| 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 standard power rule integration, finding a constant using a boundary condition, solving a quadratic, and finding a tangent equation. All techniques are routine for this level, though the multi-part structure and fractional powers require careful execution. Slightly easier than average due to being purely procedural with no conceptual challenges. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(x) = \int (3x^{\frac{1}{2}} - 4x^{-\frac{1}{2}}) \, dx\) | ||
| \(f(x) = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} + c\) | M1 A2 | |
| \((0, 0) \therefore c = 0\) | M1 | |
| \(f(x) = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}}\) | A1 | |
| (b) \(2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} = 0\) | ||
| \(2x^{\frac{1}{2}}(x - 4) = 0\) | M1 | |
| \(x = 0\) (at O), \(4 \therefore A (4, 0)\) | A1 | |
| (c) \(x = 2 \therefore y = 2(2\sqrt{2}) - 8(\sqrt{2}) = -4\sqrt{2}\) | M1 A1 | |
| grad \(= 3\sqrt{2} - \frac{4}{\sqrt{2}} = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}\) | M1 A1 | |
| \(\therefore y + 4\sqrt{2} = \sqrt{2}(x - 2)\) | M1 | |
| \(y = \sqrt{2}x - 6\sqrt{2}\) | A1 | (13 marks) |
**(a)** $f(x) = \int (3x^{\frac{1}{2}} - 4x^{-\frac{1}{2}}) \, dx$ | |
$f(x) = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} + c$ | M1 A2 |
$(0, 0) \therefore c = 0$ | M1 |
$f(x) = 2x^{\frac{3}{2}} - 8x^{\frac{1}{2}}$ | A1 |
**(b)** $2x^{\frac{3}{2}} - 8x^{\frac{1}{2}} = 0$ | |
$2x^{\frac{1}{2}}(x - 4) = 0$ | M1 |
$x = 0$ (at O), $4 \therefore A (4, 0)$ | A1 |
**(c)** $x = 2 \therefore y = 2(2\sqrt{2}) - 8(\sqrt{2}) = -4\sqrt{2}$ | M1 A1 |
grad $= 3\sqrt{2} - \frac{4}{\sqrt{2}} = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$ | M1 A1 |
$\therefore y + 4\sqrt{2} = \sqrt{2}(x - 2)$ | M1 |
$y = \sqrt{2}x - 6\sqrt{2}$ | A1 | (13 marks)
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**Total: 75 marks**\includegraphics{figure_1}
Figure 1 shows the curve with equation $y = \text{f}(x)$.
The curve meets the $x$-axis at the origin and at the point $A$.
Given that
$$\text{f}'(x) = 3x^{\frac{1}{2}} - 4x^{-\frac{1}{2}},$$
\begin{enumerate}[label=(\alph*)]
\item find f$(x)$. [5]
\item Find the coordinates of $A$. [2]
\end{enumerate}
The point $B$ on the curve has $x$-coordinate 2.
\begin{enumerate}[label=(\alph*)]\setcounter{enumi}{2}
\item Find an equation for the tangent to the curve at $B$ in the form $y = mx + c$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q10 [13]}}