Functions as a special kind of relation

Correct answer: A. Yes, because every \\(x\\) has exactly one image

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.

Correct answer: A. (16)

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.

Correct answer: A. (R={(1,1),(1,2),(2,3)})

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

Correct answer: A. (2)

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

Correct answer: A. No, because (f(3)=9\notin{1,2,3,4,5})

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

Correct answer: A. (2) has two different images (b) and (c)

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

Correct answer: A. In identity, (x\mapsto x); in constant, every (x\mapsto 1)

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

Correct answer: A. \({0,1,2})

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.

Correct answer: A. (f(x)=4x)

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

Correct answer: A. range (={2,3,4,5}), codomain (={1,2,3,4,5})

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.

Correct answer: A. (2^{6}-2^3=56)

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.

Correct answer: A. (2^6) and (3^2)

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

Correct answer: A. For (x=3), (y=4\notin B), so no image is formed

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.

Correct answer: A. When the repeated first coordinate has the same image and uniqueness remains after removing duplicate pairs

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

Correct answer: A. Yes, duplicate ((1,2)) is the same pair

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

Correct answer: A. (f) may still be a function

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

Correct answer: A. valid function with graph ({(1,3),(2,2),(3,1)})

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.

Correct answer: A. (x=1) has images (y=-1) and (y=1)

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

Correct answer: A. (R) is a function from (A) to (B) and range (=B)

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.

Correct answer: A. It is a function from (A) to (B)

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.

Correct answer: A. Because (1) has two different images

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

Correct answer: C. (4)

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

Correct answer: B. (4^3)

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

Correct answer: B. (3^2)

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.

Correct answer: B. It is a function and it can be many-one

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.