| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Moderate -0.8 This is a straightforward C1 question requiring basic algebraic manipulation to find where y=0, then differentiation using standard power rules to find the gradient. Both parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the fractional power. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(8x - x^{\frac{5}{3}} = 0\) | M1 |
| \(x(8 - x^{\frac{2}{3}}) = 0\) | M1 A1 | |
| \(x = 0\) (at O) or \(x^{\frac{2}{3}} = 8\) | M1 A1 | |
| \(x = (\sqrt[3]{8})^2 = 4\) | M1 A1 | |
| (b) | \(\frac{dy}{dx} = 8 - \frac{5}{3}x^{\frac{2}{3}}\) | M1 A1 |
| \(\text{grad} = 8 - (\frac{5}{3} \times 8) = -12\) | M1 A1 | (7) |
(a) | $8x - x^{\frac{5}{3}} = 0$ | M1 |
| $x(8 - x^{\frac{2}{3}}) = 0$ | M1 A1 |
| $x = 0$ (at O) or $x^{\frac{2}{3}} = 8$ | M1 A1 |
| $x = (\sqrt[3]{8})^2 = 4$ | M1 A1 |
(b) | $\frac{dy}{dx} = 8 - \frac{5}{3}x^{\frac{2}{3}}$ | M1 A1 |
| $\text{grad} = 8 - (\frac{5}{3} \times 8) = -12$ | M1 A1 | (7)\includegraphics{figure_1}
Figure 1 shows the curve with equation $y = 8x - x^{\frac{3}{2}}$, $x \geq 0$.
The curve meets the $x$-axis at the origin, $O$, and at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the $x$-coordinate of $A$. [3]
\item Find the gradient of the tangent to the curve at $A$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q5 [7]}}