Functions as a special kind of relation

Each given (x) has a unique real cube root in the given (Y). The cube-root relation gives a unique image here.

The value is (1-\frac{1}{x^2+2}), so the minimum is (\frac{1}{2}) and (1) is never attained. The range is (\left[\frac{1}{2},1\right)).

There are (4) choices for the first input, (3) for the second, and (2) for the third. Total is (4\cdot3\cdot2=24).

Option (B) is a function, but in the inverse relation (1) has images (0) and (1). Hence (R^{-1}) is not a function.

The direct answer is A: the domain is R minus {-2,2}. A rational expression is defined only when its denominator is not zero. Here the denominator is |x|-2. Set it equal to zero: |x|-2=0, so |x|=2. The real solutions are x=2 and x=-2. These two inputs must be excluded, while every other real number is allowed. Hence the domain is R\{-2,2}. Option A is correct. Option B removes only 2 and forgets -2; at x=-2, the denominator is also zero. Option C removes 0, but x=0 gives denominator -2, which is perfectly valid. Option D includes only numbers less than 2, unnecessarily excluding many valid numbers such as 3 and -5, and it does not properly express the full domain. The common mistake is checking only |x|=2 as if x=2 were the sole answer; modulus always requires both signs.

The counting principle is to count all functions and subtract those that fail the required condition. Each of the four elements of A can independently be mapped to either 0 or 1 in B. Therefore the total number of functions is 2^4=16. The only function having no input with image 1 is the function that maps every element of A to 0. Thus exactly one function must be excluded. The required number is 16−1=15, so option C is correct. Counting only functions with one selected input mapped to 1 would double-count many functions, while 1 is the excluded all-zero function and 16 includes the function that never takes the value 1.

The domain includes (0), but (\frac{1}{0}) is undefined. A function needs a value at every element of its domain.

The minimum value of ((x-a)^2+b) is (b). For range ([3,\infty)), we need (b=3), while (a) may be any real value.

There are (2^6=64) total subsets and (2^3=8) functions. Hence non-function subsets are (64-8=56).

The expression under the square root can be rewritten by completing the square: \\(x^2-6x+10=(x-3)^2+1\\). Since a square is always non-negative, \\((x-3)^2\\ge0\\), so the quantity inside the root is always at least 1. It reaches 1 when \\(x=3\\). Therefore, the smallest value of the function is \\(\\sqrt1=1\\).

As x moves away from 3 in either direction, \\((x-3)^2\\) increases, so the square-root value can become arbitrarily large. Every value from 1 onward is obtained, because for any \\(y\\ge1\\), the equation \\(y=\\sqrt{(x-3)^2+1}\\) can be satisfied by a suitable real x. Hence the range is \\([1,\\infty)\\), so option A is correct. The range is not \\([0,\\infty)\\), because values below 1 cannot occur.

The absolute value gives a unique value for each input. Equal images for different inputs do not prevent a function.

The graph of a function has exactly one ordered pair for each element of the domain. Therefore the number of ordered pairs is (|A|=3).

Here (\sqrt{x}) means the principal non-negative square root, so each (x) has one image. The obtained values are ({0,1,2,3}).

Both pieces give the same value (3) at (x=2), so the rule is a valid function. Overlap is allowed only when the assigned values agree.

There are (3) choices for the common value of the first three inputs and (3^2) choices for the remaining two inputs. Total functions are (3\cdot3^2=27).

The direct answer is B. For a function, every element of the domain must have exactly one image. Use y²=2x. For x=2, y²=4, so y=2 or y=-2; thus 2 has two images. For x=8, y²=16, so y=4 or y=-4. For x=18, y²=36, so y=6 or y=-6. All these y-values are listed in the codomain, but the two outputs for each input mean the relation is not a function. Option B is correct because x=2 alone already proves failure. Option A is wrong because the relation does not satisfy uniqueness. Option C is wrong because 8 does have images, 4 and -4. Option D is wrong for the same reason as A, and its proposed range {2,4,6} also ignores the negative outputs; in any case, the relation is not a function. The exam cue is: solve for y and check whether one x gives two signs.

The expression (5-ax^2) stays non-negative for all real (x) only when (a\le0). In a square-root function the radicand must be non-negative for every input.

A function from A to B assigns one allowed value to each of the five elements of A. Since the range must be exactly {2,4}, every output must be either 2 or 4, and both values must actually occur. If we temporarily allow only 2 and 4, each of the five inputs has two choices, giving 2^5=32 functions. However, two of these functions are not acceptable: the constant function with value 2 and the constant function with value 4. Each misses one required range value.

Thus the required number is 2^5−2=32−2=30. Option B is correct. The value 16 would arise from a different counting situation, while 32 fails to remove the two constant functions. The answer correctly uses the condition that the range is exactly {2,4}, not merely a subset of it.

Since (n^3-n=n(n-1)(n+1)) is the product of three consecutive integers, it is divisible by (3). For codomain checking, focus on integrality.

For every (x), (y=7-x) is unique and lies in the given codomain. In finite sets check the image of every input.

On the given domain (f(x)=x+3), but removing (x=3) removes the value (6). The output of an excluded input may also be excluded.

There are (\binom{4}{2}=6) pairs for (f(1)<f(2)) and (4) choices for (f(3)=f(4)). Total functions are (6\cdot4=24).

For (-1\le x\le3), the value is (4), and at endpoints (x=-2) and (x=4), the value is (6). Breakpoints of modulus help find the range quickly.

There is one empty function from the empty domain, and its graph has no ordered pair. The number of functions and the number of graph pairs are different facts.

The direct answer is C. If f:A to B is a function, the definition says that every input in A is assigned exactly one output in B. Therefore, since a1 and a2 belong to A, both f(a1) and f(a2) must belong to B. The two inputs being different does not force their outputs to be different; a function may send several inputs to one common value. Option C is therefore always true. Option A is not always true: for example, a constant function can have f(a1)=f(a2). Option B is also not always true, because the outputs may or may not be equal. Option D is wrong because a1 and a2 are elements of A, not necessarily elements of B; the sets may even be different types of objects. The important lesson is that “function” guarantees one output in the codomain for each input, but it does not guarantee distinct outputs. Distinct outputs require an extra property called one-one or injective.