Functions as a special kind of relation

In option (A), both (1) and (2) have exactly one image. In exams, check the whole domain before calling a subset a function.

(f(2)=4), so the preimage of (4) is (2). In exams, identify the input from the output to find a preimage.

The governing concept is the preimage of a value under a function. The preimage of 9 consists of every element x in the domain for which f(x) = 9. Since f(x) = x² and the domain is {1,2,3,4}, evaluate the rule: f(1)=1²=1, f(2)=2²=4, f(3)=3²=9, and f(4)=4²=16. Thus the only domain element mapped to 9 is 3, so f⁻¹({9}) = {3}. Option C is therefore correct. Option A is the preimage of 1, option B is the preimage of 4, and option D is the preimage of 16. Although the equation x²=9 can have x=−3 over the integers, −3 is not in the stated domain, so it cannot be included here.

(4) is not in the domain, so (16) is not obtained. In exams, a preimage must come from the domain only.

Every domain element has exactly one image and the images are (2,4,6). In exams, check both function validity and range using the rule.

Substituting \(x=2\) into the given rule gives \(f(2)=5-2=3\). Therefore, option C is correct. The values 1, 2 and 4 do not result from applying the stated rule. Exam tip: to find a function value, substitute the given input directly into the function rule.

In option (A), every domain element has one image, but (1) and (2) both have image (a). In exams, test function and one-one separately.

Repeating the same pair does not create a new different image. In exams, distinguish between a repeated pair and a different image.

(A) contains both even and odd elements, so both images (0) and (1) are obtained. In exams, check all types of inputs.

A function has exactly one pair for every domain element, so the number of pairs will be (3). In exams, pairs in a function equal the number of domain elements.

The values are (|-2|=2), (|-1|=1), (|0|=0), (|1|=1), and (|2|=2), so the range is ({0,1,2}). In exams, write only distinct obtained images in the range.

The correct answer is B. A function from A to B must assign every element of the domain A exactly one image in B. Check the first coordinates because they represent the domain elements. In option B, 1 is paired with p, 2 is paired with q, and 3 is paired with p. Every member of A appears once and has exactly one output, so B is a function. In option A, the element 1 has two outputs, p and q, which violates the definition. In option C, the element 3 has no output, so the assignment is incomplete. In option D, the element 2 has two outputs, p and q, so it is not a function. It is acceptable for two different domain elements to have the same image; therefore 1 and 3 may both map to p. Memory cue: every input gets one arrow, while several inputs may point to the same output.

The graph of a function defined on a finite domain is the set of ordered pairs (x,f(x)) for every x in the domain. The first coordinate must come from the domain {0,1,2}, and the second coordinate is obtained by squaring it. Thus f(0)=0²=0, f(1)=1²=1, and f(2)=2²=4. The graph is therefore {(0,0),(1,1),(2,4)}, which is option A. Option B uses values from a different rule, while option C reverses the ordered pair (2,4) and incorrectly places 4 as an input. Option D treats the function as though f(2)=2 rather than 4. The codomain contains 4, so the pair (2,4) is valid.

The inverse image or preimage f⁻¹({0}) contains the elements of the domain A whose function value belongs to the set {0}. We inspect each domain element: 1 is odd, so f(1)=1; 2 is even, so f(2)=0; 3 is odd, so f(3)=1; and 4 is even, so f(4)=0. Therefore the inputs that map to 0 are 2 and 4, giving f⁻¹({0})={2,4}. Option A is correct. Option B is the preimage of {1}, option C is not a subset of the given domain, and option D incorrectly includes inputs whose images are 1. A preimage consists of domain elements, not necessarily elements of the codomain.

A function from a domain to a codomain must assign exactly one value in the codomain to every element of its domain. Therefore, the rule used to define the function must produce an allowed codomain value for each input. Here the domain is {1,2,3}, the codomain is {1,2,3,4,5}, and f(x)=2x. We obtain f(1)=2 and f(2)=4, both of which belong to the codomain. However, f(3)=2×3=6, and 6 is not an element of {1,2,3,4,5}. Thus the proposed mapping does not remain inside its stated codomain and is not well-defined as written. Option A identifies this exact failure. The fact that the outputs are even, or that the domain has three elements, is not itself a problem.

The governing concept is evaluating a function from its defining rule. For every input x, f(x) is the remainder obtained when x is divided by 3. Using the division algorithm, 4 = 3×1 + 1, where the remainder is 1. Therefore f(4)=1, making option B correct. The codomain B={0,1,2} is also consistent with this result, because a remainder after division by 3 must be one of 0, 1, or 2. Option A would occur for a multiple of 3, such as 3; option C would occur for an input such as 2, whose remainder is 2; and option D cannot be a remainder on division by 3. Thus the function rule and the codomain both support option B.

