| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Show line is tangent, verify |
| Difficulty | Standard +0.3 This is a multi-part question involving standard techniques: finding line equations from gradient and intercept, solving simultaneous equations with a quadratic, and using perpendicular gradients. While it has several parts (a-d) worth multiple marks, each step uses routine P1 methods without requiring novel insight. The algebraic manipulation is straightforward, making it slightly easier than average for a full A-level question. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{1}{2}(x+2)\) | B1 | Correct equation in any form e.g. \(y - 0 = \frac{1}{2}(x+2)\), \(y = \frac{1}{2}x + 1\), \(y = \frac{1}{2}(x--2)\). Do not allow \(l_1 = \frac{1}{2}x+1\). Apply isw if necessary |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(l_1\) intersects parabola \(\Rightarrow -\frac{1}{4}(x+2)(x-b) = \frac{1}{2}(x+2) \Rightarrow x = ...\) | M1 | Sets \(-\frac{1}{4}(x+2)(x-b) =\) their \(\frac{1}{2}(x+2)\) and makes some attempt to solve for \(x\) (or \(y\)). Condone copying errors if intention clear. Allow even if \((x+2)\) cancelling not spotted |
| \(x = b-2\) | A1 | Correct simplified \(x\) |
| \(y = \frac{1}{2}b\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(l_2\) passes through \((b, 0)\) with gradient \(-2 \Rightarrow y = ...\) | M1 | |
| \(y - 0 = -2(x - b)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y - \text{"}\frac{1}{2}b\text{"} = \text{"}-2\text{"}(x - \text{"}b-2\text{"})\) | M1 | |
| \(y - \frac{1}{2}b = -2x + 2b - 4 \Rightarrow y = -2x + \frac{5}{2}b - 4\) | A1* | Printed answer — must be correctly derived |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = -2x + 2b = -2x + \frac{5}{2}b - 4 \Rightarrow 2b = \frac{5}{2}b - 4 \Rightarrow b = ...\) or substitute \(x=b,\ y=0\) into \(y = -2x + \frac{5}{2}b - 4 \Rightarrow b = ...\) or \(-\frac{1}{4}(x+2)(x-b) = -2(x-b) \Rightarrow x = ...(6) \Rightarrow b-2=6 \Rightarrow b = ...\) | M1 | |
| \(b = 8\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses gradient of \(-2\) and point \((b, 0)\) to find equation of \(l_2\) | M1 | If using \(y = mx + c\) must proceed as far as finding \(c\) in terms of \(b\); coordinates \((b, 0)\) must be correctly placed |
| Any correct equation e.g. \(y - 0 = -2(x - b)\), \(y = -2(x-b)\), \(y = -2x + 2b\) | A1 | Apply isw if necessary |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses gradient of \(-2\) and coordinates of \(P\) in terms of \(b\) from (a) to find equation of \(l_2\) | M1 | If using \(y = mx + c\) must proceed as far as finding \(c\) in terms of \(b\); coordinates of \(P\) must be correctly placed |
| Correct equation reached from fully correct working with full marks in (a)(ii) | A1* | Dependent on full marks in (a)(ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equates intercepts of the two equations for \(l_2\) and solves for \(b\); or uses equation from (c) with \(x = b\) when \(y = 0\); or equates parabola to \(y = -2(x-b)\), solves for \(x\), equates to \(x\)-coordinate of \(P\), solves for \(b\) | M1 | Must use the given equation in (c) and their attempt using \((b, 0)\) |
| \(b = 8\) | A1 | Correct answer only in (d) scores both marks |
