| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Parameter values from curve properties |
| Difficulty | Moderate -0.3 This is a straightforward curve sketching question requiring standard techniques: reading stationary points from a graph for part (a), identifying horizontal line intersections for part (b), and using given conditions to find a cubic equation in part (c). While part (c) requires setting up simultaneous equations using f'(2)=0, f'(6)=0, and f(2)=8, this is a routine textbook exercise with no novel problem-solving required. Slightly easier than average due to all key information being explicitly provided. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2 < x < 6\) | B1 | Allow \(\{x: x>2\} \cap \{x: x<6\}\), open interval \((2,6)\); do not allow \(x>2\) or \(x<6\) separately, or closed interval \([2,6]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States either \(k > 8\) or \(k < 0\) | M1 | Condone \(k \geqslant 8\) or \(k \leqslant 0\); condone \(y \leftrightarrow k\) |
| \(\{k: k>8\} \cup \{k: k<0\}\) | A1 | Condone \(\{k<0\} \cup \{k>8\}\); must use set notation with \(\{\}\) and \(\cup\); do not allow \(k<0\) or \(k>8\) without set notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States \(y = ax(x-6)^2\) or \(f(x) = ax(x-6)^2\) | M1 | Condone \(a=1\) |
| Substitutes \((2,8)\) into \(y = ax(x-6)^2\) and attempts to find \(a\) | dM1 | Dependent on correct form |
| \(y = \frac{1}{4}x(x-6)^2\) or \(f(x) = \frac{1}{4}x(x-6)^2\) | A1 | ISW after correct answer; condone \(f(x)=\frac{1}{4}x(x-6)^2\) but not \(C = \frac{1}{4}x(x-6)^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Forms two equations in \(a,b,c\) using given points/conditions | M1 | Condone \(a=1\); \(y=ax^3+bx^2+cx+d\) is M0 until \(d=0\) |
| Forms and solves three equations including \((2,8)\) | dM1 | Calculator allowed |
| \(y = \frac{1}{4}x^3 - 3x^2 + 9x\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Sets \(f'(x) = k(x-2)(x-6)\) and integrates | M1 | Condone \(k=1\) |
| Substitutes \(x=2, y=8\) into \(f(x) = k(\ldots x^3 + \ldots x + \ldots)\) to find \(k\) | dM1 | |
| \(y = \frac{3}{4}\left(\frac{1}{3}x^3 - 4x^2 + 12x\right)\) | A1 |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 < x < 6$ | B1 | Allow $\{x: x>2\} \cap \{x: x<6\}$, open interval $(2,6)$; do not allow $x>2$ or $x<6$ separately, or closed interval $[2,6]$ |
---
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States either $k > 8$ or $k < 0$ | M1 | Condone $k \geqslant 8$ or $k \leqslant 0$; condone $y \leftrightarrow k$ |
| $\{k: k>8\} \cup \{k: k<0\}$ | A1 | Condone $\{k<0\} \cup \{k>8\}$; must use set notation with $\{\}$ and $\cup$; do not allow $k<0$ or $k>8$ without set notation |
---
# Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States $y = ax(x-6)^2$ or $f(x) = ax(x-6)^2$ | M1 | Condone $a=1$ |
| Substitutes $(2,8)$ into $y = ax(x-6)^2$ and attempts to find $a$ | dM1 | Dependent on correct form |
| $y = \frac{1}{4}x(x-6)^2$ or $f(x) = \frac{1}{4}x(x-6)^2$ | A1 | ISW after correct answer; condone $f(x)=\frac{1}{4}x(x-6)^2$ but not $C = \frac{1}{4}x(x-6)^2$ |
**Alternative I:** Using $y = ax^3 + bx^2 + cx$:
| Forms two equations in $a,b,c$ using given points/conditions | M1 | Condone $a=1$; $y=ax^3+bx^2+cx+d$ is M0 until $d=0$ |
| Forms and solves three equations including $(2,8)$ | dM1 | Calculator allowed |
| $y = \frac{1}{4}x^3 - 3x^2 + 9x$ | A1 | |
**Alternative II:** Using gradient and integrating:
| Sets $f'(x) = k(x-2)(x-6)$ and integrates | M1 | Condone $k=1$ |
| Substitutes $x=2, y=8$ into $f(x) = k(\ldots x^3 + \ldots x + \ldots)$ to find $k$ | dM1 | |
| $y = \frac{3}{4}\left(\frac{1}{3}x^3 - 4x^2 + 12x\right)$ | A1 | |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{129adfbb-98fa-4e88-b636-7b4d111f3349-12_528_812_251_628}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of a curve $C$ with equation $y = \mathrm { f } ( x )$ where $\mathrm { f } ( x )$ is a cubic expression in $X$.
The curve
\begin{itemize}
\item passes through the origin
\item has a maximum turning point at $( 2,8 )$
\item has a minimum turning point at $( 6,0 )$
\begin{enumerate}[label=(\alph*)]
\item Write down the set of values of $x$ for which
\end{itemize}
$$\mathrm { f } ^ { \prime } ( x ) < 0$$
The line with equation $y = k$, where $k$ is a constant, intersects $C$ at only one point.
\item Find the set of values of $k$, giving your answer in set notation.
\item Find the equation of $C$. You may leave your answer in factorised form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2022 Q6 [6]}}