| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Tangent with specified gradient |
| Difficulty | Standard +0.3 This is a straightforward chain rule application requiring differentiation of a composite function, finding a tangent equation at a given point, then using the parallel gradient condition to find another point. While it involves multiple steps, each is routine and the problem-solving demand is minimal—students simply apply standard procedures without novel insight. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = 3(3x-5)^2 \times 3 = 9(3x-5)^2\) | M1 |
| grad = 9 | A1 |
| \(y - 1 = 9(x-2)\) [\(y = 9x - 17\)] | M1 A1 |
| Answer | Marks |
|---|---|
| \(9(3x-5)^2 = 9\) | M1 |
| \(3x - 5 = \pm 1\) | A1 |
| \(x = 2\) (at P), \(\frac{4}{3}\) | A1 |
| \(Q(\frac{4}{3}, -1)\) | A1 |
| (7) |
**Part (a)**
$\frac{dy}{dx} = 3(3x-5)^2 \times 3 = 9(3x-5)^2$ | M1 |
grad = 9 | A1 |
$y - 1 = 9(x-2)$ [$y = 9x - 17$] | M1 A1 |
**Part (b)**
$9(3x-5)^2 = 9$ | M1 |
$3x - 5 = \pm 1$ | A1 |
$x = 2$ (at P), $\frac{4}{3}$ | A1 |
$Q(\frac{4}{3}, -1)$ | A1 |
| (7) |
---\begin{enumerate}
\item A curve has the equation $y = ( 3 x - 5 ) ^ { 3 }$.\\
(a) Find an equation for the tangent to the curve at the point $P ( 2,1 )$.
\end{enumerate}
The tangent to the curve at the point $Q$ is parallel to the tangent at $P$.\\
(b) Find the coordinates of $Q$.\\
\hfill \mbox{\textit{Edexcel C3 Q1 [7]}}