Functions as a special kind of relation

Elements of a non-empty domain cannot get any image in (\emptyset). Therefore no function can be formed.

Set A has only one element, 2. Its image can be chosen as either 5 or 6 from set B, giving two different functions. Hence, the number of functions is \(2^1 = 2\). Option 1 counts only one possible function, whereas both choices are possible. For such questions, use the formula \(n(B)^{n(A)}\) for the number of functions from A to B.

The governing concept is that a function assigns exactly one output to every input, but it may assign the same output to different inputs. Such a function is called many-one. In option A, input 4 maps to 1, input 5 also maps to 1, and input 6 maps to 2. Thus every listed input has exactly one image, while two distinct inputs share an output, so this is a valid function demonstrating the required situation. In option B, input 4 has two different outputs, 1 and 2, so it is not a function. Option C similarly gives input 5 two outputs. In option D, the stated domain element 5 has no image. Therefore A is correct.

The direct answer is A: a has two images. A function must assign exactly one output to every input in its domain. If the mapping diagram shows arrows from the single domain element a to both 1 and 2, then the same input has two different images. This violates the definition of a function, even if every arrow ends in the codomain. Option A is correct because the failure is caused by two outputs for one input. Option B is wrong: being in the codomain is allowed and is not a reason for failure. Option C is wrong because belonging to the domain is required, not forbidden. Option D is wrong because arrows do exist; in fact, there are two arrows. If an input has no arrow, that also fails the function rule, but that is not what this diagram shows. Memory cue: one input, exactly one arrow; many inputs may point to one output, but one input cannot point to two different outputs.

The governing definition says that a mapping is a function when every input in its domain is associated with exactly one output. Here a is associated with 3, b is associated with 3, and c is associated with 4. Each input has one and only one image, so the mapping is a function. It is specifically a many-one function because the distinct inputs a and b have the same output 3. Repeated outputs are allowed; the restriction concerns repeated inputs with different outputs. Therefore option A is correct. Options B and C confuse the prohibition on multiple outputs for one input with the allowed sharing of an output. Option D is also invalid because an output need not have several preimages.

In an identity function, every element of the set maps to itself, so \(f(x)=x\). Hence, \(f\) is the identity function on \(A\), also denoted by \(I_A\). It is not a constant function because a constant function maps every element to the same fixed value. Exam tip: whenever \(f(x)=x\) for every element of the domain, identify the function as an identity function.

A function from set A to set B is a special relation in which every element x of the domain A is associated with exactly one element y of B. This means that no element of A may be left without an image and no single element of A may be paired with two different elements of B. Option A states both essential requirements in concise form. Option B allows many outputs for one input and therefore violates the function rule. Option C does not describe the normal assignment from a domain, while option D omits the relation between the sets entirely. The definition does not require different inputs to have different outputs; that additional property would describe a one-one function. Hence A is correct.

The elements (1), (2), and (3) of (A) each have exactly one image in (B). It is not necessary to use every element of the codomain.

The image of (m) can be any one of the (3) elements of (B). In counting, each domain element gets as many choices as the codomain size.

For a function, every domain element must have exactly one image. So checking first components is most useful.

Applying
2
x
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to each x∈A gives 2,3,4. In rule based questions do not reverse input and output.

A relation from a set of inputs to a set of outputs is a function when every input has exactly one output. In R, the first components are 2, 4 and 6, and each appears exactly once: 2 maps to 5, 4 maps to 5, and 6 maps to 7. Therefore R is a function. It is a many-one function because two different inputs, 2 and 4, have the same image 5; this is allowed. What is not allowed is one input having two different images. Thus option B is incorrect, option C is incorrect because 2 has only one image, and option D is plainly false because R contains three ordered pairs. The repeated second component does not violate the definition.

