Real-Valued Functions, Their Domain and Range

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

A logarithmic function has all real numbers as range when its argument can take all positive values. The base (\frac{1}{3}) only changes direction, not the range.

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

Put (t=\sqrt{x-1}\ge 0), then (f=\frac{t}{t+4}). At (t=0), the value is (0), and as (t\to\infty), it approaches (1).

Correct answer: A. \(\mathbb{R}\setminus\{3,5\}\)

The denominator of a rational function cannot be zero. Thus, \(|x-4|-1\ne0\), or \(|x-4|\ne1\). Hence, \(x-4=1\) or \(x-4=-1\), giving \(x=5\) or \(x=3\). Therefore, the domain is all real numbers except 3 and 5: \(\mathbb{R}\setminus\{3,5\}\). Option B incorrectly excludes 4; at \(x=4\), the denominator is \(-1\), so the function is defined. Exam tip: for a rational function, exclude every input that makes the denominator zero.

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

The original denominator is (x-4), so (x=4) is not allowed. Even after simplification, do not add the removed point to the domain.

Correct answer: A. ( \mathbb{R}\setminus{8} )

First factor the numerator: \(x^2-16=(x-4)(x+4)\). Since the original denominator is \(x-4\), the original function is defined only when \(x\ne4\). For every allowed input, cancellation gives \(f(x)=x+4\). Thus its graph is the line \(y=x+4\), but with the point corresponding to \(x=4\) removed.

If the missing input \(x=4\) were allowed, the missing output would be \(4+4=8\). Could another input produce output 8? The equation \(x+4=8\) gives only \(x=4\), which is forbidden. Every other real output is produced by \(x=y-4\), and that input is not 4 when \(y\ne8\). Therefore the range is \(\mathbb R\setminus\{8\}\), option A.

Correct answer: A. 25

The denominator must be nonzero for the function to be defined. Setting the denominator equal to zero gives \(x^2-k=0\), or \(x^2=k\). Since the excluded values are \(x=-5\) and \(x=5\), we get \(k=5^2=25\). Option 5 is incorrect because it would exclude \(x=\pm\sqrt{5}\). Exam tip: For a rational function, exclude every value of \(x\) that makes the denominator zero.

Correct answer: A. \(4\)

Completing the square gives \(f(x)=x^2-8x+20=(x-4)^2+4\). Since \((x-4)^2\geq 0\), the minimum value of the function is \(4\), attained at \(x=4\). Hence the range is \([4,\infty)\), so \(m=4\). In an exam, rewrite a quadratic as \((x-h)^2+k\) to identify its minimum value quickly; \(8\) is only related to the coefficient of x, not the minimum.

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

Here (f(x)=-(x-3)^2+4), so the maximum is (4). A downward-opening parabola has range ((-\infty,\text{maximum}]).

Correct answer: A. ( (0,\frac{1}{4}] )

The denominator ((x+2)^2+4\ge 4), so the maximum value is (\frac{1}{4}). As the denominator grows, the value approaches (0).

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

From (y=\frac{x^2}{x^2+4x+8}), real (x) is needed in ((y-1)x^2+4yx+8y=0). The discriminant gives (0\le y\le 2), and (y=1) is also possible, so the range is ([0,2]).

Correct answer: A. ( [0,2] )

From (y=\frac{x^2+4x+8}{x^2+4}), real (x) is required in ((y-1)x^2-4x+4y-8=0). The discriminant condition (16-4(y-1)(4y-8)\ge 0) gives (0\le y\le 2).

Correct answer: A. [2, 7]

For both square roots to be real, x − 2 ≥ 0 and 7 − x ≥ 0. Together these give 2 ≤ x ≤ 7, so the domain is [2, 7]. Both endpoints are valid because one radicand is zero at each endpoint.

Correct answer: A. ( [\sqrt{5},\sqrt{10}] )

At the endpoints the value is (\sqrt{5}), and at the midpoint (x=\frac{9}{2}), the maximum is (\sqrt{10}). In symmetric questions, always check the midpoint.

