Reflexive relation

Correct answer: A. It is not onto because (0) and (1) are not attained

Step 1: The function values lie between (0) and (1). Step 2: Neither (0) nor (1) is attained for any real (x), but both are in the codomain. Step 3: The difference between open and closed intervals is crucial in onto questions.

Correct answer: A. (f) is onto

Step 1: (x^2+2x=(x+1)^2-1). Step 2: The minimum value is (-1), and the range is ([-1,\infty)). Step 3: Completing the square is a clean method to test onto for quadratics.

Correct answer: A. Yes

Step 1: (f(x)=(x+1)^2-1) and (x\ge-1). Step 2: At (x=-1), the value is (-1), and as (x) increases all larger values occur. Step 3: Use the domain to select the correct branch of the parabola.

Correct answer: A. Because negative values are not attained

Step 1: If (x<0), then (|x|+x=-x+x=0). Step 2: If (x\ge0), then (|x|+x=2x), so values are non-negative. Step 3: Use piecewise behavior to find the range.

Correct answer: A. Yes it is onto

Step 1: The range of the function is ([0,\infty)). Step 2: For (y\ge0), choose (x=\frac{y}{2}), then (f(x)=y). Step 3: Use the branch that covers the whole codomain to build the preimage.

Correct answer: A. Yes

Step 1: The minimum value of (|x-3|) is (0). Step 2: For every (y\ge0), choosing (x=3+y) gives (|x-3|=y). Step 3: For modulus functions, distance interpretation makes the range easy.

Correct answer: A. Because negative values are not attained

Step 1: An absolute value is never negative. Step 2: The codomain (\mathbb{R}) contains values like (-1), which cannot be (f(x)). Step 3: A too-large codomain can make a function non-onto.

Correct answer: A. (f) is one-one and onto

Step 1: (x^3) takes all real values. Step 2: (x^3+1) also takes all real values, because for any (y), (x=\sqrt[3]{y-1}) works. Step 3: A vertical shift of a cubic does not destroy onto behavior over (\mathbb{R}).

Correct answer: A. It is continuous and takes values from (-\infty) to (\infty)

Step 1: (x^3+x) is defined and continuous for all real (x). Step 2: As (x\to\infty), the value goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Step 3: Continuity plus end behavior shows onto over (\mathbb{R}).

Correct answer: A. Because negative values are not attained

Step 1: (x^4\ge0) and (x^2\ge0). Step 2: Therefore (f(x)\ge0), so no negative real number is an image. Step 3: For sums of even powers, focus on the minimum value and range.

Correct answer: A. Yes

Step 1: The minimum value of (x^4+x^2) is (0). Step 2: Put (t=x^2\ge0); then (f=t^2+t), increasing continuously from (0) to (\infty). Step 3: Use a helper variable to find the range of difficult polynomials.

Correct answer: A. Onto but not one-one

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.

Correct answer: A. One-one and onto

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.

Correct answer: A. No because only integer values are attained

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.

Correct answer: A. Yes

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}).

Correct answer: A. (f) is onto

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.

Correct answer: A. Because its range is ([0,1))

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.

Correct answer: A. Yes

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.

Correct answer: A. There exists at least one (a\in A) such that (f(a)=b)

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.

Correct answer: A. (f) is onto

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}).