| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Graph y = a|bx+c| + d: identify vertex and intercepts |
| Difficulty | Moderate -0.3 This is a straightforward modulus function question requiring standard techniques: sketching a V-shaped graph and solving by considering cases (x ≥ 0 and x < 0). The algebra is routine and the question structure is typical of C3/C4 textbook exercises, making it slightly easier than average. |
| 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 | Mark | Guidance |
| V shape with vertex anywhere on the \(y\)-axis, branches approximately symmetrical about the \(y\)-axis | B1 | Ignore any dashed or dotted lines |
| Intercepts at \(\left(-\frac{k}{2}, 0\right)\), \(\left(\frac{k}{2}, 0\right)\) and \(\left(0, -k\right)\) and no others | B1 | Must be crossing intercepts. Allow coordinates wrong way round if positioning correct. There must be a sketch for this mark. If ambiguity exists, sketch has precedence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x - k = \frac{1}{2}x + \frac{k}{4} \Rightarrow x = \ldots\) or \(-2x - k = \frac{1}{2}x + \frac{k}{4} \Rightarrow x = \ldots\) | M1 | Attempt to solve either equation to make \(x\) or \(k\) the subject |
| \(x = \frac{5k}{6}\) or \(x = -\frac{k}{2}\) | A1 | One correct value for \(x\). Allow equivalent fractions e.g. \(\frac{10k}{12}\), \(-\frac{2k}{4}\) etc. |
| \(x = \frac{5k}{6}\) and \(x = -\frac{k}{2}\) | A1 | Both \(x\) values correct. Allow equivalent fractions e.g. \(\frac{10k}{12}\), \(-\frac{2k}{4}\) etc. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\ | x\ | - k = \frac{1}{2}x + \frac{k}{4} \Rightarrow 4x^2 = \frac{1}{4}x^2 + \frac{5k}{4}x + \frac{25}{16}k^2 \Rightarrow 60x^2 - 20kx - 25k^2 = 0 \Rightarrow x = \ldots\) |
| \(x = \frac{5k}{6}\) or \(x = -\frac{k}{2}\) | A1 | One correct value |
| \(x = \frac{5k}{6}\) and \(x = -\frac{k}{2}\) | A1 | Both values correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x - k = \frac{1}{2}x + \frac{k}{4} \Rightarrow 60x^2 - 68kx + 15k^2 = 0 \Rightarrow x = \ldots\) | M1 | Squares both sides to obtain 3 terms each time and solves resulting 3TQ |
| \(x = \frac{5k}{6}\) | A1 | One correct value only. If this is all they do, 2 marks maximum |
## Question 12:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| V shape with vertex anywhere on the $y$-axis, branches approximately symmetrical about the $y$-axis | B1 | Ignore any dashed or dotted lines |
| Intercepts at $\left(-\frac{k}{2}, 0\right)$, $\left(\frac{k}{2}, 0\right)$ and $\left(0, -k\right)$ and no others | B1 | Must be **crossing** intercepts. Allow coordinates wrong way round if positioning correct. There must be a sketch for this mark. If ambiguity exists, sketch has precedence |
**(2 marks total)**
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x - k = \frac{1}{2}x + \frac{k}{4} \Rightarrow x = \ldots$ or $-2x - k = \frac{1}{2}x + \frac{k}{4} \Rightarrow x = \ldots$ | M1 | Attempt to solve either equation to make $x$ or $k$ the subject |
| $x = \frac{5k}{6}$ or $x = -\frac{k}{2}$ | A1 | One correct value for $x$. Allow equivalent fractions e.g. $\frac{10k}{12}$, $-\frac{2k}{4}$ etc. |
| $x = \frac{5k}{6}$ and $x = -\frac{k}{2}$ | A1 | Both $x$ values correct. Allow equivalent fractions e.g. $\frac{10k}{12}$, $-\frac{2k}{4}$ etc. |
Note: $x = -\frac{k}{2}$ must clearly be from work in (b) and not from work in (a) when attempting the sketch unless clearly stated as answer to (b).
**Alternative by squaring:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\|x\| - k = \frac{1}{2}x + \frac{k}{4} \Rightarrow 4x^2 = \frac{1}{4}x^2 + \frac{5k}{4}x + \frac{25}{16}k^2 \Rightarrow 60x^2 - 20kx - 25k^2 = 0 \Rightarrow x = \ldots$ | M1 | Adds $k$ to both sides, squares and solves to obtain a 3TQ and solves for $x$ |
| $x = \frac{5k}{6}$ or $x = -\frac{k}{2}$ | A1 | One correct value |
| $x = \frac{5k}{6}$ and $x = -\frac{k}{2}$ | A1 | Both values correct |
**Special case** (squares only one side):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x - k = \frac{1}{2}x + \frac{k}{4} \Rightarrow 60x^2 - 68kx + 15k^2 = 0 \Rightarrow x = \ldots$ | M1 | Squares both sides to obtain 3 terms each time and solves resulting 3TQ |
| $x = \frac{5k}{6}$ | A1 | One correct value only. If this is all they do, **2 marks maximum** |
**(3 marks total)**
---12. Given that $k$ is a positive constant,
\begin{enumerate}[label=(\alph*)]
\item sketch the graph with equation
$$y = 2 | x | - k$$
Show on your sketch the coordinates of each point at which the graph crosses the $x$-axis and the $y$-axis.
\item Find, in terms of $k$, the values of $x$ for which
$$2 | x | - k = \frac { 1 } { 2 } x + \frac { 1 } { 4 } k$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q12 [5]}}