Reflexive relation

Step 1: On ([-2,2]), the range of (x^2) is ([0,4]). Step 2: The codomain is the same, so it is onto, but (f(-1)=f(1)), so it is not one-one. Step 3: Even functions often give equal outputs for different inputs.

Step 1: On ([0,2]), (x^2) is strictly increasing. Step 2: Its range is ([0,4]), matching the codomain. Step 3: Restricting the square function to the non-negative branch can make it bijective.

Step 1: (\lfloor x\rfloor) is always an integer. Step 2: The codomain (\mathbb{R}) contains non-integers like (\frac{1}{2}), which are not attained. Step 3: Being defined everywhere and being onto are different ideas.

Step 1: For any (n\in\mathbb{Z}), choose (x=n), then (\lfloor x\rfloor=n). Step 2: So every integer in the codomain has a preimage. Step 3: The floor function is onto when the codomain is (\mathbb{Z}).

Step 1: (\lceil x\rceil) always gives an integer. Step 2: For any (n\in\mathbb{Z}), choosing (x=n) gives (\lceil x\rceil=n). Step 3: For the ceiling function, codomain (\mathbb{Z}) is fully covered.

Step 1: Since (x^2\ge0), the value is at least (0). Step 2: Also (\frac{x^2}{1+x^2}<1), so (1) and larger values are not attained. Step 3: Compare numerator and denominator to understand the range.

Step 1: Its range is ([0,1)). Step 2: For any (0\le y<1), (x^2=\frac{y}{1-y}) is possible, so a real (x) exists. Step 3: In equations involving (x^2), a non-negative right side gives real preimages.

Step 1: Onto means no element of the codomain is left out. Step 2: Therefore, for each (b\in B), at least one preimage (a\in A) must exist. Step 3: An onto function may have multiple preimages; one-one is a separate condition.

Step 1: We can write (x^3-6x^2+12x-5=(x-2)^3+3). Step 2: Since ((x-2)^3) takes all real values, ((x-2)^3+3) also takes all real values. Step 3: A horizontal or vertical shift of a cubic still remains onto from (\mathbb{R}) to (\mathbb{R}).