Functions as a special kind of relation

In option (A), every domain element has exactly one image. In the others, an image is missing or two images occur.

Adding (5) to an integer gives an integer and the value is unique. An infinite domain is not a problem for being a function.

There are (3) choices for the common image of the first three elements and (3) choices for the image of (4). Total functions are (3\cdot3=9).

With the whole (\mathbb{R}) as domain, negative (x)-values have no real image. Hence it is not a function on the given domain.

In a constant function, all domain elements have the same codomain element as image. It is a simple valid many-one function.

For a constant function, the common image is chosen from (B). Since (B) has (3) choices, there are (3) constant functions.

The direct answer is B: exactly one function exists. A function must assign one output to each element of its domain. When the domain is empty, there are no elements that need assignments, so the empty assignment satisfies the function rule automatically. There is also only one empty set of ordered pairs, namely the empty function. The counting formula confirms this: the number of functions from a set of size \(|A|\) to a set of size \(|B|\) is \(|B|^{|A|}\), so here it is \(2^0=1\). Option A, 0, is wrong because the empty function does exist. Option B, 1, is correct. Option C, 2, incorrectly treats the two codomain elements as two choices even though there is no domain input to receive a choice. Option D, 4, would be \(2^2\), relevant to two domain elements, not an empty domain. The important idea is that there is exactly one way to make no assignments.

In a function every element of (A) must have exactly one image. In exams first check whether any input has two images.

If (|A|=3) and (|B|=3), the number of functions is (3^3=27). Remember the formula (|B|^{|A|}) for exams.

For \(y^2=4\), both \(y=2\) and \(y=-2\) are possible. Thus, the input \(4\) has two images. A function must assign exactly one image to every input, so this relation is not a function. Exam tip: To test a relation, check whether the same first component is paired with two different second components.

The value (\sqrt{x-2}) is real only when (x\ge 2). In such questions check domain validity before the formula.

The direct answer is B: 4 functions are possible. A function from \(A\) to \(B\) gives one value in \(B=\{0,1\}\) to every element of \(A\). The values of \(p\) and \(q\) are already fixed: \(f(p)=0\) and \(f(q)=1\). Only \(r\) and \(s\) remain free. Each of these two inputs independently has 2 choices, 0 or 1. Therefore the number is \(2\times2=2^2=4\). Option A, 2, counts only one choice for each remaining input and misses the combinations. Option B, 4, is correct. Option C, 8, would correspond to three free inputs, but only two are free. Option D, 16, is the unrestricted count \(2^4\) when none of the four values is fixed; it ignores the two conditions. The four possibilities for \((f(r),f(s))\) are \((0,0),(0,1),(1,0),(1,1)\). Memory cue: fixed inputs contribute no choice; multiply the number of codomain choices for every unfixed input.

For every (x\in X), exactly one (y) is obtained. Here the reverse square relation becomes a function on the given finite sets.

Here (f(1)=4), (f(2)=3), (f(3)=2), and (f(4)=1). The range is the set of actually obtained values.

In option (C), every element of (A) appears exactly once as the first component. Many inputs may have the same image in a function.

The direct answer is B. A function must give one natural-number output for every input 1, 2, 3, and 4. Compute the values: f(1)=12, f(2)=6, f(3)=4, and f(4)=3. All four values are natural numbers, so the rule is a valid function into N. The range is the set of values actually obtained, namely {12,6,4,3}, which is the same set as {3,4,6,12}; therefore option B is correct. Option A is wrong because f(4)=12/4=3, and 3 is a natural number. Option C is wrong because {1,2,3,4} is the domain, not the range; the outputs are different numbers. Option D is wrong because f(1)=12 is defined, not undefined. The key distinction is domain versus range: domain lists allowed inputs, while range lists outputs produced by those inputs.

The expression (ax+6) stays non-negative for all real (x) only when (a=0). For a square-root function on all (\mathbb{R}), the linear part must be constant non-negative.

For (2,3,5), the value is (1), and for (1,4), the value is (0). The range is the part of the codomain that is actually obtained.

The direct answer is A: R is a function and its range is {1,2,3,4}. The relation requires x+y=5, so solve for y: y=5-x. Check every x in A. For x=1, y=4; for x=2, y=3; for x=3, y=2; for x=4, y=1. Thus R={(1,4),(2,3),(3,2),(4,1)}. Each input has exactly one output, and every output belongs to B, so R is a function. The outputs together are {1,2,3,4}, the range. Option A is correct. Option B is wrong because x=1 has only one image, 4. Option C is wrong because x=4 has image 1. Option D is wrong because the range contains y-values, not the single sum 5. Check inputs one by one.

A function is a special relation where each element of the first set maps exactly once. Every codomain element need not occur.

The direct answer is A: it is a function from \(A\) to \(B\), and its range is \(\{2,3,4\}\). Evaluate the rule for every domain element: \(f(1)=1+1=2\), \(f(2)=2+1=3\), and \(f(3)=3+1=4\). These outputs all lie in \(B=\{1,2,3,4\}\), so the mapping is a valid function. The outputs actually reached form the range, namely \(\{2,3,4\}\); the value 1 is in the codomain but is not reached. Option A is therefore correct. Option B is wrong because a function need not use every codomain element. Option C is wrong because obtaining 4 from input 3 is allowed; 4 belongs to B. Option D is wrong because it confuses codomain with range and includes 1, which is never an output. The range can be a proper subset of the codomain. Exam cue: calculate all images, check that they lie in the codomain, then list only the images actually obtained.

Only (2) has two images, so removing one of them is enough. To make a function, give each first component exactly one image.

For (f(x)=\frac{1}{x}), the value is undefined at (x=0). A function from (\mathbb{R}) must have a value for every real (x).

There are (3) choices for the common value of (f(1)=f(2)) and (3^2) choices for (3,4). Total functions are (3\cdot3^2=27).

Putting (x=0) gives (\frac{1}{2}), which is not in the codomain (\mathbb{Z}). Every value of a function must lie in the codomain.