Real-Valued Functions, Their Domain and Range

Correct answer: A. (3, 4]

Complete the square in the denominator: x² + 2x + 2 = (x + 1)² + 1. Hence the denominator is at least 1, with equality at x = −1. Its reciprocal therefore lies in (0, 1], because it is positive, reaches 1, and approaches 0 without attaining it as |x| grows. Adding 3 to every value shifts this interval to (3, 4]. The value 4 is attained when x = −1, while 3 is not attained because the reciprocal term can never be zero. Thus option A is correct. Option B incorrectly includes 3, option C ignores the addition and reciprocal behavior, and option D includes values below 3 that the positive reciprocal term cannot produce. The interval endpoints must be handled carefully.

Correct answer: B. [−2, 5]

For a real-valued square root, the radicand must be non-negative: (x + 2)(5 − x) ≥ 0. The factors are zero at x = −2 and x = 5. Testing a point between them, such as x = 0, gives a positive product, while testing a point outside, such as x = −3, gives a negative product. Therefore the product is non-negative on the closed interval −2 ≤ x ≤ 5. The endpoints are included because the radicand becomes zero there, and √0 is defined. Hence the domain is [−2, 5]. Option A reverses the sign pattern, option C wrongly excludes the valid endpoints, and option D includes values for which the radicand is negative and the square root is not real.

Correct answer: A. (−∞, 4) ∪ (6, ∞)

Because the square root is in the denominator, its radicand must be strictly positive, not merely non-negative. Thus x² − 10x + 24 > 0. Factoring gives (x − 4)(x − 6) > 0. A product of two linear factors is positive outside its two roots, so the solution is x < 4 or x > 6. The points 4 and 6 must be excluded because they make the radicand zero and would place zero in the denominator. Therefore the domain is (−∞, 4) ∪ (6, ∞). Option B incorrectly includes the endpoints, option C selects the interval where the radicand is negative, and option D excludes only the roots while incorrectly retaining all points between them.

Correct answer: A. [1, ∞)

Rewrite the expression inside the radical by completing the square: x² + 2x + 2 = (x + 1)² + 1. Since (x + 1)² ≥ 0, the radicand is at least 1. Its minimum value 1 occurs at x = −1, so the function has minimum √1 = 1. As |x| increases, (x + 1)² becomes arbitrarily large, and the square root also grows without bound. Hence every value from 1 upward is attained, giving the range [1, ∞). Option B overlooks the positive minimum inside the radical, option C incorrectly adds 1 instead of taking its square root, and option D includes negative values even though a principal square root is always non-negative. The endpoint 1 is included because it occurs at x = −1.

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

The governing concept is the ceiling function: ⌈x⌉ is the least integer greater than or equal to x. The interval begins just above -2, so x = -2 is excluded, but numbers such as -1.9 have ceiling -1. Values from -2 up to -1 have ceiling -1; values in (-1, 0] have ceiling 0; values in (0, 1] have ceiling 1; and values in (1, 2] have ceiling 2. Every listed integer is attained, while -2 cannot be attained because x is never equal to or below -2. Hence the range is {-1, 0, 1, 2}, making option A correct. Option B incorrectly includes -2, C is the domain, and D omits -1.

Correct answer: A. (1, 4]

The governing rules are that a real square root requires its radicand to be nonnegative, and a denominator cannot be zero. For 1/√(x − 1), the radicand condition is x − 1 ≥ 0, but because the square root is in the denominator, √(x − 1) must actually be positive; therefore x − 1 > 0, or x > 1. For √(4 − x), we need 4 − x ≥ 0, giving x ≤ 4. Intersecting these conditions gives 1 < x ≤ 4, or (1, 4]. Thus option A is correct. Option B wrongly includes x = 1, C wrongly excludes x = 4, and D ignores the second radical.

Correct answer: A. (-2, 4]

For the function to be real, the radicand (4 − x)/(x + 2) must be at least zero, and x = −2 must be excluded because it makes the denominator zero. The critical numbers are −2, where the expression is undefined, and 4, where the numerator is zero. Testing intervals: for x < −2, the numerator is positive and denominator negative, so the quotient is negative; for −2 < x < 4, both numerator and denominator are positive, so the quotient is positive; for x > 4, the numerator is negative and denominator positive, so it is negative. At x = 4 the quotient is zero and is allowed. Therefore the domain is (-2, 4], so option A is correct.

