Real-Valued Functions, Their Domain and Range

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

Put (t=x^2\ge 0) and (t\ne 4). Then (f=\frac{t}{t-4}), giving ((-\infty,0]) for (t\in[0,4)) and ((1,\infty)) for (t>4).

Correct answer: A. (2, 7)

The argument of the natural logarithm must be positive. The fraction (x − 2)/(7 − x) is positive only when 2 < x < 7, as a sign check at the critical points 2 and 7 shows. The endpoints are excluded because the argument is zero at 2 and undefined at 7.

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

Put (t=\sqrt{x+1}\ge 0), then (f=\frac{t}{t+2}). At (t=0), the value is (0), and as (t\to\infty), it approaches (1) but never equals (1).

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

For a real square root, the expression inside the root must be non-negative. Thus, \(1-|x-2|\ge 0\), so \(|x-2|\le 1\). This gives \(-1\le x-2\le 1\), hence \(1\le x\le 3\). Therefore, the domain is \([1,3]\). Option C excludes the endpoints, but at \(x=1\) and \(x=3\), the radicand is 0, so both values are allowed. Exam tip: For a square-root function, require the radicand to be \(\ge 0\).

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

The expression inside the square root must satisfy (\frac{x-3}{x+1}\ge 0) and (x\ne -1). A sign chart gives (x<-1) or (x\ge 3).

Correct answer: C. ( \mathbb{R}\setminus{5} )

From (y=\frac{5x-1}{x+2}), (x=\frac{-1-2y}{y-5}), so (y\ne 5). In a linear fractional function, the (y) making the denominator zero is excluded.

Correct answer: A. [−3, 1]

For the square root to be real, 2 − |x + 1| must be non-negative. Therefore |x + 1| ≤ 2, which gives −3 ≤ x ≤ 1. Both endpoints are included because the radicand is zero at each one.

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

Since \\((x-2)^2\\ge 0\\), we have \\(\\sqrt{(x-2)^2+9}\\ge 3\\). Therefore, \\(f(x)=7-\\sqrt{(x-2)^2+9}\\le 7-3=4\\). The maximum value 4 is attained at \\(x=2\\). As \\(|x|\\to\\infty\\), the square-root term tends to infinity, so \\(f(x)\\to-\\infty\\). Hence the range is \\((-\\infty,4]\\). Exam tip: first find the minimum of the radical expression; because it is subtracted, that minimum gives the maximum value of the function.

Correct answer: C. \(( -\infty,-4)\cup(1,\infty)\)

For a logarithmic function, its argument must be strictly positive. Hence, \(\frac{x+4}{x-1}>0\) is required. The critical values are \(x=-4\) and \(x=1\). A sign check shows that the fraction is positive for \(x<-4\) and for \(x>1\). At \(x=-4\), the argument becomes 0, which is not allowed for a logarithm, and at \(x=1\), the fraction is undefined. Therefore, the domain is \(( -\infty,-4)\cup(1,\infty)\). Exam tip: In logarithmic questions, first impose that the complete argument is greater than zero.

Correct answer: A. ( (0,\frac{1}{2}] )

The denominator ((x+3)^2+4\ge 4), so the maximum is (\frac{1}{2}). As the denominator grows, the value approaches (0) but never becomes (0).

Correct answer: A. \( [0,\frac{7}{2}] \)

\((x+2)(5-x)=\frac{49}{4}-\left(x-\frac{3}{2}\right)^2\) has maximum \(\frac{49}{4}\). The square root gives maximum \(\frac{7}{2}\) and minimum (0).

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

Put (t=|x-1|\ge 0), then (f=\frac{t}{t+3}). At (t=0), the value is (0), and as (t\to\infty), it approaches (1).

Correct answer: B. ( [4,13] )

The direct answer is B: the range is \\( [4,13] \\). Complete the square: \\(f(x)=x^2-4x+8=(x-2)^2+4\\). Since a square is at least zero, the minimum is 4, attained at x=2, which lies in [0,5]. For the maximum on this interval, check the endpoints: \\(f(0)=8\\) and \\(f(5)=25-20+8=13\\). Therefore the largest value is 13, and the continuous parabola takes every value between 4 and 13. B is correct. A stops at 8 and misses values such as 10 and 13. C wrongly includes values below the minimum 4. D starts at 8 and misses the minimum and intermediate values. The vertex is inside the interval, so it must be checked; then endpoints determine the maximum. Memory cue: rewrite a quadratic as \\( (x-h)^2+k\\) to see its minimum.

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

The direct answer is A: (1,3]. Put t=x². Since x is real, t≥0. Then f(x)=(t+3)/(t+1)=1+2/(t+1). Because t≥0, t+1≥1, so 0<2/(t+1)≤2. Hence 1<f(x)≤3. The value 3 occurs at t=0, that is, x=0. The value 1 is never reached because 2/(t+1) is always positive, although it approaches 0 as |x| becomes very large. Option A is correct. Option B includes 1, which is impossible. Option C wrongly says values begin at 3 and extend upward. Option D contains values below 1, but the expression is always greater than 1. Memory cue: when a positive fraction is added to 1, the result is greater than 1; inspect whether the endpoint is included.

