| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Topic | Modulus function |
| Type | Solve |linear| > constant (greater than) |
| Difficulty | Easy -1.2 This is a straightforward question on composite functions and absolute value inequalities. Part (i) requires simple recognition that g(x) = |x|. Part (ii) is immediate recall that absolute value outputs are non-negative. Part (iii) is a standard textbook exercise on solving absolute value inequalities by considering two cases. All parts are routine with no problem-solving insight required, making this easier than average. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities1.02u Functions: definition and vocabulary (domain, range, mapping) |
You are given that $gf(x) = |3x - 1|$ for $x \in \mathbb{R}$.
\begin{enumerate}[label=(\roman*)]
\item Given that $f(x) = 3x - 1$, express $g(x)$ in terms of $x$. [1]
\item State the range of $gf(x)$. [1]
\item Solve the inequality $|3x - 1| > 1$. [3]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Pure 2023 Q1 [5]}}