Onto function

Step 1: To test onto nature, put (f(x)=y). Step 2: From (5x-11=y), we get (x=\frac{y+11}{5}), which is real for every real (y). Step 3: For a linear function, solving (x) in terms of (y) is the easiest method.

Step 1: If (k=0), the function becomes the constant (3), so it cannot give all real values. Step 2: If (k\ne0), for any (y), (x=\frac{y-3}{k}) exists. Step 3: A linear function from (\mathbb{R}) to (\mathbb{R}) is onto exactly when its slope is non-zero.

Step 1: (x^2-4x+7=(x-2)^2+3). Step 2: Its minimum value is (3), so the range is ([3,\infty)), equal to the codomain. Step 3: For a quadratic function, complete the square to find the range quickly.

Step 1: (f(x)=(x-2)^2+3), so the range is ([3,\infty)). Step 2: The codomain ([0,\infty)) contains (1), but it is not in the range. Step 3: For onto nature, every codomain value must be obtained, not just some values.

Step 1: Since ((x+2)^2\ge0), (5-(x+2)^2\le5). Step 2: For every (y\le5), take (x=-2+\sqrt{5-y}), then (f(x)=y). Step 3: A downward-opening quadratic can be onto a suitable codomain.

Step 1: (|3x-6|\ge0), so (f(x)\ge-4). Step 2: For every (y\ge-4), set (3x-6=y+4), giving (x=\frac{y+10}{3}). Step 3: The range of a modulus of a linear expression starts from its minimum value.

Step 1: Put (f(x)=y), so (\ln(x-2)=y). Step 2: Then (x-2=e^y), so (x=2+e^y), which is always greater than (2). Step 3: A logarithmic function gives all real values when its inside expression is positive.

Step 1: The range of (e^{2x}) is ((0,\infty)). Step 2: Adding (4) gives range ((4,\infty)), which equals the codomain. Step 3: In exponential functions, the open endpoint is never included.

Step 1: (e^{2x}) is always positive and never (0). Step 2: Hence (e^{2x}+4) is always greater than (4), but (4) is included in the codomain. Step 3: In onto questions, always check open and closed endpoints.

Step 1: Take any codomain value (y\ne0). Step 2: From (\frac{2}{x-3}=y), we get (x=3+\frac{2}{y}), which is not (3). Step 3: Reciprocal-type functions are onto when the codomain is chosen correctly.

Step 1: Put (\frac{2x+3}{x-1}=y). Step 2: Then (2x+3=xy-y), so (x(y-2)=y+3) and (x=\frac{y+3}{y-2}). Step 3: Since (y=2) is not in the codomain, this preimage is valid.

Step 1: (x^4\ge0), so (x^4+1\ge1). Step 2: For every (y\ge1), take (x=\sqrt[4]{y-1}), then (f(x)=y). Step 3: An even-power function can be onto a suitable non-negative codomain.

Step 1: The minimum value of (x^4+1) is (1). Step 2: The codomain (\mathbb{R}) contains (0), but it is not in the range. Step 3: For a real codomain, always check values below the minimum range.

Step 1: (x^3-3x) 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 continuous cubic polynomial takes all real values.

Step 1: On the given domain, the minimum value occurs at (x=1) and is (-2). Step 2: Therefore, a real codomain value like (-3) cannot be obtained. Step 3: Changing the domain can change the onto nature of the same formula.

Step 1: (\cos0=1) and (\cos\pi=-1). Step 2: On ([0,\pi]), (\cos x) decreases continuously and takes all intermediate values. Step 3: For trigonometric functions, remember the range on the chosen interval.

Step 1: On ([0,\frac{\pi}{2}]), the range of (\sin x) is ([0,1]). Step 2: The codomain ([-1,1]) contains negative values, which are not obtained. Step 3: In onto questions, immediately check the effect of domain restriction.

Step 1: On this standard interval, (\tan x) takes all real values. Step 2: For any (y\in\mathbb{R}), (x=\tan^{-1}y) lies in this interval. Step 3: Remember the onto interval of (\tan x).

Step 1: The range of (\cos x) is ([-1,1]). Step 2: The range of (2\cos x) becomes ([-2,2]), equal to the codomain. Step 3: Multiplication scales the trigonometric range accordingly.

Step 1: The codomain (B) has elements (p,q,r). Step 2: All three appear among the given images. Step 3: For finite sets, if every codomain element appears as an image, the function is onto.

Step 1: The listed images include (a,b,c). Step 2: The codomain element (d) is not obtained from any domain element. Step 3: If even one codomain element is missed, the function is not onto.

Step 1: The total number of functions is (2^4=16). Step 2: There are (2) non-onto functions, where all elements go to only one element of (B). Step 3: Hence the number of onto functions is (16-2=14).

Step 1: For finite sets of equal size, an onto function makes every codomain element appear exactly once. Step 2: This is a permutation of (3) elements, so the number is (3!=6). Step 3: When (|A|=|B|), an onto function is also bijective.

Step 1: In an onto function, every codomain element needs a preimage. Step 2: Only (2) domain elements cannot cover (5) codomain elements. Step 3: If (|A|<|B|), the number of onto functions is immediately (0).

Step 1: (f(A)) is the range of the function. Step 2: If a function is onto, its range equals its codomain. Step 3: Hence (f(A)=B={2,4,6,8}).