📝 Partial fractions linear factors (33 MCQs)

📖 Explanation: The denominator consists of distinct linear factors (x-2) and (x+3). According to the linear factor rule, each distinct linear factor contributes one term with a constant numerator. Therefore, $\frac{5x+1}{(x-2)(x+3)} = \frac{A}{x-2} + \frac{B}{x+3}$. Options B and C incorrectly use linear numerators for linear denominators, while option D fails to decompose the fraction.

📖 Explanation: The error is that the student combined the terms for the repeated linear factor (x-1)^2. The correct form for a repeated linear factor (x-1)^2 is $\frac{A}{x-1} + \frac{B}{(x-1)^2}$. Writing it as $\frac{A}{x-1} + \frac{B}{x-1}$ suggests two distinct factors and results in an incorrect algebraic form when solving for coefficients. This common error stems from not recognizing that the denominators for repeated factors must be distinct powers.

📖 Explanation: The integral $\int \frac{dS}{200 - S}$ has a denominator with a single linear factor. Its partial fraction decomposition is simply $\frac{A}{200-S}$. The equation separates to $\frac{dS}{200-S} = \frac{dt}{50}$, integrating to $-\ln|200-S| = \frac{t}{50} + C$. This is a direct Easy of the linear factor rule where the denominator has no repeated or quadratic factors, simplifying the integration process for a mixing problem.

📖 Explanation: Substituting x = -1 eliminates the term with B, leaving $4(-1)-5 = A(-1-2) \Rightarrow -9 = -3A \Rightarrow A=3$. Substituting x = 2 eliminates A, giving $8-5 = B(3) \Rightarrow B=1$. This is the most efficient method for distinct linear factors because it directly solves for each constant without solving a system of equations. The other options are either more time-consuming or incorrect.

📖 Explanation: Clearing denominators: $s+3 = A(s+2) + B(s-1)$. Substitute s=1: $4 = A(3) \Rightarrow A = 4/3$. Substitute s=-2: $1 = B(-3) \Rightarrow B = -1/3$. This decomposition is crucial in control systems and signal processing for finding inverse Laplace transforms and analyzing system response. The constants represent the residues at the poles s=1 and s=-2.

📖 Explanation: The cover-up method directly computes the coefficient for a specific linear factor. Covering (x-1) and substituting x=1 isolates the coefficient A for the term $\frac{A}{x-1}$. This works because all other terms in the decomposition either become zero at x=1 or have the factor (x-1) in the denominator, which is cancelled. The result 10/3 is indeed A. This method is efficient but only works for distinct linear factors.

📖 Explanation: The decomposition is $\frac{1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$. Clearing denominators gives $1 = A(x-2) + B(x-1)$. Substituting x=1 gives $1 = A(-1) \Rightarrow A = -1$. Substituting x=2 gives $1 = B(1) \Rightarrow B = 1$. Therefore, $\frac{1}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{1}{x-2} = \frac{1}{x-2} - \frac{1}{x-1}$. Option A is correct. This is a fundamental example of the linear factor rule.

📖 Explanation: The factor (x-1)^2 is a repeated linear factor, so it contributes two terms: $\frac{A}{x-1} + \frac{B}{(x-1)^2}$. The distinct linear factor (x+2) contributes one term $\frac{C}{x+2}$. Thus, the correct decomposition is $\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$. Option B incorrectly uses a linear numerator for the repeated factor; option C uses an unnecessary linear numerator; and option D incorrectly combines the repeated factors.

📖 Explanation: According to the linear factor rule, each distinct linear factor (ax+b)^m contributes m terms. The factor (x-1)^3 contributes 3 terms: $\frac{A_1}{x-1} + \frac{A_2}{(x-1)^2} + \frac{A_3}{(x-1)^3}$. The factor (x+4) contributes 1 term: $\frac{B}{x+4}$. Therefore, the total number of terms is 3 + 1 = 4. This is a direct Easy of the rule for repeated linear factors.

📖 Explanation: The correct equation after clearing denominators is $3x-1 = A(x-2) + B(x+1)$. Substituting A=2 and B=1 gives $2(x-2) + 1(x+1) = 2x-4 + x+1 = 3x-3$, which is not equal to $3x-1$. This indicates that the student's coefficients are incorrect. The correct coefficients are A=2, B=1? Let's check: A=2, B=1 gives $2x-4+x+1 = 3x-3$, so it's wrong. The correct solution is A=4/3, B=5/3. This question tests the ability to verify a solution by substitution.

