Algebra of real functions

Correct answer: A. (0)

(f(-2)=0) and (g(-2)=3), so the product is (0). If one factor value is (0), the product is (0).

Correct answer: A. \( \mathbb{R}\setminus{0,3} \)

(f) needs (x\ne 3), (g) needs (x\ne 0), and divisor (g(x)\ne 0) gives (x\ne 3). Together, (x\ne 0,3).

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

The direct answer is option A: \(x-2\), with \(x\ne1\). Function division means \(\left(\frac fg\right)(x)=\frac{f(x)}{g(x)}\). Factor the numerator: \(x^2-3x+2=(x-1)(x-2)\). Hence \(\frac{(x-1)(x-2)}{x-1}=x-2\), but only where the original denominator is nonzero. Since \(g(x)=x-1\), we must exclude \(x=1\), even though cancellation simplifies the expression. Option A includes both the correct rule and restriction. Option B wrongly allows x=1. Option C has the wrong factor and sign. Option D gives the wrong simplified expression and wrongly excludes 2 instead of 1. Important lesson: simplifying a quotient does not restore a value where the original denominator was zero.

Correct answer: A. (0)

((f+g)(x)=x^2+2x=(x+1)^2-1), so the minimum is (-1). Be careful because (2x-1+x^2+1=x^2+2x).

Correct answer: C. \(g(x)\ne 0\ \forall x\in\mathbb{R}\)

Although \(f\) is defined for every real \(x\), \(\frac{f(x)}{g(x)}\) is defined only when \(g(x)\ne0\). Hence C is correct. Being one-one does not prevent \(g\) from taking value zero. In exams, solve \(g(x)=0\) first to find excluded inputs.

Correct answer: A. (1)

(x=0) is in both domains and (f(0)g(0)=(-1)(-1)=1). First check whether the given value is in the domain.

Correct answer: A. (2)

\(f(1)=2\) and \(g(1)=2\), so both ratios are (1). In quotient questions, the denominator value must be non-zero.

Correct answer: A. (f(x)=g(x)) but (x\ne 3)

(f(x)) simplifies to (x+3), but the original function is not defined at (x=3). Same formula and same function require matching domains.

Correct answer: A. 0

For the pointwise product, (fg)(x)=f(x)g(x). Hence, (fg)(x)=(|x|+x)(|x|-x)=|x|^2-x^2=0, since |x|^2=x^2 for every real x. Option B results from an incorrect expansion that treats the factors as if their terms were simply added. Exam tip: apply the identity (a+b)(a-b)=a^2-b^2 directly.

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

The first function gives (x^2-4\ne 0), so (x\ne \pm2), and the second gives (x\ne 2). The common domain is ( \mathbb{R}\setminus{-2,2} ).

Correct answer: A. \(x\in\{0,1\}\)

\((f-g)(x)=f(x)-g(x)=x^3-x^2=x^2(x-1)\). By the zero-product property, \(x^2=0\) gives \(x=0\), and \(x-1=0\) gives \(x=1\). Hence the real solutions are \(0\) and \(1\). Option B is incomplete because it omits \(x=0\). Exam tip: factor the expression first, then set each factor equal to zero.

Correct answer: A. (2x²+8)/(x²−4), x≠±2

To add the rational functions, use the common denominator (x−2)(x+2)=x²−4. Thus (f+g)(x)=(x+2)/(x−2)+(x−2)/(x+2)=[(x+2)²+(x−2)²]/(x²−4). Expanding the numerator gives (x²+4x+4)+(x²−4x+4)=2x²+8, because the middle terms cancel. Therefore (f+g)(x)=(2x²+8)/(x²−4). The original denominators require x−2≠0 and x+2≠0, so x≠2 and x≠−2, written x≠±2. Hence option A is correct. Option B has an incorrect sign in the constant term, option C loses the squared terms, and option D omits the factor 2. No cancellation removes the original domain restrictions.

Correct answer: A. \(-5\)

\((f-g)(x)=f(x)-g(x)=x^2+ax-(bx+1)=x^2+(a-b)x-1\). Comparing this with the given expression \(x^2-5x-1\), the coefficient of \(x\) gives \(a-b=-5\). Hence, option A is correct. Exam tip: compare corresponding coefficients, and do not assume that \(a\) and \(b\) individually are determined when only their difference is given.

Correct answer: A. \(\frac{x+1}{x},\ x\ne 0,-1\)

Dividing gives \(\frac{1}{x}\div\frac{1}{x+1}=\frac{x+1}{x}\). The original (g) also requires (x\ne -1).

