| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of normal line |
| Difficulty | Standard +0.2 This is a straightforward C3 question requiring standard differentiation (including chain rule for ln), finding gradient at a point, calculating perpendicular gradient for the normal, and solving dy/dx=0 for stationary points. All steps are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = 2x - 5 + \frac{2}{x}\) | M1 |
| \(x=3, y=-6\), grad \(= \frac{5}{3}\) | A1 |
| grad of normal \(= -\frac{3}{5}\) | M1 |
| \(\therefore y+6 = -\frac{3}{5}(x-3)\) | M1 |
| \(5y+30 = -3x+9\) | |
| \(3x+5y+21 = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| SP: \(2x - 5 + \frac{2}{x} = 0\) | ||
| \(2x^2 - 5x + 2 = 0\) | M1 | |
| \((2x-1)(x-2) = 0\) | M1 | |
| \(x = \frac{1}{2}, 2\) | A1 | (8) |
**(a)**
$\frac{dy}{dx} = 2x - 5 + \frac{2}{x}$ | M1 |
$x=3, y=-6$, grad $= \frac{5}{3}$ | A1 |
grad of normal $= -\frac{3}{5}$ | M1 |
$\therefore y+6 = -\frac{3}{5}(x-3)$ | M1 |
$5y+30 = -3x+9$ | |
$3x+5y+21 = 0$ | A1 |
**(b)**
SP: $2x - 5 + \frac{2}{x} = 0$ | |
$2x^2 - 5x + 2 = 0$ | M1 |
$(2x-1)(x-2) = 0$ | M1 |
$x = \frac{1}{2}, 2$ | A1 | (8)
---4. The curve $C$ has the equation $y = x ^ { 2 } - 5 x + 2 \ln \frac { x } { 3 } , x > 0$.
\begin{enumerate}[label=(\alph*)]
\item Show that the normal to $C$ at the point where $x = 3$ has the equation
$$3 x + 5 y + 21 = 0$$
\item Find the $x$-coordinates of the stationary points of $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q4 [8]}}