| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |f(x)| compared to |g(x)| with parameters: sketch then solve |
| Difficulty | Standard +0.8 This question requires understanding of modulus function transformations, careful case-by-case analysis to solve the modulus equation (splitting into 4 cases based on sign changes at x=0 and x=-5a/3), and algebraic manipulation with a parameter. While the individual techniques are C3 standard, the combination of sketching with parameters and systematic case analysis elevates this above routine exercises. |
| 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 |
|---|---|---|
| (a) \(y = | 3x + 5a | \) |
| \(y = | x | - a\) |
| (b) \(-x - a = 3x + 5a \Rightarrow x = -\frac{3}{2}a\) | M1 A1 | |
| \(-x - a = -(3x + 5a) \Rightarrow x = -2a\), \(x = -2a, -\frac{3}{2}a\) | M1 A1 | (10 marks) |
**(a)** $y = |3x + 5a|$ | B3 |
$y = |x| - a$ | B3 |
**(b)** $-x - a = 3x + 5a \Rightarrow x = -\frac{3}{2}a$ | M1 A1 |
$-x - a = -(3x + 5a) \Rightarrow x = -2a$, $x = -2a, -\frac{3}{2}a$ | M1 A1 | (10 marks)\begin{enumerate}[label=(\alph*)]
\item Sketch on the same diagram the graphs of $y = |x| - a$ and $y = |3x + 5a|$, where $a$ is a positive constant.
Show on your diagram the coordinates of any points where each graph meets the coordinate axes. [6]
\item Solve the equation
$$|x| - a = |3x + 5a|.$$ [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q6 [10]}}