Correct answer: B. ( (5,\infty) )

The direct answer is B: (5,∞). A square root requires its radicand to be non-negative, while a denominator can never be zero. For √(x-2), we need x-2≥0, so x≥2. For 1/√(x-5), the expression under the root must be positive, not merely non-negative, because the root is in the denominator: x-5>0, so x>5. Both conditions must hold at once. The stronger condition x>5 already guarantees x≥2, so the domain is (5,∞). Option A, [2,∞), incorrectly includes values from 2 through 5, where the denominator is either undefined or zero. Option B is correct because every x greater than 5 makes both terms meaningful and keeps the denominator non-zero. Option C, [5,∞), incorrectly includes x=5; then √(x-5)=0 and division by zero is impossible. Option D, (2,5), fails because the denominator square root is not real there. Common mistake: for a square root in a denominator, use a strict inequality.

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

Put (t=x^2\ge 0), then (f=\frac{3t+1}{t+1}=3-\frac{2}{t+1}). At (t=0), the value is (1), and (3) is only a limiting value.

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

Taking (t=x^2-1), we get (t\in[-1,\infty)) and (t\ne 0). Hence (\frac{1}{t}) has values ((-\infty,-1]\cup(0,\infty)).

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

The sum of distances from (-2) and (4) has minimum (6). Any (x) between these points gives the same minimum value.

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

The difference of distances from fixed points (1) and (5) lies from (-4) to (4). Remove signs interval-wise and check endpoint values.

Correct answer: A. \(f(x)=x^2+1\)

Since \(x^2\ge0\), \(x^2+1\ge1\), with minimum value 1 at \(x=0\). For every \(y\ge1\), take \(x=\sqrt{y-1}\). In contrast, \(\sqrt{x}+1\) is not defined for all real \(x\). Exam tip: check the minimum of a squared term.

Correct answer: C. \(9\)

A rational function is undefined wherever its denominator is zero. Here, \(x^2-a=0\), so \(x^2=a\). The excluded values are \(x=-3\) and \(x=3\); hence \(a=(-3)^2=3^2=9\). Therefore, the correct answer is \(9\). Exam tip: Set the denominator equal to zero to identify the values excluded from the domain.

Correct answer: A. ( [1,5] )

The direct answer is A: [1,5]. For a real square root, the expression inside it must be at least zero. Thus 6x-x^2-5≥0. Rearranging gives -x^2+6x-5≥0, or -(x^2-6x+5)≥0. Factoring gives -(x-1)(x-5)≥0. Since the quadratic opens downward, its value is non-negative between its roots, including the roots themselves. Hence 1≤x≤5, so the domain is [1,5]. Option A is correct because both endpoints make the radicand zero and are allowed. Option B, (-∞,1]∪[5,∞), reverses the sign region; outside the roots the downward quadratic is negative. Option C, (1,5), wrongly excludes x=1 and x=5, although a zero radicand gives √0=0 and is valid. Option D, all real numbers, is wrong because values such as x=0 give -5 inside the square root. A reliable method is to find the roots, test one point between them and one point outside, and remember that zero is allowed under a square root.

Correct answer: A. [0,2]

The radicand can be rewritten as \(6x-x^2-5=4-(x-3)^2\). For the function to be real, \((x-3)^2\leq 4\), giving the domain \([1,5]\). The radicand varies from 0 to 4, so taking its square root gives \(f(x)\in[0,2]\). Option C is the range of the radicand, not of the function. Exam tip: for a square-root function, first find the range of the expression under the root and then apply the square root.

Correct answer: A. ( (0,\frac{1}{2}] )

The direct answer is A: the range is \\( (0,1/2] \\). Since an absolute value is never negative, \\( |x+3|\ge0\\), so the denominator satisfies \\( |x+3|+2\ge2\\). Therefore \\(0<f(x)\le1/2\\). Equality \\(1/2\\) occurs at x=-3, so the upper endpoint is included. The denominator can become arbitrarily large as x moves far from -3, so the fraction approaches 0, but it never equals 0; hence 0 is excluded. A is correct. B incorrectly includes 0. C reverses the behavior and says values start at 1/2 and increase without bound, but the reciprocal decreases as the denominator grows. D has the wrong upper bound 2; even at the smallest denominator the value is only 1/2. Memory cue: a positive reciprocal of a quantity at least 2 lies between 0 and 1/2, with 0 usually only approached.

Correct answer: A. ( \mathbb{R}\setminus{6} )

The original domain has (x\ne 3), and simplification gives (f(x)=x+3). Hence the value (6), which would occur at (x=3), is excluded from the range.