Functions

Step 1: (f(x)=2x-5) is linear with non-zero slope, so distinct inputs give distinct outputs. Step 2: For any (y\in\mathbb{R}), (x=\frac{y+5}{2}) is real, so every real output is attained. Step 3: In exams, remember that (ax+b) with (a\neq0) is both one-one and onto from (\mathbb{R}) to (\mathbb{R}).

Step 1: Since (x^2\ge 0), we get (x^2+1\ge 1). Step 2: The codomain (\mathbb{R}) contains (0), but (f(x)=0) is impossible for real (x). Step 3: To test onto, look for a codomain value that is never produced by the function.

Step 1: ((g\circ f)(x)) means apply (f) first and then apply (g). Step 2: (g(f(x))=(3x+1)^2=9x^2+6x+1). Step 3: In composition, order matters; always simplify the inside function first.

Step 1: Let (y=\frac{x-2}{3}). Step 2: Then (3y=x-2), so (x=3y+2). Replacing (y) by (x), we get (f^{-1}(x)=3x+2). Step 3: To find an inverse, isolate (x) first and then interchange the variables.

Step 1: Onto means every element of (B) is hit by at least one element of (A). Step 2: Since (A) has (7) elements and (B) has (5), all images cannot be distinct. Step 3: For finite sets, compare cardinalities to judge one-one and onto quickly.

Step 1: If (n_1+2=n_2+2), then (n_1=n_2), so the function is one-one. Step 2: For any (m\in\mathbb{Z}), (n=m-2) is also an integer, so every integer is obtained. Step 3: On integers, adding or subtracting a fixed integer usually gives a bijection.

Step 1: For (n\in\mathbb{N}), (n+2\ge 3). Step 2: The codomain (\mathbb{N}) contains (1) and (2), but they are not values of (f(n)). Step 3: Changing the domain can change whether a function is onto.

Step 1: On ([0,\infty)), (x^2) is increasing, so two different inputs do not give the same square. Step 2: For any (y\ge0), (x=\sqrt{y}) belongs to the domain and gives (f(x)=y). Step 3: The nature of the square function depends strongly on its domain.

Step 1: (f(2)=4) and (f(-2)=4), so the function is not one-one. Step 2: Every (y\ge0) in the codomain is obtained by taking (x=\sqrt{y}). Step 3: Read the codomain carefully because it decides the onto test.

Step 1: Let (y=x^3). Step 2: Then (x=\sqrt[3]{y}), so after interchanging variables (f^{-1}(x)=\sqrt[3]{x}). Step 3: The cube function is bijective on (\mathbb{R}), so its inverse exists.

Step 1: In a rational function, the denominator must not be zero. Step 2: Solving (x-3=0) gives (x=3), so this value cannot be in the domain. Step 3: In such questions, first set the denominator equal to zero to find excluded values.

Step 1: From (y=\frac{2x+1}{x-3}), we get (yx-3y=2x+1). Step 2: Thus (x(y-2)=3y+1), so (x=\frac{3y+1}{y-2}), defined for (y\neq2). Step 3: If each allowed (y) gives a unique (x), the function is bijective.

Step 1: Suppose ((g\circ f)(a_1)=(g\circ f)(a_2)). Step 2: Then (g(f(a_1))=g(f(a_2))); since (g) is one-one, (f(a_1)=f(a_2)), and since (f) is one-one, (a_1=a_2). Step 3: The composition of two one-one functions is always one-one.

Step 1: Take any (c\in C). Since (g) is onto, there is (b\in B) such that (g(b)=c). Step 2: Since (f) is onto, there is (a\in A) such that (f(a)=b), so (g(f(a))=c). Step 3: The composition of two onto functions is onto.

Step 1: For every (x), (f(x)=5), so many inputs have the same image and the function is not one-one. Step 2: The codomain (\mathbb{R}) has many values other than (5) that are not obtained. Step 3: A constant function on a large codomain is generally neither one-one nor onto.

Step 1: For finite sets of equal size, a one-one function is automatically onto. Step 2: The number of ways to match (4) elements with (4) distinct images is (4!=24). Step 3: The number of bijections between two sets of size (n) is (n!).

Step 1: The first element has (5) choices, the second has (4), and the third has (3). Step 2: Thus the number of one-one functions is (5\times4\times3=60). Step 3: For one-one functions, choose images without repetition.

Step 1: In a one-one function, distinct elements of (A) must have distinct images. Step 2: (A) has (5) elements but (B) has only (3), so (5) distinct images are impossible. Step 3: For finite sets, if the domain is larger than the codomain, no one-one function exists.

Step 1: The value of (x^3+x) keeps increasing as (x) increases. Step 2: More formally, (f'(x)=3x^2+1>0), so the function is strictly increasing. Step 3: A strictly increasing function is one-one.

Step 1: (\sin x) always lies between (-1) and (1). Step 2: The codomain is (\mathbb{R}), which contains values like (2), but these are not obtained by (\sin x). Step 3: Remembering the range of trigonometric functions helps in onto questions.

Step 1: On the given interval, (\sin x) is strictly increasing, so it is one-one. Step 2: Its values over this interval cover all of ([-1,1]). Step 3: Restricting a trigonometric function to a suitable interval can make it bijective.

Step 1: (e^x) is strictly increasing, so it is one-one. Step 2: For every (y>0), (x=\ln y) is real and gives (e^x=y). Step 3: The exponential function becomes bijective when the codomain is ((0,\infty)).

Step 1: Let (y=\ln x). Step 2: By the definition of logarithm, (x=e^y), so (f^{-1}(x)=e^x). Step 3: (\ln x) and (e^x) are inverse functions of each other.

Step 1: Complete the square: (x^2-4x+7=(x-2)^2+3). Step 2: Since ((x-2)^2\ge0), the minimum value is (3). Step 3: Completing the square is a fast way to find the range of a quadratic function.

Step 1: (f(x)=(x-2)^2+3), so the range is ([3,\infty)). Step 2: (f(1)=4) and (f(3)=4), so the function is not one-one. Step 3: If the codomain equals the range, the function is onto.