Answer
Done below.
Work Step by Step
a) If $x^2 < 3$, then $-\sqrt{3} < x < \sqrt{3}$. Since $\sqrt{3}$ is about 1.7, the statement is true if and only if $x$ is in the set $\{-1, 0, 1\}$
b) If $x^2 > x$, then $x^2 - x > 0$ so that $x(x - 1) > 0$. This obviously fails for $x = 0, 1$, but any other integer will yield a positive value.
c) If $2x + 1 = 0$, then $x = -1/2$. Thus the truth set is just the empty set.
