Onto function

Correct answer: A. Real values like (\frac{1}{2}) are not images

Step 1: (\lfloor x\rfloor) always gives an integer. Step 2: The codomain (\mathbb{R}) contains (\frac{1}{2}), but it is never an image. Step 3: The greatest integer function has range limited to integers.

Correct answer: A. It is onto

Step 1: Take any integer (k) in the codomain. Step 2: Choosing (x=k) gives (\lfloor x\rfloor=k). Step 3: With codomain (\mathbb{Z}), the floor function becomes onto.

Correct answer: A. Yes because for every (y\in[0,4]), (x=\sqrt{y}\in[0,2])

Step 1: On the domain ([0,2]), the range of (x^2) is ([0,4]). Step 2: For every (y\in[0,4]), (\sqrt{y}) lies in the given domain. Step 3: On closed intervals, always check endpoint values while finding the range.

Correct answer: A. Because (5) is not an image

Step 1: If (x\in[0,2]), then (x^2\in[0,4]). Step 2: The codomain contains (5), but (x^2=5) gives (x=\sqrt{5}), which is not in the domain. Step 3: With restricted domains, check the maximum and minimum possible output values.

Correct answer: A. Yes because it is a continuous odd-degree polynomial and is unbounded in both directions

Step 1: (x^3-3x) is continuous. Step 2: As (x\to\infty), it goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Step 3: Even with turning points, end behavior can show onto property.

Correct answer: A. It is onto

Step 1: The range of (|x-2|) is ([0,\infty)). Step 2: For any (y\ge0), choosing (x=y+2) gives (|x-2|=y). Step 3: A horizontal shift of a modulus function changes the vertex location but not the range type.

Correct answer: A. (f) is onto

Step 1: Since (|x|\ge0), we have (|x|+2\ge2). Step 2: For every (y\ge2), choosing (x=y-2) gives the image (y). Step 3: A vertical shift of a modulus function changes its minimum value.

Correct answer: A. Because for every (y\le0), (x=\sqrt{-y}) can be taken

Step 1: (-x^2\le0) for every real (x). Step 2: If (y\le0), then (x=\sqrt{-y}) is real and (-x^2=y). Step 3: A negative quadratic becomes onto when the codomain is chosen correctly.

Correct answer: A. (1) cannot be the image of any (x)

Step 1: (-x^2) is never positive. Step 2: The codomain (\mathbb{R}) contains (1), but no real (x) satisfies (-x^2=1). Step 3: Showing one missed codomain value is enough to prove not onto.

Correct answer: A. (f) is onto

Step 1: Since (x^2\ge0), (x^2-2\ge-2). Step 2: For every (y\ge-2), choosing (x=\sqrt{y+2}) gives (f(x)=y). Step 3: For a downward or upward shift of a quadratic, identify the minimum value.

Correct answer: A. (\mathbb{R})

Step 1: (\tan x) takes all real values on its defined domain. Step 2: Therefore, if the codomain is (\mathbb{R}), no real value is missed. Step 3: Remembering standard trigonometric ranges is useful in onto questions.

Correct answer: A. Because for every \(y\in\mathbb{R}\), \(x=\tan^{-1}y\) lies in this domain

Step 1: On the principal interval \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), (\tan x) covers all real numbers. Step 2: For every real (y), \(x=\tan^{-1}y\) lies in that interval. Step 3: Inverse trigonometric values help prove onto property.

Correct answer: A. Values like (1) and (-1) are not images

Step 1: (\frac{x}{1+|x|}) always lies between (-1) and (1). Step 2: It never equals (1) or (-1), though they are in the codomain. Step 3: For rational modulus functions, check limiting endpoint values.

Correct answer: A. It is onto

Step 1: The range of this function is ((-1,1)). Step 2: For (y\ge0), take (x=\frac{y}{1-y}), and for (y<0), take (x=\frac{y}{1+y}). Step 3: For modulus-based expressions, checking cases separately is helpful.

Correct answer: A. Negative numbers and (0) are not images

Step 1: (2^x) is positive for every real (x). Step 2: Therefore (0) and negative real numbers are in the codomain but not in the range. Step 3: Remember that the natural range of exponential functions is ((0,\infty)).

Correct answer: A. (x=\log_2 y)

Step 1: To prove onto, solve (2^x=y). Step 2: Since (y>0), (x=\log_2 y) is real and gives (2^x=y). Step 3: For exponential functions, logarithms provide the required preimage.

Correct answer: A. Yes because for every (y\in\mathbb{R}), (x=\sqrt[3]{y-1}) exists

Step 1: Take any real number (y) in the codomain. Step 2: (x=\sqrt[3]{y-1}) is real and gives (x^3+1=y). Step 3: A vertical shift of a cubic function does not destroy onto property over (\mathbb{R}).

Correct answer: A. It is onto

Step 1: The range of ((x-1)^2) is ([0,\infty)). Step 2: For every (y\ge0), taking (x=1+\sqrt{y}) gives (f(x)=y). Step 3: A horizontal shift of a square function does not change its range type.

Correct answer: A. (f) is onto

Step 1: Since ((x+3)^2\ge0), (f(x)\ge4). Step 2: For (y\ge4), taking (x=-3+\sqrt{y-4}) gives (f(x)=y). Step 3: The completed-square form makes the range clear immediately.

Correct answer: A. Because (2) is not an image

Step 1: (\cos x) always lies in ([-1,1]). Step 2: The codomain (\mathbb{R}) contains (2), but (\cos x=2) is impossible. Step 3: Match the standard trigonometric range with the codomain.

Correct answer: A. Yes because every value in ([-1,1]) is obtained by (\cos x)

Step 1: The range of (\cos x) is ([-1,1]). Step 2: The codomain is the same, so no element is missed. Step 3: If the codomain equals the actual range, the function is onto.

Correct answer: A. It is onto

Step 1: Put (t=|x|), so (t\ge0) and (f(x)=t^2+t). Step 2: On ([0,\infty)), (t^2+t) takes every value from (0) to (\infty), since (t=\frac{-1+\sqrt{1+4y}}{2}) works. Step 3: For harder modulus questions, substitution is useful.

Correct answer: A. It is onto

Step 1: Put (t=|x|), so (t\ge0) and (f(x)=t^2+t+1). Step 2: It gives (1) at (t=0) and increases continuously to (\infty) without missing values. Step 3: The codomain ([1,\infty)) exactly matches this range.

Correct answer: A. (f) is onto

Step 1: When (x<0), (x+1) gives all values in ((-\infty,1)). Step 2: When (x\ge0), (x^2) gives all values in ([0,\infty)). Step 3: The union of both ranges is (\mathbb{R}), so the function is onto.

Correct answer: A. Because (0) and (1) are never images

Step 1: (e^{-x}) is positive for every real (x). Step 2: So (\frac{1}{1+e^{-x}}) always lies between (0) and (1), but never equals (0) or (1). Step 3: When endpoints are in the codomain, check whether they are actually attained.