Problems

Solution:

1. 1. 

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   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.

Solution:

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   \((a,0)\), \((b,0)\), \((b,\f(b))\) and \((a, \f(a))\) forms a rectangle.
2. There are no solutions if \(a < c < b\):
   

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   There is one solution if \(a=c\) or \(a = b\)
   

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   And there are two solution if \(c \not \in [a,b]\)
   

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   There is exactly one solution unless....
   

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   ... there are infinitely many solutions when the gradients line up perfectly, ie when \(a+b=c+d\)
   

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Solution: Case 1: \(x \leq -1\) \begin{align*} && -1-x+x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && -x-2 &= x + 2 \\ \Leftrightarrow && x = -2 \end{align*} Case \(-1 < x \leq 0\): \begin{align*} && x+1+x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && x&= x + 2 \\ \end{align*} No solutions Case \(0 < x \leq 1\): \begin{align*} && x+1-x-3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && -x&= x + 2 \\ \end{align*} No solutions Case \(1 < x \leq 2\): \begin{align*} && x+1-x+3(x-1)+2(x-2) &= x + 2 \\ \Leftrightarrow && 5x-6&= x + 2 \\ \Leftrightarrow && x = 2 \end{align*} Case \(2 < x\): \begin{align*} && x+1-x+3(x-1)-2(x-2) &= x + 2 \\ \Leftrightarrow && x+2&= x + 2 \\ \end{align*} Therefore the solutions are \(x \in \{-2\} \cup [2, \infty)\)

Solution: If you arrive between 12 to and 12 past, it will take 60 minutes + how many minutes you wait at the station. If you arrive between 12 past and 12 to, it will take 75 minutes plus waiting at the station. Let's say arrival time \(X \sim U(0,60)\) minutes past the hour, then travel time is. Let's say there are two random variables, \(X_{fast} \sim U(0,24)\) \(X_{slow} \sim U(0, 36)\). Then if you wait for a fast train your expected wait time is \(72\), if you wait for a slow time, your expected wait time is \(75 + 18 = 93\). There is a \(\frac{24}{60} = \frac{4}{10}\) chance of being in the first case, and \(\frac{6}{10}\) chance of the second, ie: \(\frac{4}{10} \cdot 72 + \frac{6}{10} \cdot 93 = \frac{846}{10} = 84.6\) expected wait time. Suppose the time the trains so the expected fraction of time waiting for the fast train is \(t\) and the slow train is \(1-t\). Then the expected time is: \begin{align*} t \l 30t + 60 \r + (1-t) \l 30(1-t) + 75 \r &= 60t^2 -75t + 105 \\ &= 60 \l t^2 - \frac{5}{4}t \r + 105 \\ &= 60 \l t - \frac{5}{8} \r^2 - ? + 105 \\ \end{align*} Threfore we should choose \(x\) such that \(t = \frac58\), which is \(~37.5\) minutes after the slower train, \(25.5\) minutes past.

Solution: The probability density function will look like a triangle with base \(20\) minutes and therefore height \(\frac{1}{10}\) per minute, ie: \begin{align*} f_T(t) &= \begin{cases} \frac{1}{100}(t+10) & \text{if } -10 \leq t \leq 0 \\ \frac{1}{100}(10-t) & \text{if } 0 \leq t \leq 10 \\ 0 & \text{otherwise} \end{cases} \end{align*} \begin{align*} \mathbb{P}(\text{catch bus}) &=0.5 \mathbb{P}(\text{bus arrives after 7:55})+0.4 \mathbb{P}(\text{bus arrives after 7:58}) + 0.1 \mathbb{P}(\text{bus arrives after 8:02}) \\ &= \frac12 \cdot \left (1 - \frac18 \right) + \frac{2}{5} \cdot \left ( 1 - \frac{4^2}{5^2} \cdot \frac{1}{2} \right) + \frac{1}{10} \cdot \frac{4^2}{5^2} \cdot \frac12 \\ &= \frac{1\,483}{2\,000} \\ &\approx 74\% \end{align*} \begin{align*} \mathbb{P}(\text{catch bus}) &= \mathbb{P}(\text{bus arrives after 7:55}) \mathbb{P}(\text{catch next bus by 9:00}) \\ &= \frac78 + \frac18 \cdot \frac12 \\ &= \frac{15}{16} \end{align*} He should expect to miss \(6.25\) buses, so missing \(5\) seems about right. (Using a binomial calculation, seeing 5 or fewer buses is ~\(40\%\) which isn't suspicious).