Functions as a special kind of relation

Correct answer: B. It is a function and its range is \(\{1,4,9\}\)

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.

Correct answer: B. No, because (f(4)=8\notin{1,2,3,4})

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

Correct answer: B. ({(0,1),(1,1),(2,3)})

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.

Correct answer: A. Because the formula is undefined at (x=2)

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

Correct answer: A. It is a function

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.

Correct answer: B. (1)

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

Correct answer: C. ({(1,a),(2,b),(3,c),(4,a),(4,b)})

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

Correct answer: A. Because (1) has preimages (1) and (-1)

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

Correct answer: B. ({2,4})

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

Correct answer: A. ({-1,1})

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.

Correct answer: A. ([-2,2])

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

Correct answer: C. (16)

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

Correct answer: B. (g(n)=2n+1)

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

Correct answer: B. (0)

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

Correct answer: C. (f(x)=\begin{cases}x^2,&x\le2\x+3,&x\ge2\end{cases})

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.

Correct answer: B. (8)

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).

Correct answer: B. ({2,4})

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.

Correct answer: C. 32

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.

Correct answer: B. ([1,\infty))

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

Correct answer: A. It will never be a function on all of (\mathbb{R})

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.

Correct answer: B. (R={(x,y):y=0 \text{ if } x<3,\ y=1 \text{ if } x\ge3})

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

Correct answer: A. It is a function and range is ({1,3,5})

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.

Correct answer: B. It is not a function because elements of (A) have no images

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

Correct answer: B. (1)

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

Correct answer: A. Because at x = 2, two different values, 3 and 4, occur

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.