| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Determine nature of stationary points |
| Difficulty | Moderate -0.3 This is a standard C2 calculus question covering routine differentiation, finding stationary points, second derivative test, and finding a normal line equation. All techniques are textbook exercises requiring no problem-solving insight, though the multi-part structure and 15 total marks make it slightly more substantial than the most basic questions. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = 4x^3 - 16x\) | M1 A1 | (2) |
| (b) \(4x^3 - 16x = 0\) | M1 | |
| \(4x(x^2 - 4) = 0\); \(x = 0, 2, -2\) | A2 (1, 0) | |
| \(y = 3, -13, -13\) | M1 A1 | (5) |
| (c) \(\frac{d^2y}{dx^2} = 12x^2 - 16\) | M1 | |
| \(x = 0\) — Max.; \(x = 2\) — Min.; \(x = -2\) — Min. | A1 ft | One of these, All three (3) |
| (d) \(x = 1: y = 1 - 8 + 3 = -4\) | B1 | |
| At \(x = 1\), \(\frac{dy}{dx} = 4 - 16 = -12\) | B1 ft | \((m)\) |
| Gradient of normal \(= -\frac{1}{m}\) | M1 | \((= \frac{1}{12})\) |
| \(y - (-4) = \frac{1}{12}(x - 1)\); \(x - 12y - 49 = 0\) | M1 A1 | (5) |
**(a)** $\frac{dy}{dx} = 4x^3 - 16x$ | M1 A1 | (2)
**(b)** $4x^3 - 16x = 0$ | M1 |
$4x(x^2 - 4) = 0$; $x = 0, 2, -2$ | A2 (1, 0) |
$y = 3, -13, -13$ | M1 A1 | (5)
**(c)** $\frac{d^2y}{dx^2} = 12x^2 - 16$ | M1 |
$x = 0$ — Max.; $x = 2$ — Min.; $x = -2$ — Min. | A1 ft | One of these, All three (3)
**(d)** $x = 1: y = 1 - 8 + 3 = -4$ | B1 |
At $x = 1$, $\frac{dy}{dx} = 4 - 16 = -12$ | B1 ft | $(m)$
Gradient of normal $= -\frac{1}{m}$ | M1 | $(= \frac{1}{12})$
$y - (-4) = \frac{1}{12}(x - 1)$; $x - 12y - 49 = 0$ | M1 A1 | (5)For the curve $C$ with equation $y = x^4 - 8x^2 + 3$,
\begin{enumerate}[label=(\alph*)]
\item find $\frac{dy}{dx}$, [2]
\item find the coordinates of each of the stationary points, [5]
\item determine the nature of each stationary point. [3]
\end{enumerate}
The point $A$, on the curve $C$, has $x$-coordinate $1$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find an equation for the normal to $C$ at $A$, giving your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [15]}}