| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| 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 +0.3 This is a straightforward multi-part question requiring standard techniques: sketching basic functions including modulus, solving a linear modulus equation by cases, and applying exponential substitution (let u = 5^(y/2)) to reduce part (c) to the same form as part (b). The substitution is heavily scaffolded by the question structure, requiring only routine algebraic manipulation and logarithms to find y. Slightly above average due to the substitution step, but well within standard P2 expectations. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02s 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 | Marks | Guidance |
| Draw V-shaped graph with vertex on positive \(x\)-axis | \*B1 | Must be straight lines |
| Draw (more or less) correct graph of \(y = x + 3\) with smaller gradient together with a V shaped graph | DB1 | And crossing \(y\)-axis above \(y\)-intercept of first graph. Intersection in the first quadrant may be implied |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Solve \(x+3=2x-1\) to obtain \(x=4\) | B1 | |
| Attempt solution of linear equation where signs of \(2x\) and \(x\) are different | M1 | |
| Obtain \(x=-\frac{2}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply non-modulus equation \((x+3)^2=(2x-1)^2\) | B1 | |
| Attempt solution of 3-term quadratic equation obtained from squaring both terms | M1 | Must have B1 |
| Obtain \(-\frac{2}{3}\) and \(4\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Apply logarithms and use power law for \(5^{\frac{1}{2}y}=k\) where \(k>0\) | M1 | Using *their* positive root from part (b). Allow M1 for \(y=2\log_5 4\) |
| Obtain \(y=1.72\) | A1 | AWRT; and no other values |
| Total | 2 |
**Question 2(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Draw V-shaped graph with vertex on positive $x$-axis | \*B1 | Must be straight lines |
| Draw (more or less) correct graph of $y = x + 3$ with smaller gradient together with a V shaped graph | DB1 | And crossing $y$-axis above $y$-intercept of first graph. Intersection in the first quadrant may be implied |
| **Total** | **2** | |
## Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Solve $x+3=2x-1$ to obtain $x=4$ | B1 | |
| Attempt solution of linear equation where signs of $2x$ and $x$ are different | M1 | |
| Obtain $x=-\frac{2}{3}$ | A1 | |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modulus equation $(x+3)^2=(2x-1)^2$ | B1 | |
| Attempt solution of 3-term quadratic equation obtained from squaring both terms | M1 | Must have B1 |
| Obtain $-\frac{2}{3}$ and $4$ | A1 | |
| **Total** | **3** | |
---
## Question 2(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Apply logarithms and use power law for $5^{\frac{1}{2}y}=k$ where $k>0$ | M1 | Using *their* positive root from part (b). Allow M1 for $y=2\log_5 4$ |
| Obtain $y=1.72$ | A1 | AWRT; and no other values |
| **Total** | **2** | |
---2
\begin{enumerate}[label=(\alph*)]
\item Sketch, on the same diagram, the graphs of $y = x + 3$ and $y = | 2 x - 1 |$.
\item Solve the equation $x + 3 = | 2 x - 1 |$.
\item Find the value of $y$ such that $5 ^ { \frac { 1 } { 2 } y } + 3 = \left| 2 \times 5 ^ { \frac { 1 } { 2 } y } - 1 \right|$. Give your answer correct to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2021 Q2 [7]}}