Functions

Step 1: For finite sets with equal size, a bijection is like a permutation. Step 2: There are (3!) ways to assign (3) distinct images. Step 3: Since (3!=6), there are (6) bijective functions.

Step 1: (\sin 0=0) and (\sin \pi=0), so it is not one-one. Step 2: The range of (\sin x) is ([-1,1]), so it is not onto (\mathbb{R}). Step 3: For trigonometric functions, check periodicity and range.

Step 1: On the given interval, (\sin x) is strictly increasing, so it is one-one. Step 2: Its range on this interval is exactly ([-1,1]). Step 3: Restricting the domain can make the same function bijective.

Step 1: (\cos 0=1) and (\cos 2\pi=1), so the function is not one-one. Step 2: The range of (\cos x) is ([-1,1]), so it is not onto (\mathbb{R}). Step 3: If codomain is (\mathbb{R}), a bounded range function is not onto.

Step 1: On \([0,\pi]\), (\cos x) is strictly decreasing, so it is one-one. Step 2: On this interval, it takes all values from (1) to (-1). Step 3: Hence it is also onto the codomain ([-1,1]).

Step 1: (f^{-1}(k)=4) means (f(4)=k). Step 2: (f(4)=2\cdot4+3=11). Step 3: In inverse function questions, translate the statement into the original function.

Step 1: First find (g(2)=2+1=3). Step 2: Now (f(g(2))=f(3)=3^2-1=8). Step 3: In composition, always evaluate the inner function first.

Step 1: Rewrite (x^2-6x+10) as ((x-3)^2+1). Step 2: Since ((x-3)^2\ge 0), the minimum value is (1). Step 3: Therefore the range is ([1,\infty)).

Step 1: (\lfloor 1.2\rfloor=1) and (\lfloor 1.8\rfloor=1), so it is not one-one. Step 2: Its range is only the integers, not all real numbers. Step 3: For the greatest integer function, check the range carefully.

Step 1: If (n_1+5=n_2+5), then (n_1=n_2), so it is one-one. Step 2: For any (m\in\mathbb{Z}), (n=m-5\in\mathbb{Z}), so every (m) is attained. Step 3: Adding a fixed integer gives a bijection on integers.

Step 1: From (2n_1=2n_2), we get (n_1=n_2), so it is one-one. Step 2: But odd integers like (1) are not of the form (2n). Step 3: On integers, (2n) produces only even integers.

Step 1: (n+1) gives different values for different (n), so it is one-one. Step 2: If (\mathbb{N}={1,2,3,\ldots}), then (1) is not of the form (n+1). Step 3: On natural numbers, always check the first element of the codomain.

Step 1: (f(0)=0) and (f(-2)=0). Step 2: Two different inputs (0) and (-2) have the same image. Step 3: To disprove one-one property, one counterexample is enough.

Step 1: (|x-2|\ge 0). Step 2: The smallest value occurs when (|x-2|=0), that is (x=2), giving minimum (3). Step 3: The range of this absolute value function starts from its minimum value.

Step 1: ((f\circ g)(x)=f(g(x))). Step 2: Substitute (g(x)=x-1) into (f), giving (f(x-1)=|x-1|). Step 3: Keep the entire inner expression inside the absolute value.

Step 1: (f(0)=0^3-3\cdot0=0). Step 2: (f(\sqrt{3})=(\sqrt{3})^3-3\sqrt{3}=3\sqrt{3}-3\sqrt{3}=0). Step 3: Two distinct inputs have the same image, so the function is not one-one.

Step 1: In an onto function, every element of (B) is the image of some element of (A). Step 2: Therefore (f(A)=B). Step 3: Hence (f(A)) has (5) elements.

Step 1: In a one-one function, distinct elements of (A) have distinct images. Step 2: Since (A) has (6) elements, there will be (6) distinct images. Step 3: Therefore (f(A)) has (6) elements.

Step 1: Since (x^2\ge 0), we have (x^2-1\ge -1). Step 2: A value like (-2) is less than (-1), so it can never be attained. Step 3: To disprove onto property, one missing codomain element is enough.

Step 1: For (x>0), the value is (\frac{x}{1+x}), which lies between (0) and (1). Step 2: For (x<0), the value is (\frac{x}{1-x}), which lies between (-1) and (0). Step 3: The values (1) and (-1) are approached but not attained, so the range is ((-1,1)).

Step 1: Let (y=\frac{2x+3}{x-1}). Step 2: From (y(x-1)=2x+3), we get (x(y-2)=y+3). Step 3: Therefore (x=\frac{y+3}{y-2}), so (f^{-1}(x)=\frac{x+3}{x-2}).

Step 1: The function is strictly increasing because the odd power terms along with (x) increase consistently. Step 2: As (x\to\infty), the function goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Step 3: Hence it is both one-one and onto on (\mathbb{R}).

Step 1: For (x<0), values lie in ((-\infty,-1)), and for (x\ge0), values lie in ([1,\infty)). Step 2: The two parts do not overlap, so the function is one-one. Step 3: But values in ((-1,1)) are missing, so it is not onto.

Step 1: We can write (x^3-6x^2+12x+1=(x-2)^3+9). Step 2: The cubic expression ((x-2)^3) is strictly increasing, so distinct inputs give distinct outputs and every real value is attained. Step 3: In exams, convert such polynomials into shifted cubic form to identify one-one and onto properties quickly.

An inverse is a function only when every element of \(B\) has exactly one preimage in \(A\). Onto ensures a preimage exists, while one-one ensures it is unique. Exam tip: “exactly one” signals that both conditions are needed.