Functions as a special kind of relation

The graph of a function f from A to B is the set of ordered pairs (x,f(x)) for all x in the domain A. A function must assign exactly one output to every domain element, so the graph contains one ordered pair for r, one for s and one for t. Since A={r,s,t} has three elements, the graph contains |A|=3 ordered pairs. The size of B does not determine the number of pairs; B may contain more, fewer, or the same number of elements, although each assigned output must belong to B. Therefore option A is correct. Option D confuses the codomain size with the domain size, while options B and C omit some of the required domain elements. The defining count comes from the domain.

Since A is empty, there is no element whose image must be assigned in B. Therefore, exactly one empty function exists. This also follows from the formula \(|B|^{|A|}=3^0=1\). Option 0 is incorrect because the number of functions is one, not zero. Exam tip: There is exactly one function from an empty domain to any codomain.

The domain contains 0 but the codomain is empty, so 0 cannot get an image. Hence no function is possible.

In option A, each domain element 1, 2, and 3 has exactly one image, so it is a function. However, 1 and 2 both have the image a; hence different inputs have the same output, so the function is not one-one. Option C is one-one, whereas B and D assign two images to the same input and therefore are not functions. Exam tip: for a function, every input must have exactly one output; for a one-one function, distinct inputs must also have distinct outputs.

In the first relation, each element of \(A\)—1, 2, and 3—has exactly one image in \(B\), so it is a function. In B and D, one input has two outputs; C leaves 3 unmapped. Exam tip: check every domain element exactly once.

Using the rule \\(f(x)=2x\\), the inputs \(x=1,2,3\) give the outputs \(2,4,6\), respectively, so it matches every ordered pair. The rule \(x+2\) matches only when \(x=2\), while \(3x\) and \(x/2\) produce different outputs. Exam tip: verify a proposed rule with every given ordered pair, not just one pair.

In every ordered pair, the second component is the square of the first: \(2^2=4\), \(3^2=9\), and \(5^2=25\). Therefore, the correct rule is \(f(x)=x^2\). Option B gives \(f(2)=4\), but \(f(3)=6\), so it does not fit all the pairs. In an exam, test the proposed rule with each given input.

Substituting \(x=a\) in \(f(x)=x+7\) gives \(f(a)=a+7\). Thus, \(a+7=12\), so \(a=5\). The value 12 is \(f(a)\), not \(a\). Exam tip: replace \(x\) with the given input variable before solving the equation.

To find h(2), substitute 2 for x: h(2)=4×2−1=8−1=7. Therefore, 7 is correct. Option 8 results from forgetting to subtract 1. Exam tip: substitute the given input first, then perform the multiplication and subtraction carefully.

Substitute x = 2 into the function: p(2) = 2³ − 1 = 8 − 1 = 7. Therefore, option A is correct.

Two different y values for the same x break the function condition. Always check repeated inputs in a table.

For a function, every element of the domain must have exactly one image in the codomain. Here, 1 maps to 4, 2 maps to 5, and 3 maps to 6, so exactly one arrow leaves each domain element; hence the relation is a function. Option B is incorrect because no domain element has either zero or more than one image. Exam tip: When checking a mapping, start by examining every element of the domain.

A mapping diagram represents a function only when every input has exactly one arrow leading to one output. In the given diagram, the input p points to both 1 and 2. Thus p has two different images, which violates the defining condition of a function. The input q has one image, 3, but that does not repair the failure caused by p. Therefore option A is correct. It is not a constant function because the outputs are not all the same, and it is not an identity function because the displayed input and output labels do not show p mapping to itself or q mapping to itself. Option D is false because a mapping is not automatically a function; the one-output-per-input rule must be checked.

Every pair in the graph of a function has the form \((x,f(x))\). Here, \(f(4)=4+1=5\), so \((4,5)\) is correct. In option B, the function value is incorrect; in option C, the input and output are reversed; and for option D, \(f(6)=7\), not 8. Exam tip: substitute the first coordinate into the function and check whether it gives the second coordinate.

