| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Prove curve has no turning points |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question requiring basic differentiation, substitution into simultaneous equations, and analyzing the discriminant of a quadratic. Part (a) involves routine algebraic manipulation, while part (b) requires showing dy/dx has no real roots—a standard technique. Slightly easier than average due to clear structure and standard methods. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=2,\ y=5 \Rightarrow 5=8a-12+6+b\) | M1 | Substitutes \(x=2,\ y=5\) into \(y=ax^3-3x^2+3x+b\) (condone slips) |
| \(\frac{dy}{dx}=3ax^2-6x+3\) AND \(x=2,\ \frac{dy}{dx}=7 \Rightarrow 7=12a-12+3\) | M1 | Substitutes \(x=2,\ \frac{dy}{dx}=7\) into \(\frac{dy}{dx}=3ax^2-6x+3\) (\(\frac{dy}{dx}\) must be correct) |
| Solves \(11=8a+b\) and \(7=12a-9 \Rightarrow a=\frac{4}{3},\ b=\frac{1}{3}\) | A1 A1 | A1: \(a=\frac{4}{3}\) or exact equivalent. A1: \(b=\frac{1}{3}\) or exact equivalent. |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(\frac{dy}{dx}=3ax^2-6x+3=0\) with their value of \(a\) | M1 | May be implied by later working. |
| \(4x^2-6x+3=0\) and attempts \(b^2-4ac\) | dM1 | Attempts discriminant or roots via formula. |
| \(b^2-4ac=-12<0\) hence there are no turning points | A1* | Must achieve \(4x^2-6x+3=0\), \(b^2-4ac=-12\), state \(-12<0\) and conclude no turning points. Full marks available with incorrect \(b=\frac{1}{3}\) but not with incorrect \(a=\frac{4}{3}\). |
| Alt (b): Complete the square: \(4x^2-6x+3=4\left(x-\frac{3}{4}\right)^2+\frac{3}{4}\) | A1* | States \(\frac{dy}{dx}>0\) (for all \(x\)) hence no turning points. |
| (3) |
## Question 10:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=2,\ y=5 \Rightarrow 5=8a-12+6+b$ | M1 | Substitutes $x=2,\ y=5$ into $y=ax^3-3x^2+3x+b$ (condone slips) |
| $\frac{dy}{dx}=3ax^2-6x+3$ AND $x=2,\ \frac{dy}{dx}=7 \Rightarrow 7=12a-12+3$ | M1 | Substitutes $x=2,\ \frac{dy}{dx}=7$ into $\frac{dy}{dx}=3ax^2-6x+3$ ($\frac{dy}{dx}$ must be correct) |
| Solves $11=8a+b$ and $7=12a-9 \Rightarrow a=\frac{4}{3},\ b=\frac{1}{3}$ | A1 A1 | A1: $a=\frac{4}{3}$ or exact equivalent. A1: $b=\frac{1}{3}$ or exact equivalent. |
| | **(4)** | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $\frac{dy}{dx}=3ax^2-6x+3=0$ with their value of $a$ | M1 | May be implied by later working. |
| $4x^2-6x+3=0$ and attempts $b^2-4ac$ | dM1 | Attempts discriminant or roots via formula. |
| $b^2-4ac=-12<0$ hence there are no turning points | A1* | Must achieve $4x^2-6x+3=0$, $b^2-4ac=-12$, state $-12<0$ and conclude **no turning points**. Full marks available with incorrect $b=\frac{1}{3}$ but not with incorrect $a=\frac{4}{3}$. |
| **Alt (b):** Complete the square: $4x^2-6x+3=4\left(x-\frac{3}{4}\right)^2+\frac{3}{4}$ | A1* | States $\frac{dy}{dx}>0$ (for all $x$) hence **no turning points**. |
| | **(3)** | |
The images you've shared appear to be blank pages (pages 27 and 28) from a Pearson Education document, containing no mark scheme content — only the PMT watermark, page numbers, and the Pearson company registration footer on page 27.
There is no mark scheme content to extract from these pages. Could you share the pages that actually contain the mark scheme questions and answers?10. The curve $C$ has equation
$$y = a x ^ { 3 } - 3 x ^ { 2 } + 3 x + b$$
where $a$ and $b$ are constants.
Given that
\begin{itemize}
\item the point $( 2,5 )$ lies on $C$
\item the gradient of the curve at $( 2,5 )$ is 7
\begin{enumerate}[label=(\alph*)]
\item find the value of $a$ and the value of $b$.
\item Prove that $C$ has no turning points.\\
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2019 Q10 [7]}}