Algebra of real functions

Correct answer: A. ( \mathbb{R}-{0} )

((fg)(x)=x^2\cdot\frac{1}{x}=x), but (x=0) is not in the domain. Hence every non-zero real number is a solution.

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

The direct answer is option A: \(((-\infty,-2]\cup[2,\infty))\setminus\{5\}\). For \(f(x)=\sqrt{x^2-4}\), require \(x^2-4\ge0\). Factoring gives \((x-2)(x+2)\ge0\), which holds when \(x\le-2\) or \(x\ge2\). For \(g(x)=1/(x-5)\), require \(x-5\ne0\), so \(x\ne5\). The domain of a sum is the intersection of the two domains, so remove 5 from the two outer intervals. Option A states this exactly. Option B uses the interval between -2 and 2, but there the square-root radicand is negative except at endpoints. Option C correctly uses the outer regions but wrongly excludes the endpoints, where the radicand is zero and the square root is defined. Option D excludes only 5 and wrongly includes numbers where \(f\) is not real. Memory cue: solve the square-root inequality first, then remove denominator zeros.

Correct answer: A. ( \mathbb{R}-{3} )

(f(x)=x+3) only for (x\neq3), so (f-g=0) on that domain. Domain is essential when comparing algebraic expressions as functions.

Correct answer: A. (0)

((f-g)(x)=x^2-2x+1=(x-1)^2), whose minimum is (0). Completing the square is useful in such questions.

Correct answer: A. \(\frac{3}{4}\)

\((f-g)(x)=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\). Therefore the minimum is \(\frac{3}{4}\).

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

The denominator (x^2+1) is never (0), so no real number is excluded. Only real zeros of the denominator are removed.

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

Since (x^2+1>0) for every real (x), both (f) and (g) are defined on all (\mathbb{R}). Hence (fg=1) on all (\mathbb{R}).

Correct answer: A. ( \frac{2x}{x^2-1} )

(\frac{1}{x-1}+\frac{1}{x+1}=\frac{(x+1)+(x-1)}{x^2-1}=\frac{2x}{x^2-1}). Add the numerators correctly after taking the common denominator.

Correct answer: A. (4)

(f^2-g^2=(f-g)(f+g)=\frac{2}{x}\cdot2x=4), where (x\neq0). Using an identity makes the calculation shorter.

Correct answer: A. 25

Substituting x=0 gives f(0)=0²+4(0)+5=5 and g(0)=0²−4(0)+5=5. For the product function, (fg)(0)=f(0)g(0), so (fg)(0)=5×5=25. Option B is only the value of one function, not their product. Exam tip: Rewrite (fg)(a) as f(a)g(a) before substituting values.

Correct answer: A. \(x=3\)

The quotient \(\left(\frac{f}{g}\right)(x)\) is zero only when the numerator is zero and the denominator is non-zero. Here, \(f(x)=x^2-5x+6=(x-2)(x-3)\), so the numerator is zero at \(x=2\) and \(x=3\). However, \(g(2)=0\), so the quotient is undefined at \(x=2\). Therefore, the only valid zero is \(x=3\). Exam tip: For the zeros of a quotient function, first find the zeros of the numerator and then exclude any value that makes the denominator zero.

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

The quotient \\(\\frac{f}{g}\\) is formed by dividing \\(f(x)=\\sqrt{x-2}\\) by \\(g(x)=x-5\\). Its domain must satisfy two independent conditions: the radicand of the square root cannot be negative, and the denominator cannot be zero. Both restrictions must hold at the same time.

From \\(x-2\\ge0\\), we get \\(x\\ge2\\). From the denominator, \\(x-5\\ne0\\), so \\(x\\ne5\\). Combining these gives \\([2,\\infty)\\setminus\\{5\\}\\). The endpoint 2 is included because \\(\\sqrt{0}\\) is defined, while 5 is excluded because division by zero is impossible. Therefore option A is correct.

Correct answer: A. \((x-1)^2\)

Algebraic operations on functions are performed in the stated order: \((f-g+h)(x)=f(x)-g(x)+h(x)=x^2-2x+1\). Therefore, it simplifies to \((x-1)^2\). In option B, the sign of the middle term is incorrectly changed to \(+2x\). Exam tip: when subtracting a function, carefully retain the negative sign for all its terms.

