Reflexive relation

Step 1: (x^3-3x) is an odd degree polynomial. Step 2: As (x\to\infty), its value goes to (\infty), and as (x\to-\infty), its value goes to (-\infty), so all real values occur. Step 3: For odd degree continuous polynomials, check end behavior.

Step 1: The maximum value of (4-(x-2)^2) is (4). Step 2: Its range is ((-\infty,4]), equal to the codomain. Step 3: For a parabola, use the vertex to find the range quickly.

Step 1: (f(x)=(x-1)^2) and (x\ge1). Step 2: At (x=1), the value is (0), and as (x) increases all non-negative values occur. Step 3: For restricted domains, find the range using that exact domain.

Step 1: (e^x) is always positive. Step 2: For every (y>0), taking (x=\ln y) gives (f(x)=y). Step 3: For exponential functions, codomain ((0,\infty)) makes onto verification direct.

Step 1: The range of (e^x) is \((0,\infty)\). Step 2: The codomain is \(\mathbb{R}\), but values like (-1) are never attained. Step 3: If even one codomain element is missed, the function is not onto.

Step 1: The range of (\sin x) is ([-1,1]). Step 2: The codomain is also ([-1,1]), so the function is onto. Step 3: Since (\sin x) is periodic, different inputs can give the same output.

Step 1: On the given domain, (\sin x) is strictly increasing. Step 2: Its range is ([-1,1]), equal to the codomain. Step 3: On restricted trigonometric intervals, check both monotonicity and range.

Step 1: On ([0,\pi]), (\cos x) is strictly decreasing. Step 2: (\cos 0=1) and (\cos \pi=-1), so all values in ([-1,1]) occur. Step 3: A strictly monotonic function is one-one and becomes onto when its range equals the codomain.

Step 1: Values of \(\tan^{-1}x\) lie only in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). Step 2: The codomain is \(\mathbb{R}\), so many real values are missed. Step 3: Remember the principal range of inverse trigonometric functions.

Step 1: \(\tan^{-1}x\) is strictly increasing, so it is one-one. Step 2: Its range is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), equal to the codomain. Step 3: Choosing the exact range as codomain often makes a function onto.

Step 1: (2n+1) is always an odd integer. Step 2: The codomain (\mathbb{Z}) also contains even integers, which are not images. Step 3: For integer functions, parity checks are very useful for onto questions.

Step 1: (f(n)=2n+1) gives all odd integers. Step 2: For any odd (y), (n=\frac{y-1}{2}) is an integer. Step 3: Onto is proved when every target element has a preimage.

Step 1: For (n\in\mathbb{N}), (f(n)\ge4). Step 2: The codomain (\mathbb{N}) contains (1,2,3), but they are not attained. Step 3: For natural number functions, check the smallest possible image.

Step 1: The first value of (f(n)=n+3) is (4). Step 2: For every (y\ge4), (n=y-3) is a natural number. Step 3: In onto proofs, take an arbitrary (y) and find its preimage.

Step 1: In an onto function, every element of the codomain must be hit at least once. Step 2: Here the domain has (3) elements and the codomain has (4), so covering all four elements is impossible. Step 3: For finite sets, first compare the number of elements.

Step 1: For finite sets with equal size, an onto function is automatically one-one. Step 2: Hence such functions are bijections, counted by (4!). Step 3: (4!=24); for equal finite sizes, onto counting becomes permutation counting.

Step 1: Total functions are (2^3=8). Step 2: The non-onto functions are the (2) constant functions. Step 3: Onto functions are (8-2=6); for small sets, subtract the missed cases.

Step 1: Take any (z\in C). Step 2: Since (g) is onto, (g(y)=z) for some (y\in B), and since (f) is onto, (f(x)=y) for some (x\in A). Step 3: Thus ((g\circ f)(x)=z), so the composition is onto.

Step 1: Since (g\circ f) is onto, every element of (C) is ((g\circ f)(x)) for some (x). Step 2: But ((g\circ f)(x)=g(f(x))), so that element is attained by (g). Step 3: If the composition is onto, the outer function must be onto.

Step 1: If (g\circ f) is onto, (g) must be onto. Step 2: But (f) need not cover all of (B), because (g) may cover (C) using only part of (B). Step 3: In composition questions, keep the roles of inner and outer functions separate.