| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a standard C2 differentiation problem requiring students to use the condition that dy/dx = 0 at a stationary point and that the point lies on the curve. It involves routine differentiation of a polynomial, solving simultaneous equations, and finding a second stationary point. While it requires multiple steps and careful algebra, it follows a well-practiced procedure with no novel insight needed, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks |
|---|---|
| (a) \(\frac{dy}{dx} = 3x^2 + 2ax - 15\) | M1 A1 |
| SP when \(x = -1\) \(\therefore 3 - 2a - 15 = 0\) | M1 |
| \(a = -6\) | A1 |
| \(y = x^3 - 6x^2 - 15x + b\) | |
| \((-1, 12)\) on curve \(\therefore 12 = -1 - 6 + 15 + b\) | M1 |
| \(b = 4\) | A1 |
| (b) \(3x^2 - 12x - 15 = 0\) | M1 |
| \(3(x-5)(x+1) = 0\) | M1 |
| \(x = -1\) [at \((-1, 12)\)] or \(5\) | A1 |
| \(\therefore (5, -96)\) | (9 marks) |
**(a)** $\frac{dy}{dx} = 3x^2 + 2ax - 15$ | M1 A1 |
SP when $x = -1$ $\therefore 3 - 2a - 15 = 0$ | M1 |
$a = -6$ | A1 |
$y = x^3 - 6x^2 - 15x + b$ |
$(-1, 12)$ on curve $\therefore 12 = -1 - 6 + 15 + b$ | M1 |
$b = 4$ | A1 |
**(b)** $3x^2 - 12x - 15 = 0$ | M1 |
$3(x-5)(x+1) = 0$ | M1 |
$x = -1$ [at $(-1, 12)$] or $5$ | A1 |
$\therefore (5, -96)$ | (9 marks)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=(\alph*)]
\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{Edexcel C2 Q6 [9]}}