Correct answer: A. ( \frac{8}{x} )

Here (f-g=\frac{4}{x}) and (f+g=2), so (f^2-g^2=\frac{8}{x}). The identity (a^2-b^2=(a-b)(a+b)) is very useful.

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

For (f), (x>0), and for (g), (x\ge1). Their intersection is ( [1,\infty) ). For product also, take the common domain of both functions.

Correct answer: D. x = −2, 2

Answer: D, x=−2 and x=2. Since (f−g)=0, write f(x)−g(x)=0: (x^2−4)−(|x|−2)=0, so x^2−|x|−2=0. Let t=|x|. Because an absolute value is never negative, t≥0, and x^2=|x|^2=t^2. The equation becomes t^2−t−2=0=(t−2)(t+1). Hence t=2 or t=−1, but t=−1 is impossible. Thus |x|=2, which gives x=2 or x=−2. Substitution confirms both: 4−2−2=0. A includes −1, which is not a solution. B omits −2 and C omits 2 while including 1. D contains exactly both valid signs. Memory cue: |x|=a for a positive number always gives x=±a.

Correct answer: A. (5)

Adding the two functions combines their corresponding terms. We have \\(f(x)=x^2+px+q\\) and \\(g(x)=x^2-px+q\\). The terms containing \\(p\\) cancel because one is \\(px\\) and the other is \\(-px\\). The two constant terms add, so the result is \\(2x^2+2q\\).

The question states that this sum is \\(2x^2+10\\). Comparing the constant terms gives \\(2q=10\\), and dividing by 2 gives \\(q=5\\). The value of \\(p\\) is irrelevant because its terms cancel. Therefore option A is correct. Comparing coefficients of like powers is a reliable method whenever two polynomial expressions are equal for all values of the variable.

Correct answer: A. when they have same rule and same domain

(f(x)=x-1) only for (x\neq-1), while (g) is on all (\mathbb{R}). Equal functions must have both the same rule and the same domain.

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

The direct answer is option A: \(\mathbb{R}\setminus\{-1,1\}\). For \(f(x)=2x/(x^2-1)\), the denominator must be nonzero: \(x^2-1=(x-1)(x+1)\ne0\), so \(x\ne1,-1\). In \(g(x)=1/(x-1)+1/(x+1)\), both denominators also require \(x\ne1\) and \(x\ne-1\). The difference \(f-g\) is defined only on the common domain, so the allowed set is all real numbers except -1 and 1. Option A is correct. Option B incorrectly allows both denominator zeros. Option C removes 0, but 0 is allowed: the denominators are then -1 and 1. Option D gives only the forbidden points rather than the allowed domain. Simplifying the expression might create cancellation, but cancellation cannot restore values where the original functions were undefined. Memory cue: for sums or differences, intersect the original domains before simplifying.

Correct answer: A. only at (x=1)

(f-g=x^2-2x+1=(x-1)^2), which is (0) only at (x=1). A perfect square helps identify the zero quickly.

Correct answer: A. ( [-3,\infty)\setminus{2} )

The root needs (x\ge -3) and the denominator needs (x\ne 2). For a product, take the common domain of both functions.

Correct answer: A. (x+2,\ x\ne 2)

Even after simplification, the original denominator must satisfy (x-2\ne 0). Cancelled denominator values do not return to the domain.

Correct answer: A. ( [-3,3]\setminus{0} )

The root needs (9-x^2\ge 0), so (x\in[-3,3]), and (x\ne 0). The domain of the sum is the intersection.

Correct answer: A. \(2x^2-5x+15\)

For an algebraic combination of functions, \((2f-3g)(x)=2f(x)-3g(x)\). Thus, \(2f(x)=2x^2+4x\) and \(3g(x)=9x-15\). Therefore, \((2f-3g)(x)=(2x^2+4x)-(9x-15)=2x^2-5x+15\). Option C results from using the wrong sign while subtracting \(3g(x)\). Exam tip: place the entire function being subtracted in parentheses before simplifying.

Correct answer: A. \( [0,\infty\)\setminus{4} )

\(\sqrt{x}\) needs \(x\ge 0\) and the denominator needs (x\ne 4). In a quotient, the denominator function must not be zero.