| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve modulus equation then apply exponential/log substitution |
| Difficulty | Standard +0.8 Part (a) is routine graphing, part (b) is standard modulus equation solving. Part (c) requires recognizing the exponential substitution (letting u = 3^y) to transform it into the form of part (b), then solving the resulting exponential equation using logarithms. The substitution insight and multi-step connection between parts elevates this above average difficulty, though it's a structured question with clear guidance. |
| Spec | 1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Draw V-shaped graph with vertex on positive \(x\)-axis | B1 | |
| Draw (more or less) correct graph of \(y=2x+7\) with smaller gradient | B1 | And crossing \(y\)-axis above \(y\)-intercept of modulus graph |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Solve \(3x-5=2x+7\) to obtain \(x=12\) | B1 | |
| Attempt solution of linear equation where signs of \(3x\) and \(2x\) are different | M1 | \(3x-5=-2x-7\) OE |
| Obtain \(x=-\dfrac{2}{5}\) | A1 | |
| Alternative: State or imply \((3x-5)^2=(2x+7)^2\) | B1 | Must be working with \((3x-5)^2=(2x+7)^2\) |
| Attempt solution of 3-term quadratic equation | M1 | |
| Obtain \(-\dfrac{2}{5}\) and \(12\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Apply logarithms and use power law for \(3^y=k\) where \(k>0\) or correct equivalent | M1 | Using *their* positive answer from part (b) or greater accuracy; and no other values |
| Obtain \(2.26\) | A1 | |
| Total: 2 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Draw V-shaped graph with vertex on positive $x$-axis | B1 | |
| Draw (more or less) correct graph of $y=2x+7$ with smaller gradient | B1 | And crossing $y$-axis above $y$-intercept of modulus graph |
| **Total: 2** | | |
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Solve $3x-5=2x+7$ to obtain $x=12$ | B1 | |
| Attempt solution of linear equation where signs of $3x$ and $2x$ are different | M1 | $3x-5=-2x-7$ OE |
| Obtain $x=-\dfrac{2}{5}$ | A1 | |
| **Alternative:** State or imply $(3x-5)^2=(2x+7)^2$ | B1 | Must be working with $(3x-5)^2=(2x+7)^2$ |
| Attempt solution of 3-term quadratic equation | M1 | |
| Obtain $-\dfrac{2}{5}$ and $12$ | A1 | |
| **Total: 3** | | |
## Question 4(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Apply logarithms and use power law for $3^y=k$ where $k>0$ or correct equivalent | M1 | Using *their* positive answer from part (b) or greater accuracy; and no other values |
| Obtain $2.26$ | A1 | |
| **Total: 2** | | |4
\begin{enumerate}[label=(\alph*)]
\item Sketch, on the same diagram, the graphs of $y = | 3 x - 5 |$ and $y = 2 x + 7$.
\item Solve the equation $| 3 x - 5 | = 2 x + 7$.
\item Hence solve the equation $\left| 3 ^ { y + 1 } - 5 \right| = 2 \times 3 ^ { y } + 7$, giving your answer correct to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q4 [7]}}