📖 Explanation: For linear factors, the numerators in the partial fraction decomposition are always constants, regardless of whether the factor is repeated or distinct. For a factor (x-a)^m, the terms are $\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + ... + \frac{A_m}{(x-a)^m}$, where all A_i are constants. Option A correctly states this. Options B and C are incorrect because they either restrict to repeated factors or incorrectly suggest linear numerators. Option D is incorrect as the numerators are constants, not polynomials.

📖 Explanation: Partial fractions with only constant numerators occur when all factors in the denominator are linear (not quadratic) and the numerator is of lower degree than the denominator. Option A has denominator (x-1)(x+2), which consists of distinct linear factors, so it will have constant numerators. Options B and C contain irreducible quadratic factors (x^2+1), which require linear numerators (Ax+B). Option D has a repeated linear factor but still uses constant numerators. The key is that the denominator must be factorable into linear factors.

📖 Explanation: The decomposition is $\frac{4x}{(x-3)(x+1)} = \frac{3}{x-3} + \frac{1}{x+1}$. Integrating term by term gives $3\ln|x-3| + \ln|x+1| + C$. Using logarithm properties, this can be written as $\ln| (x-3)^3(x+1) | + C$. Therefore, options A and C are equivalent and both correct. This question combines the linear factor rule, integration of logarithmic functions, and logarithm properties to combine terms.

📖 Explanation: The decomposition gives $f(x) = \frac{4/3}{x-1} - \frac{1/3}{x+2}$. Evaluating at x=0: $f(0) = \frac{4/3}{-1} - \frac{1/3}{2} = -4/3 - 1/6 = -8/6 - 1/6 = -9/6 = -3/2$. This matches the original function $f(0) = \frac{0+3}{(-1)(2)} = 3/(-2) = -3/2$. This question checks understanding of the equivalence of the original and decomposed forms by evaluating at a convenient point.

📖 Explanation: Since the degree of the numerator (2) equals the degree of the denominator (2), the function is improper. Long division: $(2x^2+3) \div (x^2-1) = 2$ with remainder 5. So $\frac{2x^2+3}{x^2-1} = 2 + \frac{5}{x^2-1} = 2 + \frac{5}{(x-1)(x+1)}$. Decomposing $\frac{5}{(x-1)(x+1)} = \frac{5/2}{x-1} - \frac{5/2}{x+1}$. Thus, $2 + \frac{5/2}{x-1} - \frac{5/2}{x+1}$. Option A is incorrect as it shows 2 + 2/(x-1) + 1/(x+1). The correct answer is $2 + \frac{5/2}{x-1} - \frac{5/2}{x+1}$. This question tests the process of handling improper rational functions before partial fraction decomposition.

📖 Explanation: For a linear factor, the numerator in a partial fraction term must be a constant. The form $\frac{Ax+B}{x-1}$ is not a valid partial fraction term because the numerator is linear. Such a form would be used for a quadratic factor, not a linear one. Options B, C, and D all have constant numerators, which are valid for linear factors (with C being for a repeated linear factor). This is a common misconception where students might mistakenly use linear numerators for linear denominators.

📖 Explanation: Let $\frac{x^2+2x+3}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$. To find A+B+C, we can multiply both sides by x and take the limit as x→∞. The left side becomes $\frac{x^2+2x+3}{x^3-6x^2+11x-6} \cdot x \approx \frac{x^3}{x^3} = 1$. The right side becomes $\frac{Ax}{x-1} + \frac{Bx}{x-2} + \frac{Cx}{x-3}$, which approaches A+B+C. Therefore, A+B+C = 1. Alternatively, substitute a large value of x to approximate. This is an advanced technique using limits to find the sum of coefficients without solving for each individually.

📖 Explanation: To find A for the term $\frac{A}{x-1}$, the cover-up method involves covering up (x-1) in the original expression and evaluating the remaining expression at x=1. The remaining expression is $\frac{5x+1}{x+2}$. Substituting x=1 gives $\frac{5(1)+1}{1+2} = 6/3 = 2$. Option A is correct. Option B would be division by zero. Option C is the original expression, which is undefined at x=1. Option D incorrectly evaluates the original expression with a different factor covered. This question tests the correct Easy of the cover-up method.

