Real-Valued Functions, Their Domain and Range

Correct answer: B. ({7})

The direct answer is B: the range is the singleton set {7}. A range contains the values that the function actually produces. Here f(x)=7 for every allowed input x, so changing x never changes the output. For example, f(0)=7, f(2)=7, and f(−10)=7. Hence the only output is 7, written as {7}. Option A, \(\mathbb R\), is wrong because the function does not produce every real number. Option B is correct because it lists exactly the one attained value. Option C, [7,∞), is wrong because values such as 8 and 20 are not produced. Option D, (−∞,7], is also wrong because values below 7 are not produced. The domain may contain many numbers, but the range can contain only one number for a constant function. Memory cue: constant function means horizontal line and one output, so write the range in braces as {constant}.

Correct answer: B. \(\mathbb{R}\setminus\{-4\}\)

The function \(f(x)=\frac{2}{x+4}\) is defined for all real \(x\) except when the denominator is zero.

So we must have \(x+4\neq 0\Rightarrow x\neq -4\). Hence the domain is \(\mathbb{R}\setminus\{-4\}\).

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

For the square root, (x+6 \ge 0), so (x \ge -6). In exams the endpoint is included because square root is defined at (0).

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

Since (\sqrt{x} \ge 0), (\sqrt{x}+3 \ge 3). In exams adding (3) shifts the minimum value to (3).

Correct answer: A. R, अर्थात सभी वास्तविक संख्याएँ

The governing concept is that the domain of a real-valued rational function contains every real input for which its denominator is non-zero. Here the denominator is x² + 1. Since x² is always at least 0 for every real x, x² + 1 is always at least 1. Therefore it can never equal zero, so no real number must be excluded from the domain. Hence the domain is R, the set of all real numbers, and option A is correct. Options B and C incorrectly exclude 1 or −1, even though their denominators are 2. Option D includes only positive numbers, although zero and negative numbers are also valid inputs.

Correct answer: B. 5

Substituting \(x=3\) into the function rule gives \(f(3)=2(3)-1=6-1=5\). Therefore, option B is correct. Option C results from forgetting to subtract 1 after calculating \(2\times3\). Exam tip: substitute the given input directly into the function rule and simplify carefully.

Correct answer: B. ({2,3,6})

For inputs (0,1,2), the outputs are (2,3,6). In exams, for a finite domain, calculate the value for each input.

Correct answer: C. (2)

The denominator (x-2) must not be (0), so (x \ne 2). In exams check the denominator first in fractional functions.

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

For a real-valued square-root function, the expression inside the radical must be non-negative. Thus, \(2x+8\geq 0\), so \(2x\geq -8\) and \(x\geq -4\). Therefore, the domain is \([-4,\infty)\). Option B incorrectly includes values less than \(-4\); for example, at \(x=-5\), \(2x+8=-2\), which has no real square root. Exam tip: for a real square-root function, set the radicand greater than or equal to zero.

Correct answer: C. (\mathbb{R})

(5-x) is a linear function with non-zero slope, so it can give all real values. In exams a negative slope still gives range (\mathbb{R}).

Correct answer: A. ([1,\infty))

Since (|x-2| \ge 0), (|x-2|+1 \ge 1). In exams check the minimum of modulus where it becomes zero.

Correct answer: A. (\mathbb{R}\setminus{-3,3})

The direct answer is option A: ℝ\{−3,3}. The domain contains every real input for which the expression is defined. In f(x)=1/(x^2−9), the denominator must not be zero because division by zero is undefined. Set x^2−9=0. Then x^2=9, so x=3 or x=−3. These two real numbers must be excluded; every other real number is allowed. Therefore the domain is ℝ\{−3,3}. Option A is correct. Option B excludes only 9, but 9 is not the value of x that makes the denominator zero; at x=9 the denominator is 81−9=72. Option C, (−3,3), includes x=0, where the function is defined, but excludes many valid values outside the interval, so it is too small. Option D, [3,∞), incorrectly keeps only numbers at least 3 and also includes x=3, which is forbidden. Factor x^2−9=(x−3)(x+3) as a quick exam check: exclude the roots of the denominator.

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

For the square root, (10-2x \ge 0), so (x \le 5). In exams dividing by a negative coefficient reverses the inequality.

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

The function \(f(x)=\frac{1}{x}\) can never have the value \(0\), since \(\frac{1}{x}=0\) has no real solution for \(x\neq 0\). Also, for every nonzero real number \(y\), taking \(x=\frac{1}{y}\) gives \(f(x)=y\). Hence, the range is \(\mathbb{R}\setminus\{0\}\). Option \(\mathbb{R}\) is incorrect because it includes \(0\). Exam tip: For the reciprocal function \(\frac{1}{x}\), both domain and range exclude \(0\).

