Functions as a special kind of relation

Correct answer: A. Because at (x=1), two different values (4) and (2) occur

The input (x=1) belongs to both parts and gives different values (4) and (2). Two different outputs for one input do not define a function.

Correct answer: A. It is a function because both rules give (5) at (x=3)

At (x=3), (2\cdot3-1=5) and (3+2=5), so there is no conflict. Overlap is valid when both values agree.

Correct answer: A. (54)

There are (3\cdot3) choices for the two equal groups and (3\cdot2) choices for the unequal last pair. Total functions are (3\cdot3\cdot3\cdot2=54).

Correct answer: A. (\ 0\le a\le\frac{3}{2}) / (0\le a\le\frac{3}{2})

If (a<0), the radicand can become negative for large (|x|), and if (a\ge0), its minimum is (9-6a). Thus (9-6a\ge0) gives (0\le a\le\frac{3}{2}).

Correct answer: B. It is a function because (n^5-n) is always divisible by (5)

For an integer (n), (n^5-n) is always divisible by (5). Therefore every value lies in (\mathbb{Z}).

Correct answer: A. It is a function and range is ({4,6,8,10})

For every (x), (y=12-2x) is unique and lies in the given codomain. Listing all images is a safe method for finite sets.

Correct answer: B. (\mathbb{R}-{0})

On the given domain (f(x)=x+2), but removing (x=-2) removes the value (0). After simplification, note the image of the excluded input.

Correct answer: A. (40)

There are (10) choices for the triple ((f(1),f(2),f(3))) and (4) choices for (f(4)=f(5)). Total is (10\cdot4=40).

Correct answer: A. ([6,8])

For (-2\le x\le4), the value is (6), and at endpoints (x=-3,5), it is (8). Check modulus breakpoints and interval endpoints.

Correct answer: A. (1) and (0)

There is one empty function from the empty domain, but its graph has no ordered pair. Keep the count and graph size separate.

Correct answer: A. (f(C)\subseteq B)

All elements of (C) lie in (A), and their images lie in the codomain (B). Hence (f(C)\subseteq B) is always true.

Correct answer: A. Because (x=0) has two images

The correct answer is option A: x=0 has two images. A relation is a function from X to Y only when every element of X has exactly one image in Y. Use x^2+y^2=9 and test the allowed values. For x=0, y^2=9, so y=3 or y=-3. Thus the same input 0 is related to two outputs, and the relation is not a function. For x=3, 9+y^2=9, so y=0 only; therefore option B is wrong. For x=-3, the same calculation gives y=0, so x=-3 does have an image; option C is wrong. Option D is wrong because the presence of 0 in Y does not by itself prevent a relation from being a function. What matters is the number of images for each input. Option A correctly identifies the failure of the one-input-one-output rule. Memory cue: in a function, each x gets one and only one y; different x-values may share a y, but one x cannot have two y-values.

Correct answer: A. \(-6<a<6\)

The function is defined for all real \(x\) only if the denominator never becomes zero. So we need \(x^2+ax+9\ne 0\) for all \(x\). A quadratic \(x^2+ax+9\) never equals \(0\) for real \(x\) exactly when its discriminant is negative: \(a^2-4\cdot 1\cdot 9=a^2-36<0\). Thus \(a^2<36\Rightarrow -6<a<6\).

Correct answer: C. 35

We must choose the 4 elements of the 7-element domain that will be mapped to 1. This can be done in \(\binom{7}{4}=35\) ways. The remaining 3 elements must be mapped to 0, so they introduce no additional choices. Hence the number of functions is 35. Exam tip: For a function from an \(n\)-element set to \(\{0,1\}\), if exactly \(r\) domain elements map to 1, the count is \(\binom{n}{r}\).

Correct answer: A. (f^{-1}) is a function and (f^{-1}=f)

The rule sends (x) to (7-x), and applying the same rule again returns (x). Therefore the inverse relation is the same function.

Correct answer: A. (10)

On the given domain (f(x)=x+5), but removing (x=5) removes the value (10). The image of the excluded input is removed from the range.

Correct answer: A. When each of (1,2,3,4,5) appears exactly once as a first component

With exactly (5) pairs, a function needs one image for every element of (A). Repetition of second components is allowed.

Correct answer: B. (4\sqrt{2})

By symmetry the maximum occurs at (x=8), and the value is (2\sqrt{8}=4\sqrt{2}). For sums of square roots, check the balanced point.

Correct answer: A. (2^{12}) and (3^4)

Total relations are (2^{|A||B|}=2^{12}), and total functions are (|B|^{|A|}=3^4). The formulas for relations and functions are different.

Correct answer: A. \(\left{\frac{7}{2},5,-1,\frac{1}{2}\right}\)

At the given inputs, the values are \(\frac{7}{2},5,-1,\frac{1}{2}\). For a finite domain, direct substitution is fastest.

Correct answer: C. (48)

There are (2^6=64) total functions, and the opposite case (f(2)=b), (f(5)=a) gives (2^4=16) functions. Hence (64-16=48).

Correct answer: A. ([7,\infty))

The direct answer is option A: [7,∞). The expression is f(x)=|x+5|+|x−2|. Absolute value means distance on the number line: |x+5| is the distance from x to −5, and |x−2| is the distance from x to 2. The distance between −5 and 2 is 7. For every x between −5 and 2, the two distances add to exactly 7. Outside that interval, moving farther away increases the total distance, so values larger than 7 occur. Thus the minimum is 7 and the range is [7,∞). Option A is correct because 7 is included and every larger value is possible. Option B, [0,∞), wrongly assumes the sum can be zero; both distances cannot be zero because x cannot equal −5 and 2 simultaneously. Option C, {7}, includes only the minimum and ignores larger values outside the interval. Option D contains values below 7, which are impossible. The exam idea is that the sum of distances from a point to two fixed points is at least the distance between those points.

Correct answer: A. It is a function

Each given (x) has a unique real seventh root in (Y). The inverse of an odd power is unique over real numbers.

Correct answer: B. No, because (f(4)=6\notin{0,1,2,3,4})

The direct answer is B: no, this is not a valid function from the stated domain to the stated codomain. A function must assign every input exactly one output, and that output must belong to the codomain. Check the inputs one by one: f(0)=0^2−3(0)+2=2, f(1)=1−3+2=0, f(2)=4−6+2=0, f(3)=9−9+2=2, and f(4)=16−12+2=6. Although each input has one value, 6 is not in {0,1,2,3,4}; therefore the mapping fails as a function into the given codomain. Option A is wrong because not all values lie in the codomain. Option B is correct because f(4)=6 is outside it. Option C is wrong because f(1)=0 is allowed and does not cause a problem. Option D is wrong: a polynomial can be evaluated on a finite set, but its outputs must still lie in the declared codomain. Memory cue: check every input and then check whether every output belongs to the codomain.

Correct answer: A. Because (x=2) has two images

The direct answer is A. A relation is a function only when every domain element has exactly one image. For x=2, substitute into |x−2y|=2: |2−2y|=2. This means 2−2y=2 or 2−2y=−2. The first equation gives y=0, and the second gives y=2. Thus the same input x=2 has two images, 0 and 2, so the relation is not a function. Option A is correct for precisely this reason. Option B is wrong because x=0 does have images: |0−2y|=2 gives y=1 or y=−1, both in the codomain. Option C is wrong because x=4 also has images: |4−2y|=2 gives y=1 or y=3. Option D is irrelevant and wrong; the mere presence of y=3 in the codomain does not violate the function rule. The problem is two outputs for one input, not the size of the codomain. Memory cue: a function allows many inputs to share one output, but never one input to have two outputs.