Real-Valued Functions, Their Domain and Range

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

The direct answer is option A, \([0,1)\). For every real number \(x\), we have \(x^2\ge0\). Therefore the numerator is non-negative and the denominator \(x^2+1\) is always positive, so the function cannot be negative. When \(x=0\), \(f(0)=0\), so zero is included. Also, \(x^2<x^2+1\), and division by the positive denominator gives \(\frac{x^2}{x^2+1}<1\). Thus the value is always at least 0 but always less than 1. The value 1 is never reached: solving \(\frac{x^2}{x^2+1}=1\) would give \(x^2=x^2+1\), which is impossible. Option A is correct. Option B, \((0,1]\), wrongly excludes the attainable value 0 and includes the unattainable value 1. Option C, \([0,1]\), wrongly includes 1. Option D, \(\mathbb R\), is impossible because the function never gives negative values or values at least 1. Memory cue: zero may be included when the numerator can be zero; check separately whether the endpoint 1 is attained.

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

The square root requires \(x\ge 0\), so the domain is \([0,\infty)\). At \(x=0\), \(f(0)=1\), so 1 is included in the range. As \(x\to\infty\), \(f(x)\to 0\), but the function never actually equals 0 for any finite \(x\). Hence the range is \((0,1]\). Exam tip: first impose the input restrictions, then check endpoint values and limiting values to determine the range.

Correct answer: A. ( (-\infty,1]\cup[3,\infty) )

Since (x^2-4x+3=(x-1)(x-3)), it must be (\ge0). For an upward quadratic, the outside intervals are selected.

Correct answer: A. (-4, 4)

For a real-valued expression, the radicand 16 − x² must be non-negative. However, the square root is in the denominator, so it cannot be zero; the required condition is therefore 16 − x² > 0. Solving gives x² < 16, or equivalently −4 < x < 4. Thus both endpoints are excluded: at x = 4 or x = −4 the denominator becomes zero, and division by zero is undefined. Every number strictly between −4 and 4 makes the denominator positive and the function real and defined. Hence option A, (−4, 4), is correct. Option B incorrectly includes the zero-denominator endpoints, while C describes the outside region and D excludes only the endpoints but includes values where the radicand is negative.

Correct answer: A. ( [1,3] )

Since (\cos x\in[-1,1]), (2+\cos x\in[1,3]), so (f(x)\in[1,3]). The order may reverse when taking reciprocals.

Correct answer: A. ( [3,7] )

Since (\sin x\in[-1,1]), (-2\sin x\in[-2,2]), giving total range ([3,7]). Apply multiplication and addition step by step.

Correct answer: A. ( \mathbb{R}\setminus{\frac{\pi}{2}+n\pi:n\in\mathbb{Z}} )

Since (\tan x=\frac{\sin x}{\cos x}), we need (\cos x\ne0). For (\tan x), odd multiples around (\frac{\pi}{2}) are excluded.

Correct answer: A. (0, ∞) \ {1}

The logarithm log x is defined for positive x, so the first restriction is x > 0. Since log x appears in the denominator, it must also be nonzero. For the usual logarithm base greater than 1, log x = 0 exactly when x = 1. Therefore both conditions together give x > 0 and x ≠ 1, or (0, ∞) \ {1}. Option A is correct. Option B forgets that the denominator cannot be zero. Option C incorrectly allows negative and zero inputs, where log x is not real or not defined. Option D keeps only x > 1 and wrongly removes the valid interval 0 < x < 1; on that interval the logarithm is negative but nonzero, so the reciprocal is perfectly defined.

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

The square root requires (\log_3(x-2)\ge0), giving (x-2\ge1). The condition (x-2>0) is already satisfied.

Correct answer: A. ( [1,\infty)\setminus{5} )

For (\sqrt{x-1}), (x\ge1), and for the denominator (x\ne5). The endpoint (x=1) is valid because the denominator is not zero.

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

The denominator ((x-3)^2+1\ge1), so the maximum is (\frac{1}{1}=1), and (0) is not attained. Completing the square gives the bounds clearly.

Correct answer: A. [0, 6]

Each radicand must be non-negative: x ≥ 0 and 6 − x ≥ 0, so x ≤ 6. Their intersection is [0, 6]. Both endpoints are included because a square root of zero is defined.

Correct answer: A. (2\sqrt{3})

By symmetry the maximum occurs at (x=3), and the value is (2\sqrt{3}). In such forms try making the two terms equal.

Correct answer: A. ( [4,25] )

On this interval, the minimum of (|x|) is (2) and the maximum is (5). On a negative domain (x^2) may decrease, but values remain positive.

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

From (y=\frac{x}{x^2+1}), we get (yx^2-x+y=0), and discriminant (\ge0) gives (|y|\le\frac{1}{2}). The discriminant method is useful for range.

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

The denominator is ((x+1)^2+1), and (f(x)=1+\frac{3}{(x+1)^2+1}). Hence values are greater than (1) and up to (\frac{5}{2}).

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

Since (|x|+1\ge1), the maximum is (1), and (0) is not attained. As the denominator grows without bound, the fraction approaches (0).

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

For \(f(x)=\sqrt{|x|-3}\) to be defined, we need
\(|x|-3\ge 0\Rightarrow |x|\ge 3\).

So \(x\le -3\) or \(x\ge 3\). Hence the domain is \(( -\infty,-3]\cup[3,\infty)\).

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

The difference of distances from two fixed points cannot exceed their distance (3). Therefore the range is ([-3,3]).

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

Since (\sqrt{x-2}) is in the denominator, (x>2), and (\sqrt{8-x}) gives (x\le8). Because of the denominator (2) is excluded but (8) remains.

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

For every real \(x\), \(1+x^2>0\), so the domain is \(\mathbb{R}\). Put \(t=x^2\ge0\); then \(f=\frac{t}{1+t}\). It can be 0 but can never equal 1, so the range is \([0,1)\). Exam tip: first check where the denominator is zero.

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

Here (f(x)=1-\frac{2}{x^2+1}), so the minimum is (-1), and the upper bound (1) is not attained. Rewriting the expression quickly gives the range.

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

For a square root to be real, each radicand must be non-negative. Thus, \(x-1\ge 0\) gives \(x\ge 1\), and \(x-4\ge 0\) gives \(x\ge 4\). The common solution is \(x\ge 4\), so the domain is \([4,\infty)\). Option B considers only the first square root, while option D incorrectly excludes \(x=4\), where both square roots are defined. Exam tip: for multiple square roots, take the intersection of all radicand conditions.

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

The domain is (x\ge4), and at (x=4) the value is (\sqrt{3}). Both terms increase, so this is the minimum.

Correct answer: A. 1/4

The key concept is completing the square and then using the fact that a positive fraction is largest when its denominator is smallest. Rewrite the denominator as x² + 2x + 5 = (x + 1)² + 4. Since (x + 1)² ≥ 0 for every real x, the denominator is at least 4, and its minimum value 4 occurs at x = −1. Consequently, f(x) = 1/[(x + 1)² + 4] ≤ 1/4, with equality at x = −1. Therefore the maximum value is 1/4, so option A is correct. Option B would result from incorrectly treating the constant 5 as the minimum denominator; options C and D are larger than the permitted reciprocal and do not follow from the denominator bound.