Algebra of real functions

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

Here, \((fg)(x)\) denotes the product of the two functions: \((fg)(x)=f(x)g(x)=(x-2)(x+2)=x^2-4\). Therefore, \((fg)(x)+4=x^2-4+4=x^2\), so option A is correct. Option B results from adding 4 to the product without accounting for the \(-4\) term. Exam tip: apply the identity \((a-b)(a+b)=a^2-b^2\).

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

The sum of two functions is found by adding their values pointwise: \((f+g)(x)=f(x)+g(x)=(x^2+1)+(2x-3)=x^2+2x-2\). Option B has the wrong sign for the linear term and an incorrect constant term. In exams, combine like powers separately and check the signs carefully.

Correct answer: A. \(x^2-3x-10\)

Here, \((fg)(x)\) denotes the product of the two functions: \((fg)(x)=f(x)g(x)=(x+2)(x-5)=x^2-5x+2x-10=x^2-3x-10\). Option D, \(2x-3\), is \(f(x)+g(x)\), the sum rather than the product. Exam tip: distinguish \((fg)(x)\), which means a product, from \(f(g(x))\), which means composition.

Correct answer: A. (x+3)

\(\left(\frac{f}{g}\right)(x)=\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3\), where \(x \neq 3\). Always check that the denominator is not zero.

Correct answer: A. (x \in \mathbb{R}, x \neq 2, x \neq -4)

The denominators (x-2) and (x+4) must not be zero, so (x \neq 2) and (x \neq -4). In rational functions, check denominators first.

Correct answer: A. 9

For the difference of two functions, \((f-g)(x)=f(x)-g(x)=2x^2-(x+1)\). At \(x=-2\), \(f(-2)=2(-2)^2=8\) and \(g(-2)=-2+1=-1\). Therefore, \((f-g)(-2)=8-(-1)=9\). The distractor 7 results from mishandling the negative value of \(g(-2)\). Exam tip: evaluate both functions separately before subtracting.

Correct answer: A. 15

First, \(f(2)=2^2-1=3\) and \(g(2)=2(2)+1=5\). Therefore, \((fg)(2)=f(2)g(2)=3\times5=15\). Here, \((fg)\) denotes the product of the two functions, not their composition. In an exam, carefully distinguish \((fg)(x)\) from \((f\circ g)(x)\).

Correct answer: A. Both are equal

The direct answer is A: \((g+f)(x)=(f+g)(x)\). Function addition is defined pointwise: \((f+g)(x)=f(x)+g(x)\). Thus \((g+f)(x)=g(x)+f(x)\). Ordinary addition is commutative, so changing the order does not change the sum. Substituting the given functions gives \((f+g)(x)=(4x-1)+(x^2+2)=x^2+4x+1\), and \((g+f)(x)=(x^2+2)+(4x-1)=x^2+4x+1\). Option A is correct. Option B is wrong because the expressions are equal, not opposites. Option C is wrong because both functions are defined for every real number, so their domains are the same. Option D is wrong because equality holds for every real \(x\), not only at zero. Memory cue: in addition, order can be exchanged.

Correct answer: A. \(x^2-36\)

The product of two functions is defined by \((fg)(x)=f(x)g(x)\). Hence, \((fg)(x)=(x+6)(x-6)=x^2-6^2=x^2-36\). Option B incorrectly suggests a sum-of-squares pattern, whereas this is a product of conjugate binomials. Exam tip: recognize the identity \((a+b)(a-b)=a^2-b^2\).

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

In the original (f(x)), (x-1 \neq 0), so (x \neq 1), even if simplification occurs after multiplication. Decide the domain from original expressions.

Correct answer: A. (2x, x \neq 0)

((fg)(x)=\frac{2}{x}\cdot x^2=2x), but (x \neq 0) because of the original denominator. Keep original restrictions after simplification.

