| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Find curve from gradient |
| Difficulty | Moderate -0.3 This is a straightforward C1 integration question requiring algebraic simplification (splitting the fraction), basic integration of power functions, and using an initial condition to find the constant. Part (a) is simple differentiation. The techniques are routine for C1 level, though the multi-step nature and 9 total marks make it slightly more substantial than the most basic exercises. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = 1 - 4x^{-3}\) | B1 | |
| \(\frac{d^2y}{dx^2} = 12x^{-4}\) | M1 A1 | |
| (b) \(y = \int (1 - 4x^{-3}) \, dx\) | ||
| \(y = x + 2x^{-2} + c\) | M1 A2 | |
| \(x = -1, y = 0 \therefore 0 = -1 + 2 + c\) | ||
| \(c = -1\) | M1 | |
| \(y = x + 2x^{-2} - 1\) | ||
| when \(x = 2, y = 2 + \frac{1}{2} - 1 = \frac{3}{2}\) | M1 A1 | (9) |
**(a)** $\frac{dy}{dx} = 1 - 4x^{-3}$ | B1 |
$\frac{d^2y}{dx^2} = 12x^{-4}$ | M1 A1 |
**(b)** $y = \int (1 - 4x^{-3}) \, dx$ | |
$y = x + 2x^{-2} + c$ | M1 A2 |
$x = -1, y = 0 \therefore 0 = -1 + 2 + c$ | |
$c = -1$ | M1 |
$y = x + 2x^{-2} - 1$ | |
when $x = 2, y = 2 + \frac{1}{2} - 1 = \frac{3}{2}$ | M1 A1 | (9)Given that
$$\frac{dy}{dx} = \frac{x^3 - 4}{x^2}, \quad x \neq 0,$$
\begin{enumerate}[label=(\alph*)]
\item find $\frac{d^2y}{dx^2}$. [3]
\end{enumerate}
Given also that $y = 0$ when $x = -1$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the value of $y$ when $x = 2$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q8 [9]}}