Real-Valued Functions, Their Domain and Range

Correct answer: A. (−1, 5]

The fraction under the square root must be non-negative, and x ≠ −1 because the denominator cannot be zero. A sign check at the critical points −1 and 5 shows the fraction is non-negative on (−1, 5]. The endpoint 5 is included because the fraction is zero there; −1 is excluded.

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

Since \(x^2\ge0\), we get \(x^2+1\ge1\) for every real \(x\). Hence the domain is \(\mathbb R\) and the range is \([1,\infty)\). Option B wrongly includes 0. Exam tip: a square root always denotes the non-negative principal value.

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

If (y=\frac{x}{x^2+1}), then the discriminant of (yx^2-x+y=0) must satisfy (1-4y^2\ge 0). Thus (-\frac{1}{2}\le y\le \frac{1}{2}).

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

Since \(x>0\), \(f(x)-2=x+\frac{1}{x}-2=\frac{(x-1)^2}{x}\ge 0\). Thus \(f(x)\ge 2\), with equality at \(x=1\), so 2 is included in the range. Also, \(f(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\), and every value \(y\ge2\) is attained. Therefore, the range is \([2,\infty)\). Exam tip: for \(x+\frac1x\) with \(x>0\), the minimum 2 can also be found directly using AM-GM. Option B is incorrect because the value 2 is attained at \(x=1\).

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

For negative (x), put (x=-t), (t>0), then (f=-(t+\frac{4}{t})\le -4). Hence the range is ((-\infty,-4]).

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

If (y=\frac{x-2}{x+5}), then (x=\frac{-5y-2}{y-1}), so (y\ne 1). Hence the range is (\mathbb{R}-{1}).

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

For real values, the expression inside each square root must be non-negative. From \(x+2\ge 0\), we get \(x\ge -2\), and from \(6-x\ge 0\), we get \(x\le 6\). Hence, the intersection of the two conditions is \([-2,6]\), so option A is correct. Option C wrongly excludes \(-2\) and \(6\); at these endpoints, one square root is zero, which is valid. Exam tip: for a function involving square roots, require every radicand to be \(\ge 0\) and intersect the resulting conditions.

Correct answer: C. \(4\)

For square roots, \(x+2\ge 0\) and \(6-x\ge 0\Rightarrow -2\le x\le 6\). The expression is symmetric: \(x+2\) and \(6-x\) add to \(8\). The maximum occurs when the two radicands are equal: \(x+2=6-x\Rightarrow 2x=4\Rightarrow x=2\). Then \(f(2)=\sqrt{2+2}+\sqrt{6-2}=\sqrt{4}+\sqrt{4}=2+2=4\).

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

Because the square root is in the denominator, its radicand must be strictly positive: \(4-x^2>0\). Thus \(x^2<4\), which gives \(-2<x<2\). Therefore, the domain is \((-2,2)\). Option B is incorrect because at \(x=\pm2\), the denominator becomes zero. Exam tip: when a square root appears in the denominator, impose the strict condition radicand \(>0\).

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

The maximum value of (\sqrt{4-x^2}) is (2), so the minimum value of (f) is (\frac{1}{2}). Near the endpoints, the value grows without bound.

Correct answer: A. [3, 7)

The domain is found by applying separate square-root and denominator conditions. The numerator √(x − 3) is real when x − 3 ≥ 0, so x ≥ 3. The denominator √(7 − x) must also be real, requiring 7 − x ≥ 0, but it cannot equal zero because it is a denominator. Hence the stronger condition is 7 − x > 0, or x < 7. Combining x ≥ 3 with x < 7 gives [3, 7). The left endpoint 3 is included because the numerator becomes √0 while the denominator is √4, which is nonzero. The right endpoint 7 is excluded because the denominator becomes zero. Therefore option A is correct; B reverses endpoint inclusion, C incorrectly includes a zero denominator, and D violates the radicand conditions.

Correct answer: A. (2)

Since (x^2-6x+11=(x-3)^2+2), the minimum value is (2). Completing the square is the fastest method here.

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

(f(x)=-2(x-2)^2+9), so the maximum value is (9). A downward-opening parabola has range ((-\infty,9]).

