| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring completion of the square to find the minimum point, finding the equation of a line through the origin and a known point, solving a quadratic equation, and writing inequalities to define a region. All techniques are standard P1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02i Represent inequalities: graphically on coordinate plane1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x^2 - 5x + 13 = (x-2.5)^2 - 2.5^2 + 13 = (x-2.5)^2 + 6.75\) | M1A1 | For attempting to complete the square. Look for \((x-2.5)^2\). \((x-2.5)^2+6.75\) or correctly extracting \(x=2.5\) |
| Coordinates \(M = (2.5, 6.75)\) | A1 | \(M=(2.5, 6.75)\) or \(\left(\frac{5}{2}, \frac{27}{4}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts equation of \(l\) using \(M\): \(y = \frac{6.75}{2.5}x\) \((y=2.7x)\) | M1 | Uses their \(M\) to find equation for \(l\). Look for correct attempt at gradient: \(y = \frac{"6.75"}{" 2.5"}x\) |
| Attempts to solve \(y = 2.7x\) with \(y = x^2 - 5x + 13\) | M1 | Depends on having attempted (not necessarily correct) use of \(O\) and \(M\) to find equation of \(l\) |
| \(\Rightarrow 2.7x = x^2-5x+13 \Rightarrow x^2-7.7x+13=0 \Rightarrow (x-2.5)(x-5.2)=0\) | Correct answers following correct simplified quadratic fine for method | |
| \(x = 5.2\) oe | A1 | \(x=5.2\) or equivalent such as \(x=\frac{26}{5}\) |
| Coordinates \(N = (5.2, 14.04)\) | dM1 A1 | dM1: substitutes \(x=5.2\) into either equation. A1: \(N=(5.2,14.04)\) oe such as \(\left(\frac{26}{5},\frac{351}{25}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| States two of: \(y < x^2-5x+13\), \(y > 2.7x\), \(0 \leq x < 2.5\) | M1 | Also allow first two combined, or loose inequalities. Allow \(\leq\) or \(<\) on left hand end of \(0\leq x < 2.5\) |
| States all three of: \(y < x^2-5x+13\), \(y > 2.7x\), \(0 \leq x < 2.5\) | A1ft | Follow through on their \(2.7x\) from (b) and 2.5. Use of \(R\) instead of \(y\) is M0 |
# Question 3:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $x^2 - 5x + 13 = (x-2.5)^2 - 2.5^2 + 13 = (x-2.5)^2 + 6.75$ | M1A1 | For attempting to complete the square. Look for $(x-2.5)^2$. $(x-2.5)^2+6.75$ or correctly extracting $x=2.5$ |
| Coordinates $M = (2.5, 6.75)$ | A1 | $M=(2.5, 6.75)$ or $\left(\frac{5}{2}, \frac{27}{4}\right)$ |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts equation of $l$ using $M$: $y = \frac{6.75}{2.5}x$ $(y=2.7x)$ | M1 | Uses their $M$ to find equation for $l$. Look for correct attempt at gradient: $y = \frac{"6.75"}{" 2.5"}x$ |
| Attempts to solve $y = 2.7x$ with $y = x^2 - 5x + 13$ | M1 | Depends on having attempted (not necessarily correct) use of $O$ and $M$ to find equation of $l$ |
| $\Rightarrow 2.7x = x^2-5x+13 \Rightarrow x^2-7.7x+13=0 \Rightarrow (x-2.5)(x-5.2)=0$ | | Correct answers following correct simplified quadratic fine for method |
| $x = 5.2$ oe | A1 | $x=5.2$ or equivalent such as $x=\frac{26}{5}$ |
| Coordinates $N = (5.2, 14.04)$ | dM1 A1 | dM1: substitutes $x=5.2$ into either equation. A1: $N=(5.2,14.04)$ oe such as $\left(\frac{26}{5},\frac{351}{25}\right)$ |
## Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| States two of: $y < x^2-5x+13$, $y > 2.7x$, $0 \leq x < 2.5$ | M1 | Also allow first two combined, or loose inequalities. Allow $\leq$ or $<$ on left hand end of $0\leq x < 2.5$ |
| States all three of: $y < x^2-5x+13$, $y > 2.7x$, $0 \leq x < 2.5$ | A1ft | Follow through on their $2.7x$ from (b) and 2.5. Use of $R$ instead of $y$ is M0 |
---3. In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{50ec901b-b6b6-4b72-85bd-a084f313c99b-06_583_588_395_680}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve $C$ with equation $y = x ^ { 2 } - 5 x + 13$
The point $M$ is the minimum point of $C$.
The straight line $l$ passes through the origin $O$ and intersects $C$ at the points $M$ and $N$ as shown.
Find, showing your working,
\begin{enumerate}[label=(\alph*)]
\item the coordinates of $M$,
\item the coordinates of $N$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{50ec901b-b6b6-4b72-85bd-a084f313c99b-06_531_561_1793_680}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the curve $C$ and the line $l$. The finite region $R$, shown shaded in Figure 3, is bounded by $C , l$ and the $y$-axis.
\item Use inequalities to define the region $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2019 Q3 [10]}}