Algebra of real functions

Correct answer: A. \(a^2+2a\)

The sum of two functions is defined by \((f+g)(x)=f(x)+g(x)\). Hence, \((f+g)(a)=f(a)+g(a)=a^2+2a\). Options B and C incorrectly mix \(x\) and \(a\), while option D does not represent the sum. Exam tip: when evaluating a function at \(a\), replace every occurrence of \(x\) with \(a\).

Correct answer: A. (0)

The direct answer is option A, 0. The notation \\((fg)(0)\\) means the product of the two function values at 0: \\(f(0)g(0)\\). First calculate \\(f(0)=0^2-2(0)=0\\). Next calculate \\(g(0)=0^2+2(0)=0\\). Therefore \\((fg)(0)=f(0)g(0)=0\cdot0=0\\). Option A is correct. Option B, 4, might arise from confusing the signs or substituting another value, but it is not the product at zero. Option C, -4, is also not obtained by direct substitution and may come from an incorrect sign calculation. Option D, 1, has no basis in either function value. The product rule is simple: evaluate each function at the given input, then multiply. Since both values are zero, the product must be zero. Remember that \\(fg\\) here means multiplication, not composition.

Correct answer: A. \(x \in \mathbb{R}, x \neq 1\)

(f(x)) needs \(x \neq 1\), and \(f(x)=\frac{1}{x-1}\) is never zero. So the only restriction is \(x \neq 1\).

Correct answer: A. (3)

\(\left(\frac{f}{g}\right)(x)=\frac{(x-4)(x+4)}{x+4}=x-4\), so at \(x=-1\) the value is (-5). Check directly as \(\frac{-15}{3}=-5\).

Correct answer: A. (-5)

\(\frac{x^2-16}{x+4}=\frac{(x-4)(x+4)}{x+4}=x-4\), hence \(-1-4=-5\). Substitute carefully after simplification.

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

For the square root, (9-x^2 \geq 0), so (-3 \leq x \leq 3). The linear function gives no extra restriction.

Correct answer: A. \(g\circ f:A\to C\), where \((g\circ f)(x)=g(f(x))\)

Applying \(f\) first sends \(x\in A\) to \(f(x)\in B\). Applying \(g\) next gives \(g(f(x))\in C\), so the composite is \(g\circ f\). Exam tip: in a composition, the function on the right acts first.

Correct answer: A. (4)

\(\frac{x^2+3x+2}{x+1}=\frac{(x+1)(x+2)}{x+1}=x+2\), so at \(x=2\) the value is (4). Factorisation saves time.

Correct answer: A. (1)

(f(1)=1) and (g(1)=3), so ((f+g)(1)-(fg)(1)=4-3=1). In such questions, find individual values first.

Correct answer: A. No real solution

((f-g)(x)=x^2-2x+2=(x-1)^2+1), which is never (0). Completing the square quickly shows real solutions.

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

\((g-f)(x)=g(x)-f(x)=(x^2+1)-(x^2-1)=2\) for every real \(x\). Thus, the difference is a constant function, whose range is the singleton set \(\{2\}\). \(\{0\}\) is incorrect because the constant terms give \(1-(-1)=2\). Exam tip: simplify \((g-f)(x)\) completely before determining its range.

Correct answer: A. 3

The product function is (fg)(x) = f(x)g(x) = (x + 1)x² = x³ + x². The highest power of x is 3, so the degree is 3. Option B gives the degree of g(x) alone, not that of the product. In such questions, multiply the functions first and then identify the highest power of x.

Correct answer: A. (0) and (2)

The direct answer is option A, 0 and 2. The sum function is found by adding corresponding values: \\( (f+g)(x)=f(x)+g(x)=(x^2-5x)+3x=x^2-2x\\). A zero is an input for which the function value is zero, so set \\(x^2-2x=0\\). Factor: \\(x(x-2)=0\\). By the zero-product property, either \\(x=0\\) or \\(x-2=0\\), giving \\(x=2\\). Hence the zeroes are 0 and 2, option A. Option B, 0 and 5, incorrectly uses the zero of the original f term without combining it with g. Option C, 2 and 5, misses the zero x=0. Option D, only 2, misses one valid solution. Always form the requested sum first; do not find zeroes of f or g separately. Checking gives \\(0^2-2(0)=0\\) and \\(2^2-2(2)=0\\).