Correct answer: A. {-4, 0, 8}

The governing concept is that the range is the set of output values produced by applying the function to every member of the stated domain. For x = −1, f(−1) = 4(−1) = −4. For x = 0, f(0) = 0. For x = 2, f(2) = 8. Collecting these distinct outputs gives the range {−4, 0, 8}. Therefore option A is correct. Option B merely repeats the domain rather than calculating outputs. Option C uses 4 for the image of −1, which ignores the negative sign, and option D incorrectly gives 4 as the image of 0. Each input has been checked, so no further value belongs in the range.

Correct answer: B. ([0,\infty))

(\sqrt{x^2}=|x|), so the output is never negative. In exams treat (\sqrt{x^2}) as (|x|), not simply (x).

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

The square root is in the denominator, so (4-x>0), giving (x<4). In exams do not include equality for a denominator square root.

Correct answer: A. [7, ∞)

The governing concept is the real-existence condition for a square root: its radicand must be non-negative. In this function the radicand is x − 7, so we require x − 7 ≥ 0. Adding 7 to both sides gives x ≥ 7. Thus every real number from 7 onward is allowed, including 7 itself because √0 is defined. The added constant +2 is outside the radical; it changes output values but does not impose any new restriction on x. Therefore the domain is [7, ∞), making option A correct. Option B confuses the outside constant with the input restriction, option C reverses the inequality, and option D incorrectly includes values less than 7.

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

The minimum value of (\sqrt{x-7}) is (0), so (f(x)\ge 2). In exams the outside added (2) shifts the range upward.

Correct answer: B. (\mathbb{R}\setminus{-1})

The direct answer is option B: ℝ\{−1}. The domain is the set of inputs for which the original expression is meaningful. In f(x)=(x+1)/(x+1), the denominator x+1 cannot equal zero. Solving x+1=0 gives x=−1, so −1 must be excluded. For every other real x, the fraction is defined and simplifies numerically to 1, although the original function still has a missing point at −1. Therefore the domain is ℝ\{−1}. Option A, ℝ, is wrong because it includes −1, where the denominator is zero. Option B is correct because it removes exactly that forbidden input. Option C, {1}, is wrong because it gives only one input; many other real inputs are allowed. Option D, (−1,∞), is wrong because values less than −1, such as −2, are also valid; only −1 is forbidden. A crucial exam warning is never to cancel the common factor before recording the original denominator restriction. The simplified expression is 1, but the original function is not defined at −1.

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

The direct answer is A: the range is \([3,\infty)\). Start with the basic fact \(|x|\ge0\) for every real x. Multiplying by 2 gives \(2|x|\ge0\), and adding 3 gives \(2|x|+3\ge3\). Thus no output is below 3. The minimum 3 is reached at x=0, because f(0)=2|0|+3=3. As |x| can take every non-negative value, 2|x| can take every non-negative value, so all values from 3 upward are attained. Option A is correct. Option B, [0,∞), has an incorrect lower endpoint; outputs between 0 and 3 cannot occur. Option C, (−∞,3], gives the wrong direction and includes impossible negative values. Option D, all real numbers, is wrong for the same reason: the output is never less than 3. The square brackets matter because 3 is actually obtained. Memory cue: absolute value is never negative; positive scaling keeps that fact, and +3 moves the minimum to 3.

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

The correct answer is option A: [4,∞). A square root is real only when its radicand is non-negative. Therefore require 3x-12≥0. Add 12 to both sides: 3x≥12. Divide by positive 3, so x≥4. Equality is allowed because √0=0, so x=4 belongs to the domain. Hence the domain is [4,∞). Option A correctly includes 4 and all larger values. Option B, (-∞,4], reverses the direction and would include values such as x=0, for which the radicand is -12 and the square root is not real. Option C, (4,∞), wrongly excludes x=4 even though f(4)=0 is defined. Option D, ℝ\{4}, wrongly includes values below 4 and excludes the valid endpoint 4. No denominator restriction exists here; the only restriction comes from the square root. Memory cue: for √(expression), set the expression ≥0; include the zero endpoint.

Correct answer: B. ( (-\infty,9])

Since ((x-1)^2 \ge 0), (9-(x-1)^2 \le 9). In exams identify the maximum value of a downward parabola.

Correct answer: B. [2, 5]

The governing concept is the image of a closed interval under a linear function. Since f(x) = x + 2 has positive slope 1, it is increasing throughout the domain [0, 3]. Thus the least output is obtained at the left endpoint: f(0) = 0 + 2 = 2. The greatest output is obtained at the right endpoint: f(3) = 3 + 2 = 5. Because both endpoints are included in the domain and the function is defined at them, the endpoint values 2 and 5 are included in the range. Therefore the range is [2, 5], making option B correct. Option A ignores the addition of 2, option C incorrectly excludes the endpoints, and option D uses an incorrect lower endpoint.

Correct answer: A. ((-2,8))

The positive multiplier (2) changes the open interval ((-1,4)) to ((-2,8)). In exams keep open endpoints open.