| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| 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 stationary points question requiring routine differentiation of polynomials, solving a simple equation, and applying the second derivative test. All techniques are standard C1 procedures with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part nature and requirement to interpret increasing functions. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 4x^3 + 32\) | M1 | Attempt to differentiate (one term correct). "+ C" is A0 |
| A1 | Completely correct | |
| \(4x^3 + 32 = 0\) | M1 | Sets their \(\frac{dy}{dx} = 0\) (can be implied) |
| \(x = -2\) | A1 | Correct value for \(x\) (not \(\pm 2\)) www |
| \(y = -48\) | A1 FT [5] | Correct value of \(y\) for their single non-zero value of \(x\). e.g. \((2, 80)\), \((4, 384)\), \((-4, 128)\), \((8, 4352)\), \((-8, 3840)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = 12x^2\) | M1 | Correct method for determining nature of a stationary point. e.g. evaluating second derivative at \(x = -2\) and stating a conclusion. Evaluating \(\frac{dy}{dx}\) either side of \(x = -2\). Evaluating \(y\) either side of \(x = -2\) |
| When \(x = -2\), \(\frac{d^2y}{dx^2} > 0\) so minimum pt | A1 [2] | Fully correct for \(x = -2\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x > -2\) | B1 FT [1] | ft from single \(x\) value in (i) consistent with (ii). Do not accept \(x \geq -2\) |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 4x^3 + 32$ | M1 | Attempt to differentiate (one term correct). "+ C" is A0 |
| | A1 | Completely correct |
| $4x^3 + 32 = 0$ | M1 | Sets their $\frac{dy}{dx} = 0$ (can be implied) |
| $x = -2$ | A1 | Correct value for $x$ (**not** $\pm 2$) **www** |
| $y = -48$ | A1 FT [5] | Correct value of $y$ for their single non-zero value of $x$. e.g. $(2, 80)$, $(4, 384)$, $(-4, 128)$, $(8, 4352)$, $(-8, 3840)$ |
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## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2} = 12x^2$ | M1 | Correct method for determining nature of a stationary point. e.g. evaluating second derivative at $x = -2$ and stating a conclusion. Evaluating $\frac{dy}{dx}$ either side of $x = -2$. Evaluating $y$ either side of $x = -2$ |
| When $x = -2$, $\frac{d^2y}{dx^2} > 0$ so minimum pt | A1 [2] | Fully correct for $x = -2$ only |
---
## Question 8(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x > -2$ | B1 FT [1] | ft from single $x$ value in (i) consistent with (ii). **Do not accept** $x \geq -2$ |
---8 (i) Find the coordinates of the stationary point on the curve $y = x ^ { 4 } + 32 x$.\\
(ii) Determine whether this stationary point is a maximum or a minimum.\\
(iii) For what values of $x$ does $x ^ { 4 } + 32 x$ increase as $x$ increases?
\hfill \mbox{\textit{OCR C1 2012 Q8 [8]}}