| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Curve properties and tangent/normal |
| Difficulty | Moderate -0.3 This is a straightforward C1 integration question requiring: (a) evaluating dy/dx at x=1 and using point-slope form, (b) integrating a power function and using the boundary condition to find the constant, (c) solving a quadratic equation. All techniques are routine for C1 level with no novel problem-solving required, though the multi-part structure and algebraic manipulation in part (c) elevate it slightly above the most basic exercises. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\text{grad} = 8 - 2 = 6\) | B1 | |
| \(\therefore y - 1 = 6(x - 1)\) | M1 | |
| \(y = 6x - 5\) | A1 | |
| (b) \(y = \int (8x - \frac{2}{x^3}) \, dx\) | ||
| \(y = 4x^2 + x^{-2} + c\) | M1 A2 | |
| \((1, 1) \therefore 1 = 4 + 1 + c\) | ||
| \(c = -4\) | M1 | |
| \(y = 4x^2 + x^{-2} - 4\) | A1 | |
| (c) \(4x^2 + x^{-2} - 4 = 0\) | ||
| \(4x^4 - 4x^2 + 1 = 0\) | M1 | |
| \((2x^2 - 1)^2 = 0\) | M1 | |
| \(x^2 = \frac{1}{2}\) | ||
| \(x = \pm\frac{1}{\sqrt{2}}\) | A1 | |
| \(x = \pm\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \pm\frac{1}{2}\sqrt{2}\) | M1 A1 | (13) |
**(a)** $\text{grad} = 8 - 2 = 6$ | B1 |
$\therefore y - 1 = 6(x - 1)$ | M1 |
$y = 6x - 5$ | A1 |
**(b)** $y = \int (8x - \frac{2}{x^3}) \, dx$ |
$y = 4x^2 + x^{-2} + c$ | M1 A2 |
$(1, 1) \therefore 1 = 4 + 1 + c$ |
$c = -4$ | M1 |
$y = 4x^2 + x^{-2} - 4$ | A1 |
**(c)** $4x^2 + x^{-2} - 4 = 0$ |
$4x^4 - 4x^2 + 1 = 0$ | M1 |
$(2x^2 - 1)^2 = 0$ | M1 |
$x^2 = \frac{1}{2}$ |
$x = \pm\frac{1}{\sqrt{2}}$ | A1 |
$x = \pm\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \pm\frac{1}{2}\sqrt{2}$ | M1 A1 | (13)The curve $C$ has the equation $y = f(x)$.
Given that
$$\frac{dy}{dx} = 8x - \frac{2}{x^3}, \quad x \neq 0,$$
and that the point $P(1, 1)$ lies on $C$,
\begin{enumerate}[label=(\alph*)]
\item find an equation for the tangent to $C$ at $P$ in the form $y = mx + c$, [3]
\item find an equation for $C$, [5]
\item find the $x$-coordinates of the points where $C$ meets the $x$-axis, giving your answers in the form $k\sqrt{2}$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q10 [13]}}