SPS SPS FM Pure 2024 January — Question 2 6 marks

Exam BoardSPS
ModuleSPS FM Pure (SPS FM Pure)
Year2024
SessionJanuary
Marks6
TopicModulus function
TypeSolve |f(x)| compared to |g(x)| with parameters: equation or inequality only
DifficultyStandard +0.3 This is a modulus inequality problem requiring case analysis and graph transformations. Part (a) involves splitting into cases based on the critical point x = 3k/2, solving linear inequalities, and expressing in set notation—standard Further Maths technique but more involved than typical A-level. Part (b) requires understanding function transformations (horizontal stretch and vertical stretch/reflection) applied to the vertex of a modulus function—straightforward once part (a) is understood. Overall slightly above average difficulty due to algebraic manipulation with parameters and multiple steps, but uses well-practiced techniques.
Spec1.02g Inequalities: linear and quadratic in single variable1.02h Express solutions: using 'and', 'or', set and interval notation1.02l Modulus function: notation, relations, equations and inequalities1.02w Graph transformations: simple transformations of f(x)

  1. Find, in terms of \(k\), the set of values of \(x\) for which $$k - |2x - 3k| > x - k$$ giving your answer in set notation. [4]
  2. Find, in terms of \(k\), the coordinates of the minimum point of the graph with equation $$y = 3 - 5f\left(\frac{1}{2}x\right)$$ where $$f(x) = k - |2x - 3k|$$ [2]

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $k$, the set of values of $x$ for which
$$k - |2x - 3k| > x - k$$
giving your answer in set notation. [4]

\item Find, in terms of $k$, the coordinates of the minimum point of the graph with equation
$$y = 3 - 5f\left(\frac{1}{2}x\right)$$
where
$$f(x) = k - |2x - 3k|$$ [2]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM Pure 2024 Q2 [6]}}