Correct answer: A. \(x\in\{-1,0,1\}\)

The condition \((f-g)(x)=0\) gives \(x^2-|x|=0\), or \(x^2=|x|\). Let \(t=|x|\ge 0\). Then \(t^2=t\), so \(t=0\) or \(t=1\). Hence \(x=0\) or \(x=\pm1\), giving \(x\in\{-1,0,1\}\). Option B is incorrect because it omits \(x=-1\). Exam tip: for equations involving a modulus, substitute \(t=|x|\) with \(t\ge0\) before solving.

Correct answer: A. (\frac{3}{2})

((fg)(x)=\sqrt{(x+1)(2-x)}), and the quadratic inside has maximum (\frac{9}{4}). Therefore the maximum is (\frac{3}{2}).

Correct answer: A. (-\frac{9}{4})

((f-g)(x)=x^2-7x+11=(x-\frac{7}{2})^2-\frac{5}{4}), so the minimum is (-\frac{5}{4}). Check options carefully.

Correct answer: A. ( \mathbb{R}\setminus{-2,0} )

The first denominator gives (x\ne -2), and the second gives (x\ne 0). The domain of the sum keeps both restrictions.

Correct answer: A. (x=-2)

The direct answer is option A, \(x=-2\). Function addition is performed pointwise: \((f+g)(x)=f(x)+g(x)\). Substituting the given functions gives \(x^2+4x+4\). This is the perfect square \((x+2)^2\), because \((x+2)^2=x^2+4x+4\). A square is zero only when its base is zero, so \(x+2=0\), giving \(x=-2\). Option A is therefore correct. Option B, \(x=2\), gives \((4+8+4)=16\), not zero. Option C, \(x=0\), gives 4, not zero. Option D, \(x=-4\), gives \(16-16+4=4\), not zero. The word zero means the output of the combined function, not necessarily the input. Memory cue: look for \(x^2+2ax+a^2=(x+a)^2\), then set the bracket equal to zero.

Correct answer: A. \(\frac{-4x}{x^2-1},\ x\ne\pm1\)

\((f-g)(x)=\frac{x-1}{x+1}-\frac{x+1}{x-1}\). Taking the common denominator gives the numerator \((x-1)^2-(x+1)^2\). Expanding, this is \((x^2-2x+1)-(x^2+2x+1)=-4x\), while the denominator is \((x+1)(x-1)=x^2-1\). Hence the simplified form is \(\frac{-4x}{x^2-1}\). The domain excludes \(x=1\) and \(x=-1\), where the original functions are undefined. Exam tip: retain the minus sign before the entire second numerator; losing it leads to option B.

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

For \(f(x)=\sqrt{x-1}\), we need \(x\ge 1\). However, in the quotient \(\frac{x-3}{\sqrt{x-1}}\), the denominator must not be zero. At \(x=1\), the denominator is zero, so the actual condition is \(x>1\), giving the domain \((1,\infty)\). Option B incorrectly includes \(x=1\). Exam tip: For a quotient involving a square root, apply both the square-root condition and the non-zero-denominator condition.

Correct answer: A. (-4)

(f(1)=-1) and (g(1)=4), so ((fg)(1)=-4). For a product function, multiply values, do not add formulas.

Correct answer: A. \(\{1\}\)

The product is \((fg)(x)=f(x)g(x)=\frac{1}{x^2+1}(x^2+1)=1\). Since \(x^2+1\) is never zero, the product is defined for every real \(x\), but its value is always 1. Hence, its range is only \(\{1\}\). Exam tip: simplify the product first and check whether it becomes a constant function.

Correct answer: A. ( [0,1] )

The direct answer is option A: \([0,1]\). A square root is defined over the reals only when its radicand is nonnegative. For \(f(x)=\sqrt{x}\), require \(x\ge0\). For \(g(x)=\sqrt{1-x}\), require \(1-x\ge0\), so \(x\le1\). Both functions must exist because they are being added. The intersection of \(x\ge0\) and \(x\le1\) is \(0\le x\le1\), or \([0,1]\). Option A is correct and includes endpoints because both radicands are allowed to be zero. Option B wrongly excludes 0 and 1. Option C includes values greater than 1, where \(\sqrt{1-x}\) is not real. Option D includes negative values, where \(\sqrt{x}\) is not real. For sums, intersect the individual domains.

Correct answer: A. p=−2±2√5

The governing concept is the minimum value of an upward-opening quadratic. Add the functions first: (f+g)(x)=x²+px+4+2x+1=x²+(p+2)x+5. Since the coefficient of x² is positive, the quadratic has a minimum. For that minimum to equal zero, the graph must touch the x-axis at its vertex, so the discriminant must be zero. Therefore (p+2)²−4(1)(5)=0, giving (p+2)²=20. Taking both square roots yields p+2=±2√5, hence p=−2±2√5. This is option A. Option B omits the shift by −2, option C gives only the shift and ignores the square-root possibilities, and option D uses an incorrect value from the discriminant equation.