Algebra of real functions

Correct answer: A. (1)

\(\frac{f}{g}=\frac{(x-2)^2}{x-2}=x-2\) where (x\ne 2), so at \(x=3\) the value is (1). First note the restriction then substitute.

Correct answer: A. (x^2+3)

(g(x)=(f+g)(x)-f(x)=x^2+3x+5-(3x+2)=x^2+3). To find the unknown function subtract the known function from the given sum.

Correct answer: A. (x+1), (x\ne 1)

(f(x)=\frac{x^2-1}{x-1}=x+1), but (x\ne 1) because (g(x)\ne 0). While finding an unknown factor check the division restriction.

Correct answer: A. 7

First, f(1)=1²=1 and g(1)=2(1)+1=3. Therefore, (f+g)(1)=f(1)+g(1)=1+3=4, while (fg)(1)=f(1)g(1)=1×3=3. Hence, the required value is 4+3=7. Exam tip: (f+g)(x) denotes the sum of the function values, whereas (fg)(x) denotes their pointwise product; do not confuse the two operations.

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

The first function needs (x\ne -2) and the second needs (x\ne 2). The domain of the sum is obtained by excluding both restrictions.

Correct answer: A. \(-4)

The direct answer is option A, \(-4\). A product function is evaluated by multiplying the separate function values: \((fg)(x)=f(x)g(x)\). Put \(x=0\) into the first function: \(f(0)=0^2+2=2\). Put \(x=0\) into the second function: \(g(0)=0^2-2=-2\). Now multiply: \((fg)(0)=2\times(-2)=-4\). Therefore option A is correct. Option B, 0, would be obtained only if one factor were zero, but the factors are 2 and \(-2\), not zero. Option C, 4, ignores the negative sign in \(g(0)\), because \(2\times2=4\) but the second value is actually \(-2\). Option D, \(-2\), uses only one value and does not perform the required multiplication. The important distinction is that \((fg)(0)\) means product, not composition; composition would be written differently. Memory cue: evaluate each function first, then multiply.

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

\(\frac{g}{f}=\frac{x^2-4}{x+2}=\frac{(x-2)(x+2)}{x+2}=x-2\), but (x\ne -2). Exclude the zero of the cancelled factor too.

Correct answer: A. (x=\pm 1)

The direct answer is option A: \(x=\pm1\). To find where a function is zero, set its expression equal to zero. Here \((f-g)(x)=f(x)-g(x)=2x^2-(x^2+1)=x^2-1\). We need \(x^2-1=0\). Rearranging gives \(x^2=1\), so taking both square roots gives \(x=1\) or \(x=-1\), written \(x=\pm1\). Option A is correct. Option B, \(x=0\), is not a zero because \((f-g)(0)=0-1=-1\). Option C, \(x=\pm2\), is not correct because substituting either value gives \(x^2-1=4-1=3\), not zero. Option D is wrong because real solutions do exist, namely 1 and \(-1\). The symbols \(\pm\) are important because both positive and negative numbers have square 1. Memory cue: for zeros, write the function equal to zero before solving.

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

\((f+g)(x)=f(x)+g(x)=(x^2+x)+(x^2-x)=2x^2\). Since \(2x^2\geq 0\) for every real \(x\), and the minimum value \(0\) is attained at \(x=0\), the range is \([0,\infty)\). Option D is incorrect because the minimum is 0, not 1. Exam tip: a positive multiple of \(x^2\) has a range beginning at its minimum value, 0.

Correct answer: A. \(\mathbb{R}-{0,1})

((f\circ g)(x)=\frac{1}{x-1}) needs (x\ne 1) and ((g\circ f)(x)=\frac{1}{x}-1) needs (x\ne 0). In the sum both restrictions combine.

Correct answer: A. (4x)

(f(g(x))=(x+1)^2) and (f(h(x))=(x-1)^2), so the difference is (4x). The difference of squares formula is fast here.

Correct answer: A. \(\frac{2x}{x-3}), (x\ne 3)

(f(x)+1=\frac{x+3}{x-3}+1=\frac{2x}{x-3}), where (x\ne 3). Make a common denominator when adding a constant.

Correct answer: A. 0

By the definition of the sum of two functions, (f+g)(x)=f(x)+g(x)=(2x−5)+(5−2x)=0. Therefore, the result is the zero function, whose value is 0 for every x. Option B equals 2f(x), not f(x)+g(x). Exam tip: combine the variable terms and constant terms separately when adding functions.

