Functions as a special kind of relation

In a function each element of (A) is related to exactly one element of (B). In exams focus on the phrase exactly one.

For a function, every element of \(A\) must have exactly one image in \(B\). In option A, 1 maps to 3 and 2 maps to 4, so it is a function. In option B, 1 has two images; in option C, 1 has no image; and in option D, 2 has two images. Exam tip: Check that every element of the domain appears exactly once as the first component of an ordered pair.

The governing definition is that a function from A to B must assign exactly one image in B to every element of A. In the given relation, the first component 1 occurs in both (1,2) and (1,3), so the same input 1 is assigned two different outputs, 2 and 3. This violates the uniqueness condition for images. Although the element 2 does have the image 4, that does not repair the violation caused by 1. The codomain B is non-empty and all listed pairs are not identical, so C and D are plainly irrelevant. Thus option A correctly identifies the reason that R is a relation but not a function.

Each of (a), (b), and (c) has exactly one image. Two different elements may have the same image in a function.

For a function, every element of the domain (input) must be associated with exactly one element of the codomain (output), so statement A is always true. Statements B and C describe additional one-to-one behavior, which is not required for every function; different inputs may have the same output. Statement D is false because no input in the domain can be left without an associated output. Exam tip: To test a mapping, check that exactly one arrow starts from each input.

In the function \(f:A\to B\), \(A\) is the domain and \(B\) is the codomain. The set of actual output values is the range, which may be a subset of the codomain. Exam tip: the set written after the arrow is the codomain.

Using the rule \(g(x)=2x\), substitute \(x=3\): \(g(3)=2\times3=6\). Therefore, 6 is correct. The value 3 is the input, not the output of the function. Exam tip: to evaluate a function, replace its variable with the given input and simplify.

Direct answer: Option A, only one arrow leaves each domain element. A function from a domain to a codomain assigns exactly one output to every input. In a mapping diagram, this means every element of the domain must have one and only one outgoing arrow. The outputs may be the same for different inputs, and some codomain elements may receive no arrow. Option A states the essential rule. Option B is not allowed because one input would have two different outputs, so the assignment would not be single-valued. Option C is not allowed because an input with no arrow has no assigned output, so the rule is incomplete. Option D describes a surjective or onto condition, not the basic definition of every function; a function need not cover the whole codomain. For example, two domain elements may both point to one codomain element and the function can still be valid. Memory cue: one input, exactly one output; many inputs may share an output.

Each element of A has exactly one image in B: 1 maps to a, 2 maps to a, and 3 maps to b. Therefore, the relation is a function. The fact that 1 and 2 have the same image does not violate the definition; it only means the function is many-one, not one-one. Exam tip: for a function, every domain element must have exactly one image.

For a relation to be a function, every element of the domain A must have exactly one image. Here, 1 maps to 2 and 2 maps to 3, but 3 has no image. Therefore, R is not a function from A to A. Option B is incorrect because 1 has only one image, namely 2. Exam tip: To test a function, check that every domain element appears exactly once as the first component of an ordered pair.

A relation is a collection of ordered pairs that connects elements of one set with elements of another set. A function is a special relation with an additional rule: every element of the domain must have exactly one image in the codomain. Because every function satisfies the general definition of a relation, every function is necessarily a relation.

The reverse statement is not true. A relation may assign two different outputs to one input, or may fail to assign an output to some domain element, and then it is not a function. Being one-one is also not required for every function; different inputs may have the same output. Therefore option A is correct. The other choices either reverse the definition or impose conditions that do not apply to all functions.

The ordered pair (1, 2) shows that the function assigns the value 2 to the input 1. Therefore, f(1) = 2.

In the ordered pair (0,1), the first component 0 is the input and the second component 1 is its image. Therefore, h(0) = 1. Exam tip: To find h(a), locate the ordered pair (a,b) whose first component is a; the second component is the answer.

In a function, every element of the domain must have exactly one image. In option A, the input 1 has two different images, 5 and 6, so the relation is not a function. In the other options, both 1 and 2 have exactly one image each. Exam tip: To test a relation quickly, check whether any first coordinate is paired with more than one second coordinate.

