| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Range of k, line not intersecting curve |
| Difficulty | Moderate -0.3 This is a standard discriminant problem requiring substitution to form a quadratic, then applying b²-4ac < 0 for no intersection. Parts (ii) and (iii) involve routine coordinate geometry (solving simultaneous equations and finding perpendicular bisector). While multi-step, all techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02f 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 |
|---|---|---|
| 10(i) | 2x+ 12 =k−x or y=2 ( k− y )+ 12 → 3 term quadratic. | |
| x k− y | *M1 | Attempt to eliminate y (or x) to form a 3 term quadratic. |
| Answer | Marks | Guidance |
|---|---|---|
| Use of b² − 4ac → k² − 144 < 0 | DM1 | Using the discriminant, allow ⩽ , = 0; expect 12 and −12 |
| − 12 < k < 12 | A1 | Do NOT accept ⩽ . Separate statements OK. |
| Answer | Marks | Guidance |
|---|---|---|
| 10(ii) | Using k = 15 in their 3 term quadratic | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| x = 1,4 or y = 11, 14 | A1 | Either pair of x or y values correct.. |
| (1, 14) and (4, 11) | A1 | Both pairs of coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| 10(iii) | Gradient of AB = −1 → Perpendicular gradient = +1 | B1FT |
| Answer | Marks | Guidance |
|---|---|---|
| Finding their midpoint using their (1, 14) and (4, 11) | M1 | Expect (2½, 12½) |
| Equation: y – 12½ = (x – 2½) [y = x + 10] | A1 | Accept correct unsimplified and isw |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 10:
--- 10(i) ---
10(i) | 2x+ 12 =k−x or y=2 ( k− y )+ 12 → 3 term quadratic.
x k− y | *M1 | Attempt to eliminate y (or x) to form a 3 term quadratic.
Expect 3x² − kx + 12 or 3y2 – 5ky + (2k2 + 12) (= 0)
Use of b² − 4ac → k² − 144 < 0 | DM1 | Using the discriminant, allow ⩽ , = 0; expect 12 and −12
− 12 < k < 12 | A1 | Do NOT accept ⩽ . Separate statements OK.
3
--- 10(ii) ---
10(ii) | Using k = 15 in their 3 term quadratic | M1 | From (i) or restart. Expect 3x² − 15x + 12 or 3y2 – 75y + 462
(= 0)
x = 1,4 or y = 11, 14 | A1 | Either pair of x or y values correct..
(1, 14) and (4, 11) | A1 | Both pairs of coordinates
3
--- 10(iii) ---
10(iii) | Gradient of AB = −1 → Perpendicular gradient = +1 | B1FT | Use of m m =−1 to give +1 or ft from their A and B.
1 2
Finding their midpoint using their (1, 14) and (4, 11) | M1 | Expect (2½, 12½)
Equation: y – 12½ = (x – 2½) [y = x + 10] | A1 | Accept correct unsimplified and isw
3
Question | Answer | Marks | GuidanceThe equation of a curve is $y = 2x + \frac{12}{x}$ and the equation of a line is $y + x = k$, where $k$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Find the set of values of $k$ for which the line does not meet the curve. [3]
\end{enumerate}
In the case where $k = 15$, the curve intersects the line at points $A$ and $B$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{1}
\item Find the coordinates of $A$ and $B$. [3]
\item Find the equation of the perpendicular bisector of the line joining $A$ and $B$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2018 Q10 [9]}}