $B$ where $x = 4$.\\
(i) Given that $B$ is a stationary point of the curve, find the value of the constant $a$.\\
(ii) Determine whether the stationary point $B$ is a maximum point or a minimum point.\\
(iii) Find the $x$-coordinate of the other stationary point of the curve.
\hfill \mbox{\textit{OCR C1 2015 Q9 [10]}}