Correct answer: A. \([-5,5]\setminus\{3\}\)

The expression under the square root must be non-negative, so \(25-x^2\ge 0\), which gives \(-5\le x\le 5\). Thus the initial domain is \([-5,5]\). Also, the denominator \(x-3\) cannot be zero, so \(x=3\) must be excluded. Therefore, the domain is \([-5,5]\setminus\{3\}\). Option B is incorrect because the endpoints \(-5\) and \(5\) are allowed. Exam tip: impose \(\text{radicand}\ge 0\) and denominator \(\ne 0\).

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

In the given interval, (\sin x) attains both maximum (1) and minimum (-1). For trigonometric ranges, check key angles.

Correct answer: A. \( \mathbb{R}\setminus\left{\frac{3\pi}{2}+2n\pi:n\in\mathbb{Z}\right} \)

The denominator must not be zero, so (1+\sin x\ne 0). When \(\sin x=-1\), \(x=\frac{3\pi}{2}+2n\pi\) is excluded.

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

The direct answer is option A: the range is [1,3]. Start with the basic fact -1\le\sin x\le1 for every real x. Adding 2 throughout gives 1\le2+\sin x\le3. Thus f(x) can never be below 1 or above 3, and both endpoints occur: sin x=-1 gives f(x)=1, while sin x=1 gives f(x)=3. Option A is correct. Option B is only the range of sin x and forgets the added 2. Option C misses the value 1 and is too narrow. Option D claims every real value is possible, but sine is bounded. Exam cue: adding a number shifts both ends of a range by that number.

Correct answer: A. \([0,\pi]\)

For a real-valued square root, we require \(\sin x\ge 0\). Within \([0,2\pi]\), sine is non-negative on \([0,\pi]\). The endpoints are included because \(\sin 0=\sin\pi=0\). Exam tip: for a square-root function, first impose that the expression inside the root is at least zero, then intersect the result with the given interval.

Correct answer: A. \( \left[0,\frac{\pi}{2}\right]\cup\left[\frac{3\pi}{2},2\pi\right] \)

The direct answer is A: the domain is \\( [0,\pi/2]\cup[3\pi/2,2\pi] \\). For a real square root, the quantity inside must satisfy \\(\cos x\ge0\\). On the unit circle, cosine is the x-coordinate: it is non-negative in the first and fourth quadrants. Within \\([0,2\pi]\\), these are \\([0,\pi/2]\\) and \\([3\pi/2,2\pi]\\), including endpoints where cosine is 0. A is correct. B describes the second and third quadrants, where cosine is non-positive and is negative except at endpoints, so the square root is generally not real. C includes [0,π], but cosine is negative between π/2 and π. D includes the entire interval and therefore includes points such as π where \\(\cos\pi=-1\\). Memory cue: for \\(\sqrt{\cos x}\\), keep the right half of the unit circle where cosine is non-negative.

Correct answer: A. ( {1,2,3,4} )

Here (x+2\in[1,5)), so the greatest integer values are (1,2,3,4). The value (5) does not occur because the upper endpoint is open.

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

The fractional part of any real number is at least (0) and less than (1). When (2x) is an integer, the value is (0).

Correct answer: A. ( (0,\frac{1}{3}] )

Since (\sqrt{x^2+1}\ge 1), the denominator is at least (3). The maximum is (\frac{1}{3}), and the value approaches (0).

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

The denominator is always greater than \(|x|\), so \(\left|\frac{x}{\sqrt{x^2+9}}\right|<1\). For large \(|x|\), the value approaches \(\pm 1\) but never equals them.

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

The denominator is (\sqrt{(x-2)^2+4}), which is greater than (|x-2|). Hence the ratio stays strictly between (-1) and (1).

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

Here (f(x)=\frac{x}{3}+\frac{3}{x}). For positive (x), the minimum is (2), and for negative (x), the maximum is (-2).