Functions as a special kind of relation

The direct answer is A: the ordered pairs are {(1,4),(2,5),(3,6)}. The rule says y=x+3 and x can be 1, 2, or 3. Step 1: for x=1, y=1+3=4, giving (1,4). Step 2: for x=2, y=2+3=5, giving (2,5). Step 3: for x=3, y=3+3=6, giving (3,6). Option A lists all three pairs correctly. Option B keeps y equal to 3 and ignores the rule y=x+3. Option C reverses the order of each pair; (x,y) must be written with x first. Option D adds 4 instead of 3. The useful method is to substitute every allowed x and write the result as (x,y).

From (y^2=1), both (y=1) and (y=-1) are obtained. In exams, if one input has two images, it is not a function.

For every (x\in A), (x^2\in B) and the image is exactly one. In exams, check whether the rule's images lie in the codomain.

(2^1=2), (2^2=4), and (2^3=8), so the range is ({2,4,8}). In exams, substitute every input in an exponential rule.

(f(1)=-1) is not in codomain (B), so it is not a function from (A) to (B). In exams, every output must lie in the codomain.

(f(2)=0), (f(3)=1), and (f(4)=2), so the range is ({0,1,2}). In exams, changing the domain changes the range too.

In option (A), different domain elements have different images. In exams, for a one-one function, images should not repeat.

(d) is an element of codomain (B), but no domain element maps to it. In exams, the codomain may contain extra elements.

There are (2) choices from (B) for each of the (4) elements of (A), so there are (2^4=16) functions. In exams, the base is the size of the codomain.

(A\times B) has (4\times 2=8) pairs, so total relations are (2^8=256). In exams, count subsets of the Cartesian product for relations.

Every domain element has exactly one image and the images are ({0,1,4}). In exams, identify domain and range separately.

(f(1),f(2),f(3)) give (3,6,9), not (12). In exams, do not put an unattained codomain element in the range.

Each domain element can map to either (0) or (1), so there are (2) choices for each. In exams, multiply these choices for the total count.

Only (3) in domain (A) has no image, so adding ((3,5)) gives every element exactly one image. In exams, first identify the missing domain element.

Each element of domain (A) has exactly one image. In exams, repetition of the same image does not make a function invalid.

In a function every domain element must have exactly one image but (3) is missing. In exams, compare first components with the whole domain.

A relation from A to B is a function when every element of the domain A has exactly one image in B. Inspect the ordered pairs by their first components. The element 1 is paired only with 5, and 3 is paired only with 5; both satisfy the uniqueness requirement. However, the element 2 appears in two ordered pairs: (2,6) and (2,7). Thus 2 is assigned two different images, 6 and 7, which violates the function condition. The element 5 is a second component, or image, not a domain element being tested for uniqueness. Therefore option B is correct. A function may allow different inputs to share one output, as 1 and 3 both map to 5; it only forbids one input from having multiple outputs.

Each domain element has (4) choices in (B), so total functions are (4^3=64). In exams, use (n(B)^{n(A)}).

In a constant function all domain elements map to one chosen element of (B). Therefore the number of constant functions is (3).

(A\times B) has (3\times 4=12) pairs, so relations are (2^{12}=4096). In exams, relations are counted as subsets.

Putting (x=0,1,2,3) gives images (3,4,5,6). In exams, substitute all domain values to find the range.

The direct answer is C: {4,5,6}. The range of a function is the set of distinct output values, meaning the second components of its ordered pairs. The given pairs are (1,4), (2,4), (3,5), and (4,6). Their outputs are 4, 4, 5, and 6. A set does not repeat an element, so the repeated 4 is written only once. Hence the range is {4,5,6}. Option A, {1,2,3,4}, lists the input values, so it is the domain, not the range. Option B, {4,4,5,6}, lists the outputs but repeats 4; it is not written as a proper set of distinct elements. Option C correctly gives the distinct outputs. Option D, {1,4,5,6}, mixes an input value, 1, with output values and therefore does not represent the range. Notice that 4 appears as both an output and an input in different roles, but its role in the range is determined by its position as a second component. Memory cue: domain means first coordinates; range means distinct second coordinates actually obtained.

Only the image of (3) is missing, so adding ((3,1)) is correct. In exams, first identify the missing domain element.

(2) has two images and after removing ((2,c)) every first component has exactly one image. In exams, remove one pair from the repeated first component.

In the reversed relation, each of (3,5,7) has exactly one image. In exams, test the reversed relation separately by the function condition.