IB Mathematics: Analysis and Approaches Quiz: Partial Fractions
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Partial FractionsQuestion 1 of 20

Let f(x)=1x2−3x+2f(x) = \frac{1}{x^2-3x+2}. The partial fraction decomposition of f(x)f(x) is Ax−2+Bx−1\frac{A}{x-2} + \frac{B}{x-1}. Which statement correctly describes the behavior of the graph of y=f(x)y=f(x) near its vertical asymptotes?

The function approaches +∞\infty on both sides of each asymptote.
The function approaches +∞\infty as x→1−x \to 1^{-} and +∞\infty as x→2+x \to 2^{+}.
The function approaches -∞\infty as x→1−x \to 1^{-} and -∞\infty as x→2+x \to 2^{+}.
The function approaches +∞\infty as x→1+x \to 1^{+} and -∞\infty as x→2+x \to 2^{+}.
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Partial Fractions

Practice Partial Fractions in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Partial Fractions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let f(x)=1x2−3x+2f(x) = \frac{1}{x^2-3x+2}. The partial fraction decomposition of f(x)f(x) is Ax−2+Bx−1\frac{A}{x-2} + \frac{B}{x-1}. Which statement correctly describes the behavior of the graph of y=f(x)y=f(x) near its vertical asymptotes?

  1. The function approaches +∞\infty on both sides of each asymptote.
  2. The function approaches +∞\infty as x→1−x \to 1^{-} and +∞\infty as x→2+x \to 2^{+}. (correct answer)
  3. The function approaches -∞\infty as x→1−x \to 1^{-} and -∞\infty as x→2+x \to 2^{+}.
  4. The function approaches +∞\infty as x→1+x \to 1^{+} and -∞\infty as x→2+x \to 2^{+}.
Explanation: First, find the partial fraction decomposition. The denominator is x2−3x+2=(x−1)(x−2)x^2-3x+2 = (x-1)(x-2). The vertical asymptotes are at x=1x=1 and x=2x=2. 1(x−2)(x−1)=Ax−2+Bx−1\frac{1}{(x-2)(x-1)} = \frac{A}{x-2} + \frac{B}{x-1} 1=A(x−1)+B(x−2)1 = A(x-1) + B(x-2) Let x=2x=2: 1=A(2−1)  ⟹  A=11 = A(2-1) \implies A=1. Let x=1x=1: 1=B(1−2)  ⟹  B=−11 = B(1-2) \implies B=-1. So, f(x)=1x−2−1x−1f(x) = \frac{1}{x-2} - \frac{1}{x-1}. The behavior near the asymptotes is determined by the signs of the coefficients A and B. Near x=1x=1, the term −1x−1-\frac{1}{x-1} dominates. As x→1−x \to 1^{-} (e.g., x=0.99), x−1x-1 is a small negative number. So −1x−1-\frac{1}{x-1} is a large positive number. Thus, f(x)→+∞f(x) \to +\infty. As x→1+x \to 1^{+} (e.g., x=1.01), x−1x-1 is a small positive number. So −1x−1-\frac{1}{x-1} is a large negative number. Thus, f(x)→−∞f(x) \to -\infty. Near x=2x=2, the term 1x−2\frac{1}{x-2} dominates. As x→2−x \to 2^{-} (e.g., x=1.99), x−2x-2 is a small negative number. So 1x−2\frac{1}{x-2} is a large negative number. Thus, f(x)→−∞f(x) \to -\infty. As x→2+x \to 2^{+} (e.g., x=2.01), x−2x-2 is a small positive number. So 1x−2\frac{1}{x-2} is a large positive number. Thus, f(x)→+∞f(x) \to +\infty. Checking the options: B) The function approaches +∞\infty as x→1−x \to 1^{-} (correct) and +∞\infty as x→2+x \to 2^{+} (correct). This statement is true.

Question 2

The expression kx−1(x−3)(x+2)\frac{kx-1}{(x-3)(x+2)} is written as Ax−3+Bx+2\frac{A}{x-3} + \frac{B}{x+2}. Given that A=2A=2, find the value of kk.

