| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Determine constant from stationary point condition |
| Difficulty | Moderate -0.3 This is a straightforward integration problem requiring students to use the condition that dy/dx = 0 at a turning point to find k, then integrate to find the original function. While it involves multiple steps (finding k, integrating, using initial condition, finding y-intercept), each step follows standard procedures with no novel insight required. Slightly easier than average due to clear structure and routine techniques. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(2(3)^3 - 9(3)^2 + 5(3) + k = 0 \Rightarrow k = \ldots\) | M1 | Substitutes \(x=3\) into given derivative, sets \(=0\), solves for \(k\) |
| \(54 - 81 + 15 + k = 0 \Rightarrow k = 12^*\) or \(-12 + k = 0 \Rightarrow k = 12^*\) | A1\* | Obtains \(k=12\) with no errors and sufficient working shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(2(3)^3 - 9(3)^2 + 5(3) + 12 = 0\) | M1 | Substitutes \(x=3\) and \(k=12\) into derivative |
| \(54 - 81 + 15 + 12 = 0\) hence \(k=12^*\) | A1\* | Correct work obtaining 0 with minimal conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Divide \(2x^3 - 9x^2 + 5x + k\) by \((x-3)\), proceed to remainder \(= 0\); requires at least \(2x^2 \pm 3x\) | M1 | Attempts division by \((x-3)\) proceeding as far as remainder \(= 0\) |
| \(k - 12 = 0 \Rightarrow k = 12\) | A1\* | Obtains \(k=12\); division must be correct but allow \(k-12\) or \(0\) as "remainder" |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int(2x^3 - 9x^2 + 5x + 12)\,\mathrm{d}x \to x^4 \pm \ldots x^3 \pm \ldots x^2 \pm \ldots x \pm \ldots\) | M1 | Attempts to integrate; evidence from at least 2 terms: \(2x^3 \to \ldots x^4\), \(-9x^2 \to \ldots x^3\), \(5x \to \ldots x^2\), or \(12 \to \ldots x\) |
| \(\frac{1}{2}(3)^4 - 3(3)^3 + \frac{5}{2}(3)^2 + 12(3) + c = -10 \Rightarrow c = \ldots\) | dM1 | Substitutes \(x=3\) into integrated expression including constant of integration, sets equal to \(\pm 10\), finds \(c\). Depends on previous mark |
| \((0,\ -28)\) | A1 | Condone \(-28\) or \(y=-28\) but not just \(c=-28\). No other values or points |
## Question 5:
### Part (a):
| $2(3)^3 - 9(3)^2 + 5(3) + k = 0 \Rightarrow k = \ldots$ | **M1** | Substitutes $x=3$ into given derivative, sets $=0$, solves for $k$ |
| $54 - 81 + 15 + k = 0 \Rightarrow k = 12^*$ or $-12 + k = 0 \Rightarrow k = 12^*$ | **A1\*** | Obtains $k=12$ with no errors and sufficient working shown |
**Alternative by verification:**
| $2(3)^3 - 9(3)^2 + 5(3) + 12 = 0$ | **M1** | Substitutes $x=3$ and $k=12$ into derivative |
| $54 - 81 + 15 + 12 = 0$ hence $k=12^*$ | **A1\*** | Correct work obtaining 0 with minimal conclusion |
**Alternative using algebraic division:**
| Divide $2x^3 - 9x^2 + 5x + k$ by $(x-3)$, proceed to remainder $= 0$; requires at least $2x^2 \pm 3x$ | **M1** | Attempts division by $(x-3)$ proceeding as far as remainder $= 0$ |
| $k - 12 = 0 \Rightarrow k = 12$ | **A1\*** | Obtains $k=12$; division must be correct but allow $k-12$ or $0$ as "remainder" |
### Part (b):
| $\int(2x^3 - 9x^2 + 5x + 12)\,\mathrm{d}x \to x^4 \pm \ldots x^3 \pm \ldots x^2 \pm \ldots x \pm \ldots$ | **M1** | Attempts to integrate; evidence from at least 2 terms: $2x^3 \to \ldots x^4$, $-9x^2 \to \ldots x^3$, $5x \to \ldots x^2$, or $12 \to \ldots x$ |
| $\frac{1}{2}(3)^4 - 3(3)^3 + \frac{5}{2}(3)^2 + 12(3) + c = -10 \Rightarrow c = \ldots$ | **dM1** | Substitutes $x=3$ into integrated expression including constant of integration, sets equal to $\pm 10$, finds $c$. Depends on previous mark |
| $(0,\ -28)$ | **A1** | Condone $-28$ or $y=-28$ but not just $c=-28$. No other values or points |
---\begin{enumerate}
\item The curve $C$ has equation $y = \mathrm { f } ( x )$
\end{enumerate}
The curve
\begin{itemize}
\item passes through the point $P ( 3 , - 10 )$
\item has a turning point at $P$
\end{itemize}
Given that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 x ^ { 3 } - 9 x ^ { 2 } + 5 x + k$$
where $k$ is a constant,\\
(a) show that $k = 12$\\
(b) Hence find the coordinates of the point where $C$ crosses the $y$-axis.
\hfill \mbox{\textit{Edexcel Paper 2 2023 Q5 [5]}}