| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch two |linear| functions and solve related equation/inequality |
| Difficulty | Moderate -0.8 This is a straightforward modulus sketching question requiring students to identify V-shapes with vertices at x=3a/2 and x=-2a, then find their intersection by solving |2x-3a|=|2x+4a|. It involves routine application of modulus graph techniques with minimal problem-solving beyond standard procedures. |
| Spec | 1.02s Modulus graphs: sketch graph of |ax+b| |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Draw two V-shaped graphs, one with vertex on negative \(x\)-axis, the other with vertex on positive \(x\)-axis | M1 | |
| Show graphs in correct relation to each other, with lines (roughly) parallel as appropriate | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State coordinates \((-2a, 0)\), \((\frac{3}{2}a, 0)\), \((0, 4a)\), \((0, 3a)\) | B1 | Accept values inserted at correct points on axes |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Square both sides to obtain linear equation or inequality and attempt solution OR solve linear equation \(2x + 4a = -(2x - 3a)\) OR equivalent OR use symmetry of diagram to determine \(x\)-coordinate of point of intersection | M1 | |
| Obtain value \(-\frac{1}{4}a\) and no other | A1 | |
| State answer \(x > -\frac{1}{4}a\) | A1 | |
| Total | 3 |
## Question 3:
**Part 3(a)(i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Draw two V-shaped graphs, one with vertex on negative $x$-axis, the other with vertex on positive $x$-axis | M1 | |
| Show graphs in correct relation to each other, with lines (roughly) parallel as appropriate | A1 | |
| **Total** | **2** | |
**Part 3(a)(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| State coordinates $(-2a, 0)$, $(\frac{3}{2}a, 0)$, $(0, 4a)$, $(0, 3a)$ | B1 | Accept values inserted at correct points on axes |
| **Total** | **1** | |
**Part 3(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Square both sides to obtain linear equation or inequality and attempt solution **OR** solve linear equation $2x + 4a = -(2x - 3a)$ **OR** equivalent **OR** use symmetry of diagram to determine $x$-coordinate of point of intersection | M1 | |
| Obtain value $-\frac{1}{4}a$ and no other | A1 | |
| State answer $x > -\frac{1}{4}a$ | A1 | |
| **Total** | **3** | |
---3 It is given that $a$ is a positive constant.\\
(a) (i) Sketch on a single diagram the graphs of $y = | 2 x - 3 a |$ and $y = | 2 x + 4 a |$.\\
(ii) State the coordinates of each of the points where each graph meets an axis.\\
(b) Solve the inequality $| 2 x - 3 a | < | 2 x + 4 a |$.\\
\hfill \mbox{\textit{CAIE P2 2020 Q3 [6]}}