Official solution
f(x) = ex -- x, g(x) = x2 -- x
f(g(x)) = e(x2 - x) - (x2 - x)
If f(g(x)) is increasing function
(f (g(x)))x = \[{e^{\left( {{x^2} - x} \right)}} \times (2x - 1) - 2x + 1\]
\[ \Rightarrow \mathop {(2x - 1)}\limits_A \mathop {[{e^{\left( {{x^2} - x} \right)}} - 1]}\limits_B \]
A \& B are either both positive or negative
for (f (g(x)))' \[ \ge \] 0,
\[x \in \left[ {0,{1 \over 2}} \right] \cup \left[ {1,\infty } \right)\]