Romanian Mathematical Olympiad

The continuity at $x=1$ implies $a_1+b_1=a_2+b_2$. If $a_1=a_2$, then $b_1=b_2$ and we can take $m_1=a_1=a_2$, $n_1=b_1=b_2$, $m_2=n_2=0$, $\epsilon =\pm 1$. Otherwise, we search $A$ such that $m_{2}x+n_2=A(x-1)$. The required form can be obtained if $m_1-\epsilon |A|=a_1$, $n_1+\epsilon |A|=b_1$ and $m_1+\epsilon |A|=a_2$, $n_1-\epsilon |A|=b_2$, that is $$m_1=\frac{a_1+a_2}{2},n_1=\frac{b_1+b_2}{2},2\epsilon |A|=a_2-a_1=b_1-b_2.$$ Choosing $\epsilon =\mathrm{sign}(a_2-a_1)$, $m_2=\frac{1}{2}(a_1-a_2)$ and $n_2=\frac{1}{2}(a_2-a_1)$, we get $$f(x)=\frac{a_1+a_2}{2}x+\frac{b_1+b_2}{2}+\mathrm{sign}(a_2-a_1)\left|\frac{a_1-a_2}{2}x+\frac{a_2-a_1}{2}\right|.$$