| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch y=|linear| and y=linear with unknown constants, then solve |
| Difficulty | Moderate -0.3 This is a straightforward modulus function question requiring standard techniques: sketching V-shaped graphs with given transformations and solving a modulus equation by considering two cases. The algebra is routine and the question structure is typical for C3, making it slightly easier than average but not trivial due to the parametric form and need for careful case analysis. |
| 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 function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| V shape in correct position touching negative \(x\)-axis | B1 | Correct orientation; could be tick shape |
| \((-a/2, 0)\) and \((0, a)\) | B1 | Accept \(-a/2\) and \(a\) marked on correct axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Translation down of previous V shape | B1 ft | ft from (i), or correct position if starts again |
| \(((b-a)/2, 0)\) and \((-(a+b)/2, 0)\) | B1, B1 | May be shown on wrong parts of \(x\)-axis or interchanged |
| Completely correct graph with \(y\)-intercept at \((0, a-b)\) | B1 | Graph must have no negative \(y\) intercept; must be two \(x\)-intercepts, one positive and one negative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2x+a) - b = \frac{1}{3}x \rightarrow \frac{5}{3}x = b - a\) | M1 | Attempts first \(+\)ve solution correctly using \((2x+a)\) |
| \(x = \frac{3}{5}(b-a)\) | A1 | Any equivalent e.g. \(x = \frac{3}{5}b - \frac{3}{5}a\) |
| \(-(2x+a) - b = \frac{1}{3}x \rightarrow -2x - \frac{1}{3}x = a+b\) | M1 | Attempts second \(-\)ve solution correctly using \(-(2x+a)\) |
| \(x = -\frac{3}{7}(a+b)\) | A1 | Any equivalent e.g. \(x = -\frac{3}{7}a - \frac{3}{7}b\) |
# Question 6:
## Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| V shape in correct position touching negative $x$-axis | B1 | Correct orientation; could be tick shape |
| $(-a/2, 0)$ **and** $(0, a)$ | B1 | Accept $-a/2$ and $a$ marked on correct axes |
## Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Translation down of previous V shape | B1 ft | ft from (i), or correct position if starts again |
| $((b-a)/2, 0)$ and $(-(a+b)/2, 0)$ | B1, B1 | May be shown on wrong parts of $x$-axis or interchanged |
| Completely correct graph with $y$-intercept at $(0, a-b)$ | B1 | Graph must have no negative $y$ intercept; must be **two** $x$-intercepts, one positive and one negative | (6) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2x+a) - b = \frac{1}{3}x \rightarrow \frac{5}{3}x = b - a$ | M1 | Attempts first $+$ve solution correctly using $(2x+a)$ |
| $x = \frac{3}{5}(b-a)$ | A1 | Any equivalent e.g. $x = \frac{3}{5}b - \frac{3}{5}a$ |
| $-(2x+a) - b = \frac{1}{3}x \rightarrow -2x - \frac{1}{3}x = a+b$ | M1 | Attempts second $-$ve solution correctly using $-(2x+a)$ |
| $x = -\frac{3}{7}(a+b)$ | A1 | Any equivalent e.g. $x = -\frac{3}{7}a - \frac{3}{7}b$ | (4) [10] |
---\begin{enumerate}
\item Given that $a$ and $b$ are constants and that $0 < a < b$,\\
(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.\\
(b) Solve, for $x$, the equation
$$| 2 x + a | - b = \frac { 1 } { 3 } x$$
giving any answers in terms of $a$ and $b$.
\hfill \mbox{\textit{Edexcel C3 2014 Q6 [10]}}