Functions as a special kind of relation

Each element of \(A\) has exactly one image in \(B\) under \(y=x^2\): \(1\mapsto1\), \(2\mapsto4\), and \(3\mapsto9\). Hence, the relation is a function and its range is \(\{1,4,9\}\). The codomain is the complete set \(B=\{1,4,9,16\}\), so option A is incorrect. The fact that \(16\) has no preimage does not disqualify the relation from being a function. Exam tip: The range is always a subset of the codomain, and the codomain may contain elements absent from the range.

Here (f(4)=16-12+4=8), which is not in the codomain. For finite domains, checking all values is the safe method.

In option (B), both (0) and (1) have image (1), and every input has exactly one image. A common image does not prevent a function.

Even though simplification looks like (x+2), the original formula is undefined at (x=2). When deciding domain, check the original denominator.

Each (x) has a unique real cube root lying in the given (Y). A cube-root relation does not give two values like a square-root relation.

For the range to be exactly ({1}), every input must have image (1). So only one constant function is possible.

In option (C), (4) has two images (a) and (b). Having all first components is not enough; uniqueness is also required.

In the inverse relation, (1) is related to both (1) and (-1). Two images for one input make the inverse relation not a function.

The value (0) is produced by even inputs, namely (2) and (4). A preimage contains elements of the domain, not of the codomain.

For negative (x), the value is (-1), and for positive (x), it is (1). Since (x=0) is not in the domain, there is no issue.

For real values, (4-x^2\ge0), so (-2\le x\le2). In square-root questions keep the radicand non-negative.

The image of (a) is fixed, so (b,c) have (4^2=16) choices. In restricted functions count the free inputs.

For every natural (n), (2n+1) is a natural number. The other options do not give values in (\mathbb{N}) for all (n).

Since (0\ge0), the second rule applies and (f(0)=0^2=0). In piecewise functions read the boundary symbol carefully.

In option (C), (x=2) belongs to both rules and gives values (4) and (5). Two values for one input do not define a function.

A function is a subset of (A\times B) where one pair is chosen for each element of (A) from (2) choices. Total is (2^3=8).

Here (f(1)=4), (f(2)=2), (f(3)=2), and (f(4)=4), so the range is ({2,4}). For a piecewise rule list every input value.

The counting principle is different for arbitrary relations and functions. A relation from A to B is any subset of A×B, so with 5×2=10 ordered pairs there are 2¹⁰ relations. For a function, however, each of the 5 elements of A must independently choose exactly one of the 2 elements of B as its image. By the multiplication principle, the number of functions is 2×2×2×2×2 = 2⁵ = 32. Therefore option C is correct. Option D is the number of all relations, not functions; option A and option B do not follow from the function-counting formula. The fact that several domain elements may select the same codomain element does not reduce this count.

Since (x^2+2x+2=(x+1)^2+1), the minimum value is (1). Completing the square is useful for finding range.

The denominator (x^2-4) is zero at (x=\pm2), so the original formula is undefined there. A zero denominator cannot be fully removed for all real inputs.

Option (B) assigns exactly one value to every (x). The other options miss some inputs or give multiple values.

The direct answer is A: it is a function and its range is {1,3,5}. The domain is {1,2,3}, and the rule is f(x)=2x-1. Calculate each value: f(1)=2(1)-1=1, f(2)=2(2)-1=3, and f(3)=2(3)-1=5. Every value lies in the codomain {1,2,3,4,5}, so the rule defines a function. The set of values actually obtained is the range {1,3,5}; the codomain need not be completely used. Option A is correct. Option B is wrong because an unused codomain element does not stop a rule from being a function. Option C confuses domain or codomain with range and includes 2, which is not obtained. Option D is wrong because f(3)=5 is allowed and is inside the codomain. Remember: a function need not cover every codomain element.

When (A\ne\emptyset), the empty relation gives no image to any element of (A). A function requires an image for every input.

From the empty domain to any set, there is exactly one empty function. The formula (|B|^{|A|}=3^0=1) gives the same result.

A piecewise rule defines a function only when every input has one well-defined output. Here the first condition is x ≤ 2, while the second is x ≥ 2. Their domains overlap at x = 2, so both formulas must be applied there. The first formula gives 2 + 1 = 3, whereas the second gives 2² = 4. Thus the same input is assigned two different outputs, violating the uniqueness requirement for a function. Therefore option A is correct. There is no issue at x = 0 or x = 1 because the first formula gives real values, and f(3) = 9 is a valid single output under the second formula. If one condition had been x < 2, the overlap would disappear.