I repeatedly tried to solve it but by no means I could match my answer with the given answer: $5$
The problem goes like this:
$y=f(x)$ is differentiable function and $g(x)= f(x - x^2).$
If $y=g(x)$ has local maxima at $x=1/2,$ but the absolute maximum exists at some other point, then minimum number of solutions of $g'(x)=0$ is ________
Note that
$$ g(1/2) = f(1/2 - 1/4) = f(1/4).$$
Thus $f$ has a local maximum at $1/4$. Solving
$$x- x^2 = \frac 14$$
gives $x = 1/2$ as the only (double) root.
Let $c\neq 1/2$ be the point where $g$ attains an absolute maximum. Now solving
$$x-x^2 =c-c^2$$
gives two distinct solutions, $x = c$ and $x = c'$ (it cannot be a double root) and either
$$c< 1/2 < c' \text{or } c'<1/2 < c.$$
Thus $g$ has at least three local maximum, at $c, 1/2, c'$. It is clear that it has at least $5$ critical value.