$f:[0,1]\to\mathbb{R}$ monotonic function with $f(1/4)f(3/4)<0$. Suppose $\sup\{x\in[0,1]:f(x)<0\}=\alpha$. Which of the following statements are true?
$1$ $f(\alpha)<0$
$2$ if $f$ is increasing then $f(\alpha)\leq0$
$3$ if $f$ is continuous and $1/4<\alpha<3/4$ then $f(\alpha)=0$
$4$ if $f$ is decreasing then $f(\alpha)<0$
My Try:
Since $\alpha$ is supremum of this set $f(\alpha)\geq0$. First option not true in general. By intermediate value theorem third option is true.
Since $\alpha$ is supremum and $f$ is increasing, $\alpha-\varepsilon<x\implies f(\alpha-\varepsilon)\leq f(x)<0$ for all $\varepsilon>0$ hence $f(\alpha)\geq0$
Since $\alpha$ is supremum and $f$ is decreasing, $\alpha-\varepsilon<x\implies 0>f(\alpha-\varepsilon)\geq f(x)$ for all $\varepsilon>0$ hence $f(\alpha)<0$
Second option is false and fourth is true.
Are my arguments correct? I have some doubts over my arguments for second and fourth options . Could anybody clarify?