Onto function

Correct answer: A. Onto

Step 1: For onto, every (y\in\mathbb{R}) must have some preimage (x\in\mathbb{R}). Step 2: From (y=x^3-1), we get (x=\sqrt[3]{y+1}), which is real for every real (y). Step 3: In exams, cubic functions often cover all real values.

Correct answer: B. No because numbers less than (4) are not obtained

Step 1: Since (x^2\ge 0), we have (x^2+4\ge 4). Step 2: The codomain is (\mathbb{R}), but the range is only ([4,\infty)). Step 3: Always compare the range with the codomain for onto questions.

Correct answer: A. (f) is onto

Step 1: Since (x^2\ge 0), (f(x)=x^2+2\ge 2). Step 2: For every (y\ge 2), we can take (x=\sqrt{y-2}). Step 3: A quadratic function can become onto when the codomain matches its actual range.

Correct answer: B. Not onto

Step 1: (f(n)=n+1) gives values (2,3,4,\ldots). Step 2: The codomain (\mathbb{N}) contains (1), but (1) has no preimage. Step 3: In natural-number questions, checking the first few codomain values is very helpful.

Correct answer: A. Because for every (y\in\mathbb{Z}), (x=y+5) is an integer

Step 1: Take any target value (y\in\mathbb{Z}). Step 2: From (y=x-5), we get (x=y+5), which is an integer. Step 3: For linear functions, solving for (x) is a quick way to test onto.

Correct answer: B. Not onto because negative values are not obtained

Step 1: (e^x) is always greater than (0). Step 2: The codomain (\mathbb{R}) contains negative numbers and (0), which are not in the range. Step 3: Remembering the range of exponential functions helps in onto questions.

Correct answer: A. Onto

Step 1: The range of (e^x) is ((0,\infty)). Step 2: The codomain is also ((0,\infty)), so for every (y>0), (x=\ln y) exists. Step 3: Never ignore the given codomain in onto problems.

Correct answer: B. It is not onto because range is ([-1,1])

Step 1: (\sin x) always lies between (-1) and (1). Step 2: The codomain (\mathbb{R}) has values like (2), which are never obtained. Step 3: First identify the bounded range of trigonometric functions.

Correct answer: A. Onto

Step 1: The range of (\cos x) is ([-1,1]). Step 2: The codomain is also ([-1,1]), so every codomain value is attained. Step 3: A function is onto when its codomain equals its actual range.

Correct answer: B. Not onto because negative values are not obtained

Step 1: (|x|) is always (0) or positive. Step 2: The codomain (\mathbb{R}) contains values like (-3), which cannot be obtained from (|x|). Step 3: Negative target values quickly show that the modulus function is not onto over (\mathbb{R}).

Correct answer: A. Onto

Step 1: The range of (|x|) is ([0,\infty)). Step 2: For every (y\ge 0), taking (x=y) gives (|x|=y). Step 3: To make the modulus function onto, choose codomain ([0,\infty)).

Correct answer: A. Onto

Step 1: For any (y\in\mathbb{R}), solve (y=2x+7). Step 2: (x=\frac{y-7}{2}) is real for every real (y). Step 3: A non-zero-slope linear function from (\mathbb{R}) to (\mathbb{R}) is onto.

Correct answer: B. Not onto

Step 1: The constant function (f(x)=5) always gives only (5). Step 2: Other codomain values such as (4) and (6) are not obtained. Step 3: If the range has only one value and the codomain is larger, the function is not onto.

Correct answer: A. Onto

Step 1: The function value is (5) for every (x). Step 2: The codomain also contains only (5), so every codomain value is attained. Step 3: A constant function can be onto if the codomain has exactly that one value.

Correct answer: B. No because (c) has no preimage

Step 1: For onto, every element of (B) must be obtained at least once. Step 2: Here (a) and (b) are obtained, but (c) is not. Step 3: For finite sets, mark each codomain element and check whether all are hit.

Correct answer: A. Onto

Step 1: The codomain (B) has elements (p) and (q). Step 2: Both (p) and (q) are obtained, so no codomain element is missed. Step 3: Onto does not require distinct preimages, only that every codomain element is hit.

Correct answer: A. For every (b\in B), there exists (a\in A) such that (f(a)=b)

Step 1: Onto is a condition about every element of the codomain. Step 2: If every (b\in B) is the image of some (a\in A), then range equals (B). Step 3: In definition questions, identify the statement that starts from the codomain.

Correct answer: A. Range (=) codomain

Step 1: In an onto function, no codomain element is left out. Step 2: Hence the range of the function equals the whole codomain. Step 3: The shortest test for onto is to compare range and codomain.

Correct answer: B. Not onto

Step 1: (x^2-6x+10=(x-3)^2+1). Step 2: Therefore the minimum value is (1), and the range is ([1,\infty)). Step 3: Completing the square quickly reveals whether a quadratic is onto.

Correct answer: A. Onto

Step 1: (f(x)=(x-3)^2+1). Step 2: Its range is ([1,\infty)), which equals the given codomain. Step 3: The same formula can give a different answer when the codomain changes.

Correct answer: A. Onto

Step 1: (x^3+x) is a continuous increasing function because (3x^2+1>0). Step 2: As (x\to \infty), the value goes to (\infty), and as (x\to -\infty), it goes to (-\infty). Step 3: Such cubic functions cover all real values.

Correct answer: B. Not onto

Step 1: (x^4\ge 0). Step 2: Therefore (x^4+1\ge 1), so values less than (1) are not obtained. Step 3: For even-power functions, check the minimum value to test onto.

Correct answer: A. Onto

Step 1: Take any codomain value (y\ge 0). Step 2: Choosing (x=\sqrt{y}) gives (x\in[0,\infty)) and (x^2=y). Step 3: Restricting the domain can make the square function onto for a suitable codomain.

Correct answer: B. Not onto because negative values are not obtained

Step 1: Even when (x\ge 0), (x^2\ge 0). Step 2: The codomain (\mathbb{R}) contains negative values that are not obtained. Step 3: Getting some values is not enough; the whole codomain must be covered.

Correct answer: A. It is not defined on all of (\mathbb{R})

Step 1: (\tan x) is not defined at (x=\frac{\pi}{2}+n\pi). Step 2: So it cannot directly be treated as a function from (\mathbb{R}) to (\mathbb{R}). Step 3: Before testing onto, check whether the rule is defined on the entire domain.