| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Quadratic in disguise, intersection count |
| Difficulty | Standard +0.3 This is a multi-part question involving curve sketching, algebraic manipulation, and discriminant analysis. Part (a) is routine (finding intercepts and sketching a cubic). Part (b) is straightforward algebraic substitution. Part (c) requires recognizing the quartic as a quadratic in x², then applying the discriminant condition for two positive real solutions—a standard technique but requiring some insight. Overall slightly easier than average due to clear scaffolding and standard methods. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Negative cubic curve with one maximum and one minimum | B1 | Negative cubic in any position, must have one max and one min |
| Curve passing through (not touching) the origin | B1 | Positive or negative cubic passing through origin; origin does not need to be labelled |
| Cuts \(x\)-axis at \(x = 2\) and \(x = -2\) | B1 | Must cut so extends beyond \(x\)-axis from above or below, not a turning point. Condone \((0,2)\) and \((0,-2)\) if in correct positions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x(4-x^2) = \frac{A}{x} \Rightarrow 4x^2 - x^4 = A \Rightarrow x^4 - 4x^2 + A = 0^*\) | B1* | Must show sufficient working with at least one intermediate step between \(x(4-x^2) = \frac{A}{x}\) and printed answer. As minimum accept: \(x(4-x^2) = \frac{A}{x} \Rightarrow 4x - x^3 = \frac{A}{x} \Rightarrow x^4 - 4x^2 + A = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A > 0\) | B1 | For \(A > 0\) as one boundary. \(x > 0\) is B0 |
| \(b^2 = 4ac \Rightarrow 16 = 4A \Rightarrow A = ...\) | M1 | Attempts \(b^2 = 4ac\) or equivalent e.g. \(b^2 - 4ac = 0\) with \(a=1,\ b=-4,\ c=A\). Condone use of inequality |
| \(0 < A < 4\) | A1 | Allow equivalents e.g. "\(A > 0\) and \(A < 4\)", "\((0,4)\)" but not "\(A > 0\) or \(A < 4\)". Correct answer with no working scores 3/3. Partially correct e.g. \(0 < A \leqslant 4\) scores B1M1(implied)A0 |
## Question 8:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Negative cubic curve with one maximum and one minimum | B1 | Negative cubic in any position, must have one max and one min |
| Curve passing through (not touching) the origin | B1 | Positive or negative cubic passing through origin; origin does not need to be labelled |
| Cuts $x$-axis at $x = 2$ and $x = -2$ | B1 | Must cut so extends beyond $x$-axis from above or below, not a turning point. Condone $(0,2)$ and $(0,-2)$ if in correct positions |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x(4-x^2) = \frac{A}{x} \Rightarrow 4x^2 - x^4 = A \Rightarrow x^4 - 4x^2 + A = 0^*$ | B1* | Must show sufficient working with at least one intermediate step between $x(4-x^2) = \frac{A}{x}$ and printed answer. As minimum accept: $x(4-x^2) = \frac{A}{x} \Rightarrow 4x - x^3 = \frac{A}{x} \Rightarrow x^4 - 4x^2 + A = 0$ |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $A > 0$ | B1 | For $A > 0$ as one boundary. $x > 0$ is B0 |
| $b^2 = 4ac \Rightarrow 16 = 4A \Rightarrow A = ...$ | M1 | Attempts $b^2 = 4ac$ or equivalent e.g. $b^2 - 4ac = 0$ with $a=1,\ b=-4,\ c=A$. Condone use of inequality |
| $0 < A < 4$ | A1 | Allow equivalents e.g. "$A > 0$ and $A < 4$", "$(0,4)$" but not "$A > 0$ or $A < 4$". Correct answer with no working scores 3/3. Partially correct e.g. $0 < A \leqslant 4$ scores B1M1(implied)A0 |
---\begin{enumerate}
\item The curve $C _ { 1 }$ has equation
\end{enumerate}
$$y = x \left( 4 - x ^ { 2 } \right)$$
(a) Sketch the graph of $C _ { 1 }$ showing the coordinates of any points of intersection with the coordinate axes.
The curve $C _ { 2 }$ has equation $y = \frac { A } { x }$ where $A$ is a constant.\\
(b) Show that the $x$ coordinates of the points of intersection of $C _ { 1 }$ and $C _ { 2 }$ satisfy the equation
$$x ^ { 4 } - 4 x ^ { 2 } + A = 0$$
(c) Hence find the range of possible values of $A$ for which $C _ { 1 }$ meets $C _ { 2 }$ at 4 distinct points.
\hfill \mbox{\textit{Edexcel P1 2024 Q8 [7]}}