| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular bisector of segment |
| Difficulty | Moderate -0.3 Part (i) is straightforward gradient calculation with algebraic simplification to show independence from k. Part (ii) requires finding the midpoint, perpendicular gradient, and forming an equation—all standard coordinate geometry techniques. The algebra is slightly involved but follows routine procedures with no novel insight required, making this slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks |
|---|---|
| 6(i) | Carry out relevant method to find A and B such that |
| Answer | Marks |
|---|---|
| 4− y2 2+ y 2− y | M1 |
| Answer | Marks |
|---|---|
| 4 | A1 |
| Total: | 2 |
| Answer | Marks |
|---|---|
| 6(ii) | Separate variables correctly and integrate at least one side to obtain one of the terms |
| a ln x, b ln (2 + y) or c ln (2 – y) | M1 |
| Obtain term ln x | B1 |
| Answer | Marks |
|---|---|
| 4 4 | A1FT |
| Answer | Marks |
|---|---|
| least two terms of the form a ln x, b ln (2 + y) and c ln (2 – y) | M1 |
| Answer | Marks |
|---|---|
| 4 4 4 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3x4 +1 | A1 | |
| Total: | 6 | |
| Question | Answer | Marks |
Question 6:
--- 6(i) ---
6(i) | Carry out relevant method to find A and B such that
1 A B
≡ +
4− y2 2+ y 2− y | M1
1
Obtain A = B =
4 | A1
Total: | 2
--- 6(ii) ---
6(ii) | Separate variables correctly and integrate at least one side to obtain one of the terms
a ln x, b ln (2 + y) or c ln (2 – y) | M1
Obtain term ln x | B1
Integrate and obtain terms 1 ln ( 2+ y ) − 1 ln ( 2− y )
4 4 | A1FT
Use x = 1 and y = 1 to evaluate a constant, or as limits, in a solution containing at
least two terms of the form a ln x, b ln (2 + y) and c ln (2 – y) | M1
Obtain a correct solution in any form, e.g.
ln x = 1 ln ( 2+ y ) − 1 ln ( 2− y ) − 1 ln3
4 4 4 | A1
( )
2 3x4 −1
Rearrange as ( ) , or equivalent
3x4 +1 | A1
Total: | 6
Question | Answer | MarksThe coordinates of points $A$ and $B$ are $(-3k - 1, k + 3)$ and $(k + 3, 3k + 5)$ respectively, where $k$ is a constant $(k \neq -1)$.
\begin{enumerate}[label=(\roman*)]
\item Find and simplify the gradient of $AB$, showing that it is independent of $k$. [2]
\item Find and simplify the equation of the perpendicular bisector of $AB$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2018 Q6 [7]}}