Real-Valued Functions, Their Domain and Range

Correct answer: A. ( [0,4) )

For (\sqrt{x}), (x\ge 0), and for the denominator square root, (4-x>0) is needed. Together, (0\le x<4).

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

The direct answer is option A, \(\mathbb R\setminus\{2\}\). The domain contains all real inputs for which the formula is defined. A fraction is undefined when its denominator is zero. Here the original denominator is \(x-2\), so we require \(x-2\ne0\), which means \(x\ne2\). Thus every real number except 2 belongs to the domain. Although \(x^2-4=(x-2)(x+2)\), cancelling gives \(f(x)=x+2\) only for \(x\ne2\). At \(x=2\), the original expression is \(0/0\), undefined; cancellation cannot restore that missing input. Option A is correct. Option B incorrectly includes 2. Option C incorrectly removes -2 as well, even though the original denominator at -2 is -4 and is non-zero. Option D includes only numbers less than 2 and wrongly omits all numbers greater than 2. Memory cue: for a rational function, inspect the original denominator before cancelling factors.

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

For a real square root, the expression under the radical must be non-negative. Hence \(|x|-3\ge 0\), so \(|x|\ge 3\). Therefore, \(x\le -3\) or \(x\ge 3\), giving the domain \(( -\infty,-3]\cup[3,\infty)\). Option B wrongly includes values between \(-3\) and \(3\), where \(|x|-3<0\). Exam tip: For real functions involving a square root, first set the radicand greater than or equal to zero.

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

The denominator must not be zero. Solving \(|x|-2=0\) gives \(|x|=2\), so \(x=-2\) or \(x=2\) makes the function undefined. All other real numbers are allowed; hence the domain is \(\mathbb{R}\setminus\{-2,2\}\).

Correct answer: A. \([0,1)\)

Let \(t=|x|\). Then \(t\ge 0\) and \(f(x)=\frac{t}{1+t}\). At \(x=0\), we get \(t=0\) and hence \(f(x)=0\). For every finite \(t\), \(\frac{t}{1+t}<1\), so the value 1 is never attained, although the function approaches 1 as \(|x|\) increases. Therefore, the range is \([0,1)\). Option B is incorrect because 0 is included, while option C is incorrect because 1 is excluded. Exam tip: for rational functions, check both limiting values and whether each endpoint is actually attained.

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

For positive (x), the value is at least (2), and for negative (x), it is at most (-2). Check both cases separately.

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

The direct answer is A: [0,∞). First simplify the expression: x^2-4x+4=(x-2)^2, so f(x)=√((x-2)^2)=|x-2|. An absolute value can never be negative. Its smallest value is 0, reached when x=2. As x moves away from 2, |x-2| can become any positive number; for example, choosing x=2+t gives f(x)=t for every t≥0. Therefore the range is [0,∞). Option A is correct because it includes zero and every non-negative real value. Option B, (0,∞), wrongly excludes zero, even though f(2)=0. Option C, [2,∞), wrongly says the minimum is 2; the actual minimum is 0. Option D, all real numbers, is wrong because the square-root output is never negative. Remember that √(u^2)=|u|, not simply u, and an absolute value has a V-shaped graph with lowest point on the x-axis. Exam cue: whenever a square root contains a perfect square, simplify it using an absolute value and then identify the least possible output.

Correct answer: A. ( \left[\frac{1}{4},\infty\right) )

The maximum of (\sqrt{16-x^2}) is (4), so the minimum fraction is (\frac{1}{4}). Near the endpoints, the denominator tends to (0).

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

The greatest integer function gives the largest integer not greater than (x). Since (4) is not included, the value (4) will not occur.

Correct answer: A. ( [0,1) )

The fractional part is always at least (0) and less than (1). At integer (x), the value is (0).

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

The expression inside the square root must satisfy (\frac{x-1}{x+2}\ge 0) and (x\ne -2). A sign chart gives (x<-2) or (x\ge 1).

Correct answer: A. \([-3,5)\)

For a real square root, the radicand must be non-negative, and its denominator must be nonzero. The critical points are \(-3\), where the numerator is zero, and \(5\), where the denominator is zero. Sign analysis gives \(\frac{x+3}{5-x}\ge 0\) for \(-3\le x<5\). Thus \(-3\) is included, but \(5\) is excluded, so the domain is \([-3,5)\).

Correct answer: A. 7

The expression under a square root must be non-negative, so \(x-a\ge 0\), which gives \(x\ge a\). Therefore, the domain of \(f(x)=\sqrt{x-a}\) is \([a,\infty)\). Comparing it with the given domain \([7,\infty)\), we get \(a=7\). Choosing \(-7\) would give the domain \([-7,\infty)\), so it is incorrect. Exam tip: The domain of \(\sqrt{x-c}\) is directly \([c,\infty)\).