Direct answer: Option A, \(\{(2,1),(4,5)\}\). For a function from \(A=\{2,4\}\) to \(B=\{1,3,5\}\), every element of A must occur exactly once as a first component, and each second component must belong to B. In option A, 2 is mapped to 1 and 4 is mapped to 5; both outputs are in B, and each input has exactly one image. Option B is invalid because input 2 is assigned both 1 and 3, while input 4 has no image. Option C is invalid because input 2 has no image. Option D is invalid because input 4 is assigned two different outputs, 5 and 3, while input 2 has one. The fact that 3 is unused in option A does not matter; a function need not use every codomain element. Memory cue: check the first coordinates—each domain element must appear once and only once.

In a constant function, every domain element has the same image or function value; therefore, all domain elements are associated with one fixed element of the codomain. Hence, option A is correct. Option B represents different function values, which is not the defining property of a constant function. Exam tip: the graph of a constant function is a horizontal line.

The function value is (7) for every input, so it is a constant function. In a constant function all images are the same.

By definition, an identity function satisfies \(I_A(x)=x\). Thus, every element of the set \(A\) is mapped to itself. The image is not necessarily \(0\) or \(1\); it remains the input element \(x\). Exam tip: in an identity function, input and output are always the same.

In an identity function, every element of A maps to itself, so f(x) = x. Hence all ordered pairs must be of the form (x, x), giving {(1, 1), (2, 2), (3, 3)}. In option B the elements map cyclically, while options C and D do not map every element to itself. Exam tip: for an identity function, check that the two entries in every ordered pair are equal.

Direct answer: Option A, the set of ordered pairs \(\{(x,f(x)):x\in A\}\). A function records an input and its corresponding output together. For every \(x\) in the domain, the graph contains the ordered pair \((x,f(x))\). The first coordinate tells us the input, and the second tells us the output. Option A is therefore the correct mathematical description. Option B is wrong because the domain elements alone do not tell us their outputs and are not the graph. Option C is wrong for the same reason: codomain elements alone omit the input-output pairing. Option D is clearly wrong because a function graph is not limited to the numbers 0 and 1; its pairs depend on the function and its domain. For example, if \(f(2)=5\), the graph contains \((2,5)\), not just an isolated input or output. Memory cue: a graph of a function is a collection of input-output pairs.

In option A, the input x = 2 is associated with two different outputs, y = 5 and y = 6. Thus, it violates the defining condition of a function: every input must have exactly one output. In option B, different x-values share the same y-value, which is allowed, so it is a function. Exam tip: check whether any x-value is paired with more than one y-value.

The governing concept is the Cartesian-product definition of a relation. A relation from A to B consists of ordered pairs whose first component belongs to A and whose second component belongs to B. The set of every possible such ordered pair is A × B. Therefore any relation from A to B, including a function, must be a subset of A × B. A function adds the extra requirement that every element of A appears as a first component exactly once with one image in B; it does not change the product containing the relation. B × A reverses the order of the components, while union and intersection are sets of elements rather than the appropriate collection of ordered pairs.

The governing concept is the requirement that a function assign exactly one image to every element of its domain. The domain here is A = {1,2}, so it has |A| = 2 elements. A function from A to B must therefore contain one ordered pair beginning with 1 and one ordered pair beginning with 2, giving exactly two ordered pairs. The codomain B determines the possible choices of images, not the number of pairs in a completed function. There are 3 choices for the image of each input, so six different functions can be formed in total, but each individual function still has two pairs. Thus D confuses the number of functions with the number of pairs, while B and C do not follow from the domain size.

In a function, each element of the domain is associated with exactly one element of the codomain. Thus, 4 domain elements produce 4 ordered pairs. The number of elements in the codomain does not affect this answer. Exam tip: The number of ordered pairs in a function equals the number of elements in its domain.

Each element (p) and (q) has (2) choices, so total functions are (2^2=4). Use (|B|^{|A|}) for counting.