Correct answer: A. (\frac{-5}{x+1}, x \neq -1)

With the same denominator, subtract numerators: (\frac{x-2-(x+3)}{x+1}=\frac{-5}{x+1}). Because of denominator (x+1), (x \neq -1).

Correct answer: A. \(8x-1\)

For a linear combination of functions, multiply each entire function by its coefficient: \((3f+2g)(x)=3f(x)+2g(x)=3(2x+1)+2(x-2)=6x+3+2x-4=8x-1\). Hence, option A is correct. Option B results from an incorrect sign in the constant term. In the exam, distribute each coefficient across every term of the corresponding function.

Correct answer: A. (3)

\(\frac{x^2+6x+9}{x+3}=\frac{(x+3)^2}{x+3}=x+3\), hence the value at \(x=0\) is (3). The perfect square identity is useful.

Correct answer: A. \(x>1\)

In \(f(x)\), \(\sqrt{x-1}\) is in the denominator. Thus, the radicand must satisfy \(x-1\geq 0\), and the denominator must not be zero, so \(x-1\neq 0\). Therefore, \(x>1\). Since \(g(x)=x\) is defined for all real numbers, the domain of \((f+g)\) is the intersection of the two domains, namely \((1,\infty)\). Exam tip: For a square root in the denominator, require the expression under the root to be strictly positive.

Correct answer: A. (2)

((f-g)(x)=(x^2+1)-(x^2-1)=2), so at any (x) the value is (2). Simplifying first reduces calculation.

Correct answer: A. \(2x\)

The difference of two functions is \((f-g)(x)=f(x)-g(x)\). Thus, \((x^2+x+1)-(x^2-x+1)=x^2+x+1-x^2+x-1=2x\). Option B results from failing to change the sign of the \(-x\) term when subtracting the entire function \(g(x)\). Exam tip: when a function is preceded by a minus sign, change the sign of every term inside its brackets.

Correct answer: A. 6

The product of two real-valued functions is defined by \((fg)(x)=f(x)g(x)\). Thus, \((fg)(x)=\frac{x+2}{x-3}(x-3)=x+2\), with the restriction \(x\neq 3\). Therefore, \((fg)(4)=4+2=6\). Option 1 is only the value of \(g(4)\), while option 4 incorrectly treats \(x=4\) as outside the domain. Exam tip: write \((fg)(x)=f(x)g(x)\) first and check the original domain restriction before cancelling factors.

Correct answer: A. (x, x \geq 0)

((fg)(x)=\sqrt{x}\cdot\sqrt{x}=x), but both square roots require (x \geq 0). The simplified form does not change the original domain.

Correct answer: A. (1)

\(\frac{x^2-3x+2}{x-1}=\frac{(x-1)(x-2)}{x-1}=x-2\), so at \(x=3\) the value is (1). Factorisation makes division simple.

Correct answer: A. \(16-x^2\)

For the product of two functions, \((fg)(x)=f(x)g(x)\). Thus, \((fg)(x)=(x+4)(4-x)=4^2-x^2=16-x^2\), using the identity \((a+b)(a-b)=a^2-b^2\). Option B has the sign reversed. In an exam, first check whether \(fg\) denotes the product of functions or their composition.

Correct answer: A. \(3x-9\)

By the definition of the difference of two functions, \((f-g)(x)=f(x)-g(x)\). Therefore, \((4x-7)-(x+2)=4x-7-x-2=3x-9\), so option A is correct. Option C results from combining the constant terms incorrectly. Exam tip: when subtracting a bracket, change the sign of every term inside it.

Correct answer: A. (x+3)

The direct answer is A: \(x+3\), with \(x\ne3\). By definition, \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2-9}{x-3}\). Factor the numerator as a difference of squares: \(x^2-9=(x-3)(x+3)\). Therefore \(\frac{(x-3)(x+3)}{x-3}=x+3\), but cancellation is allowed only when \(x-3\ne0\), so \(x=3\) remains excluded. Option A is correct. Option B merely shows a cancelled factor and is not the simplified quotient. Option C incorrectly changes subtraction into addition. Option D, 1, would result only if numerator and denominator were identical, which they are not. Also, the simplified rule does not restore the missing value at \(x=3\). The key cue is: factor first, cancel common factors, and keep denominator restrictions.