| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 6 |
| Topic | Stationary points and optimisation |
| Type | Determine constant from stationary point condition |
| Difficulty | Standard +0.3 This is a straightforward stationary points question requiring differentiation and solving simultaneous equations. Part (i) is routine application of dy/dx=0, while part (ii) adds the standard condition d²y/dx²=0 for inflection. The algebra is simple and the method is textbook standard, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 4ax^3 + 3bx^2 - 2\) | M1 | Attempt to differentiate – all powers reduced by 1 |
| A1 | Correct first derivative | |
| \(4a(2)^3 + 3b(2)^2 - 2 = 0 \Rightarrow 16a + 6b = 1\) | A1 | AG – sufficient working must be shown to establish given result |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = 12ax^2 + 6bx\) | B1FT | Correct second derivative following through from their first derivative |
| \(16a + 6b = 1\) and \(4a + b = 0 \Rightarrow a = \ldots\) and \(b = \ldots\) | M1 | Formulate two equations in \(a\) and \(b\) and attempt to solve for both \(a\) and \(b\) |
| \(a = -\frac{1}{8}, b = \frac{1}{2}\) | A1 | Both values correct |
### (i)
$\frac{dy}{dx} = 4ax^3 + 3bx^2 - 2$ | M1 | Attempt to differentiate – all powers reduced by 1
| A1 | Correct first derivative
$4a(2)^3 + 3b(2)^2 - 2 = 0 \Rightarrow 16a + 6b = 1$ | A1 | AG – sufficient working must be shown to establish given result
**[3 marks total]**
### (ii)
$\frac{d^2y}{dx^2} = 12ax^2 + 6bx$ | B1FT | Correct second derivative following through from their first derivative
$16a + 6b = 1$ and $4a + b = 0 \Rightarrow a = \ldots$ and $b = \ldots$ | M1 | Formulate two equations in $a$ and $b$ and attempt to solve for both $a$ and $b$
$a = -\frac{1}{8}, b = \frac{1}{2}$ | A1 | Both values correct
**[3 marks total]**
---2 A curve has equation $y = a x ^ { 4 } + b x ^ { 3 } - 2 x + 3$.\\
(i) Given that the curve has a stationary point where $x = 2$, show that $16 a + 6 b = 1$.\\
(ii) Given also that this stationary point is a point of inflection, determine the values of $a$ and $b$.
\hfill \mbox{\textit{OCR H240/03 2018 Q2 [6]}}