  1. 1
  2. 3
  3. \frac{11}{3} (correct answer)
  4. \frac{9}{2}
Explanation: We start with the identity kx−1=A(x+2)+B(x−3)kx-1 = A(x+2) + B(x-3). We can find an expression for A using the cover-up method. Let x=3x=3: k(3)−1=A(3+2)+B(0)  ⟹  3k−1=5Ak(3)-1 = A(3+2) + B(0) \implies 3k-1 = 5A. So, A=3k−15A = \frac{3k-1}{5}. We are given that A=2A=2. 2=3k−152 = \frac{3k-1}{5} 10=3k−110 = 3k-1 11=3k11 = 3k k=113k = \frac{11}{3}. Alternatively, we can use simultaneous equations. We are given A=2A=2. Equating coefficients of xx: k=A+B=2+Bk = A+B = 2+B. Equating constant terms: −1=2A−3B=2(2)−3B=4−3B-1 = 2A-3B = 2(2)-3B = 4-3B. From the constant terms: −5=−3B  ⟹  B=5/3-5 = -3B \implies B=5/3. Then, substitute B back into the equation for k: k=2+5/3=6/3+5/3=11/3k = 2 + 5/3 = 6/3 + 5/3 = 11/3. Distractor Rationale: A) 1: Results from an error in the cover-up denominator, e.g., using 3−2=13-2=1 instead of 3−(−2)=53-(-2)=5. B) 3: Results from an algebraic slip, e.g., 3k−1=10  ⟹  3k=9  ⟹  k=33k-1=10 \implies 3k=9 \implies k=3. D) 9/2: Results from confusing A and B, setting the expression for B equal to 2.

Question 3

Given that 2x+32x2−x−1=A2x+1+Bx−1\frac{2x+3}{2x^2-x-1} = \frac{A}{2x+1} + \frac{B}{x-1}, find the value of BB.

  1. -\frac{4}{3}
  2. -\frac{1}{3}
  3. \frac{5}{3} (correct answer)
  4. 5
Explanation: First, factor the denominator: 2x2−x−1=(2x+1)(x−1)2x^2-x-1 = (2x+1)(x-1). So we have the identity 2x+3=A(x−1)+B(2x+1)2x+3 = A(x-1) + B(2x+1). To find B, we choose a value of xx that makes the term with A equal to zero. Let x=1x=1. 2(1)+3=A(1−1)+B(2(1)+1)  ⟹  5=A(0)+B(3)  ⟹  5=3B  ⟹  B=532(1)+3 = A(1-1) + B(2(1)+1) \implies 5 = A(0) + B(3) \implies 5 = 3B \implies B = \frac{5}{3}. For completeness, to find A, let x=−1/2x=-1/2. 2(−1/2)+3=A(−1/2−1)+B(0)  ⟹  −1+3=A(−3/2)  ⟹  2=−32A  ⟹  A=−432(-1/2)+3 = A(-1/2-1) + B(0) \implies -1+3 = A(-3/2) \implies 2 = -\frac{3}{2}A \implies A = -\frac{4}{3}. Distractor Rationale: A) -4/3: This is the value of A. B) -1/3: A possible arithmetic error. D) 5: Forgetting to divide by 3 when solving for B.

Question 4

Let f(x)=3−x(1−2x)(1+x)f(x) = \frac{3-x}{(1-2x)(1+x)}. When f(x)f(x) is expanded as a series in ascending powers of xx, find the coefficient of x2x^2.

  1. 5
  2. 7
  3. 8 (correct answer)
  4. \frac{16}{3}
Explanation: First, decompose f(x)f(x) into partial fractions: 3−x(1−2x)(1+x)=A1−2x+B1+x\frac{3-x}{(1-2x)(1+x)} = \frac{A}{1-2x} + \frac{B}{1+x} 3−x=A(1+x)+B(1−2x)3-x = A(1+x) + B(1-2x) Let x=1/2x=1/2: 3−1/2=A(1+1/2)  ⟹  5/2=A(3/2)  ⟹  A=5/33-1/2 = A(1+1/2) \implies 5/2 = A(3/2) \implies A=5/3. Let x=−1x=-1: 3−(−1)=B(1−2(−1))  ⟹  4=B(3)  ⟹  B=4/33-(-1) = B(1-2(-1)) \implies 4 = B(3) \implies B=4/3. So, f(x)=53(1−2x)−1+43(1+x)−1f(x) = \frac{5}{3}(1-2x)^{-1} + \frac{4}{3}(1+x)^{-1}. Now, use the binomial expansion (1+y)n=1+ny+n(n−1)2y2+...(1+y)^n = 1+ny+\frac{n(n-1)}{2}y^2+.... (1−2x)−1=1+(−1)(−2x)+(−1)(−2)2(−2x)2+...=1+2x+4x2+...(1-2x)^{-1} = 1 + (-1)(-2x) + \frac{(-1)(-2)}{2}(-2x)^2 + ... = 1+2x+4x^2+... (1+x)−1=1+(−1)(x)+(−1)(−2)2(x)2+...=1−x+x2+...(1+x)^{-1} = 1 + (-1)(x) + \frac{(-1)(-2)}{2}(x)^2 + ... = 1-x+x^2+... Substitute these back into the expression for f(x)f(x): f(x)=53(1+2x+4x2+...)+43(1−x+x2+...)f(x) = \frac{5}{3}(1+2x+4x^2+...) + \frac{4}{3}(1-x+x^2+...) The coefficient of x2x^2 is 53(4)+43(1)=203+43=243=8\frac{5}{3}(4) + \frac{4}{3}(1) = \frac{20}{3} + \frac{4}{3} = \frac{24}{3} = 8. Distractor Rationale: A) 5: This is the sum of the x2x^2 coefficients (4+1) without multiplying by A and B. B) 7: This results from swapping A and B: 43(4)+53(1)=16+53=7\frac{4}{3}(4) + \frac{5}{3}(1) = \frac{16+5}{3} = 7. D) 16/3: This results from a sign error in the expansion of (1+x)−1(1+x)^{-1}, for example, getting a coefficient of -1 for x2x^2. Then 53(4)+43(−1)=20−43=163\frac{5}{3}(4) + \frac{4}{3}(-1) = \frac{20-4}{3} = \frac{16}{3}.

