| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Rational curve intersections |
| Difficulty | Moderate -0.3 This C1 question requires sketching two standard curves (a parabola and a reciprocal function) and interpreting intersections graphically. While it involves multiple steps and graphical reasoning, the individual components are routine: completing the square for the parabola, recognizing the reciprocal curve shape, and counting intersections. The conceptual demand is modest for A-level, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Graphs of \(y = x^2 - 4x\) and \(y = -\frac{1}{x}\) | B2 B2 | |
| (b) 3 solutions | B1 | |
| \(x^2 - 4x + \frac{1}{x} = 0 \Rightarrow x^2 - 4x = -\frac{1}{x}\) | B1 | |
| and the graphs of \(y = x^2 - 4x\) and \(y = -\frac{1}{x}\) intersect at 3 points | B1 | (6) |
**(a)** Graphs of $y = x^2 - 4x$ and $y = -\frac{1}{x}$ | B2 B2 |
**(b)** 3 solutions | B1 |
$x^2 - 4x + \frac{1}{x} = 0 \Rightarrow x^2 - 4x = -\frac{1}{x}$ | B1 |
and the graphs of $y = x^2 - 4x$ and $y = -\frac{1}{x}$ intersect at 3 points | B1 | (6)\begin{enumerate}[label=(\alph*)]
\item Sketch on the same diagram the curves $y = x^2 - 4x$ and $y = -\frac{1}{x}$. [4]
\item State, with a reason, the number of real solutions to the equation
$$x^2 - 4x + \frac{1}{x} = 0.$$ [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q4 [6]}}