Functions

Step 1: An absolute value is always (0) or positive. Step 2: The minimum of (|x-2|) is (0), which occurs when (x-2=0). Step 3: Therefore, the function is minimum at (x=2).

Since \(f(-1)=0\) and \(f(0)=0\) although \(-1\ne0\), the function is not one-one. As this cubic attains values from \(-\infty\) to \(+\infty\), it is onto \(\mathbb{R}\). Exam tip: equal outputs for distinct inputs disprove injectivity.

Step 1: ((f\circ f)(x)=f(2x+3)=2(2x+3)+3=4x+9). Step 2: (4x+9=19), so (4x=10). Step 3: The calculated value is (x=\frac{5}{2}), so the options should include (\frac{5}{2}).

Step 1: ((f\circ f)(x)=f(2x+3)=2(2x+3)+3=4x+9). Step 2: (4x+9=17), so (4x=8). Step 3: Therefore, (x=2).

Step 1: The value of (\sin x) always lies between (-1) and (1). Step 2: The codomain is (\mathbb{R}), which includes values like (2) that are not attained. Step 3: For onto checking, compare the range with the codomain.

Step 1: A permutation function means a one-one and onto function from (A) to (A). Step 2: The number of ways to arrange (4) elements in (4) positions is (4!). Step 3: Hence there are (24) such functions.

Step 1: The expression inside the square root must satisfy (9-x^2\geq 0). Step 2: This gives (x^2\leq 9), meaning (-3\leq x\leq 3). Step 3: When square roots and squares appear together, solve the inequality carefully.

Step 1: (2^x) is an increasing function, so it is one-one. Step 2: Its value is always positive, so no negative real number is attained. Step 3: Knowing the range of an exponential function helps in checking onto.

Step 1: (\ln x) is defined only for positive (x). Step 2: (x=0) and negative (x) are not included in the real domain. Step 3: For logarithmic functions, the inside expression must be (>0).

Step 1: (f(2)=b) and (f(3)=b), so two distinct elements have the same image; the function is not one-one. Step 2: (c) is not the image of any element, so the function is not onto. Step 3: For finite sets, list the images to test both properties.

Step 1: (f(0)=0). Step 2: (f(\sqrt{3})=(\sqrt{3})^3-3\sqrt{3}=3\sqrt{3}-3\sqrt{3}=0), while (0\neq \sqrt{3}). Step 3: If two distinct inputs give the same image, the function is not one-one.

Step 1: Let (y=\frac{2x+3}{x-1}). Step 2: Then (yx-y=2x+3), so (x(y-2)=y+3) and (x=\frac{y+3}{y-2}). Step 3: Replacing (y) by (x), we get (f^{-1}(x)=\frac{x+3}{x-2}).

Step 1: Suppose (f(a_1)=f(a_2)). Step 2: Then (g(f(a_1))=g(f(a_2))), so ((g\circ f)(a_1)=(g\circ f)(a_2)). Step 3: Since (g\circ f) is one-one, (a_1=a_2); therefore (f) is one-one.

Step 1: Put \(y=\frac{3x-4}{2x+1}\) and solve for (x). Step 2: \(y(2x+1)=3x-4\), so \(x(2y-3)=-(y+4)\). Step 3: If \(2y-3=0\), (x) is not obtained; hence \(\frac{3}{2}\) is not in the range.

Step 1: Total functions are (3^4=81). Step 2: Remove functions that miss at least one element: (3\cdot2^4-3\cdot1^4=48-3=45). Step 3: Onto functions are (81-45=36).

Step 1: (f(x)=x^2-6x+10=(x-3)^2+1). Step 2: From (y=(x-3)^2+1), we get ((x-3)^2=y-1). Step 3: Since the domain is ([3,\infty)), (x-3\geq0), so the inverse is (3+\sqrt{x-1}).

Step 1: In (g\circ f), (f) acts first and then (g). Step 2: To reverse the process, apply (g^{-1}) first and then (f^{-1}). Step 3: Hence the order reverses, so ((g\circ f)^{-1}=f^{-1}\circ g^{-1}).

Step 1: If (n_1\neq n_2), then (n_1+1\neq n_2+1), so the function is one-one. Step 2: The codomain (\mathbb{N}) contains (1), but there is no (n\in\mathbb{N}) such that (n+1=1). Step 3: Therefore, the function is one-one but not onto.

Step 1: To disprove one-one, it is enough to find two different inputs with the same output. Step 2: (f(0)=1) and (f(\sqrt{3})=1), while (0\neq\sqrt{3}). Step 3: For polynomial functions, one clear counterexample quickly settles one-one behaviour.

Step 1: (x^3+x+5) is strictly increasing, so different inputs give different outputs. Step 2: As (x\to\infty), the value goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Step 3: A strictly increasing function covering all real values is bijective.