Correct answer: A. [-5, 2)

The radicand must satisfy (x + 5)/(2 − x) ≥ 0, and the denominator requires x ≠ 2. The critical values are x = −5, where the numerator is zero, and x = 2, where the expression is undefined. On x < −5, the numerator is negative and denominator positive, so the quotient is negative. On −5 < x < 2, both are positive, so the quotient is positive. For x > 2, the numerator is positive but denominator negative, giving a negative quotient. The value x = −5 is allowed because it produces a zero radicand; x = 2 is excluded. Hence the domain is [-5, 2), which is option A. B reverses the sign intervals, C includes an undefined point, and D excludes a valid endpoint.

Correct answer: A. [-4,3]

For a real-valued square root, its radicand must be nonnegative. Hence 12−x−x^2 ≥ 0. Rearranging gives x^2+x−12 ≤ 0, and factoring gives (x+4)(x−3) ≤ 0. A product of these two linear factors is nonpositive between the roots, including both roots because equality makes the radicand zero and √0 is defined. Therefore −4≤x≤3, so the domain is [-4,3], option A. Option C wrongly excludes the endpoints. Option B selects the outside intervals, where the radicand is negative, and option D includes values for which the square root is not real.

Correct answer: A. (-∞,-2) ∪ (4,∞)

Two restrictions must be applied because the square root is in the denominator. First, the radicand must be nonnegative for a real square root; second, the denominator cannot be zero. Together these require x^2−2x−8 > 0. Factoring gives (x−4)(x+2)>0. A product of two factors is positive outside its roots, so x<−2 or x>4. The endpoints x=−2 and x=4 are excluded because the radicand becomes zero and the denominator would be zero. Thus the domain is (-∞,-2)∪(4,∞), option A. Option B incorrectly includes the undefined endpoints, option C chooses the interval where the expression is negative, and option D omits only the roots while wrongly including points between them.

Correct answer: A. (-∞,-7] ∪ [3,∞)

The radicand of a real square root must be at least zero. Therefore |x+2|−5 ≥ 0, which is equivalent to |x+2| ≥ 5. For an absolute-value inequality of the form |u|≥a, we have u≤−a or u≥a. Hence x+2≤−5 or x+2≥5, giving x≤−7 or x≥3. The boundary values are included because the radicand equals zero there. Consequently the domain is (-∞,-7]∪[3,∞), option A. Option B and option C describe the central interval, where |x+2|<5 and the radicand is negative. Option D includes all real numbers, including values that do not produce a real square root.

Correct answer: A. (-2,6]

The domain is obtained by intersecting the restrictions from both terms. In the first term, √(x+2) is in the denominator, so its radicand must be strictly positive: x+2>0, giving x>−2. Equality is forbidden because it would make the denominator zero. In the second term, √(6−x) is an ordinary square root, so 6−x≥0, giving x≤6; equality is allowed because √0 exists. Combining these conditions produces −2<x≤6, or (-2,6]. Therefore option A is correct. Option B incorrectly includes −2, option C incorrectly excludes the valid endpoint 6, and option D ignores the first term’s restriction. The intersection, not a union, is required because both terms must be defined simultaneously.

Correct answer: A. {1, 2, 3}

The relevant idea is the range of the ceiling function on a restricted interval. Since 0<x≤1, multiplying by 3 gives 0<3x≤3. When 0<3x≤1, ⌈3x⌉=1; when 1<3x≤2, the ceiling is 2; and when 2<3x≤3, it is 3. All three outputs actually occur: for example, x=1/3 gives 1, x=1/2 gives 2, and x=1 gives 3. The value 0 never occurs because 3x is always positive, even when x approaches zero from the right. The output is always an integer, so the range is the set {1,2,3}, not a continuous interval. Therefore option A is correct. Option B incorrectly includes 0, option C omits 3, and option D describes an interval rather than the discrete output set.

