The function $f(x) = \sqrt{3} \sin 2x - \cos 2x + 4$ is one-one in the interval

If $f: R \rightarrow R$ is defined by $f(x)=x-[x]-\frac{1}{2}$ for $x \in R$, where $[x]$ is the greatest integer not exceeding $x$, then $\{x \in R: f(x)=\frac{1}{2}\}$ is equal to :

Consider the sets $A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 = 25\}$,$B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + 9y^2 = 144\}$,$C = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : x^2 + y^2 \leq 4\}$,and $D = A \cap B$. The total number of one-one functions from the set $D$ to the set $C$ is:

If $f: R \rightarrow R$,then the function $f(x) = x|x|$ is:

Statement-$I$: Let $f : R \rightarrow R$ be a function such that $f(x) = x^3 + x^2 + 3x + \sin x$. Then $f$ is a one-one function.
Statement-$II$: $f(x)$ is a decreasing function.

If $f(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$ and $g(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$,then the function $(f - g)$ is: