The Problem: Suppose we know the following about a random variable $X$: $$P(X<1)=P(X>2)=0,$$ $$P(X=1)=P(X=2),$$ $$P(a<X<b)=\frac{b^3-a^3}{14}\quad\text{for }1\leq a<b\leq2.$$ Find the cumulative distribution function of $X$.
My Thoughts: By the given conditions and additivity, we have \begin{equation}\begin{split} 1&=P(X=1)+P(1<X<2)+P(X=2)\\ &=P(X=1)+\frac{1}{2}+P(X=2), \end{split}\end{equation} so that $P(X=1)=P(X=2)=\dfrac{1}{4}.$ Now, if $P(X=t)=\delta>0$ for some $t\in(1,2)$, then \begin{equation}\begin{split} P(1<X<2) &=P(1<X<t)+P(X=t)+P(t<X<2)\\ &=\frac{t^3-1}{14}+\delta+\frac{2^3-t^3}{14}\\ &=P(1<X<2)+\delta, \end{split}\end{equation} so that $P(X=t)=0$ for all $1<t<2$, whence $P(1<X\leq t)=P(1<X<t)$ for $1<t<2$. Therefore, $$F_X(t)=\begin{cases} 0&\text{if }t<1\\ \dfrac{1}{4}+\dfrac{t^3-1}{14}&\text{if }1\leq t<2\\ 1&\text{if }t\geq2. \end{cases}$$
Do you agree with my execution? Any feedback is much appreciated and welcomed. Thank you very much for your time.