| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a straightforward P1 question requiring standard differentiation (including x^{-1}), finding a normal equation, and solving a simultaneous equation. All steps are routine textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.07m Tangents and normals: gradient and equations |
| Answer | Marks |
|---|---|
| 9(a) | 18 |
| Answer | Marks |
|---|---|
| 2 | B1 |
| Answer | Marks |
|---|---|
| dx x2 | M1 A1 |
| Answer | Marks |
|---|---|
| dx (cid:169)2(cid:185) | dM1 |
| States or uses y – 3 = –2(x – 2) or y = –2x + c with their (2, 3) | ddM1 |
| to deduce that y = –2x + 7 | A1* |
| Answer | Marks |
|---|---|
| (b) | 18 |
| Answer | Marks |
|---|---|
| (cid:169) 2 (cid:185) | M1 A1 |
| (2x − (cid:28)(cid:12)(cid:11)x − 2(cid:12) (cid:32) (cid:19) (cid:86)(cid:82) x = or (y – 3) (y + 2) = 0 so y = | dM1 |
| Answer | Marks |
|---|---|
| (cid:169)2 (cid:185) | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 9:
--- 9(a) ---
9(a) | 18
Substitutes x = 2 intoy(cid:32)20(cid:16)4(cid:117)2(cid:16) and gets 3
2 | B1
dy 18
(cid:32)(cid:16)4(cid:14)
dx x2 | M1 A1
dy (cid:167)1(cid:183)
Substitute x(cid:32)2(cid:159) (cid:32) then finds negative reciprocal ((cid:16)2)
(cid:168) (cid:184)
dx (cid:169)2(cid:185) | dM1
States or uses y – 3 = –2(x – 2) or y = –2x + c with their (2, 3) | ddM1
to deduce that y = –2x + 7 | A1*
(6)
(b) | 18
Put 20(cid:16)4x(cid:16) (cid:32)(cid:16)2x(cid:14)7 and simplify to give 2x2 – 13x + 18 = 0
x
(cid:167)7(cid:16) y(cid:183) 18
Or put y (cid:32)20(cid:16)4 (cid:168) (cid:184) (cid:16) to give y2 − y – 6 = 0
(cid:169) 2 (cid:185) (cid:167)7(cid:16) y(cid:183)
(cid:168) (cid:184)
(cid:169) 2 (cid:185) | M1 A1
(2x − (cid:28)(cid:12)(cid:11)x − 2(cid:12) (cid:32) (cid:19) (cid:86)(cid:82) x = or (y – 3) (y + 2) = 0 so y = | dM1
(cid:167) 9 (cid:183)
(cid:168) ,(cid:16)2(cid:184)
(cid:169)2 (cid:185) | A1 A1
(5)
(11 marks)
Notes:
(a)
B1: Substitutes x = 2 into expression for y and gets 3 cao (must be in part (a) and must use
curve equation – not line equation). This must be seen to be substituted.
M1: For an attempt to differentiate the negative power with x−1 to x−2.
dy 18
A1: Correct expression for (cid:32)(cid:16)4(cid:14)
dx x2
dM1: Dependent on first M1 substitutes x = 2 into their derivative to obtain a numerical gradient
1
and find negative reciprocal or (cid:86)(cid:87)(cid:68)(cid:87)(cid:72)(cid:86) (cid:87)(cid:75)(cid:68)(cid:87) −2× (cid:32)(cid:16)1
2
Alternative 1
1 1
dM1: Dependent on first M1. Finds equation of line using changed gradient (not their but (cid:16)
2 2
2 (cid:82)(cid:85) −2(cid:12) e.g. y – (cid:398)3(cid:398)(cid:32) – (cid:398)2(cid:398)(cid:11)x –2) or y(cid:32) (cid:398)−2(cid:398) x (cid:14) (cid:70) (cid:68)(cid:81)(cid:71) (cid:88)(cid:86)(cid:72) (cid:82)(cid:73) (cid:11)2(cid:15) (cid:398)3(cid:398)(cid:12) (cid:87)(cid:82) (cid:73)(cid:76)(cid:81)(cid:71) c =
A1*: cso. This is a given answer y= −2x + 7 obtained with no errors seen and equation should be
stated.
Alternative 2 – checking given answer
dM1: Uses given equation of line and checks that (2, 3) lies on the line.
A1*: cso. This is a given answer y (cid:32) −2x + 7 so statement that normal and line have the same
gradient and pass through the same point must be stated.
Question 9 notes continued
Question | Scheme | Marks\includegraphics{figure_3}
A sketch of part of the curve $C$ with equation
$$y = 20 - 4x - \frac{18}{x}, \quad x > 0$$
is shown in Figure 3.
Point $A$ lies on $C$ and has $x$ coordinate equal to 2
\begin{enumerate}[label=(\alph*)]
\item Show that the equation of the normal to $C$ at $A$ is $y = -2x + 7$. [6]
\end{enumerate}
The normal to $C$ at $A$ meets $C$ again at the point $B$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Use algebra to find the coordinates of $B$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2018 Q9 [11]}}