Algebra of real functions

For (\sqrt{x}), (x\ge0), and for (\frac{1}{x-4}), (x\neq4). Hence the domain is ( [0,\infty)-{4} ).

Although (fg=1), the original functions do not allow (x=-2) and (x=1). It is wrong to enlarge the domain after simplification.

(\frac{x^2-1}{x-1}=x+1), but (x=1) makes the original denominator zero. After cancellation, remember the removed point.

When (x\ge2), (|x-2|=x-2), so (f-g=0). For modulus functions, making cases is the safest method.

Here,
(fg)(x)
denotes the product of the two functions:
(fg)(x)=f(x)g(x)=x^2(2x+1)
. Thus,
(fg)(-2)=(-2)^2[2(-2)+1]=4(-3)=-12
. Option C is only the value of
g(-2)
, whereas the question asks for the product of
f(-2)
and
g(-2)
. Exam tip: distinguish
(fg)
, which means product, from
f\circ g
, which means composition.

For (\frac{1}{\sqrt{x-1}}), (x>1), and for (\sqrt{4-x}), (x\le4). Thus the common domain is ( (1,4] ).

Adding and subtracting the functions gives (f + g)(x) = 2x² and (f − g)(x) = 2. Therefore, ((f + g)/(f − g))(x) = 2x²/2 = x², so option A is correct. The denominator is 2, so the quotient is defined for every real x.

The first function needs (x\neq-1), and the second needs (x\neq2). The domain of the sum is the common part of these conditions.

Subtract the known function from the sum: \(g(x)=(f+g)(x)-f(x)=(x^2+x)-(3x-2)=x^2-2x+2\). Hence, option A is correct. Option D results from failing to subtract the \(3x\) term correctly. Exam tip: For the sum of functions, subtract the known function term by term to obtain the unknown function.

Since \((f-g)(x)=f(x)-g(x)\), we get \(g(x)=f(x)-(f-g)(x)\). Substituting the given functions, \(g(x)=(x^2+3x)-(2x^2-x)=-x^2+4x\), so option A is correct. Option B results from reversing the overall sign. Exam tip: when \(f-g\) is given, use \(g=f-(f-g)\) before substituting.

(\frac{(x+2)(x-2)}{x^2-4}=1), but (h(x)\neq0) gives (x\neq\pm2). The original denominator restriction remains after simplification.

For (\sqrt{9-x^2}), (-3\le x\le3), and for (\frac{1}{x}), (x\neq0). Hence the domain is ( [-3,3]-{0} ).

Zeros of both denominators must be removed, so (x=\pm1,\pm2) are excluded. For subtraction also, take the intersection of domains.

((f-g)(x)=x^3-x=x(x-1)(x+1)), so the zeros are (-1,0,1). For a zero set, equate the function to (0).

First form the quotient while respecting its domain. Since x²−4x+4 = (x−2)², we have (f/g)(x) = (x−2)²/(x−2) = x−2, but this cancellation is valid only for x ≠ 2 because the original denominator g(x) is zero at x = 2. A zero of a function must be an allowed domain value at which the function itself equals zero. The simplified expression x−2 would be zero at x = 2, but x = 2 is excluded from the original quotient, so the graph has a removable hole there rather than a zero. Consequently the quotient has no zero in its domain, and option A is correct. Option B ignores the domain restriction; C does not make x−2 zero; and D is false because the quotient is not zero for every allowed x.

The expression is (x^2-|x|). Put (t=|x|\ge0), then (t^2-t) has minimum (-\frac{1}{4}). For hard modulus questions, using (t) is helpful.

The governing concept is addition of functions together with a domain check. By definition, (f+g)(x)=f(x)+g(x), provided both functions are defined at x. At x=0, f(0)=(0+1)/(0−1)=1/(−1)=−1, while g(0)=(0−1)/(0+1)=(−1)/1=−1. Hence (f+g)(0)=−1+(−1)=−2, so option A is correct. The value is defined because f excludes x=1, g excludes x=−1, and x=0 is allowed for both. Option B would incorrectly cancel or combine terms, option C ignores the negative signs, and option D incorrectly treats zero as a forbidden denominator value.

For a quotient of real-valued functions, both square roots must be real and the denominator must be nonzero. The numerator g(x)=√(x−1) is real when x−1≥0, so x≥1. The denominator f(x)=√(x+3) is real when x+3≥0, so x≥−3; additionally, f(x) cannot equal zero, which would exclude x=−3. Combining all conditions gives x≥1. For every x in [1,∞), x+3 is at least 4, so the denominator is automatically positive and never zero. Therefore the domain is [1,∞), option A. Option D forgets the numerator restriction, B uses an incorrect lower bound, and C subtracts a value that is already outside [1,∞).

The denominator is (g(x)), so (\sqrt{x-1}\neq0) and (x>1) are required. In division, the denominator radical cannot be zero.

\((f+g)(1)=f(1)+g(1)=(1+a)+(2-a)=3\). The parameter \(a\) cancels, so \((f+g)(1)\) is always 3 and can never equal 7. Hence, no real value of \(a\) satisfies the condition. Exam tip: simplify the sum of the function values first; if the parameter cancels, check whether the given condition is consistent.

To subtract the two functions, subtract their corresponding expressions. The x^2 terms cancel because they have equal coefficients, and the constant terms also cancel. The remaining terms contain k, so comparing the coefficient of x with the given expression determines k directly.

We calculate f(x)-g(x)=(x^2+kx+1)-(x^2-kx+1). Changing signs in the second bracket gives x^2+kx+1-x^2+kx-1=2kx. The question states that this difference equals 6x, so 2kx=6x for every x. Comparing coefficients gives 2k=6 and therefore k=3. Option A is correct. The value -3 would produce -6x, not 6x.

\((af+bg)(x)=a f(x)+b g(x)=a(x+1)+b(x-1)=(a+b)x+(a-b)\). Comparing this with \(5x+1\) gives \(a+b=5\) and \(a-b=1\). Therefore, the correct answer is 5. In such questions, compare coefficients of corresponding powers, especially the coefficient of \(x\).

For a linear combination of functions, \((2f-3g)(x)=2f(x)-3g(x)\). Therefore, \(2(x^2-2x)-3(3x-1)=2x^2-4x-9x+3=2x^2-13x+3\). Hence, option A is correct. In option C, the \(x\)-terms from \(-3g(x)\) have not been combined correctly. Exam tip: Keep the entire function in parentheses when distributing a negative coefficient.

(\frac{x}{x-1}-\frac{x}{x+1}=\frac{x[(x+1)-(x-1)]}{x^2-1}=\frac{2x}{x^2-1}). Keep brackets correct while taking a common denominator.

The equation (x^2+\frac{1}{x}=2) gives (x^3-2x+1=0=(x-1)(x^2+x-1)). Since (x\neq0), count all valid real roots carefully.