| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary points coordinates |
| Difficulty | Standard +0.8 This question requires finding the second derivative to verify inflection point (routine), then using the distance formula combined with solving a cubic equation to find k. The integration, root-finding, and algebraic manipulation across multiple steps elevates this above standard textbook exercises, though the techniques themselves are A-level standard. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07f Convexity/concavity: points of inflection1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f''(x)=-4-6x=0\) | M1 | Sets \(f''(x)=0\) |
| Criteria 1: \(f''(x)=0 \Rightarrow x=\frac{4}{-6}\Rightarrow x=-\frac{2}{3}\), or verifies \(f''\!\left(-\frac{2}{3}\right)=0\) | ||
| Criteria 2: Either checks \(f''(-0.7)=0.2>0\), \(f''(-0.6)=-0.4<0\); or \(f'''\!\left(-\frac{2}{3}\right)=-6\neq 0\) | ||
| At least one of Criteria 1 or Criteria 2 | B1 | AO 2.4 |
| Both Criteria 1 and Criteria 2, and concludes \(C\) has a point of inflection at \(x=-\frac{2}{3}\) | A1 | AO 2.1 |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x)=kx-2x^2-x^3\ \{+c\}\) | M1, A1 | Integrates \(f'(x)\) correctly |
| \(f(0)=0 \Rightarrow c=0 \Rightarrow f(x)=kx-2x^2-x^3\) | A1 | Uses \((0,0)\) to find \(c=0\); AO 2.2a |
| \(f(x)=0 \Rightarrow x(k-2x-x^2)=0 \Rightarrow x=0\) or \(k-2x-x^2=0\) | ||
| \(x^2+2x-k=0 \Rightarrow (x+1)^2-1-k=0\) | M1 | Completes the square or uses quadratic formula; AO 2.1 |
| \(x=-1\pm\sqrt{k+1}\) | A1 | |
| \(AB=\left(-1+\sqrt{k+1}\right)-\left(-1-\sqrt{k+1}\right)=4\sqrt{2}\) | M1 | Uses \(AB=4\sqrt{2}\); AO 2.1 |
| \(2\sqrt{k+1}=4\sqrt{2}\Rightarrow k=7\) | A1 | |
| Total: (7) | (10 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to find \(f''\left(-\frac{2}{3}\right)\) | M1 | Attempts to find \(f''(x)\) and sets result equal to 0 |
| See scheme | B1 | |
| See scheme | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integrates \(f'(x)\) to give \(f(x) = \pm kx \pm \alpha x^2 \pm \beta x^3\), \(\alpha, \beta \neq 0\) | M1 | With or without constant of integration |
| \(f(x) = kx - 2x^2 - x^3\) | A1 | With or without constant of integration |
| Finds \(f(x) = kx - 2x^2 - x^3 + c\); deduces \(c = 0\) since curve passes through origin; gives \(k - 2x - x^2 = 0\) or \(x^2 + 2x - k = 0\) | A1 | References \(y = f(x)\) passing through origin to deduce \(c = 0\) |
| Uses valid method to solve quadratic to give \(x\) in terms of \(k\) | M1 | |
| \(x = -1 \pm \sqrt{k+1}\) | A1 | Correct roots for \(x\) in terms of \(k\) |
| Applies \(AB = 4\sqrt{2}\) on \(x = -1 \pm \sqrt{k+1}\) to find \(k\) | M1 | Complete method to find \(k\) |
| \(k = 7\) | A1 | From correct solution only |
## Question 11:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f''(x)=-4-6x=0$ | M1 | Sets $f''(x)=0$ |
| **Criteria 1:** $f''(x)=0 \Rightarrow x=\frac{4}{-6}\Rightarrow x=-\frac{2}{3}$, or verifies $f''\!\left(-\frac{2}{3}\right)=0$ | | |
| **Criteria 2:** Either checks $f''(-0.7)=0.2>0$, $f''(-0.6)=-0.4<0$; or $f'''\!\left(-\frac{2}{3}\right)=-6\neq 0$ | | |
| At least one of Criteria 1 or Criteria 2 | B1 | AO 2.4 |
| Both Criteria 1 and Criteria 2, **and** concludes $C$ has a point of inflection at $x=-\frac{2}{3}$ | A1 | AO 2.1 |
| **Total: (3)** | | |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x)=kx-2x^2-x^3\ \{+c\}$ | M1, A1 | Integrates $f'(x)$ correctly |
| $f(0)=0 \Rightarrow c=0 \Rightarrow f(x)=kx-2x^2-x^3$ | A1 | Uses $(0,0)$ to find $c=0$; AO 2.2a |
| $f(x)=0 \Rightarrow x(k-2x-x^2)=0 \Rightarrow x=0$ or $k-2x-x^2=0$ | | |
| $x^2+2x-k=0 \Rightarrow (x+1)^2-1-k=0$ | M1 | Completes the square or uses quadratic formula; AO 2.1 |
| $x=-1\pm\sqrt{k+1}$ | A1 | |
| $AB=\left(-1+\sqrt{k+1}\right)-\left(-1-\sqrt{k+1}\right)=4\sqrt{2}$ | M1 | Uses $AB=4\sqrt{2}$; AO 2.1 |
| $2\sqrt{k+1}=4\sqrt{2}\Rightarrow k=7$ | A1 | |
| **Total: (7)** | **(10 marks)** | |
## Question 11:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to find $f''\left(-\frac{2}{3}\right)$ | M1 | Attempts to find $f''(x)$ and sets result equal to 0 |
| See scheme | B1 | |
| See scheme | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrates $f'(x)$ to give $f(x) = \pm kx \pm \alpha x^2 \pm \beta x^3$, $\alpha, \beta \neq 0$ | M1 | With or without constant of integration |
| $f(x) = kx - 2x^2 - x^3$ | A1 | With or without constant of integration |
| Finds $f(x) = kx - 2x^2 - x^3 + c$; deduces $c = 0$ since curve passes through origin; gives $k - 2x - x^2 = 0$ or $x^2 + 2x - k = 0$ | A1 | References $y = f(x)$ passing through origin to deduce $c = 0$ |
| Uses valid method to solve quadratic to give $x$ in terms of $k$ | M1 | |
| $x = -1 \pm \sqrt{k+1}$ | A1 | Correct roots for $x$ in terms of $k$ |
| Applies $AB = 4\sqrt{2}$ on $x = -1 \pm \sqrt{k+1}$ to find $k$ | M1 | Complete method to find $k$ |
| $k = 7$ | A1 | From correct solution only |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{96e004d9-c6b6-474b-9b67-06e1771c609e-22_760_1182_248_443}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows a sketch of the curve $C$ with equation $y = \mathrm { f } ( x )$.\\
The curve $C$ crosses the $x$-axis at the origin, $O$, and at the points $A$ and $B$ as shown in Figure 5.
Given that
$$f ^ { \prime } ( x ) = k - 4 x - 3 x ^ { 2 }$$
where $k$ is constant,
\begin{enumerate}[label=(\alph*)]
\item show that $C$ has a point of inflection at $x = - \frac { 2 } { 3 }$
Given also that the distance $A B = 4 \sqrt { 2 }$
\item find, showing your working, the integer value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 Q11 [10]}}