| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 9 |
| 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 standard C1 stationary points question requiring routine differentiation, substituting given conditions to form simultaneous equations, and solving. While it involves multiple steps (differentiate, use point on curve, use dy/dx=0, solve system, find second stationary point), all techniques are straightforward applications of core methods with no conceptual challenges or novel insights required. Slightly easier than average due to its predictable structure. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = 3x^2 + 2ax - 15\) | M1 A1 |
| SP when \(x = -1\) ∴ \(3 - 2a - 15 = 0\), \(a = -6\) | M1 A1 |
| \(y = x^3 - 6x^2 - 15x + b\); \((-1, 12)\) on curve ∴ \(12 = -1 - 6 + 15 + b\), \(b = 4\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(3x^2 - 12x - 15 = 0\) | M1 |
| \(3(x - 5)(x + 1) = 0\) | M1 |
| \(x = -1\) [at \((-1, 12)\)] or \(5\) ∴ \((5, -96)\) | A1 |
| (9) |
## (i)
$\frac{dy}{dx} = 3x^2 + 2ax - 15$ | M1 A1
SP when $x = -1$ ∴ $3 - 2a - 15 = 0$, $a = -6$ | M1 A1
$y = x^3 - 6x^2 - 15x + b$; $(-1, 12)$ on curve ∴ $12 = -1 - 6 + 15 + b$, $b = 4$ | M1 A1
## (ii)
$3x^2 - 12x - 15 = 0$ | M1
$3(x - 5)(x + 1) = 0$ | M1
$x = -1$ [at $(-1, 12)$] or $5$ ∴ $(5, -96)$ | A1
| (9)
---A curve has the equation
$$y = x^3 + ax^2 - 15x + b,$$
where $a$ and $b$ are constants.
Given that the curve is stationary at the point $(-1, 12)$,
\begin{enumerate}[label=(\roman*)]
\item find the values of $a$ and $b$, [6]
\item find the coordinates of the other stationary point of the curve. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 Q7 [9]}}