Real-Valued Functions, Their Domain and Range

Correct answer: A. 4

Substituting \(x=1\), \(f(1)=|1-2|+|1+2|=|-1|+|3|=1+3=4\). The value 2 results from considering only one modulus term, so it is incorrect. Exam tip: evaluate each modulus term separately before combining the results.

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

Option \(f(x)=x^2\) is correct because \(x^2\geq0\) for every real \(x\), and it equals 0 at \(x=0\). In contrast, \(x^2+1\) has minimum value 1. Exam tip: find extrema to identify a function’s range.

Correct answer: A. \(x\ge -3\)

For a square root to be real, the expression inside the radical must be non-negative: \(2x+6\ge 0\). Thus, \(2x\ge -6\), giving \(x\ge -3\). Option B reverses the inequality, while option D includes real numbers for which the radicand is negative. Exam tip: For a square-root function, set the radicand greater than or equal to zero to find the domain.

Correct answer: A. 4

Substituting \(x=4\), we get \(f(4)=|4-2|+|4-6|=|2|+|-2|=2+2=4\). Hence, 4 is correct. Option B results from considering only one modulus term, whereas both terms must be added. Exam tip: The modulus of a negative number is its positive magnitude.

Correct answer: A. 0

Substitute x = −1: f(−1) = |−1 + 4| − |−1 − 2| = |3| − |−3| = 3 − 3 = 0. Therefore, the correct answer is 0.

Correct answer: A. \(x=-3\) और \(x=-5\)

To find the zeros, set \(f(x)=0\): \(x^2+8x+15=(x+3)(x+5)=0\). Hence, \(x=-3\) or \(x=-5\), so option A is correct. In option B, the signs of both zeros are incorrect. Exam tip: For a factorised quadratic, set each linear factor equal to zero to find its zeros quickly.

Correct answer: A. 3

Because the square root is defined for \(x\ge 1\), we have \(\sqrt{x-1}\ge 0\). Hence \(f(x)=3-\sqrt{x-1}\le 3\). At \(x=1\), \(\sqrt{x-1}=0\), so \(f(1)=3\); therefore, the maximum value is 3. The value 1 may occur for another input, but it is not the maximum. Exam tip: For an expression of the form \(a-\sqrt{g(x)}\), find its maximum by making the square-root term as small as possible, usually 0.

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

For the square root to be defined, \\(x+1\\ge 0\\), and \\(\\sqrt{x+1}\\ge 0\\). Thus, \\(2\\sqrt{x+1}\\ge 0\\), so \\(f(x)=2\\sqrt{x+1}-5\\ge -5\\). At \\(x=-1\\), the function equals \\(-5\\), and as \\(x\\) increases, the function values grow without bound. Therefore, the range is \\([-5,\\infty)\\). Option C ignores the downward shift by \\(-5\\). Exam tip: Find the minimum value of the square-root part first, then account for any vertical translation.

Correct answer: A. \(( -\infty,6)\cup(6,\infty)\)

The denominator requires \(x\ne -4\). Since \(\frac{1}{x+4}\) is never zero, \(f(x)\) can never equal 6. Every other real value \(y\) is attainable: setting \(y=\frac{1}{x+4}+6\) gives \(x=\frac{1}{y-6}-4\), which is defined whenever \(y\ne6\). Therefore, the range is \(( -\infty,6)\cup(6,\infty)\).

Correct answer: A. [0, 1)

For real x, the fractional part is {x} = x − ⌊x⌋, so it is always at least 0 and less than 1. The value 0 occurs for integers, but 1 is never attained. Hence, the range is [0, 1).

Correct answer: A. 0.75

Substituting \(x=5.75\) gives \(x-2=5.75-2=3.75\). The fractional part of \(3.75\) is \(0.75\), so \(f(5.75)=0.75\). Option B, \(3.75\), is only the intermediate value before taking its fractional part. Exam tip: subtract the integer part from a number to obtain its fractional part.

