Real-Valued Functions, Their Domain and Range

Correct answer: A. {1,2}

The governing concept is the domain of a relation. The domain consists of all first components a for which there exists a second component b in A satisfying a+2=b. Test the possible elements of A: if a=1, then b=3, which belongs to A, so (1,3) is valid. If a=2, then b=4, also in A, so (2,4) is valid. If a=3, then b=5, which is outside A; if a=4, then b=6, also outside A. Thus the first components that actually occur are 1 and 2, giving domain(R)={1,2}. Option A is correct. Option B is the range, while C includes first components that produce invalid second components.

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

For the elements of \(A\), we get \(f(1)=1\), \(f(2)=0\), and \(f(3)=1\). Thus, the only values actually obtained are 0 and 1, so the range of \(f\) is \(\{0,1\}\). Option C is the codomain \(B\), not the range, because 2 is not attained by the function. Exam tip: To find a range, evaluate the function for every element of the domain and list only the distinct outputs.

Correct answer: A. {1, 3, 5}

Evaluate the function for every element of A: f(1) = 2(1) − 1 = 1, f(2) = 3, and f(3) = 5. Hence, the range of f is {1, 3, 5}. Option B is the codomain, not the range; the range contains only the values actually attained by the function. Exam tip: apply the function rule to every element of the given domain.

Correct answer: B. \(\{2,4,6,8\}\)

The domain of a relation or function is the set of first components of its ordered pairs. Here, the first components are 2, 4, 6, and 8, so the domain is \(\{2,4,6,8\}\). Option A is the set of second components, which represents the range. Exam tip: take the first entry from every ordered pair to find the domain.

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

For every element of the domain, \(f(1)=4\), \(f(2)=3\), \(f(3)=2\), and \(f(4)=1\). Thus, the set of all images is \(\{1,2,3,4\}\), which is the range of the function. Option C omits 1, while option B omits 4. Exam tip: To find the range, evaluate the function for every element of the domain and collect the distinct output values.

Correct answer: A. {−1,0,3}

The range is the set of distinct output values produced by applying the function to every element of its specified domain. Calculate f(x)=x²−1 for each x in A. We get f(-2)=4−1=3, f(-1)=1−1=0, f(0)=0−1=-1, f(1)=1−1=0, and f(2)=4−1=3. Thus the list of outputs is 3,0,-1,0,3. Repeated values are written only once in a set, so the range is {-1,0,3}. Therefore option A is correct. Option B lists squares without subtracting 1, option C contains values not generated by the given domain, and option D incorrectly assumes every integer is obtained even though the domain is finite.

Correct answer: B. It is a function and its range is B

The governing ideas are the definition of a function and the distinction between codomain and range. Every element of A has exactly one assigned image: 1 and 2 map to a, while 3 and 4 map to b. Repeated images are permitted; they simply make the function many-one rather than one-one. Since both a and b actually occur as outputs, the range is {a,b}, which equals the stated codomain B. Thus option B is correct. Option A incorrectly treats repeated outputs as invalid, option C omits the output b, and option D wrongly assumes that the domain and codomain must have equal sizes.

Correct answer: B. Domain = ℝ − {−3, 3} and range = ℝ − {0}

The governing concept is that the denominator of a real rational function can never be zero. Here x² − 9 = 0 gives x² = 9, so x = −3 or x = 3. These two inputs must be excluded from the domain; every other real number makes the denominator nonzero, so the natural domain is ℝ − {−3, 3}. Also, the function can never equal zero because its numerator is 1, so its actual range is ℝ − {0}. Thus option B correctly gives both the restricted domain and the range. Option A incorrectly includes undefined inputs, option C excludes 0 unnecessarily rather than excluding ±3, and option D is far too restrictive.

Correct answer: A. [3, ∞)

The governing property is |t| ≥ 0 for every real t, with equality only when t = 0. Therefore |x − 2| + 3 ≥ 3 for every real x. The minimum value 3 is actually attained when x − 2 = 0, that is, when x = 2. As x moves away from 2, the absolute-value term can become any nonnegative real number, so the function can produce every value from 3 upward. Consequently, the range is [3, ∞), including 3. Option B wrongly excludes the attained minimum, option C reverses the direction of the possible values, and option D includes values below 3 that the formula can never produce. Thus option A is correct.

Correct answer: A. [4, ∞)

To determine the range, rewrite the quadratic by completing the square: f(x) = x² + 2x + 5 = (x + 1)² + 4. For every real x, the square (x + 1)² is at least zero. Therefore f(x) is at least 4, and equality occurs when x + 1 = 0, or x = -1. As x moves away from -1, the square can take every nonnegative real value, so the function assumes every value from 4 upward. Hence the range is [4, ∞), which is option A. Option B overlooks the lowering effect of the linear term, option C reverses the direction of the bound, and option D incorrectly claims that values below 4 can occur.

Correct answer: A. (1, 3]

The governing idea is to determine all possible values of a real-valued function. Rewrite the expression as f(x) = 1 + 6/(x² + 3). Since x² ≥ 0, the denominator satisfies x² + 3 ≥ 3, so 0 < 6/(x² + 3) ≤ 2. Therefore 1 < f(x) ≤ 3. The value 3 is attained when x = 0, because f(0) = 9/3 = 3. The value 1 is never attained for any finite real x, since 6/(x² + 3) is always positive, although it approaches 0 as |x| increases. Hence the range is (1, 3]. Option B incorrectly includes 1, while C and D do not describe the actual bounds.

