Onto function

Correct answer: A. It is onto

Step 1: This function always gives values between (-1) and (1). Step 2: Every value in the open interval ((-1,1)) is obtained for some real (x). Step 3: Since the endpoints are open, obtaining (-1) and (1) is not required.

Correct answer: A. \(x=y^2+1\) / (x=y^2+1)

Step 1: Take any codomain value (y\ge0). Step 2: If (x=y^2+1), then (x\ge1) and (\sqrt{x-1}=y). Step 3: For a square-root function, square the target value to find a preimage.

Correct answer: A. It is onto

Step 1: The cube root is defined for every real number. Step 2: For any (y\in\mathbb{R}), take (x=y^3-8), then (\sqrt[3]{x+8}=y). Step 3: A cube-root form gives all real values on (\mathbb{R}).

Correct answer: A. Because values less than (2) are not obtained

Step 1: Since (x\ge0), (x+8\ge8). Step 2: Hence (\sqrt[3]{x+8}\ge2), so a real value like (1) is not obtained. Step 3: Even a cube-root function may have a restricted range if the domain is restricted.

Correct answer: A. It is onto

Step 1: (\lfloor x-3\rfloor) gives integer values. Step 2: For any (k\in\mathbb{Z}), take (x=k+3), then (\lfloor x-3\rfloor=k). Step 3: A floor function can be onto an integer codomain.

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

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

Correct answer: A. It is onto

Step 1: (\lceil 2x\rceil) gives integer values. Step 2: For any (k\in\mathbb{Z}), take (x=\frac{k}{2}), then (\lceil 2x\rceil=k). Step 3: The linear expression inside the ceiling can produce all required integers.

Correct answer: A. It is onto

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

Correct answer: A. Because (2) has no preimage

Step 1: The function gives only values (0) and (1). Step 2: The codomain contains (2), but no (x) gives (f(x)=2). Step 3: A missed codomain value destroys onto nature.

Correct answer: A. Because an integer like (2) is not obtained

Step 1: (n^3) is always a perfect cube. Step 2: The codomain (\mathbb{Z}) contains (2), but no integer (n) satisfies (n^3=2). Step 3: On an integer domain, a power function does not necessarily give all integers.

Correct answer: A. It is onto

Step 1: (\sqrt{(x+5)^2}=|x+5|). Step 2: The range of (|x+5|) is ([0,\infty)), equal to the codomain. Step 3: Recognize (\sqrt{u^2}) as a modulus expression.

Correct answer: A. Because negative values are not obtained

Step 1: (\sqrt{(x+5)^2}=|x+5|) is always non-negative. Step 2: The codomain (\mathbb{R}) contains negative values, which are not obtained. Step 3: A function with a non-negative range is not onto the whole (\mathbb{R}).

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

Step 1: (x^3+x+1) is a continuous cubic polynomial. Step 2: As (x\to-\infty), it tends to (-\infty), and as (x\to\infty), it tends to (\infty). Step 3: By the intermediate value idea, it obtains every real value.

Correct answer: A. It is onto

Step 1: (x^6\ge0), so (x^6+2\ge2). Step 2: For every (y\ge2), take (x=\sqrt[6]{y-2}), then (f(x)=y). Step 3: An even-power function is onto its correctly shifted codomain.

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

Step 1: The minimum value of (x^6+2) is (2). Step 2: The codomain (\mathbb{R}) contains (0), but no real (x) can give it. Step 3: For even-power polynomials, codomain values below the minimum are missed.

Correct answer: A. It is onto

Step 1: (|x|\ge0), so (e^{|x|}\ge1). Step 2: For every (y\ge1), taking (x=\ln y) gives the needed absolute value and (e^{|x|}=y). Step 3: A modulus with an exponential gives range ([1,\infty)).

Correct answer: A. Because it also gives value (1), which is outside the codomain

Step 1: At (x=0), (e^{|x|}=1). Step 2: But (1) is not included in the codomain ((1,\infty)), so the mapping is not properly a function into this codomain. Step 3: Before checking onto nature, verify that all function values lie in the codomain.

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

Step 1: (\frac{x^2-1}{x^2+1}=1-\frac{2}{x^2+1}). Step 2: This value is always less than (1), so (1) is never obtained. Step 3: For rational functions, checking endpoint values is very important.

Correct answer: A. It is onto

Step 1: At (x=0), the value (-1) is obtained. Step 2: As (x) grows large, the value approaches (1) but never reaches it, so the range is ([-1,1)). Step 3: Approaching a boundary and attaining it are different ideas.

Correct answer: A. It will be onto

Step 1: Take any element (c) of (C). Step 2: Since (g) is onto, some (b\in B) has (g(b)=c); since (f) is onto, some (a\in A) has (f(a)=b). Step 3: Therefore ((g\circ f)(a)=c), so the composition is onto.

Correct answer: A. (g) is onto

Step 1: Take any (c\in C). Step 2: Since (g\circ f) is onto, there is some (a\in A) such that (g(f(a))=c). Here (f(a)\in B), so (c) has a preimage under (g). Step 3: Hence (g) must be onto.

Correct answer: A. It is onto

Step 1: The range of (\tan^{-1}x) is ((-\frac{\pi}{2},\frac{\pi}{2})). Step 2: For any (y) in this interval, take (x=\tan y), then (f(x)=y). Step 3: Remember the standard ranges of inverse trigonometric functions.

Correct answer: A. Because (-\frac{\pi}{2}) and (\frac{\pi}{2}) are not obtained

Step 1: The range of (\tan^{-1}x) is the open interval ((-\frac{\pi}{2},\frac{\pi}{2})). Step 2: The codomain includes both endpoint values, but the function never attains them. Step 3: The difference between open and closed endpoints is very important in onto questions.

Correct answer: A. \(x=-1+\sqrt{y+1}\) / (x=-1+\sqrt{y+1})

Step 1: Write (x^2+2x=y) as ((x+1)^2=y+1). Step 2: Taking (x=-1+\sqrt{y+1}) keeps (x\ge0) and gives function value (y). Step 3: Complete the square to find a preimage in quadratic forms.

Correct answer: A. Because negative values are not obtained

Step 1: When (x\ge0), (x^2+2x\ge0). Step 2: The codomain (\mathbb{R}) contains negative values such as (-1), but they are not in the range. Step 3: When the domain is restricted, the range can also be restricted.