| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch y=|linear| and y=linear with unknown constants, then solve |
| Difficulty | Standard +0.3 This question requires sketching modulus functions and solving a modulus equation with parameters. Part (a) is routine sketching with straightforward intercept calculations. Part (b) involves using the given condition at x=0 to find b, then solving the resulting equation—standard algebraic manipulation but slightly elevated by the parametric context and need to consider cases for the modulus. |
| Spec | 1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| V shape on \(x\)-axis or coordinates \(\left(\frac{1}{2}a, 0\right)\) and \((0,a)\) shown | B1 | V shape anywhere on \(x\)-axis or correct intercept coordinates |
| V shape on positive \(x\)-axis at \(\left(\frac{1}{2}a,0\right)\), cutting \(y\)-axis at \((0,a)\), in quadrants 1 and 2 | B1 | Accept \(\frac{1}{2}a\) and \(a\) on correct axes; condone dotted reflection line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| V shape translated up by \((0, a+b)\) | B1ft | Follow through on (a)(i); allow U shapes and non-symmetrical |
| Correct shape, position and \((0, a+b)\) shown, vertex in quadrant 1 | B1 | V shape in quadrants 1 and 2, cutting \(y\)-axis at \((0,a+b)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses \(a + b = 8\) | B1 | Not scored for just \( |
| Solves \( | 2x-a | +b = \frac{3}{2}x+8\): \((2x-a)+b = \frac{3}{2}x+8 \Rightarrow kc = f(a,b)\) |
| Combines \(kc = f(a,b)\) with \(a+b=8 \Rightarrow c = 4a\) | dM1, A1 | Both equations must have been correct; \(c=4a\) only |
# Question 6:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| V shape on $x$-axis **or** coordinates $\left(\frac{1}{2}a, 0\right)$ and $(0,a)$ shown | B1 | V shape anywhere on $x$-axis or correct intercept coordinates |
| V shape on positive $x$-axis at $\left(\frac{1}{2}a,0\right)$, cutting $y$-axis at $(0,a)$, in quadrants 1 and 2 | B1 | Accept $\frac{1}{2}a$ and $a$ on correct axes; condone dotted reflection line |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| V shape translated up by $(0, a+b)$ | B1ft | Follow through on (a)(i); allow U shapes and non-symmetrical |
| Correct shape, position and $(0, a+b)$ shown, vertex in quadrant 1 | B1 | V shape in quadrants 1 and 2, cutting $y$-axis at $(0,a+b)$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $a + b = 8$ | B1 | Not scored for just $|0-a|+b=8$ |
| Solves $|2x-a|+b = \frac{3}{2}x+8$: $(2x-a)+b = \frac{3}{2}x+8 \Rightarrow kc = f(a,b)$ | M1 | Understanding of modulus; signs of $2x$ and $a$ must differ |
| Combines $kc = f(a,b)$ with $a+b=8 \Rightarrow c = 4a$ | dM1, A1 | Both equations must have been correct; $c=4a$ **only** |
---\begin{enumerate}
\item Given that $a$ and $b$ are positive constants,\\
(a) on separate diagrams, sketch the graph with equation\\
(i) $y = | 2 x - a |$\\
(ii) $y = | 2 x - a | + b$
\end{enumerate}
Show, on each sketch, the coordinates of each point at which the graph crosses or meets the axes.
Given that the equation
$$| 2 x - a | + b = \frac { 3 } { 2 } x + 8$$
has a solution at $x = 0$ and a solution at $x = c$,\\
(b) find $c$ in terms of $a$.
\hfill \mbox{\textit{Edexcel C3 2017 Q6 [8]}}