| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Finding quadratic constants from real-world trajectory or context |
| Difficulty | Standard +0.8 This is a cubic function problem requiring algebraic manipulation, differentiation, and proof. Students must set up a general cubic with constraints, use the stationary point condition (g'(2)=0 and g(2)=9), solve simultaneous equations, and prove the nature of the stationary point. It requires more problem-solving and synthesis than typical AS questions, though the individual techniques are standard. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(g(x) = ax^3 + bx^2 + ax\) | B1 | Seen or implied |
| \((2,9)\) on curve \(\Rightarrow 9 = 8a + 4b + 2a\) | M1 | Uses \(x=2, y=9\) |
| \(\Rightarrow 10a + 4b = 9\) | A1 | |
| \(g'(2) = 0 \Rightarrow 0 = 12a + 4b + a\) | M1 | Differentiates and uses \(g'(2)=0\) |
| \(\Rightarrow 13a + 4b = 0\) | A1 | |
| Solve simultaneously \(\Rightarrow a, b\) | dM1 | Dependent on both M marks |
| \(g(x) = -3x^3 + \dfrac{39}{4}x^2 - 3x\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(g''(x) = -18x + \dfrac{39}{2}\), substitute \(x = 2\) | M1 | Award for \(g''(x) = Ax + B\) with \(x=2\) substituted; or first derivative / function values either side of \(x=2\) |
| \(g''(2) = -\dfrac{33}{2} < 0\), hence maximum | A1 | Must reference sign or state less than zero |
# Question 14:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g(x) = ax^3 + bx^2 + ax$ | B1 | Seen or implied |
| $(2,9)$ on curve $\Rightarrow 9 = 8a + 4b + 2a$ | M1 | Uses $x=2, y=9$ |
| $\Rightarrow 10a + 4b = 9$ | A1 | |
| $g'(2) = 0 \Rightarrow 0 = 12a + 4b + a$ | M1 | Differentiates and uses $g'(2)=0$ |
| $\Rightarrow 13a + 4b = 0$ | A1 | |
| Solve simultaneously $\Rightarrow a, b$ | dM1 | Dependent on both M marks |
| $g(x) = -3x^3 + \dfrac{39}{4}x^2 - 3x$ | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g''(x) = -18x + \dfrac{39}{2}$, substitute $x = 2$ | M1 | Award for $g''(x) = Ax + B$ with $x=2$ substituted; or first derivative / function values either side of $x=2$ |
| $g''(2) = -\dfrac{33}{2} < 0$, hence maximum | A1 | Must reference sign or state less than zero |\begin{enumerate}
\item A curve has equation $y = \mathrm { g } ( x )$.
\end{enumerate}
Given that
\begin{itemize}
\item $\mathrm { g } ( x )$ is a cubic expression in which the coefficient of $x ^ { 3 }$ is equal to the coefficient of $x$
\item the curve with equation $y = \mathrm { g } ( x )$ passes through the origin
\item the curve with equation $y = \mathrm { g } ( x )$ has a stationary point at $( 2,9 )$\\
(a) find $\mathrm { g } ( x )$,\\
(b) prove that the stationary point at $( 2,9 )$ is a maximum.
\end{itemize}
\hfill \mbox{\textit{Edexcel AS Paper 1 2020 Q14 [9]}}