Onto function

Correct answer: A. Because (0) is not the image of any (x)

Step 1: Since (x^2\ge0), we get (x^2+1\ge1). Step 2: So (0) and all real numbers less than (1) are not in the range. Step 3: To disprove onto, it is enough to show one missed codomain element.

Correct answer: A. Yes because for (m\in\mathbb{Z}), (n=m-5) can be taken

Step 1: Take any codomain element (m\in\mathbb{Z}). Step 2: (n=m-5) is also an integer and (f(n)=m). Step 3: For translation functions on integers, test onto by writing the preimage directly.

Correct answer: A. Because (1) and (2) are missed

Step 1: For (n\in\mathbb{N}), (n+2\ge3). Step 2: The codomain (\mathbb{N}) contains (1) and (2), but they are not images. Step 3: In functions on natural numbers, quickly check the first few codomain values.

Correct answer: A. It is onto because for every (y\ge0), (x=\sqrt{y}) exists

Step 1: Here the codomain is only ([0,\infty)). Step 2: For every (y) in this codomain, (\sqrt{y}) lies in the domain and (f(\sqrt{y})=y). Step 3: The same formula may give a different answer when the codomain changes.

Correct answer: A. Yes because for every (y\le3), (x=\sqrt{3-y}) can be taken

Step 1: The maximum value of (3-x^2) is (3). Step 2: If (y\le3), then (x=\sqrt{3-y}) is real and (f(x)=y). Step 3: A function is onto when its range equals its codomain.

Correct answer: A. Because for any (y\in\mathbb{R}), (x=\frac{y+7}{2}) exists

Step 1: Take any real number (y) from the codomain. Step 2: Solving (y=2x-7) gives (x=\frac{y+7}{2}), which is real. Step 3: A non-constant linear function from real numbers to real numbers is onto.

Correct answer: A. Zero and negative real numbers are not images

Step 1: (e^x) is positive for every real (x). Step 2: The codomain (\mathbb{R}) contains (0) and negative numbers, but they are not obtained. Step 3: For exponential functions, compare the range carefully with the codomain.

Correct answer: A. (f) is onto

Step 1: The codomain is ((0,\infty)). Step 2: For every (y>0), (x=\ln y) is real and (e^x=y). Step 3: An exponential function can be onto when its codomain is chosen as its natural range.

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

Step 1: The range of (\sin x) is ([-1,1]). Step 2: The codomain is also ([-1,1]), so no codomain element is missed. Step 3: In trigonometric functions, use the standard range directly.

Correct answer: A. Yes because for every (y\ge1), (x=\sqrt{y-1}) can be taken

Step 1: The minimum value of (x^2+1) is (1). Step 2: For every (y\ge1), (x=\sqrt{y-1}) is real and gives (f(x)=y). Step 3: Use the minimum or maximum value to match the codomain.

Correct answer: A. (f) is onto

Step 1: For any real (y), \(x=\sqrt[3]{y}\) is real. Step 2: Then \(f(x)=\left(\sqrt[3]{y}\right)^3=y\). Step 3: Simple odd-power polynomial functions often cover the whole real codomain.

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

Step 1: (x^4\ge0) for every real (x). Step 2: So a negative codomain element such as (-1) is not an image. Step 3: For even powers, check negative codomain values first.

Correct answer: A. It is onto

Step 1: Put (y=\frac{2x+1}{x-1}), which gives (x=\frac{y+1}{y-2}). Step 2: Since the codomain has (y\ne2), this (x) is defined and also (x\ne1). Step 3: For rational functions, solve for (x) in terms of (y) to test onto.

Correct answer: A. Because (0) is in the codomain but is not an image

Step 1: (\frac{1}{x}) can never be equal to (0). Step 2: The codomain (\mathbb{R}) includes (0), so one element is missed. Step 3: For reciprocal-type functions, always check whether (0) can be an output.

Correct answer: A. It is onto

Step 1: Take any (y\ne0) in the codomain. Step 2: Choosing (x=\frac{1}{y}) gives (x\ne0) and (f(x)=y). Step 3: Removing (0) from the codomain makes this reciprocal function onto.

Correct answer: A. (f) is onto but not one-one

Step 1: All three codomain elements (a,b,c) appear as images. Step 2: But (2) and (3) both map to (b), so the function is not one-one. Step 3: For finite sets, listing images is the easiest test.

Correct answer: A. No because the domain has fewer elements than the codomain

Step 1: In an onto function, every codomain element must be hit at least once. Step 2: (A) has only (3) elements, while (B) has (4) elements. Step 3: For finite sets, the domain size cannot be less than the codomain size for an onto function.

Correct answer: A. (3^5-3\cdot2^5+3)

Step 1: The total number of functions is (3^5). Step 2: Subtract functions missing at least one codomain element: (\binom{3}{1}2^5), then add those missing two: (\binom{3}{2}1^5). Step 3: Inclusion-exclusion is the key tool for counting onto functions.

Correct answer: A. Because negative real numbers are not images

Step 1: (|x|) is always (0) or positive. Step 2: The codomain (\mathbb{R}) contains negative numbers, but they are not in the range. Step 3: For modulus functions, quickly check negative codomain values.

Correct answer: A. (f) is onto but not one-one

Step 1: For every (y\ge0), taking (x=y) gives (|x|=y). Step 2: But (|a|=|-a|), so different domain elements can have the same image. Step 3: Onto and one-one properties must be checked separately.

Correct answer: A. It is continuous and (f(x)\to\infty) as (x\to\infty), (f(x)\to-\infty) as (x\to-\infty)

Step 1: (x^3+x) is a continuous polynomial. Step 2: For very large positive (x), its value is very large positive, and for very large negative (x), its value is very large negative. Step 3: The end behavior of a continuous odd-degree polynomial supports onto over (\mathbb{R}).

Correct answer: A. Its minimum value is (3) and (2) is not an image

Step 1: Write (x^2-4x+7=(x-2)^2+3). Step 2: Its range is ([3,\infty)), so (2) in the codomain is not an image. Step 3: Completing the square quickly gives the range of a quadratic function.

Correct answer: A. It is onto

Step 1: (f(x)=(x-2)^2+3), so the range is ([3,\infty)). Step 2: The codomain is also ([3,\infty)), hence every codomain element is an image. Step 3: When the codomain equals the range, the function is onto.

Correct answer: A. Yes because for every (y\in(0,1]), (x=\sqrt{\frac{1}{y}-1}) can be taken

Step 1: (\frac{1}{1+x^2}) takes values greater than (0) and up to (1). Step 2: For any (0<y\le1), (\frac{1}{y}-1\ge0), so a suitable (x) exists. Step 3: In such questions, identify the range interval first.

Correct answer: A. Because (1) is never an image

Step 1: (\frac{x^2}{1+x^2}<1) for every real (x). Step 2: (0) is obtained at (x=0), but (1) is not obtained though it is in the codomain. Step 3: When a variable occurs in both numerator and denominator, check endpoint values carefully.