Correct answer: A. (1, 4]

Complete the square in the denominator: x² − 2x + 2 = (x − 1)² + 1. Therefore the denominator is at least 1 and is always positive. Let d = (x − 1)² + 1, so d ≥ 1. Then 3/d lies in (0, 3], because it equals 3 when d = 1, which occurs at x = 1, and it approaches 0 as |x| becomes large but never equals 0. Adding 1 shifts every value upward by 1, giving the range (1, 4]. The upper endpoint 4 is included at x = 1, while the lower endpoint 1 is excluded because the positive fraction can never be zero. Hence option A is correct. Option B incorrectly includes 1, C omits values between 1 and 4, and D includes impossible negative outputs.

Correct answer: A. ([\frac{2}{5},1))

Let (t=x^2\ge 0), then (f=\frac{t+2}{t+5}). It is (\frac{2}{5}) at (t=0) and approaches (1) but never equals (1).

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

The denominator of a rational function cannot be zero. Since \(x^2-1=(x-1)(x+1)\), the function is undefined at \(x=1\) and \(x=-1\). Hence its domain is \(\mathbb{R}\setminus\{-1,1\}\). In an exam, set the denominator equal to zero first and exclude every resulting value.

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

If (t=x^2\ge 0) and (t\ne 1), then (y=\frac{t+1}{t-1}). Using (t=\frac{y+1}{y-1}\ge 0) gives the range ((-\infty,-1]\cup(1,\infty)).

Correct answer: A. [0, 2]

For the square root to be real, 2x − x² ≥ 0. Factoring gives x(2 − x) ≥ 0, which holds for 0 ≤ x ≤ 2. Therefore, the domain is [0, 2], including both endpoints where the radicand is zero.

Correct answer: A. [0,1]

Rewrite the radicand as \(2x-x^2=1-(x-1)^2\). For the square root to be defined, \(2x-x^2\ge 0\), giving the domain \([0,2]\). The radicand has maximum value \(1\) at \(x=1\), and minimum value \(0\) at \(x=0,2\). Therefore, the function values range from \(0\) to \(1\), so the range is \([0,1]\). Option B is incorrect because a square root cannot be negative, while option D wrongly excludes \(0\). Exam tip: For a square-root function, first find the minimum and maximum values of the radicand.

Correct answer: A. (\mathbb{R}-{-1,1})

Both denominators must be non-zero, so (x\ne 1) and (x\ne -1). In sums of rational functions, check every denominator.

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

The expression inside the square root is \(x^2+6x+10=(x+3)^2+1\), whose minimum value is 1. Hence \(\sqrt{x^2+6x+10}\ge 1\), so \(0<f(x)\le 1\). The value 1 is attained at x=-3, but f(x) can never be 0; therefore, the range is \((0,1]\). Exam tip: For reciprocal square-root functions, first find the minimum value of the expression under the radical.

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

The denominator \(\sqrt{x^2+1}\) is always positive because \(x^2+1\ge1\). Also, \(x^2+1>x^2\), so taking positive square roots gives \(\sqrt{x^2+1}>|x|\). Consequently, the absolute value of the fraction is strictly less than 1. This places every function value strictly between \(-1\) and \(1\).

More precisely, \(|f(x)|=|x|/\sqrt{x^2+1}<1\), so \(-1<f(x)<1\). The endpoints cannot occur for any finite real \(x\), although the values approach them as \(x\) becomes very large positive or negative. Every number strictly between these endpoints is attained, for example by solving the equation for \(x\). Hence the range is \((-1,1)\), option A.

Correct answer: A. [0, 1]

Both radicands must be non-negative: x ≥ 0 and 1 − x ≥ 0, so x ≤ 1. The intersection is [0, 1]. The endpoints are included because a square root of zero is defined.

Correct answer: A. ([1,\sqrt{2}])

On ([0,1]), the minimum at endpoints is (1) and the maximum at (x=\frac{1}{2}) is (\sqrt{2}). Hence the range is ([1,\sqrt{2}]).

Correct answer: A. ([2,\infty)-{3})

For the square root (x\ge 2), and for the denominator (x\ne -3,3) are required. But (-3) is already outside (x\ge 2), so only (3) is removed.