| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (extended problem with normals, stationary points, or further geometry) |
| Difficulty | Moderate -0.3 This is a straightforward integration question requiring standard techniques (integrating polynomials and x^{-2}), finding a constant using given coordinates, then basic tangent line work. Part (c) requires solving a quadratic discriminant condition, adding slight challenge. Overall slightly easier than average due to routine methods and clear structure, though the discriminant step elevates it above pure recall. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(y = \int (3 - \frac{2}{x^2}) \, dx\) | ||
| \(y = 3x + 2x^{-1} + c\) | M1 A2 | |
| \((2, 6) \therefore 6 = 6 + 1 + c\) | ||
| \(c = -1\) | M1 | |
| \(y = 3x + 2x^{-1} - 1\) | A1 | |
| (b) \(\text{grad} = 3 - \frac{1}{2} = \frac{5}{2}\) | M1 A1 | |
| \(y - 6 = \frac{5}{2}(x - 2)\) | M1 | |
| \(2y - 12 = 5x - 10\) | ||
| \(5x - 2y + 2 = 0\) | A1 | |
| (c) \(3x + 2x^{-1} - 1 = x + 3\) | ||
| \(3x + 2 - x = x^2 + 3x\) | M1 | |
| \(x^2 - 2x + 1 = 0\) | ||
| \((x - 1)^2 = 0\), repeated root \(\therefore\) tangent | M1 A1 | (12) |
(a) $y = \int (3 - \frac{2}{x^2}) \, dx$ | |
$y = 3x + 2x^{-1} + c$ | M1 A2 |
$(2, 6) \therefore 6 = 6 + 1 + c$ | |
$c = -1$ | M1 |
$y = 3x + 2x^{-1} - 1$ | A1 |
(b) $\text{grad} = 3 - \frac{1}{2} = \frac{5}{2}$ | M1 A1 |
$y - 6 = \frac{5}{2}(x - 2)$ | M1 |
$2y - 12 = 5x - 10$ | |
$5x - 2y + 2 = 0$ | A1 |
(c) $3x + 2x^{-1} - 1 = x + 3$ | |
$3x + 2 - x = x^2 + 3x$ | M1 |
$x^2 - 2x + 1 = 0$ | |
$(x - 1)^2 = 0$, repeated root $\therefore$ tangent | M1 A1 | (12)
**Total: (75)**10. The curve $C$ has the equation $y = \mathrm { f } ( x )$.
Given that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 - \frac { 2 } { x ^ { 2 } } , \quad x \neq 0 ,$$
and that the point $A$ on $C$ has coordinates (2, 6),
\begin{enumerate}[label=(\alph*)]
\item find an equation for $C$,
\item find an equation for the tangent to $C$ at $A$, giving your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers,
\item show that the line $y = x + 3$ is also a tangent to $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q10 [12]}}