For a relation from A to B to be a function, each element of the domain A must be associated with exactly one element of B. In option A, the input 5 maps to 1, but the input 7 maps to both 2 and 3. Since one input has two different images, this relation is not a function. In option B, both inputs have exactly one image, even though they share the same output 1; that is a valid many-one function. Options C and D also assign one output to each of 5 and 7. The key rule concerns repeated first components with different second components, not repeated second components. Therefore option A is the only correct answer.

From the pair (−1,1), the image of −1 is 1. Read negative inputs the same way as positive inputs.

To count functions from a finite set A to a finite set B, use the rule |B|^|A|. Every element of the domain A must independently choose one image from the codomain B. Here |A|=4 and |B|=2, so the number of choices is 2^4=2×2×2×2=16. Equivalently, the images of 1, 2, 3 and 4 can each be selected as either 0 or 1, producing sixteen different assignments. Option B would result from using an incorrect exponent or partial counting, while option D counts only the available codomain elements rather than complete functions. Thus option A is correct. The formula counts all functions, including constant, one-one where possible, and many-one assignments.

Each of the 3 elements of A can be mapped independently to any of the 4 elements of B. Hence, the total number of functions is \(|B|^{|A|}=4^3=64\). The expression \(3^4\) reverses the roles of the domain and codomain. Exam tip: The number of functions from A to B is always \(|B|^{|A|}\).

The codomain has only 9, so every domain element maps to 9. No domain element should be left out.

In an identity function, every element of the set maps to itself, so each ordered pair has the form (x, x). Hence, {(2,2), (3,3), (4,4)} is correct. Option B gives a cyclic mapping, option C is a constant function, and option D interchanges 2 and 4. Exam tip: In an identity function, the input and output are equal for every element.

An identity function maps every element to itself, so \(I_A(x)=x\). Therefore, \(I_A(8)=8\). Although 9 is also an element of set \(A\), it is not the image of 8 under the identity function. Exam tip: for an identity function, the input and output are always the same.

The governing concept is the definition of a constant function. A function assigns exactly one output to each input in its domain, and it is called constant when that output is the same for every input. Since f(x) = 6 for every x in A, every element of the domain is mapped to the fixed value 6. Therefore f is a constant function, so option A is correct. It is not an identity function, because an identity function satisfies f(x) = x, not f(x) = 6 for all x. The rule clearly defines a function, so option C is false. Nothing states that A is empty, so option D is also unjustified; even if A had particular elements, they would all map to 6.

The range of \(f\) is the set of second coordinates of its ordered pairs. Here all ordered pairs have second component \(0\). So the range is exactly \(\{0\}\).

In a function, \(f(x)\) is the second component of the ordered pair \((x,y)\) whose first component is \(x\). Here, the pair \((20,2)\) is given, so \(f(20)=2\). Option B is the input 20, not its image. Exam tip: To find \(f(a)\), locate the ordered pair whose first component is \(a\).

Substituting \(x=5\) in the given function gives \(f(5)=\frac{2}{5+1}=\frac{2}{6}=\frac{1}{3}\). Option B, \(\frac{2}{5}\), results from omitting the \(+1\) in the denominator. In an exam, write the complete denominator in parentheses before substituting the value.

The direct answer is A: {(1,5),(3,7)}. A function from A to another set must assign exactly one output to every element of its domain. Here the domain is A = {1,2,3}, so 1, 2, and 3 must all appear as first components. In option A, first components are only 1 and 3; 2 is missing. Thus this relation is not a function from A because the domain element 2 has no image. Option A is correct. Option B contains first components 1, 2, and 3, and each appears once, so it is a function. Option C also includes all three domain elements; two different inputs may have the same output, so it is still a function. Option D includes 1, 2, and 3 exactly once, so it is also a function. The common mistake is thinking that outputs must be different. They need not be different; only each input must have one and only one output. Memory cue: check the first coordinates, cover every domain element exactly once.

In the first option, both 1 and 2 have exactly one image. In the other options, an input is missing or mapped twice.