Functions as a special kind of relation

To determine whether R is a function from A to B, every element of the domain A must be associated with exactly one element of B. Apply y=x+1: for x=1, y=2; for x=2, y=3; and for x=3, y=4. All three outputs belong to B={2,3,4,5}, and each input in A has exactly one output. Thus R={(1,2),(2,3),(3,4)} is a function from A to B. It is not empty, so option D is false. It is not option A because no input has missing or multiple outputs. The relation need not be described only in the reverse direction; although some elements of B may have no preimage, that does not prevent it from being a function from A to B. Hence option B is correct.

Using (y=x^2), each (x) gets exactly one image. Even in formula-based relations, check the whole domain.

For a relation to be a function, every element of the domain \(A\) must have exactly one image in \(B\). In option (D), no ordered pair is given with 3 as the first component, so the domain element \(3\) has no image. Hence, it is not a function. In options (A), (B), and (C), 1, 2, and 3 each occur exactly once as first components. Exam tip: To test a function, check that every domain element appears exactly once as the first component.

The element (2) is linked to both (3) and (4), so the single-image condition fails. Having all first elements is not enough.

A function is a special relation where each domain element has exactly one image. This is the key point in definition-based questions.

The number of functions is (2^3=8) and the number of relations is (2^{3\cdot2}=64). Keep the order of the ratio in mind.

To check one-one, look for different inputs with the same output. \(f(2)=2^2=4\) and \(f(-2)=(-2)^2=4\). Since \(2\neq -2\) but \(f(2)=f(-2)\), the function is not one-one. Hence option (B) is correct.

In option (B), each of (1,2,3) appears exactly once as a first element. This identifies a valid function.

The direct answer is C: the range is \(\{1,2\}\). The range means the set of output values that the function actually produces; it is not automatically the whole codomain. We are told \(f(a)=1\), \(f(b)=2\), and \(f(c)=2\). Collect the distinct outputs: 1 and 2. Thus \(\operatorname{Range}(f)=\{1,2\}\), even though 2 occurs twice in the mapping. Option A is incomplete because it leaves out the output 2. Option B is incomplete because it leaves out the output 1. Option C is correct because it contains both actual images, without repeating 2. Option D is the domain \(A=\{a,b,c\}\), not the range; those symbols are inputs, not outputs. The codomain is \(B=\{1,2\}\), and here the range happens to equal the codomain, but the two ideas should still be kept separate. Memory cue: domain means inputs, codomain means allowed outputs, and range means outputs actually reached.

For the same (x), two different (y)-values occur, so it is not a function. In square relations, check positive and negative roots.

A function from A to B assigns one output in B to every input in A. The condition f(1)=2 fixes the output of the first input, so there is no choice left for 1. The values of f(2) and f(3) are still free. Each of these two inputs can independently be assigned any one of the three elements 1, 2, or 3 in B.

Therefore the number of possible functions is \\(3\\) choices for f(2) multiplied by \\(3\\) choices for f(3), giving \\(3^2=9\\). The fixed value does not reduce the choices for the other inputs beyond removing the choice for f(1). Hence option C, 9, is correct. The value 27 would count all functions without the fixed condition, while 6 does not follow from the independent choices.

At (x=2), the denominator becomes zero, so the value is undefined. A function must have a value at every domain element.

A function is a special type of relation with the single-image condition. This basic definition is asked often in exams.

The graph of a function consists of ordered pairs \((x,f(x))\). Here, \(f(1)=0\), \(f(2)=1\), and \(f(3)=2\), so the graph is \(\{(1,0),(2,1),(3,2)\}\). Option B represents \(f(x)=x\), option C reverses the roles of the input and output values, and option D assigns two outputs to the same input 1, so it is not the graph of a function. Exam tip: evaluate the function at every domain element and write each result as \((x,f(x))\).

The codomain is the given set (B), while the range is the set of actual images. Here (8) is in the codomain but not in the range.

Two functions are equal when their domain, codomain, and values at every element are the same. The given condition gives (f=g).

(y=3x-5) gives exactly one real (y) for every real (x). The other options give zero or multiple (y)-values for some (x).

There are (2) choices for (f(1)=f(2)), and (2) choices each for (3) and (4). Total functions are (2\cdot2\cdot2=8).

For every real (x), (|x|) has one real value. Different inputs with the same image do not make a function invalid.

(f(1)) is fixed, (f(2)) has (2) choices, and (f(3)) has (3) choices. Hence total functions are (2\cdot3=6).

If any domain element has no image, it is not a function. Same images or extra codomain elements are usually not a problem.

The codomain (B) does not contain (4), so the image of (4) goes outside. In a function, each image must lie in the codomain.

The positive factors of (4) are (1,2,4), so (f(4)=3). In rule-based functions, first understand the rule clearly.

Every real (x) has a unique real cube root. Hence this relation is a function from (\mathbb{R}) to (\mathbb{R}).

There are (2^3=8) total functions, and two constant functions have range only ({0}) or ({1}). Hence (8-2=6) functions remain.