A function from a domain to a codomain must assign exactly one codomain value to every domain element. Here the relation is defined by y=x+2. If any natural number x is selected, the rule produces the single value x+2; this value is also natural, so it belongs to the stated codomain N. For example, 1 maps to 3, 2 maps to 4, and so on. The fact that y is larger than x does not violate the definition, and an infinite domain is completely acceptable. There are no two different outputs for the same x. Hence option A is correct; the other options confuse size or finiteness with the one-output requirement.

Answer: option D, (3,1). A graph of f with domain A contains an ordered pair (x,f(x)) for each x in A. We test the first coordinate of every option. For x=0, f(0)=0²−3(0)+2=2, so (0,2) is present. For x=1, f(1)=1−3+2=0, so (1,0) is present. For x=2, f(2)=4−6+2=0, so (2,0) is present. For x=3, f(3)=9−9+2=2, so the graph contains (3,2), not (3,1). Hence D is the only pair that does not belong. The domain tells us which x-values may be used; the output must still equal the stated formula. A common error is checking only whether x belongs to A and forgetting to calculate y=f(x).

Two functions are equal only when their domains, codomains, and corresponding output values are all the same. Equality cannot be decided from the displayed formula alone. Both functions here have the same domain, {1,2,3}, and the same rule, x+1. Their actual values are f(1)=g(1)=2, f(2)=g(2)=3, and f(3)=g(3)=4, so their ranges are both {2,3,4}. Nevertheless, f has codomain {2,3,4}, whereas g has codomain {2,3,4,5}. Since the codomains differ, the functions themselves are different even though the formula and range agree. Therefore option B is correct. Option A overlooks the role of the codomain, option C confuses range with codomain, and option D is false because the domains are identical.

A function requires every element of A to have exactly one image in B; different inputs are allowed to share the same image. Using congruence modulo 2, 1 and 3 are odd, so their corresponding value is 1 in B, while 2 is even, so its corresponding value is 0. Therefore the relation contains (1,1), (2,0), and (3,1). Each element of A appears once as a first component, so R is a function. Its set of actual outputs is {0,1}, which is also B. Hence option B is correct. Option A confuses repeated outputs with failure of the function rule, option C names the domain rather than the range, and option D overlooks the image (2,0).

The function criterion concerns each input having exactly one output, not different inputs having different outputs. Evaluating y = |x| gives |−2| = 2, |−1| = 1, |0| = 0, |1| = 1, and |2| = 2. Thus every member of the stated domain has one and only one image, and all images belong to the stated codomain {0,1,2}. The distinct output values are 0, 1, and 2, so the range is {0,1,2}. Therefore option A is correct. Option B incorrectly treats repeated outputs as forbidden, option C is false because 2 maps to 2, and option D confuses the domain with the range.

For a mapping f: A → B to be a function, every element of A must be assigned exactly one value, and that value must belong to the codomain B. Although the rule x² gives a single value for each input, f(1)=1 and f(2)=4, while 4 is not an element of B={1,2,3}. Thus the mapping does not stay inside the declared codomain and cannot be a function from A to B. Option A identifies the first decisive failure. Option B is not a failure because 1 belongs to B, option C merely reports a value outside B without stating the correct input condition, and option D is irrelevant to the function requirement.

For every x in the domain [−2,2], the rule y=x² assigns exactly one real output, so the relation is a function. To determine the range, note that the smallest value of x² on this interval is 0, attained at x=0, and the largest value is 4, attained at x=−2 or x=2. Because x varies continuously through the whole interval, every value between 0 and 4 is obtained. Hence the range is [0,4], and option A is correct. Equal outputs at x=−2 and x=2 do not violate the function rule. Option C confuses domain and range, while option D mistakes a property of the outputs for a reason that would disqualify a function.

To test whether R is a function from X to Y, each fixed x in X must be related to exactly one y in Y. The condition x=|y| gives the pairs (0,0), (1,−1), (1,1), (2,−2), and (2,2). In particular, the single input x=1 is related to two different outputs, −1 and 1. Therefore R fails the uniqueness requirement and is not a function from X to Y. Option B states this precise defect. Option A checks only existence and ignores uniqueness, option C is false because x=0 has y=0, and option D focuses on |y| without considering that one x can correspond to two y-values.

A function from Z to Z must assign exactly one integer output to every integer input. The expression can be written as f(n) = n(n + 1)/2. Among any two consecutive integers n and n + 1, one is even, so their product is divisible by 2. Consequently, n(n + 1)/2 is an integer for every n in Z, including negative integers and zero. The rule is therefore defined for every domain element and its value lies in the stated codomain, so it is a valid function. Option B gives this reason. A fraction symbol does not automatically mean a non-integer result; for n = 1 the value is 1. Also, f(0) = 0 is perfectly allowed, and there is no restriction to positive inputs.