| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch then find derivative/gradient/tangent |
| Difficulty | Moderate -0.3 This is a slightly below-average C1 question. Part (a) requires identifying roots and basic curve sketching (a repeated root means 'touches'). Part (b) involves finding dy/dx using the product rule, evaluating at point A, and verifying the given tangent equation—all standard techniques with clear guidance from the question structure. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Sketch showing curve with \(y\)-intercept at origin, \(x\)-intercept at \((2,0)\) and point \(B\) at \((3,0)\) | B3 | |
| (b) \(y = (2-x)(9-6x+x^2)\) | M1 | |
| \(y = 18 - 12x + 2x^2 - 9x + 6x^2 - x^3\) | M1 | |
| \(y = 18 - 21x + 8x^2 - x^3\) | A1 | |
| \(\frac{dy}{dx} = -21 + 16x - 3x^2\) | M1 A1 | |
| At \((2,0)\): \(\text{grad} = -21 + 32 - 12 = -1\) | M1 | |
| Tangent equation: \(y - 0 = -1(x - 2)\), so \(x + y = 2\) | A1 | (10) |
**(a)** Sketch showing curve with $y$-intercept at origin, $x$-intercept at $(2,0)$ and point $B$ at $(3,0)$ | B3 |
**(b)** $y = (2-x)(9-6x+x^2)$ | M1 |
$y = 18 - 12x + 2x^2 - 9x + 6x^2 - x^3$ | M1 |
$y = 18 - 21x + 8x^2 - x^3$ | A1 |
$\frac{dy}{dx} = -21 + 16x - 3x^2$ | M1 A1 |
At $(2,0)$: $\text{grad} = -21 + 32 - 12 = -1$ | M1 |
Tangent equation: $y - 0 = -1(x - 2)$, so $x + y = 2$ | A1 | (10)5. The curve $C$ with equation $y = ( 2 - x ) ( 3 - x ) ^ { 2 }$ crosses the $x$-axis at the point $A$ and touches the $x$-axis at the point $B$.
\begin{enumerate}[label=(\alph*)]
\item Sketch the curve $C$, showing the coordinates of $A$ and $B$.
\item Show that the tangent to $C$ at $A$ has the equation
$$x + y = 2 .$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q5 [10]}}