Problems

Solution: Since \(\int f(x) \, dx = 1\), and \(f(x)\) is a triangle with base \(b-a\), it must have height \(\frac{2}{b-a}\) in order to have the desired area. Since \(g(a) = 0, g(c) = \frac{2}{b-a}\), \(g(x) = A(x-a)\) and \(\frac{2}{b-a} = A (c-a) \Rightarrow g(x) = \frac{2(x-a)}{(b-a)(c-a)}\) as required. Similarly, \(h(x) = B(x-b)\) and \(\frac{2}{b-a} = B(c-b) \Rightarrow h(x) = \frac{2(b-x)}{(b-a)(b-c)}\) The mean of the distribution will be: \begin{align*} \int_a^b xf(x) \, dx &= \int_a^c xg(x) \, dx + \int_c^b xh(x) \, dx \\ &= \frac{2}{(b-a)(c-a)} \int_a^c x(x-a) dx + \frac{2}{(b-a)(b-c)} \int_c^b x(b-x) \, dx \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \left [ \frac{x^3}{3} - a\frac{x^2}{2} \right ]_a^c + \frac{1}{b-c} \left [ b\frac{x^2}{2} - \frac{x^3}{3} \right ]_c^b\r \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \l \frac{c^3}{3} - a\frac{c^2}{2} - \frac{a^3}{3} + \frac{a^3}{2} \r + \frac{1}{b-c} \l \frac{b^3}{2} - \frac{b^3}{3} - \frac{bc^2}{2} + \frac{c^3}{3} \r \r \\ &= \frac{2}{(b-a)} \l \l \frac{c^2+ac+a^2}{3} - \frac{a(a+c)}{2} \r +\l \frac{b(b+c)}{2} - \frac{b^2+bc+c^2}{3} \r\r \\ &= \frac{2}{(b-a)} \l \frac{2c^2+2ac+2a^2}{6} - \frac{3a^2+3ac}{6} + \frac{3b^2+3bc}{6} - \frac{2b^2+2bc+2c^2}{6} \r \\ &= \frac{2}{(b-a)} \l \frac{-a^2+b^2-ac+bc}{6} \r \\ &= \frac{a+b+c}{3} \\ \end{align*} The median \(M\) satisfies: \begin{align*} && \int_a^M f(x) \, dx &= \frac12 \\ \end{align*} The left hand triangle will have area: \(\frac{c-a}{b-a}\) which will be \(\geq \frac12\) if \(c \geq \frac{a+b}{2}\). In this case we need \begin{align*} && \frac{(M-a)^2}{(b-a)(c-a)} &= \frac12 \\ \Rightarrow && M &= a + \sqrt{\frac12 (b-a)(c-a)} \end{align*} Otherwise, we need: \begin{align*} && \frac{(b-M)^2}{(b-a)(b-c)} &= \frac12 \\ \Rightarrow && M &= b - \sqrt{\frac12 (b-a)(b-c)} \end{align*} These are consistent, if \(c = \frac{b+a}{2}\)

Solution: \begin{align*} && a^{2}x+ay+z & =a^{2} \tag{1}\\ && ax+y+bz & =1 \tag{2}\\ && a^{2}bx+y+bz & =b \tag{3} \\ \\ (1) - a(2): && (1-ba)z &= a^2-a \\ \Rightarrow && z &= \frac{a^2-a}{1-ab} \tag{if \(ab \neq 1\)} \\ \\ (2) - (3): && (a-a^2b)x &= b - 1 \\ \Rightarrow && x &= \frac{b-1}{a(1-ab)} \tag{if \(a \neq 0, ab \neq 1\)} \\ \\ b(1) - (3): && (ab-1)y &= a^2 - b^2 \\ \Rightarrow && y &= \frac{a^2-b^2}{ab-1} \end{align*} Let's consider the cases where \(a = 0\), then \begin{align*} && z &= 0 \\ && y + bz &= 1 \\ && y+bz &= b \\ \Rightarrow && y &= 1 = b \end{align*} So if \(a = 0\) then \(b = 1\) and \(x \in \mathbb{R}, y = 1, z = 0\). If \(a \neq 0, ab = 1\), then \begin{align*} && a^2 x + ay + z &= a^2 \\ && ax + y + \frac1{a}z &= a \\ && ax + y + \frac{1}{a}z &= b \\ \end{align*} The last two equations imply \(a = b = \pm 1\). \(a = 1 \Rightarrow x+y+z = 1\), so we have a lot of solutions. \(a = -1 \Rightarrow x -y +z = 1\) so again, lots of solutions. Conclusion: If \(ab \neq 1, a \neq 0\), we have: \[ (x,y,z) = \left (\frac{b-1}{a(1-ab)}, \frac{a^2-b^2}{ab-1}, \frac{a^2-a}{1-ab} \right)\] If \(a = 0\) then \(b = 1\) and we have: \((x,y,z) = (t, 1, 0)\). If \(ab = 1\) then \(a = 1\) or \(a = -1\). If \(a = 1\) then \((x,y,z) = (t, s, 1-t-s)\) If \(a = -1\) then \((x,y,z) = (t,s,1-t+s)\)