📖 Explanation: The partial fraction form makes it easy to find the limit: $\lim_{t \to \infty} x(t) = \lim_{t \to \infty} \left( \frac{5/3}{t-1} - \frac{1/3}{t+2} \right) = 0 - 0 = 0$. This matches the original function where the degree of the numerator (1) is less than the degree of the denominator (2), so the limit is 0. This question demonstrates the utility of partial fractions in analyzing asymptotic behavior of functions by separating terms that tend to zero.

📖 Explanation: The denominator has factors x and (x-1)^2. The decomposition is $\frac{1}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$. Clearing denominators: $1 = A(x-1)^2 + Bx(x-1) + Cx$. Substitute x=0: $1 = A(1) \Rightarrow A=1$. Substitute x=1: $1 = C(1) \Rightarrow C=1$. To find B, equate coefficients of x^2: $0 = A + B \Rightarrow B = -1$. Thus, $\frac{1}{x} - \frac{1}{x-1} + \frac{1}{(x-1)^2}$. Wait, option A is $\frac{1}{x} + \frac{1}{x-1} - \frac{1}{(x-1)^2}$, which is incorrect. Let's re-evaluate: A=1, C=1, B=-1 gives $\frac{1}{x} - \frac{1}{x-1} + \frac{1}{(x-1)^2}$. None of the options match exactly. The correct decomposition is $\frac{1}{x} - \frac{1}{x-1} + \frac{1}{(x-1)^2}$. Option B is close but has a negative sign error for the last term. The correct answer is B. Let's check: B gives $\frac{1}{x} - \frac{1}{x-1} + \frac{1}{(x-1)^2} = \frac{(x-1)^2 - x(x-1) + x}{x(x-1)^2} = \frac{x^2-2x+1 - x^2 + x + x}{x(x-1)^2} = \frac{1}{x(x-1)^2}$. Yes, B is correct.

📖 Explanation: The integral is $\int_{0}^{1} \left( \frac{4/3}{x-1} - \frac{1/3}{x+2} \right) dx = \frac{4}{3}[\ln|x-1|]_{0}^{1} - \frac{1}{3}[\ln|x+2|]_{0}^{1}$. However, the first integral is improper because of the singularity at x=1. Evaluating the improper integral: $\lim_{b \to 1^-} \frac{4}{3}\ln|b-1| - \frac{4}{3}\ln|0-1| - \frac{1}{3}\ln 3 + \frac{1}{3}\ln 2$. The term $\frac{4}{3}\ln|b-1|$ diverges to -∞, so the integral diverges. Therefore, none of the options are correct. The question is flawed. However, if the integral was from 0 to 1/2, it would be $\frac{4}{3}\ln(1/2) - \frac{1}{3}\ln(5/2)$. The correct answer should be 'diverges'. But option A gives a finite value. This is an Medium question: recognize the improper integral and the divergence.

📖 Explanation: The denominator has a linear factor (x-1) and an irreducible quadratic factor (x^2+2x+2). The linear factor contributes one term with a constant numerator: $\frac{A}{x-1}$. The quadratic factor contributes one term with a linear numerator: $\frac{Bx+C}{x^2+2x+2}$. Thus, there are 2 terms total. This question combines the rules for linear and quadratic factors, requiring understanding of both types of factor contributions.

📖 Explanation: The original function and its partial fraction decomposition are algebraic identities that hold for all values of x where both sides are defined. The only points where they are undefined are the roots of the denominator: x=1 and x=-2. For all other real numbers, the equivalence holds. This addresses a common misconception that partial fraction decompositions might have restricted domains beyond the points of discontinuity. They are actually equivalent functions on their common domain.

📖 Explanation: The correct sequence is to first decompose the rational function into partial fractions, which yields terms of the form $\frac{A}{x+1} + \frac{B}{x-2}$. Then, each term is integrated using the formula $\int \frac{1}{x-a} dx = \ln|x-a| + C$. Option A correctly describes this process. Option B is incorrect because the derivative of the denominator is not the numerator. Option C is incorrect because the integration of 1/x terms involves logarithms. Option D is unrelated.

📖 Explanation: Clearing denominators: $2x+5 = A(x+3) + B(x-1) = (A+B)x + (3A - B)$. Equating coefficients gives: coefficient of x: A+B = 2; constant term: 3A - B = 5. Option A correctly represents this system. Option B has the wrong sign for the constant term equation. Option C incorrectly sets the coefficients. Option D has the wrong constant term equation. This tests the ability to correctly set up and solve a system of equations for partial fraction coefficients.

