Real-Valued Functions, Their Domain and Range

Correct answer: B. ( \mathbb{R}\setminus{3} )

The original denominator is (x-3), so (x=3) is not allowed. Even after simplification, do not add the removed point to the domain.

Correct answer: A. \([-2,8]\)

For both square roots to be real, each radicand must be non-negative. From \(x+2\ge 0\), we get \(x\ge -2\), and from \(8-x\ge 0\), we get \(x\le 8\). Hence, the intersection is \([-2,8]\). At \(x=-2\) and \(x=8\), one square root becomes zero, so both endpoints are included; therefore, option B is incorrect. Exam tip: For radical functions, require every radicand to be \(\ge 0\) and intersect all resulting conditions.

Correct answer: A. \(2\sqrt{5}\)

For real values, \(x+2\ge 0\) and \(8-x\ge 0\), so the domain is \(-2\le x\le 8\). By the Cauchy–Schwarz inequality, \((\sqrt{x+2}+\sqrt{8-x})^2\le 2[(x+2)+(8-x)]=20\). Hence \(f(x)\le 2\sqrt{5}\). Equality occurs when \(\sqrt{x+2}=\sqrt{8-x}\), which gives \(x=3\). Therefore, the maximum value is \(2\sqrt{5}\). Exam tip: first determine the domain of radical expressions, then check the equality condition for the inequality used.

Correct answer: A. ( \sqrt{10} )

The square roots require x+2≥0 and 8−x≥0, so the domain is the closed interval [−2,8]. Let a=x+2 and b=8−x. Then a,b are nonnegative and a+b=10. By the inequality (√a+√b)²=a+b+2√(ab)≥a+b=10, we obtain f(x)≥√10. Equality occurs when ab=0, which happens at x=−2 or x=8. Therefore the minimum is √10.

Option A is correct. The value 2√5 is larger than √10 and is actually the maximum, reached when the two radicands are equal. The value 0 is impossible because both square roots cannot vanish at the same x. The supplied explanation gives the correct endpoint idea, although the inequality makes the minimum argument more explicit.

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

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

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

Since \(|3x|\le \frac{x^2+9}{2}\), \(\left|\frac{3x}{x^2+9}\right|\le \frac{1}{2}\). Equality occurs at \(x=\pm 3\).

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

Since \(\sqrt{x^2-4}\) is in the denominator, its radicand must be non-negative and the denominator must also be nonzero. Hence, \(x^2-4>0\), so \(x^2>4\), giving \(x<-2\) or \(x>2\). Therefore, the domain is \((-\infty,-2)\cup(2,\infty)\). Option B is incorrect because it includes \(x=\pm2\), where the denominator becomes zero. Exam tip: when a square root occurs in the denominator, require the expression under the root to be strictly positive.

Correct answer: A. \([-3,3]\setminus\{-1\}\)

The radicand must satisfy \(9-x^2\ge 0\), giving \(-3\le x\le 3\). The denominator must also be nonzero, so exclude \(x=-1\). Thus the domain is \([-3,3]\setminus\{-1\}\).

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

Since (\sqrt{x}\ge 0), the denominator is at least (1). At (x=0), the value is (1), and as (x\to\infty), it approaches (0).

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

Complete the square: \(x^2+2x+2=(x+1)^2+1\), so the logarithm's argument has minimum value \(1\), attained at \(x=-1\), and is unbounded above. Since \(\log_2\) is increasing, the function has minimum value \(\log_2 1=0\) and grows without bound. Its range is \([0,\infty)\).

Correct answer: A. \(\{0,1,2,3\}\)

Since \(x\in[0,2)\), we have \(2x\in[0,4)\). Therefore, the possible integer values of \(\lfloor 2x\rfloor\) are \(0,1,2,3\), and each of them is attained. For example, the value is \(0\) when \(x\in[0,\tfrac12)\), \(1\) when \(x\in[\tfrac12,1)\), and so on. The endpoint \(x=2\) is excluded, so \(2x=4\) and the value \(4\) cannot occur; hence option B is incorrect. Thus, the range is \(\{0,1,2,3\}\). Exam tip: For the greatest-integer function, check carefully whether the endpoints of the input interval are included.

