Functions as a special kind of relation

For a mapping f: A → B to be a function, every element of the domain A must have exactly one image, and that image must belong to the codomain B. Evaluate the rule on the four domain elements: f(1) = 1 - 4 + 5 = 2, f(2) = 4 - 8 + 5 = 1, f(3) = 9 - 12 + 5 = 2, and f(4) = 16 - 16 + 5 = 5. The first three outputs are in B, but 5 is not in {1,2,3,4}. Thus the stated mapping to B is not a function. Option C is correct. Repeated outputs, such as f(1) = f(3) = 2, are allowed; the problem is the output f(4) = 5 outside the codomain.

The relation condition |x - y| = 1 means that y must be one unit away from x. For x = 1, the possible codomain values are y = 2, so it has one image. For x = 2, both y = 1 and y = 3 satisfy the condition, giving two images. For x = 3, y = 2 and y = 4 are possible, again showing the relation may have multiple outputs. A function must assign exactly one output to every input in its domain. The two images of x = 2 alone are enough to disqualify the relation, so option B is correct. The presence of 4 in Y is not itself a problem, and x = 1 does have an image.

A relation from A to B is a function precisely when each element of A occurs as the first component of exactly one ordered pair. It is acceptable for different inputs to share the same output; this is why options A, B, and D are functions even though some second components repeat. In option C, the input 0 appears twice: it is paired with p and also with r. Thus 0 has two distinct images, while the other inputs have one image each. This violates the uniqueness condition, so option C is the relation that is not a function. The fact that every element of A appears at least once is necessary but not sufficient; it must appear exactly once.

The relation represents selected points on the circle x² + y² = 4. To test whether it is a function from X to Y, inspect the possible y-values for each x in X. When x = 0, the equation becomes y² = 4, so y = 2 or y = -2. Both values belong to Y, meaning the single input 0 has two distinct images. This violates the defining rule that a function must assign exactly one output to each input. Hence option A is correct. For x = 2 and x = -2, the equation gives y² = 0 and therefore only y = 0, so those inputs do not create the same problem. The presence of 0 in Y is valid and is not a reason for failure.

The governing definition says that a function from A to B must assign exactly one element of B to every element of A. In this relation, (x,y) is included whenever x−y is odd. Consider x=1: y=2 gives 1−2=−1, which is odd, and y=4 gives 1−4=−3, which is also odd. Therefore x=1 has at least two different images, 2 and 4. This violates the uniqueness requirement, so R is not a function and option A is correct. Option B is false because x=2 is related to y=1 and y=3. Option C is false because B is not empty, while option D is irrelevant: equal domain and codomain sets do not guarantee that a relation is a function.

For f:A→B to be a valid function, every element of A must have exactly one image and each image must belong to the codomain B. The polynomial rule is single-valued, so calculate its outputs and check the codomain. We get f(0)=0−0+1=1, f(1)=1−2+1=0, f(2)=4−4+1=1, and f(3)=9−6+1=4. Although each input has one numerical output, 4 is not in the stated codomain {0,1,2,3}. Therefore this rule does not define a function from the given A to the given B, and option B is correct. Option C is not an issue because 0 belongs to B; option D ignores the essential codomain requirement.

A relation from A to B is a function when every element of A occurs as a first component exactly once, with one and only one corresponding element of B. Options A, B and D each contain five ordered pairs, and the first components 1, 2, 3, 4 and 5 occur once each. They therefore satisfy the function condition, even if several different inputs have the same output. In option C, the first component 4 occurs twice: (4,r) and (4,q). Hence 4 has two images, r and q, so the uniqueness requirement fails. The repeated output for different inputs in the other options is allowed; a function need not be one-one. It is also enough that each input has one image; not every element of B must be used. Therefore option C is the only correct choice.

For a function from A to B, each input x in A must be paired with exactly one output y in B. The relation includes a pair whenever x + y is odd. Take x = 1. Since 1 + 2 = 3 is odd and 1 + 4 = 5 is also odd, both (1,2) and (1,4) belong to R. Thus the same input 1 has at least two images, 2 and 4, which violates the uniqueness condition. In fact, every odd x pairs with all even elements of B, and every even x pairs with all odd elements, so the relation generally has multiple outputs for each input. Option B is false because x = 2 has odd images such as 1, 3 and 5. Options C and D do not affect the function definition. Therefore option A gives the correct reason.

A piecewise-defined function uses the rule whose condition includes the given input. At x = 4, the first rule 3x − 2 cannot be used because it applies only when x < 4. The second rule applies when x ≥ 4, and it includes 4. Therefore f(4) = 4² − 10 = 16 − 10 = 6. The value approached from the left would be 3(4) − 2 = 10, but that does not determine f(4), because the first branch excludes the boundary. The second branch explicitly assigns the value at 4. Hence option A is correct. Option B is the left-hand expression evaluated at 4, not the function’s actual assigned value; option C results from an incorrect calculation or rule choice, and option D is false because the second branch defines the function at x = 4.

The governing concept is the real-valued domain of a square-root function. A square root is defined over the real numbers only when its radicand is non-negative. Here the radicand is x+4, so we require x+4≥0. Subtracting 4 from both sides gives x≥−4. The endpoint is included because √0 is defined and equals 0; therefore x=−4 is allowed. The vertical shift by −2 changes the output values but does not restrict the inputs. Hence option A is correct. Option B reverses the sign incorrectly, option C uses the wrong direction of inequality, and option D wrongly excludes the valid endpoint.

In a function, every element of the domain A must have one and only one image in the codomain B. Option B describes a stronger situation similar to a one-one and onto mapping, which is not required for every function. Exam tip: if even one element of A has two different images, the relation is not a function.