The direct answer is A: 3. In an ordered pair (input, output), the first component is the input and the second component is its image. The given function contains (1,8), (2,8), and (3,9). To find whose image is 9, locate the pair whose second component is 9. That pair is (3,9), so f(3) = 9 and the required input is 3. Option A is correct. Option B, 1, is not correct because the pair with first component 1 is (1,8), so the image of 1 is 8. Option C, 2, is not correct because (2,8) shows that the image of 2 is also 8. Option D, 8, is not the required input; it is an output value, and the question asks whose image is 9. This illustrates that different inputs may have the same image: both 1 and 2 map to 8, while 3 maps to 9. Memory cue: in (x,y), x goes to y; to find the input for a known output, search the second coordinate and read the first.

The element 3 of domain A is not the first component in any pair. Every domain element must have an image in a function.

For the function \(f:A\to B\), the value obtained for \(x\in A\), namely \(f(x)\), is called the image of \(x\), and it belongs to the codomain \(B\). The domain \(A\) is the set of inputs, whereas the codomain \(B\) is the set of possible outputs. Exam tip: distinguish the image of an input from its preimage.

In function terminology, if f(x)=y, then x is an input from the domain that maps to the output y. Such an input is called a preimage, or inverse image, of y under f. Therefore option A is correct. The codomain is the whole target set B, not an individual input or output. The range is the set of outputs actually attained by the function, so it is also a set rather than the name for x. An empty set describes the absence of elements and has no role in this statement. A value y may have one preimage, several preimages, or none, depending on the function and the chosen output; when f(x)=y is explicitly given, x is certainly a preimage of y.

Each domain element has 2 choices, so the total number of functions is 2
2
=4. Multiply the choices in counting.

In a constant function, every element of A must have the same image in B. In option A, the images of 1, 2, and 3 are all a, so it is a constant function. Options B and C have different images, so they are not constant. Option D is not a function because 1 has two images and 2 has no image. Exam tip: To identify a constant function, check whether the second components of all ordered pairs are identical.

The governing concept is the uniqueness condition for a function. A relation from a domain to a codomain is a function only if every input, or first component, is associated with exactly one output, or image. In R, the input 2 occurs in both (2, 2) and (2, 3), so the same input has two different images, 2 and 3. This violates the definition of a function, making option A correct. The pair (1, 1) is acceptable because input 1 has only one image. Having 3 as a second component is not a problem, and the fact that the relation has three pairs does not by itself decide whether it is a function. The failure is specifically the repeated input with different outputs.

The relation is given in set-builder form by R={(x,y): y=x+2, x∈A}, with A={2,3,4}. To list its ordered pairs, substitute each allowed value of x into the rule y=x+2. For x=2, y=4, giving (2,4); for x=3, y=5, giving (3,5); and for x=4, y=6, giving (4,6). Hence R={(2,4),(3,5),(4,6)}, so option A is correct. Option B would represent y=x, while option C reverses the order of the coordinates. Option D adds 3 instead of 2 to each x. The first coordinate must come from A, and the second coordinate must be calculated using the stated rule.

Here, \(f(-1)=(-1)^2=1\) and \(f(1)=1^2=1\). Thus, two different inputs can have the same output; such a function is many-one rather than one-one. Option B is incorrect because each input in a function must have exactly one output. Exam tip: to test whether a relation is a function, check that no single input is assigned two different outputs.

Checking first components is most important while testing a function. No first component should be linked with two different second components.

The direct answer is A: each x in A has exactly one image in B. A relation is a collection of ordered pairs connecting elements of two sets. A function is a relation with one extra strict rule: every element of the domain must be connected to one and only one element of the codomain. Thus, every function is automatically a relation, but a general relation may fail the uniqueness rule. Option A is correct because it states both required ideas: every input has exactly one output. Option B is wrong because an input cannot have many outputs in a function; that describes a relation that is not a function. Option C is wrong because functions are made of ordered pairs. Option D is wrong because the domain need not be empty; it may contain many elements. Remember: one input, exactly one output.