Functions as a special kind of relation

Each element of \\(A\\) has exactly one image determined by its parity: an even number maps to \\(0\\), while an odd number maps to \\(1\\). Thus, every \\(x\\in A\\) is associated with exactly one \\(y\\in B\\), so \\(R\\) is a function. Different inputs are allowed to have the same output, so option B is incorrect. Exam tip: To test a function, verify that every domain element has exactly one image.

From the ordered pairs, (f(3)=7) and (f(4)=9), so the sum is (16). While reading function values, the input is the first coordinate.

It is not empty, but (1) has two images and (3) has no image. In hard MCQs, check both conditions separately.

At (x=2), the denominator becomes (0), so (f(2)) is not defined. In a rational function, the denominator must not be zero.

The output is unique, but (f(3)=9) is not in the codomain. For a function, every image must also lie in the codomain.

One input (2) is related to two outputs, so it is not a function. In function checks, first notice repeated first coordinates.

Identity maps each input to itself, while a constant function sends every input to the same value. Both can be valid functions.

The outputs are (2,1,0,1,2), so the distinct range is ({0,1,2}). In a set, order and repetition are not important.

In every pair, the second coordinate is (4) times the first coordinate. To identify a rule, verify it on all given values.

The actual outputs are (2,3,4,5), while the codomain is the given set (B). The range is always a subset of the codomain.

The direct answer is A: 2^6-2^3=56 relations are not functions. A has 3 elements and B has 2, so A×B has 3×2=6 ordered pairs. Every relation from A to B is any subset of these 6 pairs, giving 2^6=64 relations. To form a function, each of the 3 elements of A must choose exactly one of the 2 elements of B, giving 2×2×2=2^3=8 functions. Hence the relations that are not functions are 64-8=56. Option A is correct. Option B gives only the number of functions, not the number of non-functions. Option C gives all relations and fails to subtract functions. Option D uses an incorrect counting idea. Remember: non-functions = all relations - functions.

(|A\times B|=6), so relations are (2^6) and functions are (3^2). Keep relation and function counting formulas separate.

The required output for (x=3) is (4), which is not in the codomain, so every domain element is not mapped. A codomain restriction can change the relation.

A duplicate ordered pair is not a separate element in a set. If the same input gives the same output, uniqueness is not broken.

In sets, a repeated pair is not counted, so each input has a unique image. Understand duplicate pair and different image separately.

Two different inputs having the same image does not violate the function rule. Violation occurs when one input has two different images.

The direct answer is A. For a function from A to A, every input in A must produce exactly one output that is also in A. Calculate step by step: f(1)=4-1=3, f(2)=4-2=2, and f(3)=4-3=1. Thus the graph is {(1,3),(2,2),(3,1)}. Each input has one output, and all outputs 1, 2, and 3 belong to A, so the rule defines a valid function from A to A. Option A is correct. Option B is wrong because f(1)=3 is not a problem; 3 belongs to A and a function may map 1 to 3. Option C is wrong because f(2)=2 is also completely valid; an input may map to itself. Option D is wrong because the relation is not empty: it contains the three calculated ordered pairs. The exam method is to check both conditions: unique output and output remaining inside the stated codomain.

For (x=1), two possible (y)-values exist, so uniqueness fails. Read an absolute value relation with direction carefully.

Every (x\in A) has a unique image in (1,4,9,16), and all codomain elements are images. In this case, range and codomain are equal.

The direct answer is A: \(R\) is a function from \(A\) to \(B\). For a function from \(A\) to \(B\), every element of the domain \(A\) must have exactly one image in \(B\). Check the first entries: 1 is paired with 4, 2 with 5, and 3 with 4. Thus every member of \(A\) appears once as an input, and every second entry is in \(B=\{4,5\}\). Repetition of an output is allowed; different inputs may have the same image. Therefore A is correct. Option B is wrong because 4 appearing twice is not a problem: both 1 and 3 may map to 4. Option C is wrong because 5 need not appear more than once; a codomain element may be unused or used once or several times. Option D is wrong because the stated mapping is from A to B, not only from B to A, and B has no obligation to supply the inputs. Exam cue: inspect first components and demand exactly one pair for each domain element.

In a function, one first element cannot have two different images. In exams, watch for repeated first elements.

A function has exactly one pair for every element of the domain. Hence the number of pairs is (|A|).

Each domain element has (4) choices in (B), so total functions are (4^3). Remember the formula: (|B|^{|A|}).

For a function, each element of (A) gets one image in (B), so there are (3^2) functions. Do not confuse this with all relations.

Different first elements may have the same image, so a many-one function is valid. The common mistake is treating the same image as an error.