If $f(x) = \begin{cases} [x], & -3 < x \leq -1 \\ |x|, & -1 < x < 1 \\ |[x]|, & 1 \leq x < 3 \end{cases}$ then the set $\{x : f(x) \geq 0\}$ is equal to

If a set $A$ has $n$ elements,then the number of functions defined from $A$ to $A$ that are not one-one is

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:

Show that the function $f : R \rightarrow R$ given by $f(x) = x^{3}$ is injective.

The function $f:R \to R$ defined by $f(x) = (x - 1)(x - 2)(x - 3)$ 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.