Functions as a special kind of relation

Correct answer: A. (R={(1,4),(2,4),(3,5)})

Every (x\in A) has exactly one image in (B). In exams, first check presence and uniqueness for each domain element.

Correct answer: A. (2^4=16)

Each of the (4) elements of (A) has (2) choices in (B), so total functions are (2^4=16). Remember the formula: functions from (A) to (B) (=|B|^{|A|}).

Correct answer: A. (R) is a function

A function needs exactly one output for each input; different inputs may have the same output. Common mistake: do not treat same image as a violation of function rule.

Correct answer: C. (R={(1,p),(2,q),(2,r),(3,p)})

The direct answer is C. A function must assign every element of the domain exactly one image in the codomain. In option C, the input 1 is paired with p, input 3 with p, but input 2 is paired with both q and r. Since q and r are different outputs, one input has two images, so this relation is not a function. Option A is a function: 1, 2, and 3 each have exactly one image, namely p, q, and r. Option B is also a function: different inputs are allowed to have the same output p; a function does not require different outputs. Option C fails the one-output rule because of (2,q) and (2,r). Option D is a function because each domain element has one image, even though the order of the outputs is reversed. The useful test is to inspect the first coordinates: every domain element must occur once with only one second coordinate.

Correct answer: A. (f(x)=2x+1)

Every ordered pair satisfies (y=2x+1). While identifying a rule from pairs, test the rule on all pairs.

Correct answer: A. \({1,3,5,7})

Putting (x=0,1,2,3) gives outputs (1,3,5,7). The range is the set of actual images, not necessarily the whole codomain.

Correct answer: A. Every (a\in A) has exactly one arrow to an element of (B)

In a function, exactly one arrow must start from each domain element. In diagrams, check both arrow direction and arrow count.

Correct answer: A. (4\in A) has no image

Every element of (A) must have an image, but (4) is missing. An unused element of the codomain does not make a function invalid.

Correct answer: A. For (x=3), (y=0,1,2) are possible

For (x=3), there are many images, so uniqueness fails. In inequality relations, count possible outputs for one input.

Correct answer: D. \((4,0))

(f(4)=1), so ((4,0)) is not in the relation. To test an ordered pair, put the first coordinate into the rule.

Correct answer: A. For every (a\in A), there is exactly one (b\in B) such that ((a,b)\in R)

The condition for a function is on domain elements, not codomain elements. In definition questions, the words exactly one are most important.

Correct answer: A. (3)

For (x=1,2,3), pairs are ((1,1),(2,4),(3,9)). In rule-based relations, substitute domain values systematically.

Correct answer: A. (3)

In any function, the number of ordered pairs equals the number of elements in the domain. Here (|A|=3), so there are (3) pairs.

Correct answer: A. It is a constant function

Every input has image (2), so it is a constant function. Repeated output does not make a function invalid.

Correct answer: A. \(R=\{(1,1),(2,2),(3,3),(4,4)\}\)

In an identity function, every \(x\in A\) maps to itself, so \(I_A(x)=x\). In option A, the mappings are \(1\to1, 2\to2, 3\to3\), and \(4\to4\); hence it is the identity function. Option C is a constant function, while B and D map at least some elements to different elements. Exam tip: for an identity function, check whether every ordered pair has equal components, \((x,y)\) with \(x=y\).

Correct answer: A. \({1,2,3})

The domain is the set of first coordinates, and here (A={1,2,3}). Distinguish ordered pairs from the domain.

Correct answer: A. For every (a\in A), (f(a)\in B) is unique

In a function, every domain element has a unique image in the codomain. The whole codomain need not become the range.

Correct answer: A. \({-4,-3,0,5})

The outputs are (5,0,-3,-4,-3), so distinct values ({-4,-3,0,5}) form the range. Repeated values are not written in a set.

Correct answer: A. (3^2=9)

Here (|A|=2) and (|B|=3), so functions (=3^2=9). In counting, keep codomain size as base and domain size as exponent.

Correct answer: B. (y^2=x) with (x>0)

In (y^2=x), for (x=4), both (y=2) and (y=-2) are possible. Two (y)-values for one (x) do not define a function.

Correct answer: A. \({(-1,1),(0,0),(1,1),(2,2)})

For absolute value, (|-1|=1), (|0|=0), (|1|=1), and (|2|=2). In an ordered pair, the first coordinate remains the input.

Correct answer: A. It is a function

For every (x\in A), (y=5-x) is unique and lies in (A). For an equation relation, check uniqueness to decide function status.

Correct answer: A. For (x=2), both (y=1) and (y=2) are possible

(x=2) gets two images, so uniqueness fails. In divisibility relations, check all divisors in the codomain for one input.

Correct answer: A. \({(1,3),(2,4),(3,5)})

The direct answer is A: the relation is \(\{(1,3),(2,4),(3,5)\}\). A function rule tells us to take each input from the domain and calculate exactly one output. Here \(f(x)=x+2\). Step by step, \(f(1)=1+2=3\), \(f(2)=2+2=4\), and \(f(3)=3+2=5\). All outputs belong to \(B\), so this is a valid relation from \(A\) to \(B\). Option A is correct because it contains precisely these three ordered pairs. Option B uses \(x+1\), since it gives 2, 3, and 4. Option C reverses the direction of the pairs and does not have the required domain elements as first entries. Option D gives 4, 5, and 6, which corresponds to \(x+3\), not \(x+2\). Remember: substitute every domain value into the formula, and write the input first and its image second.

Correct answer: A. \({2,4,6})

The outputs are (2,4,6), so the smallest codomain can be the set of these images. The codomain must contain all possible outputs.