| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find tangent to polynomial curve |
| Difficulty | Moderate -0.3 This is a straightforward C1 question involving basic curve sketching, differentiation using the chain rule, and finding tangent equations. Part (d) requires finding where a parallel tangent occurs by solving f'(x)=3, which adds mild problem-solving but remains routine for C1 standard. Slightly easier than average due to the simple cubic function and clear step-by-step structure. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| (a) Sketch with \((-2, 0)\) and \((0, 8)\) marked, correct cubic shape | B3 | |
| (b) \(\text{f}(x) = (x+2)(x^2+4x+4)\) | M1 | |
| \(\text{f}(x) = x^3 + 4x^2 + 4x + 2x^2 + 8x + 8 = x^3 + 6x^2 + 12x + 8\) | A1 | |
| \(\text{f}'(x) = 3x^2 + 12x + 12\) | M1 A1 | |
| (c) grad \(= 3 - 12 + 12 = 3\) | B1 | |
| \(\therefore y - 1 = 3(x+1)\) \([y = 3x + 4]\) | M1 A1 | |
| (d) grad \(m = 3\): \(3x^2 + 12x + 12 = 3\) → \(x^2 + 4x + 3 = 0\) | ||
| \((x+1)(x+3) = 0\) | M1 | |
| \(x = -1\) (at \(P\)), \(-3\) | A1 | |
| \(x = -3 \therefore y = -1\) | M1 | |
| \(\therefore y + 1 = 3(x+3)\) → \(y = 3x + 8\) | A1 | (14) |
## Question 10:
| Answer/Working | Marks | Notes |
|---|---|---|
| **(a)** Sketch with $(-2, 0)$ and $(0, 8)$ marked, correct cubic shape | B3 | |
| **(b)** $\text{f}(x) = (x+2)(x^2+4x+4)$ | M1 | |
| $\text{f}(x) = x^3 + 4x^2 + 4x + 2x^2 + 8x + 8 = x^3 + 6x^2 + 12x + 8$ | A1 | |
| $\text{f}'(x) = 3x^2 + 12x + 12$ | M1 A1 | |
| **(c)** grad $= 3 - 12 + 12 = 3$ | B1 | |
| $\therefore y - 1 = 3(x+1)$ $[y = 3x + 4]$ | M1 A1 | |
| **(d)** grad $m = 3$: $3x^2 + 12x + 12 = 3$ → $x^2 + 4x + 3 = 0$ | | |
| $(x+1)(x+3) = 0$ | M1 | |
| $x = -1$ (at $P$), $-3$ | A1 | |
| $x = -3 \therefore y = -1$ | M1 | |
| $\therefore y + 1 = 3(x+3)$ → $y = 3x + 8$ | A1 | **(14)** |
---
**Total: (75)**10. The curve $C$ has the equation $y = \mathrm { f } ( x )$ where
$$f ( x ) = ( x + 2 ) ^ { 3 }$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the curve $C$, showing the coordinates of any points of intersection with the coordinate axes.
\item Find f ${ } ^ { \prime } ( x )$.
The straight line $l$ is the tangent to $C$ at the point $P ( - 1,1 )$.
\item Find an equation for $l$.
The straight line $m$ is parallel to $l$ and is also a tangent to $C$.
\item Show that $m$ has the equation $y = 3 x + 8$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q10 [14]}}