# Question 10:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{1}{2}(x+2)$ | B1 | Correct equation in any form e.g. $y - 0 = \frac{1}{2}(x+2)$, $y = \frac{1}{2}x + 1$, $y = \frac{1}{2}(x--2)$. Do not allow $l_1 = \frac{1}{2}x+1$. Apply isw if necessary |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $l_1$ intersects parabola $\Rightarrow -\frac{1}{4}(x+2)(x-b) = \frac{1}{2}(x+2) \Rightarrow x = ...$ | M1 | Sets $-\frac{1}{4}(x+2)(x-b) =$ their $\frac{1}{2}(x+2)$ and makes some attempt to solve for $x$ (or $y$). Condone copying errors if intention clear. Allow even if $(x+2)$ cancelling not spotted |
| $x = b-2$ | A1 | Correct simplified $x$ |
| $y = \frac{1}{2}b$ | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $l_2$ passes through $(b, 0)$ with gradient $-2 \Rightarrow y = ...$ | M1 | |
| $y - 0 = -2(x - b)$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y - \text{"}\frac{1}{2}b\text{"} = \text{"}-2\text{"}(x - \text{"}b-2\text{"})$ | M1 | |
| $y - \frac{1}{2}b = -2x + 2b - 4 \Rightarrow y = -2x + \frac{5}{2}b - 4$ | A1* | Printed answer — must be correctly derived |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = -2x + 2b = -2x + \frac{5}{2}b - 4 \Rightarrow 2b = \frac{5}{2}b - 4 \Rightarrow b = ...$ **or** substitute $x=b,\ y=0$ into $y = -2x + \frac{5}{2}b - 4 \Rightarrow b = ...$ **or** $-\frac{1}{4}(x+2)(x-b) = -2(x-b) \Rightarrow x = ...(6) \Rightarrow b-2=6 \Rightarrow b = ...$ | M1 | |
| $b = 8$ | A1 | |
## Question (parts b, c, d):
---
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses gradient of $-2$ and point $(b, 0)$ to find equation of $l_2$ | M1 | If using $y = mx + c$ must proceed as far as finding $c$ in terms of $b$; coordinates $(b, 0)$ must be correctly placed |
| Any correct equation e.g. $y - 0 = -2(x - b)$, $y = -2(x-b)$, $y = -2x + 2b$ | A1 | Apply isw if necessary |
---
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses gradient of $-2$ and coordinates of $P$ in terms of $b$ from (a) to find equation of $l_2$ | M1 | If using $y = mx + c$ must proceed as far as finding $c$ in terms of $b$; coordinates of $P$ must be correctly placed |
| Correct equation reached from fully correct working with full marks in (a)(ii) | A1* | Dependent on full marks in (a)(ii) |
---
**Part (d):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equates intercepts of the two equations for $l_2$ and solves for $b$; or uses equation from (c) with $x = b$ when $y = 0$; or equates parabola to $y = -2(x-b)$, solves for $x$, equates to $x$-coordinate of $P$, solves for $b$ | M1 | Must use the **given equation** in (c) and their attempt using $(b, 0)$ |
| $b = 8$ | A1 | Correct answer only in (d) scores both marks |10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a5a5dd8b-1438-4698-929a-c5e3d5ed0694-28_903_1010_219_539}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows a sketch of the quadratic curve $C$ with equation
$$y = - \frac { 1 } { 4 } ( x + 2 ) ( x - b ) \quad \text { where } b \text { is a positive constant }$$
The line $l _ { 1 }$ also shown in Figure 5,
\begin{itemize}
\item has gradient $\frac { 1 } { 2 }$
\item intersects $C$ on the negative $x$-axis and at the point $P$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down an equation for $l _ { 1 }$
\item Find, in terms of $b$, the coordinates of $P$
\end{itemize}
Given that the line $l _ { 2 }$ is perpendicular to $l _ { 1 }$ and intersects $C$ on the positive $x$-axis,
\end{enumerate}\item find, in terms of $b$, an equation for $l _ { 2 }$
Given also that $l _ { 2 }$ intersects $C$ at the point $P$
\item show that another equation for $l _ { 2 }$ is
$$y = - 2 x + \frac { 5 b } { 2 } - 4$$
\item Hence, or otherwise, find the value of $b$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2023 Q10 [10]}}