📖 Explanation: The decomposition is $\frac{1}{(x-a)(x-b)} = \frac{1}{a-b} \left( \frac{1}{x-a} - \frac{1}{x-b} \right)$. Integrating gives $\frac{1}{a-b} (\ln|x-a| - \ln|x-b|) + C = \frac{1}{a-b} \ln \left| \frac{x-a}{x-b} \right| + C$. Option A is correct. Option B has the denominator reversed (b-a instead of a-b), which would give the negative of the correct result. Option C has the logarithm argument inverted. Option D is incorrect. This is a general formula that can be derived from the linear factor rule.

📖 Explanation: To find the original numerator, combine the fractions: $\frac{5}{x+1} + \frac{2}{2x-3} = \frac{5(2x-3) + 2(x+1)}{(x+1)(2x-3)} = \frac{10x - 15 + 2x + 2}{(x+1)(2x-3)} = \frac{12x - 13}{(x+1)(2x-3)}$. Thus, the original numerator is 12x - 13. Option A is correct. This question reverses the process, requiring the student to combine partial fractions to reconstruct the original rational function, which tests understanding of the equivalence.

📖 Explanation: The student has used a linear numerator for the linear factor (x-1), which is incorrect. The correct form should have a constant numerator for (x-1). Setting up the incorrect form: $3x+2 = (Ax+B)(x+2) + C(x-1) = Ax^2 + (2A+B)x + (2B - C)$. Equating coefficients gives A=0 from x^2, then 2A+B = B = 3, and 2B - C = 6 - C = 2, so C=4. There is a solution! A=0, B=3, C=4. The decomposition is $\frac{0x+3}{x-1} + \frac{4}{x+2} = \frac{3}{x-1} + \frac{4}{x+2}$, which is actually correct. So the error does not lead to no solution; the extra degree of freedom is resolved with A=0. This is a trick question. The correct answer is that the decomposition will still be correct with A=0. This highlights that adding an unnecessary linear numerator to a linear denominator may still yield a correct decomposition if the extra coefficient becomes zero.

📖 Explanation: Substituting x=1 in the equation $4x-1 = A(x+3) + B(x-1)$ gives $3 = A(4) \Rightarrow A = 3/4$. Substituting x=0 gives $-1 = A(3) + B(-1) = 3A - B \Rightarrow -1 = 9/4 - B \Rightarrow B = 13/4$. However, to find B more directly, one should substitute x=-3, which eliminates A and gives $-13 = B(-4) \Rightarrow B = 13/4$. Substituting x=0 is valid but requires knowing A first. The student's approach is valid but less efficient. Option A is incorrect because x=0 is sufficient to find B once A is known. The correct answer is D: the approach is correct and will yield the correct values. The question tests whether students recognize that any convenient values of x can be used, but some are more efficient than others.

📖 Explanation: For a repeated linear factor (x-a)^m, the decomposition is $\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + ... + \frac{A_m}{(x-a)^m}$, where all numerators are constants. This is a direct Easy of the linear factor rule for repeated factors. Option B is incorrect because it suggests polynomial numerators. Option C incorrectly combines the terms into a single denominator. Option D is incomplete because it doesn't account for the lower powers of (x-a).

📖 Explanation: Long division is required when the degree of the numerator is greater than or equal to the degree of the denominator. In option A, the numerator has degree 3, and the denominator has degree 2 (since (x-1)(x+2) = x^2+x-2), so long division is necessary. In options B, C, and D, the degree of the numerator is less than the degree of the denominator, so they are proper rational functions and can be decomposed directly. This question tests the prerequisite condition for applying partial fractions.

📖 Explanation: The decomposition gives $\frac{11/5}{x-2} + \frac{24/5}{x+3}$. Integrating term by term: $\frac{11}{5}\ln|x-2| + \frac{24}{5}\ln|x+3| + C$. Option A correctly matches the coefficients with the corresponding logarithmic terms. Option B has the wrong sign for the second term. Option C swaps the coefficients. Option D swaps the coefficients and the arguments of the logarithms. This tests the ability to correctly integrate partial fraction terms.

📖 Explanation: Check the student's claim: $\frac{1}{x-1} - \frac{1}{x+1} = \frac{(x+1) - (x-1)}{(x-1)(x+1)} = \frac{2}{(x-1)(x+1)}$. This is actually correct! Wait, 2/( (x-1)(x+1) ) = 1/(x-1) - 1/(x+1). Yes, the student is correct. The decomposition is valid. So the answer is A. This is a trick question where the student's claim is actually correct, testing the student's ability to verify the result rather than assuming it's wrong.