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2025 Paper 3 Q2
D: 1500.0 B: 1500.0
modulus function sequences recurrence relation curve sketching periodic sequences iterated functions fixed points piecewise linear

Let \(f(x) = 7 - 2|x|\). A sequence \(u_0, u_1, u_2, \ldots\) is defined by \(u_0 = a\) and \(u_n = f(u_{n-1})\) for \(n > 0\).

    1. Sketch, on the same axes, the graphs with equations \(y = f(x)\) and \(y = f(f(x))\).
    2. Find all solutions of the equation \(f(f(x)) = x\).
    3. Find the values of \(a\) for which the sequence \(u_0, u_1, u_2, \ldots\) has period 2.
    4. Show that, if \(a = \frac{28}{5}\), then the sequence \(u_2, u_3, u_4, \ldots\) has period 2, but neither \(u_0\) or \(u_1\) is equal to either of \(u_2\) or \(u_3\).
    1. Sketch, on the same axes, the graphs with equations \(y = f(x)\) and \(y = f(f(f(x)))\).
    2. Consider the sequence \(u_0, u_1, u_2, \ldots\) in the cases \(a = 1\) and \(a = -\tfrac79\). Hence find all the solutions of the equation \(f(f(f(x))) = x\).
    3. Find a value of \(a\) such that the sequence \(u_3, u_4, u_5, \ldots\) has period 3, but where none of \(u_0, u_1\) or \(u_2\) is equal to any of \(u_3, u_4\) or \(u_5\).

Solution:

    1. TikZ diagram
    2. If \(a = 1\) then \(u_1 = f(a) = 7-2 = 5\), \(u_2 = f(5) = -3\), \(u_3 = f(-3) = 7-6 = 1\). Therefore it must be the case that \(f(f(f(x))) = x\) for \(x = 1, 5, -3\). Similarly, if \(a = -\tfrac79\) then \(u_1 = f(-\tfrac79) = \tfrac{49}{9}\), \(u_2 = f(\tfrac{49}{9}) = -\tfrac{35}{9}\) and \(u_3 = f(-\tfrac{35}{9}) = -\tfrac79\). Therefore we must also have roots \(x = -\tfrac79, \tfrac{49}{9}, -\tfrac{35}9\). We also have the roots \(x = -7, \tfrac73\) from the first part so we have found all \(8\) roots.
    3. We need \(f(f(f(x))) = 1\) but \(f(f(x)) \neq -3, f(x) \neq 5, x \neq 1\). Suppose \(f(y) = 1 \Rightarrow 7-2|y| = 1 \Rightarrow y = \pm 3\). So \(y = 3\), ie \(f(f(x)) = 3\). Suppose \(f(z) = 3 \Rightarrow 7-2|z| = 3 \Rightarrow z = \pm 2\). Finally we need \(f(x) = \pm 2\), so say \(7-2|x| = 2 \Rightarrow x = \tfrac52\), so we have the sequence \(\tfrac52, 2, 3, 1, 5, -3, 1, \cdots\)as required.

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2003 Paper 2 Q6
D: 1600.0 B: 1500.0
integration absolute value iterated functions curve sketching piecewise functions comparison of areas trigonometric function composition

The function \(\f\) is defined by $$ \f(x)= \vert x-1 \vert\;, $$ where the domain is \({\bf R}\,\), the set of all real numbers. The function \(\g_n =\f^n\), with domain \({\bf R}\,\), so for example \(\g_3(x) = \f(\f(\f(x)))\,\). In separate diagrams, sketch graphs of \(\g_1\,\), \(\g_2\,\), \(\g_3\,\) and \(\g_4\,\). The function \(\h\) is defined by \[ \h(x) = |\sin {{{\pi}x} \over 2}|, \] where the domain is \({\bf R}\,\). Show that if \(n\) is even, \[ \int_0^n\,\big( \h(x)-\g_n(x)\big)\,\d x = \frac{2n}{\pi} -\frac{n}2\;. \]

Solution:

TikZ diagram
TikZ diagram
TikZ diagram
TikZ diagram
If \(n\) is even, and \(0 < x < n\) then \(g_n(x) = \begin{cases} \{x \} & \text{if }\lfloor x \rfloor\text{ is even} \\ 1-\{x \} & \text{if }\lfloor x \rfloor\text{ is odd} \\\end{cases}\), in other words, there are \(\frac{n}{2}\) triangles, with height \(1\) and base \(2\), giving total area of \(\frac{n}{2}\). Each section of \(|\sin (\frac{n \pi}{2})|\) will have area \(\frac{2}{\pi}\) and there will be \(n\) of them, therefore \(\frac{2n}{\pi} - \frac{n}{2}\)

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