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

For every real \(x\), \(x^2+4x+5=(x+2)^2+1>0\), so the domain is \(\mathbb{R}\). Put \(t=(x+2)^2\ge0\). Then \(f(x)=\frac{t+4}{t+1}=1+\frac{3}{t+1}\). At \(t=0\), the value \(4\) is attained. As \(t\to\infty\), the function approaches \(1\), but it never equals \(1\). Hence the range is \((1,4]\). Option B incorrectly includes \(1\), while option C excludes the attained value \(4\). Exam tip: For such rational functions, complete the square and determine the allowed range of the substituted variable.

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

Put (t=(x-1)^2\ge 0), then (f=\frac{t+4}{t+1}=1+\frac{3}{t+1}). The maximum is (4), and (1) is only a limiting value.

Correct answer: A. [1, 9]

Each radicand must be non-negative: x − 1 ≥ 0 requires x ≥ 1, and 9 − x ≥ 0 requires x ≤ 9. Their intersection is [1, 9]. The endpoints are included because a square root of zero is defined.

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

At the endpoints the value is (2\sqrt{2}), and at the midpoint (x=5), the maximum is (4). For symmetric radicals, check both endpoints and the midpoint.

Correct answer: A. ( [2,\infty)\setminus{7} )

The direct answer is option A: the domain is [2,\infty)\setminus\{7\}. A real square root is defined only when its radicand is non-negative, so x-2\ge0, giving x\ge2. Also, a denominator cannot be zero; therefore x-7\ne0, giving x\ne7. Combining both conditions gives [2,\infty) with 7 removed. Option A is correct because it includes 2 and all larger numbers except 7. Option B wrongly removes 2, although \sqrt{0} is defined. Option C includes numbers below 2, where the square root is not real. Option D wrongly excludes every x greater than 7, even though such values are allowed. Memory cue: square root means radicand \ge0; denominator means denominator \ne0.

Correct answer: A. \(( -\infty,1]\cup[5,\infty)\)

For a real square root, the expression inside the root must be non-negative. Thus, \(x^2-6x+5\ge 0\). Factoring gives \((x-1)(x-5)\ge 0\), which is true for \(x\le 1\) or \(x\ge 5\). Therefore, the domain is \(( -\infty,1]\cup[5,\infty)\). Option \([1,5]\) is incorrect because the quadratic is negative between its roots. Exam tip: for any square-root function, set the radicand greater than or equal to zero first.

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

On the domain, the inside expression starts at (0) and goes to infinity. Hence the square root range is ([0,\infty)).

Correct answer: B. 2

Completing the square gives \(f(x)=x^2+2x+3=(x+1)^2+2\). Since \((x+1)^2\geq 0\), the function has minimum value \(2\), attained at \(x=-1\). Hence its range is \([2,\\infty)\), so \(m=2\). Exam tip: For a quadratic written as \(a(x-h)^2+k\) with \(a>0\), the minimum value is \(k\). The constant term \(3\) is not the minimum value.

Correct answer: A. ( (4,5] )

The square root needs (\log_{\frac{1}{2}}(x-4)\ge 0) and (x-4>0). Since the base (\frac{1}{2}<1), (0<x-4\le 1), so the domain is ((4,5]).

Correct answer: A. ( (-\infty,2)\cup[5,\infty) )

The expression inside the square root must satisfy (\frac{x-5}{x-2}\ge 0) and (x\ne 2). A sign chart gives (x<2) or (x\ge 5).

Correct answer: A. ( \mathbb{R}\setminus{\frac{7}{2}} )

From (y=\frac{7x+4}{2x-3}), (x=\frac{3y+4}{2y-7}), so (y\ne \frac{7}{2}). In a linear fractional function, remove the impossible (y).

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

Since \(12-x-x^2=\frac{49}{4}-\left(x+\frac{1}{2}\right)^2\), the inside maximum is \(\frac{49}{4}\). Taking square root gives maximum \(\frac{7}{2}\).

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

Put (t=x^2\ge 0), then (f=\frac{t+5}{t+2}=1+\frac{3}{t+2}). The maximum is (\frac{5}{2}), and (1) is only a limiting value.

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

The argument of a logarithm must be strictly positive, so \(x^2-9>0\). Factoring gives \((x-3)(x+3)>0\), which holds for \(x<-3\) or \(x>3\). Thus the domain is \((-\infty,-3)\cup(3,\infty)\); the endpoints are excluded because the logarithm’s argument is zero there.