| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch y=|linear| and y=linear, solve inequality: numeric coefficients |
| Difficulty | Moderate -0.8 Part (a) is a routine sketch of a single modulus function requiring only knowledge of the V-shape transformation (reflection of negative part). Part (b) is a standard inequality solved by considering two cases based on the critical point x = -3/2, requiring algebraic manipulation but following a well-practiced procedure. This is below average difficulty as it's a textbook exercise with no novel insight required. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b| |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sketch of \(y = \ | 2x+3\ | \): straight lines, vertex at approximately \(\left(-\dfrac{3}{2}, 0\right)\), \(y\)-intercept at \(3\), symmetrical V-shape |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Find \(x\)-coordinate of intersection with \(y = 3x + 8\) | M1 | |
| Obtain \(x = -\frac{11}{5}\) | A1 | |
| State final answer \(x > -\frac{11}{5}\) only | A1 | \((x > -2.2)\) Do not condone \(\geqslant\) for \(>\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solve the linear inequality \(3x + 8 > -(2x+3)\), or corresponding linear equation | M1 | |
| Obtain critical value \(x = -\frac{11}{5}\) | A1 | |
| State final answer \(x > -\frac{11}{5}\) only | A1 | \((x > -2.2)\) Do not condone \(\geqslant\) for \(>\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solve the quadratic inequality \((3x+8)^2 > (2x+3)^2\), or corresponding quadratic equation | (M1) | \(5x^2 + 36x + 55\) |
| Obtain critical value \(x = -\frac{11}{5}\) | (A1) | Ignore \(-5\) if seen |
| State final answer \(x > -\frac{11}{5}\) only | (A1) | \((x > -2.2)\) Do not condone \(\geqslant\) for \(>\) |
| 3 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sketch of $y = \|2x+3\|$: straight lines, vertex at approximately $\left(-\dfrac{3}{2}, 0\right)$, $y$-intercept at $3$, symmetrical V-shape | **B1** | Show a recognisable sketch graph of $y = \|2x+3\|$. Ignore any attempt to sketch $y = 3x+8$. Straight lines. Vertex in approximately correct position on $x$-axis. Symmetry. |
| | **Total: 1** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Find $x$-coordinate of intersection with $y = 3x + 8$ | M1 | |
| Obtain $x = -\frac{11}{5}$ | A1 | |
| State final answer $x > -\frac{11}{5}$ only | A1 | $(x > -2.2)$ Do not condone $\geqslant$ for $>$ |
**Alternative Method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solve the linear inequality $3x + 8 > -(2x+3)$, or corresponding linear equation | M1 | |
| Obtain critical value $x = -\frac{11}{5}$ | A1 | |
| State final answer $x > -\frac{11}{5}$ only | A1 | $(x > -2.2)$ Do not condone $\geqslant$ for $>$ |
**Alternative Method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solve the quadratic inequality $(3x+8)^2 > (2x+3)^2$, or corresponding quadratic equation | (M1) | $5x^2 + 36x + 55$ |
| Obtain critical value $x = -\frac{11}{5}$ | (A1) | Ignore $-5$ if seen |
| State final answer $x > -\frac{11}{5}$ only | (A1) | $(x > -2.2)$ Do not condone $\geqslant$ for $>$ |
| | **3** | |
---2
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $y = | 2 x + 3 |$.
\item Solve the inequality $3 x + 8 > | 2 x + 3 |$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q2 [4]}}