Correct answer: A. (x+2)

\(\frac{x^2+4x+4}{x+2}=\frac{(x+2)^2}{x+2}=x+2\), where \(x \neq -2\). Write the zero of the denominator separately.

Correct answer: A. 10

By the definition of subtraction of functions, \((f-g)(x)=f(x)-g(x)\). Hence, \((f-g)(-5)=|-5|-(-5)=5+5=10\). Option B is incorrect because the two function values are not equal. Exam tip: \(|-5|=5\), while \(g(-5)=-5\); do not replace the latter by its absolute value.

Correct answer: A. \(1\)

For the algebraic combination of functions, \((2f-g)(x)=2f(x)-g(x)\). Therefore, \(2x^3-(2x^3-1)=2x^3-2x^3+1=1\). Option B results from handling the negative sign incorrectly. In an exam, remember that a minus sign before brackets changes the signs of all terms inside.

Correct answer: A. (x \geq 0)

(x^2) is defined for all real (x), but (\sqrt{x}) needs (x \geq 0). The domain of the sum is the intersection.

Correct answer: A. (x \geq 2)

(\sqrt{x-2}) requires (x \geq 2) and (\frac{1}{x}) requires (x \neq 0); the intersection is (x \geq 2). Identify the strongest restriction.

Correct answer: A. (-1)

\(\left(\frac{f}{g}\right)(0)=\frac{f(0)}{g(0)}=\frac{-1}{1}=-1\). In division, first check whether the denominator value is zero.

Correct answer: A. 8

By definition, (g − f)(x) = g(x) − f(x). Substituting the given functions gives (x² + 4) − (x² − 4) = x² + 4 − x² + 4 = 8. The x² terms cancel, so the difference function is the constant 8.

Correct answer: A. -x^2+3x-1

By the definition of addition and subtraction of functions, \((f+g-h)(x)=f(x)+g(x)-h(x)\). Therefore, \((2x+3)+(x-4)-x^2=-x^2+3x-1\), so option A is correct. In option B, the sign of the \(x^2\) term is incorrect because \(h(x)\) is being subtracted. Exam tip: write the complete expressions in brackets before combining like terms.

Correct answer: A. (x+1)

\(\frac{x^2+x}{x}=\frac{x(x+1)}{x}=x+1\), where \(x \neq 0\). Do not forget the restriction while cancelling a common factor.

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

((f+g)(x)=\frac{x+2}{x+2}=1), but (x+2 \neq 0). It is important to write the domain restriction with the simplified answer.

Correct answer: A. 3x² − 21x

The product function is defined by (fg)(x) = f(x)g(x). Substituting the given functions gives 3(x² − 7x) = 3x² − 21x, so option A is correct.

Correct answer: A. 2x²+2

By the definition of the sum of functions, (f+f)(x)=f(x)+f(x)=2f(x). Therefore, (f+f)(x)=2(x²+1)=2x²+2, so option A is correct. Option B incorrectly doubles only the constant term instead of the entire function. Exam tip: the sum of a function with itself is twice the function.

Correct answer: A. \(x^2-4x+4\)

By the definition of the product of functions, \((ff)(x)=f(x)\cdot f(x)\). Therefore, \((ff)(x)=(x-2)(x-2)=(x-2)^2=x^2-4x+4\), so option A is correct. Option B represents \(2f(x)\), not the product \(f(x)f(x)\). In an exam, distinguish \((ff)(x)\) from \(f(f(x))\), which denotes composition.

Correct answer: A. \(1\)

For pointwise function operations, \((f^2+g^2)(x)=[f(x)]^2+[g(x)]^2=\sin^2 x+\cos^2 x\). Using the fundamental trigonometric identity \(\sin^2 x+\cos^2 x=1\), the value is \(1\). Option C uses subtraction, while option D is the double-angle expression \(\sin 2x\). Exam tip: immediately recognize \(\sin^2 x+\cos^2 x=1\).