Correct answer: C. \(f(x)=\frac{1}{\sqrt{x}}\)

For \(\frac{1}{\sqrt{x}}\), the square root requires \(x\ge 0\), but the denominator cannot be zero. Hence only \(x>0\) is allowed. In \(\sqrt{x}\), zero is included. Exam tip: check radical and denominator restrictions separately.

Correct answer: A. (0, 1/4]

For every real number x, x^2 is non-negative, so x^2 + 4 is at least 4. Since the denominator is always positive, taking reciprocals gives 0 < 1/(x^2 + 4) ≤ 1/4. The upper endpoint is actually attained: when x = 0, y = 1/4. The lower endpoint 0 is not attained because the numerator remains 1 for every real x, although the function approaches 0 as |x| becomes arbitrarily large. Therefore the range is (0, 1/4], making option A correct. Option B incorrectly includes 0, option C lists possible denominator values, and option D ignores the bounds imposed by x^2 ≥ 0.

Correct answer: A. \(\mathbb{R}\setminus\{-2,5\}\)

The product is \((fg)(x)=\frac{3x}{(x+2)(x-5)}\), so both denominator factors must be non-zero. Thus, \(x\neq -2\) and \(x\neq 5\), giving the domain \(\mathbb{R}\setminus\{-2,5\}\). Excluding only \(-2\) or only \(5\) is incomplete. Exam tip: the domain of a product function is the intersection of the domains of the two functions.

Correct answer: A. Both \(p\) and \(q\) are defined at \(x\), and \(q(x)\ne0\)

The quotient \(\frac{p(x)}{q(x)}\) is defined only when both \(p(x)\) and \(q(x)\) exist and the denominator \(q(x)\) is not zero. Therefore, option A is correct. Having only \(p\) defined is not enough, because \(q\) must also be defined and non-zero. Exam tip: While finding the domain of a quotient function, always exclude values that make the denominator zero.

Correct answer: A. \((1,\infty)\)

For \(f(x)=\frac{1}{\sqrt{x-1}}\), the radicand must be non-negative, but the square root is in the denominator, so it cannot be zero. Thus \(x-1>0\), giving \(x>1\). Since \(g(x)=x^2\) is defined for every real number, the domain of \((f+g)\) is the intersection of the two domains: \((1,\infty)\). The endpoint \(x=1\) is excluded because it makes the denominator zero. Exam tip: when a square root occurs in a denominator, the radicand must be strictly positive.

Correct answer: A. R\{-1}

The governing domain rule for a quotient of functions is that both functions must be defined and the denominator must not be zero. Both f(x)=x^2+2x+2 and g(x)=x+1 are polynomials, so each is defined for every real x. However, (f/g)(x) is undefined when g(x)=0. Solving x+1=0 gives x=-1, so every real number except -1 belongs to the domain. The numerator does not create a restriction; even if it happened to be zero at some point, a zero numerator would be allowed as long as the denominator were nonzero. Option B incorrectly includes the point -1, option C imposes an unrelated restriction, and option D excludes the wrong value. Therefore option A is correct.

Correct answer: A. [-4,∞)\{1}

The governing concept is that the domain of a sum is the intersection of the domains of its component functions. For f(x)=√(x+4), the radicand must be nonnegative: x+4≥0, so x≥-4. For g(x)=1/(x-1), the denominator cannot vanish, so x≠1. Combining both restrictions gives [-4,∞) with the single point 1 removed. The endpoint -4 is included because the square root of zero is defined. Option B forgets the denominator restriction, option C allows values less than -4 where the square root is not real, and option D removes too much of the valid interval. Hence option A is correct.

Correct answer: A. [-2,2]

The governing concept is the domain of a product: both factors must be defined at the same real input. The linear function g(x)=x+1 is defined for every real x, so it adds no restriction. For f(x)=√(4-x^2), the radicand must satisfy 4-x^2≥0. This is equivalent to x^2≤4, hence -2≤x≤2. Therefore the product is defined exactly on the closed interval [-2,2]. The endpoints are included because the radicand is zero there, and √0 exists. Option B includes values below -2 incorrectly, option C keeps only values above 2, and option D ignores the square-root restriction. Thus option A is correct.