1.02t Solve modulus equations: graphically with modulus function

80 questions

Sort by: Default | Easiest first | Hardest first
AQA Further Paper 2 2024 June Q15
7 marks Standard +0.8
The diagram shows the line \(y = 5 - x\) \includegraphics{figure_15}
  1. On the diagram above, sketch the graph of \(y = |x^2 - 4x|\), including all parts of the graph where it intersects the line \(y = 5 - x\) (You do not need to show the coordinates of the points of intersection.) [3 marks]
  2. Find the solution of the inequality $$|x^2 - 4x| > 5 - x$$ Give your answer in an exact form. [4 marks]
SPS SPS FM Pure 2021 June Q9
6 marks Moderate -0.8
\includegraphics{figure_2} Figure 2 shows a sketch of part of the graph \(y = f(x)\), where $$f(x) = 2|3 - x| + 5, \quad x \geq 0$$
  1. State the range of \(f\) [1]
  2. Solve the equation $$f(x) = \frac{1}{2}x + 30$$ [3] Given that the equation \(f(x) = k\), where \(k\) is a constant, has two distinct roots,
  3. state the set of possible values for \(k\). [2]
SPS SPS FM Pure 2022 June Q2
9 marks Standard +0.3
\includegraphics{figure_1} Figure 1 shows part of the graph of \(y = f(x)\), \(x \in \mathbb{R}\). The graph consists of two line segments that meet at the point \((1, a)\), \(a < 0\). One line meets the \(x\)-axis at \((3, 0)\). The other line meets the \(x\)-axis at \((-1, 0)\) and the \(y\)-axis at \((0, b)\), \(b < 0\). In separate diagrams, sketch the graph with equation
  1. \(y = f(x + 1)\), [2]
  2. \(y = f(|x|)\). [2]
Indicate clearly on each sketch the coordinates of any points of intersection with the axes. Given that \(f(x) = |x - 1| - 2\), find
  1. the value of \(a\) and the value of \(b\), [2]
  2. the value of \(x\) for which \(f(x) = 5x\). [3]
Pre-U Pre-U 9794/1 2011 June Q3
3 marks Moderate -0.5
Solve the equation \(3 + 2x = |7 - 4x|\). [3]
Edexcel AEA 2014 June Q3
11 marks Standard +0.8
  1. On separate diagrams sketch the curves with the following equations. On each sketch you should mark the coordinates of the points where the curve crosses the coordinate axes.
    1. \(y = x^2 - 2x - 3\)
    2. \(y = x^2 - 2|x| - 3\)
    3. \(y = x^2 - x - |x| - 3\)
    [7]
  2. Solve the equation $$x^2 - x - |x| - 3 = x + |x|$$ [4]