Correct answer: A. 0

For algebraic operations on functions, \(g^2(x)\) means \((g(x))^2\), not the composition \(g(g(x))\). Thus, \(g^2(x)=(x+1)^2=x^2+2x+1\). Therefore, \((f-g^2)(x)=(x^2+2x+1)-(x^2+2x+1)=0\). Option B results from an incorrect subtraction of only some terms. Exam tip: distinguish the pointwise square \(g^2\) from the composition \(g\circ g\).

Correct answer: A. \(x^4-9x^2\)

The product of two real functions is defined by \((fg)(x)=f(x)g(x)\). Therefore, \((fg)(x)=(x^2-3x)(x^2+3x)\). This is of the form \((a-b)(a+b)=a^2-b^2\), so \((x^2)^2-(3x)^2=x^4-9x^2\). Option B has the wrong sign, while C and D do not result from the correct expansion. Exam tip: recognize conjugate binomials and apply the difference-of-squares identity directly.

Correct answer: A. \(\frac{4}{x^2-4}), (x\ne \pm 2)

((f-g)(x)=\frac{1}{x-2}-\frac{1}{x+2}=\frac{4}{x^2-4}), where (x\ne \pm 2). Keep restrictions while multiplying denominators.

Correct answer: A. (9)

The direct answer is option A, 9. A composition means applying the function on the right first. Thus \\(f\\circ g)(1)=f(g(1))\\). First, \\(g(1)=2\\). Then \\(f(2)=2^2+1=5\\). In the other order, \\(g\\circ f)(1)=g(f(1))\\). Now \\(f(1)=1^2+1=2\\), so \\(g(2)=2(2)=4\\). Therefore the required sum is \\(5+4=9\\). Option A works because it gives this complete calculation. Option B, 7, is not the sum of the two correct composition values. Option C, 6, is also incorrect and may result from adding or evaluating only partly. Option D, 5, is just one of the two values, not their total. Remember: in \\(f\\circ g\\), do g first, then f.

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

The direct answer is option A: \(\mathbb R\setminus\{2\}\). A quotient \((f/g)(x)\) equals \(f(x)/g(x)\), so the denominator must not be zero. Here the original denominator is \(g(x)=x-2\). Set it equal to zero: \(x-2=0\), giving \(x=2\). Therefore every real number is allowed except 2. Although \(f(x)=x^2-5x+6=(x-2)(x-3)\), cancellation may simplify the formula to \(x-3\\) only for values where the original quotient was defined; it cannot restore the missing value \(x=2\). Option A is correct. Option B wrongly excludes 3; the denominator at 3 is 1, so the quotient is defined. Option C excludes both 2 and 3, but 3 is allowed. Option D includes 2, where division by zero occurs. Memory cue: for a quotient, inspect the original denominator before cancelling factors.

Correct answer: A. (1)

Here (f^2(x)+g^2(x)=\frac{(x^2-1)^2+(2x)^2}{(x^2+1)^2}=\frac{(x^2+1)^2}{(x^2+1)^2}=1). Recognise the identity while adding squares.

Correct answer: B. (\mathbb{R}\setminus{2})

In a quotient, the denominator (g(x)) must not be zero, so (x=2) is excluded. In exams, check the domain before simplifying.

Correct answer: A. (\mathbb{R}\setminus{-3,4})

For subtraction, both functions must be defined, so (x=-3) and (x=4) are excluded. Always remove values making any denominator zero.

Correct answer: A. \(x\ge 0\)

Here, \((f-g)(x)=f(x)-g(x)=|x|-x\). For \(x\ge 0\), \(|x|=x\), so \(|x|-x=0\). For \(x<0\), \(|x|=-x\), giving \(|x|-x=-2x\), which is not zero. Hence option A is correct; option C is incomplete because every non-negative value, not just zero, is a solution. Exam tip: For modulus expressions, always check the cases \(x\ge 0\) and \(x<0\).

Correct answer: A. \(3x^2-4x+9\)

For a linear combination of functions, the coefficients multiply the entire functions: \((3f-2g)(x)=3f(x)-2g(x)=3(x^2+1)-2(2x-3)\). Simplifying gives \(3x^2+3-4x+6=3x^2-4x+9\), so option A is correct. In option C, the constant term has the wrong sign. Exam tip: distribute a negative coefficient across every term inside the parentheses.

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

Even if the product simplifies to (1), the original functions forbid (x=1) and (x=-1). The domain should not be changed after simplification.

Correct answer: A. 1

For the sum of two functions, \((f+g)(x)=f(x)+g(x)\). Thus, \(f(-2)=(-2)^2=4\) and \(g(-2)=2(-2)+1=-3\), so \((f+g)(-2)=4+(-3)=1\). The value 3 may result from an incorrect operation such as subtracting \(g(-2)\) from \(f(-2)\). Exam tip: evaluate the two functions separately before adding their values.