Functions as a special kind of relation

Every (x\in A) has exactly one image in (B). In exams, first check presence and uniqueness for each domain element.

Each of the (4) elements of (A) has (2) choices in (B), so total functions are (2^4=16). Remember the formula: functions from (A) to (B) (=|B|^{|A|}).

A function needs exactly one output for each input; different inputs may have the same output. Common mistake: do not treat same image as a violation of function rule.

The direct answer is C. A function must assign every element of the domain exactly one image in the codomain. In option C, the input 1 is paired with p, input 3 with p, but input 2 is paired with both q and r. Since q and r are different outputs, one input has two images, so this relation is not a function. Option A is a function: 1, 2, and 3 each have exactly one image, namely p, q, and r. Option B is also a function: different inputs are allowed to have the same output p; a function does not require different outputs. Option C fails the one-output rule because of (2,q) and (2,r). Option D is a function because each domain element has one image, even though the order of the outputs is reversed. The useful test is to inspect the first coordinates: every domain element must occur once with only one second coordinate.

Every ordered pair satisfies (y=2x+1). While identifying a rule from pairs, test the rule on all pairs.

Putting (x=0,1,2,3) gives outputs (1,3,5,7). The range is the set of actual images, not necessarily the whole codomain.

In a function, exactly one arrow must start from each domain element. In diagrams, check both arrow direction and arrow count.

Every element of (A) must have an image, but (4) is missing. An unused element of the codomain does not make a function invalid.

For (x=3), there are many images, so uniqueness fails. In inequality relations, count possible outputs for one input.

(f(4)=1), so ((4,0)) is not in the relation. To test an ordered pair, put the first coordinate into the rule.

The condition for a function is on domain elements, not codomain elements. In definition questions, the words exactly one are most important.

For (x=1,2,3), pairs are ((1,1),(2,4),(3,9)). In rule-based relations, substitute domain values systematically.

In any function, the number of ordered pairs equals the number of elements in the domain. Here (|A|=3), so there are (3) pairs.

Every input has image (2), so it is a constant function. Repeated output does not make a function invalid.

In an identity function, every \(x\in A\) maps to itself, so \(I_A(x)=x\). In option A, the mappings are \(1\to1, 2\to2, 3\to3\), and \(4\to4\); hence it is the identity function. Option C is a constant function, while B and D map at least some elements to different elements. Exam tip: for an identity function, check whether every ordered pair has equal components, \((x,y)\) with \(x=y\).

The domain is the set of first coordinates, and here (A={1,2,3}). Distinguish ordered pairs from the domain.

In a function, every domain element has a unique image in the codomain. The whole codomain need not become the range.

The outputs are (5,0,-3,-4,-3), so distinct values ({-4,-3,0,5}) form the range. Repeated values are not written in a set.

Here (|A|=2) and (|B|=3), so functions (=3^2=9). In counting, keep codomain size as base and domain size as exponent.

In (y^2=x), for (x=4), both (y=2) and (y=-2) are possible. Two (y)-values for one (x) do not define a function.

For absolute value, (|-1|=1), (|0|=0), (|1|=1), and (|2|=2). In an ordered pair, the first coordinate remains the input.

For every (x\in A), (y=5-x) is unique and lies in (A). For an equation relation, check uniqueness to decide function status.

(x=2) gets two images, so uniqueness fails. In divisibility relations, check all divisors in the codomain for one input.

The direct answer is A: the relation is \(\{(1,3),(2,4),(3,5)\}\). A function rule tells us to take each input from the domain and calculate exactly one output. Here \(f(x)=x+2\). Step by step, \(f(1)=1+2=3\), \(f(2)=2+2=4\), and \(f(3)=3+2=5\). All outputs belong to \(B\), so this is a valid relation from \(A\) to \(B\). Option A is correct because it contains precisely these three ordered pairs. Option B uses \(x+1\), since it gives 2, 3, and 4. Option C reverses the direction of the pairs and does not have the required domain elements as first entries. Option D gives 4, 5, and 6, which corresponds to \(x+3\), not \(x+2\). Remember: substitute every domain value into the formula, and write the input first and its image second.

The outputs are (2,4,6), so the smallest codomain can be the set of these images. The codomain must contain all possible outputs.