Functions as a special kind of relation

Substituting \(x=3\) into the function gives \(f(3)=3^2-1=9-1=8\). Therefore, option A is correct. Option C is only \(3^2\) and does not subtract 1. In such questions, substitute the value first, evaluate the power, and then perform addition or subtraction.

Substituting \(x=8\) into the rule \(g(x)=\frac{x}{2}\) gives \(g(8)=\frac{8}{2}=4\). Therefore, option A is correct. Option B leaves 8 unchanged instead of dividing by 2, while options C and D result from incorrect calculations. Exam tip: To evaluate a function, substitute the given input directly into its rule.

The governing counting principle is that a function from A to B assigns exactly one element of B to each element of A. Set A has three independent inputs: 1, 2, and 3. For each input, there are two possible output choices, 0 or 1, from set B. By the multiplication principle, the total number of assignments is 2 × 2 × 2 = 2^3 = 8, or more generally |B|^|A|. Each different assignment gives a different function, even when some outputs are repeated. Therefore option A, 8, is correct. The values 6, 5, and 3 undercount the possible assignments and do not follow the required power rule. No one-to-one or onto condition is imposed, so all assignments are included.

For a function from A to B, each of the 2 elements of A can be mapped to any of the 4 elements of B. Hence, the total number of functions is \(|B|^{|A|}=4^2=16\). The expression \(2^2\) incorrectly ignores the 4 available choices in B. Exam tip: For functions from A to B, use \(|B|^{|A|}\), with the codomain size as the base and the domain size as the exponent.

The direct answer is A: for every x in A, f(x) belongs to B. The notation f:A→B means that A is the domain, or set of allowed inputs, and B is the codomain, or set in which outputs must lie. Thus, whenever x is chosen from A, its image f(x) must be an element of B. Option A states exactly this defining requirement. Option B reverses the sets: elements of B are not automatically inputs, and f(x) is not thereby required to lie in A. Option C is wrong because the domain may contain many elements; it need not be the empty set. Option D is wrong because the codomain and domain can be different sets, even though some functions happen to use equal sets. For example, if A={1,2} and B={3,4}, a valid function maps each input in A to an output in B. Remember: in f:A→B, arrows go from A to B.

A function is a relation with a complete domain assignment. This means that every member of the domain set A must be connected to one and only one member of the codomain set B. The word “every” is essential: a function cannot leave even one input without an output. Therefore, if some element of A is not related to any element of B, the relation fails the basic existence requirement for a function. It may still be a relation, but it is not a function from A to B.

Option A is correct because every x in A must have an image under the function. It is not necessary that every y in B have an image; that condition would concern surjectivity, not the definition of a function. The other choices do not state the defining requirement. Hence the supplied answer is correct.

Two values for the same input (p) are allowed only if they are the same. Here (5\ne7), so generally it is not a function.

The governing concept is the defining condition of a function. Any set of ordered pairs is a relation, but it is a function only if each first component is associated with exactly one second component. In option A, the input 1 appears twice and is paired with two different outputs, 2 and 3. Thus R is certainly a relation, but it fails the function test because one input has more than one image. Therefore option A is correct. In option B, the first components 1 and 2 are different and each has one image. The same is true in option C for 1 and 2 and in option D for 3 and 4. Those options can therefore represent functions on their displayed domains. The issue is not repeated outputs; it is a repeated input with different outputs.

An identity function maps every element to itself, so \(I_A(x)=x\). Hence, \(I_A(5)=5\). Although 4 also belongs to the set \(A\), the identity function does not map 5 to 4. Exam tip: for an identity function, the input and output are always the same.

The governing concept is the identity function. For a set A, the identity function I_A maps every element to itself, so I_A(x)=x for every x in A. Since A={0,1,2}, the required ordered pairs are (0,0), (1,1), and (2,2). This is exactly option A. Option B cycles the elements, so the outputs differ from the inputs. Option C maps both 1 and 2 to 0, making it a constant-like mapping rather than an identity mapping. Option D interchanges 0 and 2 while leaving 1 fixed. Although the alternatives may describe relations or valid functions, none sends every element to itself. Therefore only option A represents the identity function on A.

