For any two real numbers $u$ and $v$, consider the equations $x^2 + ay = v$, and $x^2 + by = u$. Solving it yields $y = \frac{u-v}{b-a}$ and $x^2 = v - a\frac{u-v}{b-a} = \frac{bv-au}{b-a}$, so solutions exists if and only if $au \leq bv$. In particular, the problem is equivalent to $f(u) \leq f(v)$ whenever $au \leq bv$. Part (a)Immediate from $as \leq 0 \leq bt$. Part (b)The above property implies that for any two positive real numbers $u$ and $v$ such that $\frac{a}{b} \leq \frac{v}{u} \leq \frac{b}{a}$, we have $f(u) \leq f(v) \leq f(u)$ and thus $f(u) = f(v)$. Induction implies that for every natural number $n$ and every positive real number $u$ and $v$ with $\left(\frac{a}{b}\right)^n \leq \frac{v}{u} \leq \left(\frac{b}{a}\right)^n$ we have $f(u) = f(v)$. But for every positive real number $u$ and $v$, such $n$ satisfying the inequality exists, so we are done. Part (c)Consider the function below: $$ f(x) = \begin{cases} 0, & x \geq 0 \\ -2^{g(\log_{(ba^{-1})} (-x))}, & x < 0 \end{cases}, \quad \quad g(x) = \lfloor x \rfloor - \{x\} $$ By property of $g$, for any two real numbers $x$ and $y$, we have $g(x) \geq g(y)$ if and only if either $\lfloor x \rfloor > \lfloor y \rfloor$ or $\lfloor x \rfloor = \lfloor y \rfloor$ and $x \leq y$ (where $g(x) = g(y)$ if and only if $x = y$). The latter implies that we actually have $f(-1) < f\left(-\sqrt{\frac{b}{a}}\right)$ as $f(-1) = -2^{g(0)} = -1$ while $f\left(-\sqrt{\frac{b}{a}}\right) = -2^{g(0.5)} = -2^{-0.5} = -\frac{1}{2}\sqrt{2}$. Consider any two real numbers $u$ and $v$ such that $au \leq bv$. If $v \geq 0$, it's obvious that $f(u) \leq f(v) = 0$, so now suppose $v < 0$. Then, $u \leq \frac{bv}{a} < v < 0$, and thus $f(u) \leq f(v)$ is equivalent to \begin{align*} -2^{g(\log_{ba^{-1}} (-u))} &\leq -2^{g(\log_{ba^{-1}} (-v))} \\ g(\log_{ba^{-1}} (-u)) &\geq g(\log_{ba^{-1}} (-v)) \end{align*} The former of the condition $g(x) \geq g(y)$ always holds whenever $x - y \geq 1$, so it is sufficient to prove that $\log_{ba^{-1}} (-u) - \log_{ba^{-1}} (-v) \geq 1$. But this is equivalent to $\frac{u}{v} \geq \frac{b}{a}$, which since $u < 0 < a$, is equivalent to $au \leq bv$. We are done.