Onto function

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

Step 1: The definition of onto is about every element of the codomain. Step 2: For every (b\in B), at least one (a\in A) must exist such that (f(a)=b). Step 3: The existence of a preimage is the key idea in onto nature.

Correct answer: A. \(f(A)=B\) / (f(A)=B)

Step 1: (f(A)) is the range of the function. Step 2: Onto means the range is equal to the codomain. Step 3: Therefore, (f(A)=B) is the identifying condition for onto functions.

Correct answer: A. Because (1) is not obtained

Step 1: (\frac{x^2}{1+x^2}) is always less than (1). Step 2: The codomain contains (1), but no real (x) makes the value equal to (1). Step 3: In rational functions, check endpoint values separately.

Correct answer: A. It is onto

Step 1: The range of this function is ([0,1)). Step 2: (0) is obtained at (x=0), and values below (1) are obtained by suitable (x). Step 3: In an open-end codomain, obtaining (1) is not required.

Correct answer: A. It is onto

Step 1: The value of (\frac{x}{\sqrt{1+x^2}}) lies between (-1) and (1). Step 2: Every value in the open interval ((-1,1)) can be obtained for some real (x). Step 3: In bounded ranges, check whether endpoints are included or excluded.

Correct answer: A. Because (-1) and (1) are not obtained

Step 1: The function values stay between (-1) and (1), but never reach the endpoints. Step 2: The codomain ([-1,1]) includes (-1) and (1). Step 3: Missing endpoints in a closed codomain prevents onto nature.

Correct answer: A. The polynomial is continuous and goes unbounded in both directions

Step 1: (x^3-5x) is a continuous cubic polynomial. Step 2: As (x\to\infty), it goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Step 3: Such a cubic polynomial takes all real values.

Correct answer: A. Because negative values are not obtained

Step 1: Both (x^4) and (x^2) are non-negative. Step 2: Hence (x^4+x^2\ge0), and a codomain value such as (-1) is not obtained. Step 3: A sum of even powers often has a non-negative range.

Correct answer: A. It is onto

Step 1: The minimum value of (x^4+x^2) is (0), and for large (x), it becomes very large. Step 2: Since it is continuous, it takes all values above (0). Step 3: With a non-negative codomain, such a polynomial can be onto.

Correct answer: A. It is onto

Step 1: (|x-2|\ge0), so (|x-2|+1\ge1). Step 2: For every (y\ge1), take (x=2+(y-1)), then (f(x)=y). Step 3: Use the minimum value of a modulus function to identify its range.

Correct answer: A. Because a value like (0) is not obtained

Step 1: The minimum value of (|x-2|+1) is (1). Step 2: The codomain (\mathbb{R}) contains (0), but this function can never give (0). Step 3: Codomain values below the minimum stay outside the range.

Correct answer: A. It is onto

Step 1: Even elements (2,4) give (0). Step 2: Odd elements (1,3,5) give (1). Step 3: Both codomain values (0) and (1) are obtained, so the function is onto.

Correct answer: A. Because (0) has no preimage

Step 1: All domain elements (1,3,5) are odd. Step 2: Therefore, the function value is always (1), and (0) is never obtained. Step 3: If any codomain value is missed, the function is not onto.

Correct answer: A. It is onto

Step 1: (\sqrt{x^2}=|x|). Step 2: The range of (|x|) is ([0,\infty)), equal to the codomain. Step 3: Recognizing (\sqrt{x^2}) as a modulus is an easy method.

Correct answer: A. Because negative values are not obtained

Step 1: (\sqrt{x^2}=|x|). Step 2: Its value is never negative, while the codomain (\mathbb{R}) contains negative values. Step 3: Compare a modulus-type range with the real codomain carefully.

Correct answer: A. It is onto

Step 1: (\sqrt[3]{x}) can take all real values. Step 2: Adding (2) still leaves the range as (\mathbb{R}). Step 3: A vertical shift of the cube-root function does not remove onto nature.

Correct answer: A. Because a value like (0) is not obtained

Step 1: (\sqrt{x}\ge0), so (\sqrt{x}+1\ge1). Step 2: The codomain (\mathbb{R}) contains (0), but the function cannot give (0). Step 3: Always check the minimum bound of a square-root function.

Correct answer: A. It is onto

Step 1: The range of (\sqrt{x}+1) is ([1,\infty)). Step 2: For every (y\ge1), take (x=(y-1)^2), then (f(x)=y). Step 3: A square-root function is onto its correct codomain.

Correct answer: A. It is not onto

Step 1: (\lfloor x\rfloor) always gives an integer value. Step 2: The codomain (\mathbb{R}) contains non-integer real values such as (\frac{1}{2}), which are not obtained. Step 3: An integer-valued function is not onto (\mathbb{R}).

Correct answer: A. It is onto

Step 1: (\lfloor x\rfloor) can give any integer value. Step 2: For any (k\in\mathbb{Z}), take (x=k), then (\lfloor x\rfloor=k). Step 3: With codomain (\mathbb{Z}), this function is onto.

Correct answer: A. It is onto

Step 1: (\lceil x\rceil) gives integer values. Step 2: For any (k\in\mathbb{Z}), take (x=k), then (\lceil x\rceil=k). Step 3: The ceiling function is onto an integer codomain.

Correct answer: A. Because non-integer real values are not obtained

Step 1: (\lceil x\rceil) is always an integer. Step 2: The codomain (\mathbb{R}) contains non-integer values such as (\frac{3}{2}), which are not in the range. Step 3: An integer-valued function does not cover the full real codomain.

Correct answer: A. It is onto

Step 1: Taking (x=-1) gives (f(x)=0). Step 2: Taking (x=0) gives (f(x)=1). Step 3: Both codomain values are obtained, so the function is onto.

Correct answer: A. Because (0) has no preimage

Step 1: The function gives only (1) for every (x). Step 2: The codomain also contains (0), but it is not obtained from any (x). Step 3: A constant function cannot be onto a two-element codomain.

Correct answer: A. It is onto

Step 1: Since (x^2\ge0), we have (x^2-1\ge-1). Step 2: For every (y\ge-1), take (x=\sqrt{y+1}), then (f(x)=y). Step 3: If the minimum value of a quadratic matches the codomain start, onto nature is easy to verify.