Functions as a special kind of relation

Each of the 3 elements of A can be mapped independently to either of the 2 elements of B. Therefore, the total number of functions is 2 × 2 × 2 = 2³ = 8.

If \(|A|=m\) and \(|B|=n\), the number of functions from \(A\) to \(B\) is \(n^m\). Each of the 2 elements of \(A\) can be mapped independently to any of the 3 elements of \(B\), giving \(3^2=9\) functions. \(2^3\) is incorrect because it reverses the roles of the domain and codomain. Exam tip: use \(|B|^{|A|}\) for the number of functions from \(A\) to \(B\).

Two different inputs (1) and (2) map to the same output (a). Still each input has exactly one image.

In a function, each input in the domain is associated with exactly one output in the codomain. Therefore, the same input cannot have two different outputs, so statement A is false. However, two different inputs may have the same output; this is a many-to-one situation. Exam tip: to test whether a relation is a function, check that every input has exactly one output.

In the ordered pair \((c,2)\), the first component is the input \(c\), and the second component is its image or function value. Therefore, \(f(c)=2\). The value 3 is the image of \(b\), while \(a\) and \(c\) are inputs. Exam tip: To find \(f(x)\), locate the ordered pair whose first component is \(x\).

To find \(f(1)\), substitute \(x=1\) in the function: \(f(1)=3\times1+1=4\). Therefore, 4 is correct. Option B results from calculating only \(3\times1\) and omitting \(+1\). In such questions, substitute the given input carefully and follow multiplication before addition.

Substitute 2 for x: f(2) = 2² + 1 = 4 + 1 = 5. Therefore, the correct answer is 5.

To find \(f(0)\), substitute \(x=0\) in the given function: \(f(0)=0+5=5\). Therefore, the correct answer is 5. Option B is only the input \(x=0\), not the value of the function. Exam tip: Always substitute the given input in place of \(x\) when evaluating a function.

For a function, every element of the domain \(A\) must have exactly one image in \(B\). In option A, 1, 2, and 3 each occur once, with images 4, 5, and 4 respectively. In option B, 3 has no image; in C, 2 has two images; and in D, 1 has two images. Exam tip: check that every domain element appears exactly once as the first component.

In \(f:A\to B\), \(A\) is the domain and \(B\) is the codomain. Therefore, for every \(x\in A\), the function value \(f(x)\) belongs to \(B\). The set \(A\) contains the inputs, not necessarily the outputs, so option B is incorrect. Exam tip: in \(f:A\to B\), the set written after the arrow is the codomain of the function values.

Since \(x\) is an element of the domain \(A\), it is the input, or independent variable, of the function. Its image \(f(x)\) is the output. The codomain is \(B\), whereas the range is the set of actual values \(f(A)\). Exam tip: an element taken from the domain is the input, and its image is the output.

A function can assign only one output to a given input. Since both ordered pairs have the same first component \(a\), their second components must be equal; hence \(b=c\). Option D is incorrect because it represents two different outputs for the same input. Exam tip: To test whether a relation is a function, check that repeated first components never have different second components.

In a function, every x in the domain must have exactly one corresponding value f(x) in the codomain, so option A is correct. Option B violates uniqueness, while option C leaves a domain element without an image. Option D is not required for a function; it relates to an injective function, where distinct inputs have distinct outputs. Exam tip: check that each input has one and only one output.

Since \(B\) has only one element \(0\), every element of \(A\) must map to \(0\). So \(f(1)=0\), \(f(2)=0\), and \(f(3)=0\) are forced, giving exactly \(1\) function.

In a function, every element of the domain must have exactly one image. Since \(A\) is empty, there are no elements whose images need to be chosen, so exactly one empty function exists. The counting formula also gives \(|B|^{|A|}=2^0=1\). Exam tip: there is exactly one function from the empty set to any non-empty set, not zero.

Set A has only one element, 1. Its image can be any one of the three elements of B: 2, 3, or 4. Hence, the total number of functions is 3, that is, 3^1 = 3. Option 1 would represent only one particular function, not all possible functions. Exam tip: If the domain has n elements and the codomain has m elements, the number of functions is m^n.

The images of all inputs 1, 2, and 3 are 0. Since every input has the same output, the relation is a constant function. It is not an identity function because an identity function maps each input to itself. Exam tip: If the second components of all ordered pairs are equal, the function is constant.

A function must assign exactly one image to every element of its domain. In option A, 1 is mapped to a and 2 is mapped to b, so both elements of A have exactly one image. In option B, 2 has no image, while in options C and D, one element has two images. Exam tip: Check the first components of the ordered pairs to verify that every domain element appears exactly once.

A function is a relation with the extra condition of exactly one image. Hence it is called a special relation.

For each (x\in A), taking (2x) gives the correct pairs. Check every input separately while applying a rule.

The governing concept is classification of a function by its output values. A relation is a function here because each input, 1, 2, and 3, has exactly one assigned output. Every one of these inputs maps to the same value, 4. A function that has the same value for every input in its domain is called a constant function and can be written as f(x)=c. Therefore R is a constant function, so option A is correct. It is not an identity function because an identity function must satisfy f(x)=x; here f(1)=4, f(2)=4, and f(3)=4. It is not a non-function because no input has two different images, and it is not empty because three ordered pairs are present.

The governing concept is evaluation of a function represented by ordered pairs. In an ordered pair (input, output), the first component is the input and the second component is its image. To calculate f(2), locate the pair whose first component is 2. The given set contains (2,4), so the image of 2 is 4 and f(2)=4. Thus option A is correct. Option B is the input itself rather than its function value. Option C is the image of 1 because the pair (1,3) is present. Option D is another input, namely the first component of (0,2). Since the pair for input 2 is explicitly given and each input has one image, the answer is unambiguous.

The governing rule is that every element of the domain must have exactly one image in the codomain. In option A, a and b each have one image, but c is paired with both 1 and 2. Thus the same input c has two different outputs, which violates the definition of a function. Options B, C, and D each contain exactly one pair beginning with a, one beginning with b, and one beginning with c, so each can represent a function from A to B. It is permissible for different inputs to share an output, as seen in option C; the restriction concerns one input having multiple outputs. Therefore option A is the only non-function.

Direct answer: Option A, 4. The graph of a function contains exactly one ordered pair for each domain element. Here the domain is \(A=\{1,2,3,4\}\), so it has \(|A|=4\) elements. Therefore the graph contains one pair beginning with 1, one beginning with 2, one beginning with 3, and one beginning with 4. The outputs may repeat, but that does not change the number of pairs. Option A is correct. Option B, 2, is too small and would leave some domain elements without images. Option C, 8, is too large because a function cannot have two different output pairs for one input. Option D, 1, is also too small and would represent only one input, not the whole domain. The question asks for the number of pairs, not the number of distinct output values; even repeated outputs still belong to separate pairs. Memory cue: for a function, number of graph pairs equals the number of domain elements.

The governing concept is the defining test for a function from a domain: every domain element must occur as a first component exactly once, with one associated output. In option A, the domain elements 3 and 6 each appear once, paired with 9 and 12 respectively. Therefore it satisfies the stated condition. In option B, 3 has two images and 6 has none. In option C, 3 has no image, so the relation is incomplete for the domain A. In option D, 6 has two different images, 12 and 9, while 3 has one. Thus only option A gives every element of A exactly one image, irrespective of whether different domain elements might share an output in other examples.