Since (B) has only (a), every domain element maps to (a). All (1), (2), and (3) must be included.

For the rule \(f(x)=2x+1\), substituting all the given first components gives \(f(1)=3\), \(f(2)=5\), and \(f(3)=7\). Thus, it matches every ordered pair in the function. Each other option gives an incorrect value for at least one input. Exam tip: test a proposed rule with every given input, not just one.

The function gives \(1\mapsto1\), \(2\mapsto8\), and \(3\mapsto27\). These outputs are respectively \(1^3, 2^3\), and \(3^3\), so the correct rule is \(f(x)=x^3\). For instance, \(f(x)=x^2\) would give \(f(2)=4\), so option B is incorrect. Exam tip: test the proposed rule with all the ordered pairs, not just one pair.

In a function, every element of domain \(A\) must have exactly one image in codomain \(B\). In option A, 1, 2 and 3 each have one image. In B and D, one input has two images; in C, 3 has no image. Exam tip: check each input exactly once.

In the first option, (1), (2), and (3) each appear exactly once. A function needs exactly one output for each input.

In a function \(f:A\to B\), every element \(a\) of the domain has exactly one corresponding element \(f(a)\) in the codomain. Thus, \(f(a)\) cannot have zero or multiple values. Exam tip: in a mapping diagram, exactly one arrow must originate from each domain element.

In the first subset, (1) and (2) each have one image. In the others an input has two different values or is missing.

The governing concept is evaluation from an ordered-pair representation. In a pair (input, output), the first component identifies the input and the second component gives its function value. To find f(4), locate the pair whose first component is 4. The given function contains (4,8), so f(4)=8. Therefore option A is correct. The other values correspond to different roles or inputs: 4 in option B is the argument itself, 6 is the image of 3 because (3,6) is present, and 2 is the image of 1. A common error is to select the first component instead of the second or to read a neighboring pair. Since every listed input has exactly one image, the value is direct and unambiguous.

The governing concept is the domain requirement for a function. A function from A={1,2,3} to another set must assign exactly one image to every element of A. In option A, the first components are 1 and 3 only; the element 2 does not appear at all. Consequently, 2 has no image, so this relation is not a function whose domain is the whole set A. Option A is correct. In option B, each of 1, 2, and 3 appears once and receives an image. The same complete coverage occurs in options C and D. A repeated output is allowed in a function, so the fact that several elements could map to the same value would not by itself cause failure. The decisive problem in A is the missing domain element 2.

Substitute x = 5 into the function: f(5) = 2(5) − 3 = 10 − 3 = 7. Therefore, the correct answer is 7.

In an ordered pair \((x,f(x))\), the first component is the input and the second component is its image. Since \((2,c)\) is given, the preimage of \(c\) is \(2\). The elements \(1\) and \(3\) have image \(b\), while \(b\) is not an input element in this relation. Exam tip: To find the preimage of a given image, match it with the second component and read the first component of that ordered pair.

For a function \(f:A\to B\), each element \(x\) of the domain \(A\) has a unique output \(f(x)\); this output is the image of \(x\) and lies in the codomain \(B\). Option B represents the input or preimage, while option D is the ordered pair used to describe a point on the graph. Exam tip: the image is obtained by applying the function to the input, giving \(f(x)\).

If \(f(x)=y\), then \(x\) is called the preimage of \(y\). Thus, a preimage is always an element of the domain \(A\), whereas \(y\) is an element of the codomain \(B\). Exam tip: the input comes from the domain and the output belongs to the codomain.

The obtained values of the function lie in (B), so the range is a subset of (B). It need not be the whole of (B).

An empty domain has no input, so the graph has no pair. Still it is considered the empty function.