Correct answer: A. [-3,2)

The numerator and denominator impose different square-root conditions. For the numerator, x+3≥0, so x≥−3; equality is permitted because a zero numerator is valid. For the denominator, √(2−x) must be real and cannot be zero. Therefore the stronger condition is 2−x>0, giving x<2. Intersecting x≥−3 with x<2 yields −3≤x<2, or [-3,2). Hence option A is correct. The endpoint −3 is included because the numerator becomes zero while the denominator remains √5, which is nonzero. The endpoint 2 is excluded because the denominator becomes zero. Option B wrongly includes 2, option C wrongly excludes −3, and option D includes values below −3 where the numerator is not real.

Correct answer: A. \(x\ge 4\)

For a real-valued function, the expression under the square root cannot be negative. Thus, \(x-4\ge 0\), which gives \(x\ge 4\). Therefore, the domain is \([4,\infty)\). Option B reverses the inequality, while option D incorrectly includes values that make the radicand negative. Exam tip: For a square-root function, set the radicand greater than or equal to zero to find the domain.

Correct answer: A. \([3,\infty)\)

For every real \(x\), \((x+1)^2\ge 0\), so \(f(x)=(x+1)^2+3\ge 3\). At \(x=-1\), the squared term is zero and \(f(-1)=3\), so 3 is included in the range. Therefore, the range is \([3,\infty)\). Exam tip: a function of the form \((x-a)^2+b\) has minimum value \(b\) and range \([b,\infty)\).

Correct answer: A. (−∞, −1]

Since (x − 4)² ≥ 0, we have −(x − 4)² − 1 ≤ −1. Equality occurs at x = 4, so −1 is the maximum value and is included. The range is therefore (−∞, −1].

Correct answer: A. 3

For the square root to be defined, \(x-1\ge 0\), so the domain is \(x\ge 1\). Since \(\sqrt{x-1}\ge 0\), we have \(2\sqrt{x-1}+3\ge 3\). At \(x=1\), \(f(1)=3\); hence the minimum value is 3. Exam tip: For an expression of the form \(a\sqrt{g(x)}+b\) with \(a>0\), check the minimum when \(\sqrt{g(x)}=0\).

Correct answer: A. ℝ \ {2}

The denominator must not be zero. Since (x − 2)² = 0 only when x = 2, every real number except 2 is allowed. Thus, the domain is ℝ \ {2}.

Correct answer: A. -1

When \\(x<-3\\), we have \\(2x+6<2(-3)+6=0\\). Therefore, the signum function has value \\(-1\\) for this negative input, so the correct answer is A. Exam tip: the signum function equals \\(-1,0,1\\) for negative, zero and positive inputs, respectively.

Correct answer: A. 0.3

Substituting \(x=2.3\), we get \(f(2.3)=\{2.3+1\}=\{3.3\}\). The fractional part of \(3.3\) is \(0.3\), obtained after removing its integer part, \(3\). Hence, option A is correct. Exam tip: use \(\{y\}=y-\lfloor y\rfloor\); therefore, \(\{3.3\}=3.3-3=0.3\). The value \(3.3\) itself is not the fractional part.

Correct answer: A. 2

Substituting \(x=3\), we get \(f(3)=|3|-|3-2|=3-1=2\). Therefore, the correct answer is 2. The distractor 4 results from adding \(|1|\) instead of subtracting it. Exam tip: first substitute the given value into every modulus, then evaluate the absolute values.

Correct answer: A. \([-2,1)\cup(1,2]\)

For the square root, \(4-x^2\ge0\), so \(-2\le x\le2\). But the denominator cannot be zero, hence \(x=1\) is excluded. Exam tip: apply radical and denominator restrictions together.

Correct answer: A. 3

Completing the square gives \(f(x)=x^2+4x+7=(x+2)^2+3\). Since \((x+2)^2\geq 0\), its least value is 0, attained at \(x=-2\). Therefore, the minimum value of the function is \(3\). Option 7 is merely the value at \(x=0\), not the minimum. Exam tip: For a quadratic function, complete the square or use its vertex to find the minimum or maximum value.

Correct answer: A. \(x=2\) and \(x=-2\)

The modulus of a quantity is zero only when the quantity itself is zero. Thus, \(|x^2-4|=0\) requires \(x^2-4=0\), or \(x^2=4\), giving \(x=2\) or \(x=-2\). Option B incorrectly uses 4 and -4, which would correspond to \(x^2=16\). Exam tip: The zeros of \(|g(x)|\) are exactly the zeros of \(g(x)\).