Correct answer: A. 5

The denominator \(x-a\) must not be zero. Thus, the only excluded value from the domain is obtained from \(x-a=0\), which gives \(x=a\). Since the excluded value is \(5\), we have \(a=5\). Choosing \(-5\) would exclude \(-5\), not \(5\). Exam tip: For a reciprocal expression, set its denominator equal to zero to identify the value excluded from the domain.

Correct answer: A. ( (1,4] )

Put (t=(x+1)^2\ge 0), then (f=\frac{t+4}{t+1}=1+\frac{3}{t+1}). Hence the values are greater than (1) and up to (4).

Correct answer: A. ( [3,4] )

The direct answer is A: [3,4]. For a square root to be real, its inside must be non-negative: 2x−x²≥0. Rewrite it as 1−(x−1)². Since (x−1)² is at least 0, the expression 1−(x−1)² has maximum 1. On its allowed domain, it has minimum 0, so 0≤2x−x²≤1. Taking square roots gives 0≤√(2x−x²)≤1. Adding 3 throughout gives 3≤f(x)≤4. Thus the range is [3,4]. Option A is correct. Option B [0,1] is the range of the square-root part before adding 3. Option C [2,4] incorrectly includes values below 3. Option D [3,∞) incorrectly allows values larger than 4, although the square-root part never exceeds 1. Memory cue: find the range inside the root first, then apply the outside operations.

Correct answer: A. \(1\)

Complete the square in the denominator: \(x^2-6x+10=(x-3)^2+1\). Since \((x-3)^2\ge 0\), the denominator has minimum value \(1\), attained at \(x=3\). Therefore, the maximum value of \(f(x)\) is \(\frac{1}{1}=1\). For instance, \(\frac{1}{10}\) is not the maximum; it occurs only for particular values of \(x\). Exam tip: when the denominator is positive, the reciprocal is maximized when the denominator is minimized.

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

For both square roots to be real, each radicand must be non-negative. From \(x+4\ge 0\), we get \(x\ge -4\), and from \(4-x\ge 0\), we get \(x\le 4\). Their intersection is \([-4,4]\), so option A is correct. Option B wrongly excludes \(-4\) and \(4\); at each endpoint, one square root is zero and the function is still defined. Exam tip: For several square roots, write a non-negative condition for every radicand and intersect the resulting sets.

Correct answer: A. ( [2\sqrt{2},4] )

At the endpoints the value is (2\sqrt{2}), and at (x=0) the maximum is (4). For symmetric radical functions, also check the midpoint.

Correct answer: A. (−2, 2)

The argument of a real logarithm must be positive, so 4 − x² > 0. This gives x² < 4, or −2 < x < 2. The endpoints are excluded because the logarithm's argument would be zero.

Correct answer: A. (1, 5)

Each logarithm's argument must be positive. The conditions are x − 1 > 0, so x > 1, and 5 − x > 0, so x < 5. Their intersection is (1, 5); the endpoints are excluded because an argument would be zero.

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

For every real \(x\), \(|2x-1|\ge 0\), so \(f(x)\ge 3\). At \(x=\frac{1}{2}\), the absolute-value term is zero and \(f(x)=3\), so the minimum is attained. As \(|2x-1|\) can grow without bound, the range is \([3,\infty)\).

Correct answer: A. (−∞, −1] ∪ [1, ∞)

The radicand must be non-negative: x² − 1 ≥ 0. Hence x² ≥ 1, so x ≤ −1 or x ≥ 1. Therefore, the domain is (−∞, −1] ∪ [1, ∞), including the endpoints where the radicand is zero.

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

The direct answer is A: [0,∞). A square root can never be negative. First require x²−1≥0, so x≤−1 or x≥1; this is the domain. At x=1 or x=−1, f(x)=√0=0, so 0 is included. For any y≥0, choose x=√(y²+1), which belongs to the domain; then √(x²−1)=√y²=y. Therefore every non-negative real number occurs, and no negative number can occur. Option A is correct. Option B [1,∞) wrongly excludes the value 0 and all values between 0 and 1. Option C includes negative outputs, impossible for a square root. Option D contains only non-positive values and misses all positive outputs. Memory cue: √ means the principal, non-negative root; check both domain and range.

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

(|4x|\le x^2+4) because ((x-2)^2\ge 0) and ((x+2)^2\ge 0). Hence values lie from (-1) to (1), and both endpoints occur.