| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Deduce inequality solutions from sketch |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring reading roots from a graph for inequalities, polynomial expansion (routine algebra), and finding a tangent-curve intersection point. While part (c) involves several steps (finding tangent equation, solving cubic), the structure is standard and the algebraic manipulation is typical for P1 level with clear guidance provided. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| One of \(-\frac{20}{3} < x < -6\), \(x > \frac{3}{2}\) | M1 | Condone \(\leqslant \leftrightarrow <\). Allow equivalent notation e.g. \(\left(-\frac{20}{3},-6\right)\), \(\left(\frac{3}{2},\infty\right)\) |
| Both \(-\frac{20}{3} < x < -6\), \(x > \frac{3}{2}\) | A1 | Not e.g. \(-\frac{20}{3} < f(x) < -6\), \(f(x) > \frac{3}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((3x+20)(x+6)(2x-3) = (3x+20)(2x^2+9x-18) =\) | M1 | Attempts to multiply two brackets to create a quadratic before multiplying by third to form a cubic |
| \(= 6x^3 + 67x^2 + 126x - 360\) | A1 A1 | \(6x^3 + \alpha x^2 + \beta x \pm 360\) where \(\alpha\) and \(\beta\) are not both zero for first A1. Condone spurious "\(=0\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 18x^2 + 134x + 126 \Rightarrow\) Gradient at \(x=0\) is \(126\) | M1 | Attempts gradient of \(l\) by differentiating and substituting \(x=0\) |
| Equation of \(l\) is \(y = 126x - 360\) | A1ft | Follow through on their \(y = ax^3+bx^2+cx+d \Rightarrow y = cx+d\) |
| \(l\) cuts \(C\) again when \(6x^3+67x^2+126x-360 = 126x-360\) | dM1 | Sets equation of their \(l\) to their answer to (b). Depends on first M mark |
| \(6x^3+67x^2 = 0 \Rightarrow x^2(6x+67) = 0\) | ddM1 | Attempts to solve cubic of form \(ax^3+bx^2=0\) by taking out factor of \(x^2\). Depends on both previous M marks |
| \(x = -\frac{67}{6}\) | A1 | No other solutions apart from \(x=0\) which can be ignored. Ignore attempts to find \(y\) |
# Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| One of $-\frac{20}{3} < x < -6$, $x > \frac{3}{2}$ | M1 | Condone $\leqslant \leftrightarrow <$. Allow equivalent notation e.g. $\left(-\frac{20}{3},-6\right)$, $\left(\frac{3}{2},\infty\right)$ |
| Both $-\frac{20}{3} < x < -6$, $x > \frac{3}{2}$ | A1 | Not e.g. $-\frac{20}{3} < f(x) < -6$, $f(x) > \frac{3}{2}$ |
# Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(3x+20)(x+6)(2x-3) = (3x+20)(2x^2+9x-18) =$ | M1 | Attempts to multiply two brackets to create a quadratic before multiplying by third to form a cubic |
| $= 6x^3 + 67x^2 + 126x - 360$ | A1 A1 | $6x^3 + \alpha x^2 + \beta x \pm 360$ where $\alpha$ and $\beta$ are not both zero for first A1. Condone spurious "$=0$" |
# Question 10(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 18x^2 + 134x + 126 \Rightarrow$ Gradient at $x=0$ is $126$ | M1 | Attempts gradient of $l$ by differentiating and substituting $x=0$ |
| Equation of $l$ is $y = 126x - 360$ | A1ft | Follow through on their $y = ax^3+bx^2+cx+d \Rightarrow y = cx+d$ |
| $l$ cuts $C$ again when $6x^3+67x^2+126x-360 = 126x-360$ | dM1 | Sets equation of their $l$ to their answer to (b). Depends on first M mark |
| $6x^3+67x^2 = 0 \Rightarrow x^2(6x+67) = 0$ | ddM1 | Attempts to solve cubic of form $ax^3+bx^2=0$ by taking out factor of $x^2$. Depends on both previous M marks |
| $x = -\frac{67}{6}$ | A1 | No other solutions apart from $x=0$ which can be ignored. Ignore attempts to find $y$ |10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb21001f-fe68-4776-992d-ede1aae233d7-26_902_896_248_587}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve $C$ with equation $y = \mathrm { f } ( x )$, where
$$f ( x ) = ( 3 x + 20 ) ( x + 6 ) ( 2 x - 3 )$$
\begin{enumerate}[label=(\alph*)]
\item Use the given information to state the values of $x$ for which
$$f ( x ) > 0$$
\item Expand $( 3 x + 20 ) ( x + 6 ) ( 2 x - 3 )$, writing your answer as a polynomial in simplest form.
The straight line $l$ is the tangent to $C$ at the point where $C$ cuts the $y$-axis.\\
Given that $l$ cuts $C$ at the point $P$, as shown in Figure 4,
\item find, using algebra, the $x$ coordinate of $P$\\
(Solutions based on calculator technology are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2023 Q10 [10]}}