Correct answer: A. (−∞, 3]

The governing concept is the effect of the absolute-value term on the output. For every real x, |x| ≥ 0, so 3 − |x| ≤ 3. Equality occurs at x = 0, giving f(0) = 3; hence the upper endpoint is included. There is no lower bound: for any large positive number M, taking x = 3 + M gives f(x) = 3 − (3 + M) = −M. Thus arbitrarily negative values occur, while no value greater than 3 can occur. The range is (−∞, 3]. Option B reverses the direction of the bound, option C incorrectly imposes a lower bound of zero, and option D incorrectly includes values above 3.

Correct answer: A. [3, ∞) \ {5}

The domain of a real-valued expression must satisfy every restriction at the same time. For √(x − 3), the radicand must be non-negative, so x − 3 ≥ 0 and therefore x ≥ 3. The denominator x − 5 cannot be zero, so x ≠ 5. Taking the intersection of these conditions gives [3, ∞) with 5 removed, written [3, ∞) \ {5}. Thus option A is correct. Notice that x = 3 is allowed because √0 is defined, which rules out option B. Values less than 3 are excluded by the square root, so option D is too broad. Option C unnecessarily excludes all values below 5. The governing concept is intersecting the domains of the component expressions rather than treating either restriction in isolation.

Correct answer: A. ([\frac{1}{2},\infty))

The square root needs (2x-1\ge 0), so (x\ge \frac{1}{2}). Since (x=-4) is not in this interval, it need not be removed separately.

Correct answer: A. ((-\infty,-4]\cup[4,\infty))

The square root needs (x^2-16\ge 0), that is (x^2\ge 16). In exams (x^2\ge a^2) gives the outer intervals.

Correct answer: A. ((-\infty,-5)\cup(5,\infty))

The square root is in the denominator, so (x^2-25>0) is required. In exams do not include equality for a denominator square root.

Correct answer: A. (\mathbb{R}\setminus{-2,2})

The original denominator is (x^2-4=(x-2)(x+2)), so (x\ne -2,2). In exams remove zeros of the original denominator before cancellation.

Correct answer: A. ℝ \ {1, −3}

A rational function is defined for every real x except those values that make its denominator zero. Factor the denominator: x² + 2x − 3 = (x + 3)(x − 1). Hence the denominator is zero at x = −3 and x = 1. These two values must be excluded, while every other real number is allowed. Therefore the domain is ℝ \ {−3, 1}, which is the same set as ℝ \ {1, −3}; option A is correct. The numerator being zero at x = 1 does not cancel the original restriction, because the original denominator is also zero there. Option B contains the wrong excluded values, option C omits −3, and option D wrongly removes intervals instead of only two points. The governing concept is finding and excluding zeros of the original denominator.

Correct answer: A. ([1,7])

The conditions (x-1\ge 0) and (7-x\ge 0) together give (1\le x\le 7). In exams take the intersection for the combined domain.

Correct answer: A. ℝ

For this rational function, the only possible restriction is that the denominator must not be zero. Complete the square: x² + 6x + 10 = (x + 3)² + 1. Since (x + 3)² is always at least 0, the denominator is always at least 1, so it can never equal zero for any real x. Consequently, every real number is permitted and the domain is ℝ; option A is correct. Equivalently, the discriminant of the denominator is 6² − 4(1)(10) = 36 − 40 = −4, confirming that it has no real zeros. Options B and C incorrectly remove individual values, while option D incorrectly excludes x = −3 even though the denominator there is 1, not 0. The governing concept is checking denominator zeros before stating a rational function’s domain.

Correct answer: A. ((-\infty,4])

The square root needs (12-3x\ge 0), giving (x\le 4). In exams dividing by a negative coefficient reverses the inequality.

Correct answer: A. ((-\infty,2]\cup[6,\infty))

The direct answer is option A: (-\infty,2]\cup[6,\infty). A square root is real only when its radicand, the expression inside it, is non-negative. Therefore we need (x-2)(x-6)≥0. The critical values are 2 and 6. For x<2 both factors are negative, so their product is positive; at x=2 it is zero. Between 2 and 6 the factors have opposite signs, so the product is negative. For x>6 both are positive, so the product is positive; x=6 is also included. Thus the domain is the two outer intervals. Option A is correct. Option B, [2,6], contains values where the product is negative, so the square root is not real. Option C, (2,6), has the same problem and also wrongly excludes endpoints. Option D excludes only 2 and 6, but it includes forbidden values such as 3. Memory cue: for a product of two linear factors with a positive sign outside, the non-negative regions lie outside the two roots.

Correct answer: A. ([2,\infty))

(x^2-6x+11=(x-3)^2+2), so the minimum is (2). In exams completing the square gives the range quickly.

Correct answer: A. ( (-\infty,5])

Since ((x+1)^2\ge 0), (5-(x+1)^2\le 5). In exams check the maximum value of a downward parabola.

Correct answer: A. ([-2,\infty))

Direct answer: A, [-2,∞). The modulus |x-3| is a distance, so it can never be negative. Its smallest value is 0, reached when x=3. Hence f(x)=|x-3|-2 has minimum 0-2=-2, and larger values continue without bound. Therefore the range is [-2,∞). Option A is correct because -2 is included: f(3)=-2. Option B [0,∞) forgets the subtraction of 2. Option C (-∞,-2] reverses the direction and incorrectly says values can decrease without limit. Option D all real numbers includes values below -2, which are impossible. Exam cue: for |x-a|+b, the range starts at b and goes upward when the coefficient is positive.