Question 5

Let f(x)=1x2−3x+2f(x) = \frac{1}{x^2-3x+2}. The partial fraction decomposition of f(x)f(x) is Ax−2+Bx−1\frac{A}{x-2} + \frac{B}{x-1}. Which statement correctly describes the behavior of the graph of y=f(x)y=f(x) near its vertical asymptotes?

  1. The function approaches +∞\infty on both sides of each asymptote.
  2. The function approaches +∞\infty as x→1−x \to 1^{-} and +∞\infty as x→2+x \to 2^{+}. (correct answer)
  3. The function approaches -∞\infty as x→1−x \to 1^{-} and -∞\infty as x→2+x \to 2^{+}.
  4. The function approaches +∞\infty as x→1+x \to 1^{+} and -∞\infty as x→2+x \to 2^{+}.
Explanation: First, find the partial fraction decomposition. The denominator is x2−3x+2=(x−1)(x−2)x^2-3x+2 = (x-1)(x-2). The vertical asymptotes are at x=1x=1 and x=2x=2. 1(x−2)(x−1)=Ax−2+Bx−1\frac{1}{(x-2)(x-1)} = \frac{A}{x-2} + \frac{B}{x-1} 1=A(x−1)+B(x−2)1 = A(x-1) + B(x-2) Let x=2x=2: 1=A(2−1)  ⟹  A=11 = A(2-1) \implies A=1. Let x=1x=1: 1=B(1−2)  ⟹  B=−11 = B(1-2) \implies B=-1. So, f(x)=1x−2−1x−1f(x) = \frac{1}{x-2} - \frac{1}{x-1}. The behavior near the asymptotes is determined by the signs of the coefficients A and B. Near x=1x=1, the term −1x−1-\frac{1}{x-1} dominates. As x→1−x \to 1^{-} (e.g., x=0.99), x−1x-1 is a small negative number. So −1x−1-\frac{1}{x-1} is a large positive number. Thus, f(x)→+∞f(x) \to +\infty. As x→1+x \to 1^{+} (e.g., x=1.01), x−1x-1 is a small positive number. So −1x−1-\frac{1}{x-1} is a large negative number. Thus, f(x)→−∞f(x) \to -\infty. Near x=2x=2, the term 1x−2\frac{1}{x-2} dominates. As x→2−x \to 2^{-} (e.g., x=1.99), x−2x-2 is a small negative number. So 1x−2\frac{1}{x-2} is a large negative number. Thus, f(x)→−∞f(x) \to -\infty. As x→2+x \to 2^{+} (e.g., x=2.01), x−2x-2 is a small positive number. So 1x−2\frac{1}{x-2} is a large positive number. Thus, f(x)→+∞f(x) \to +\infty. Checking the options: B) The function approaches +∞\infty as x→1−x \to 1^{-} (correct) and +∞\infty as x→2+x \to 2^{+} (correct). This statement is true.

Question 6

Given k(x−a)(x+a)=Px−a+Qx+a\frac{k}{(x-a)(x+a)} = \frac{P}{x-a} + \frac{Q}{x+a}, where kk and aa are non-zero constants. Which of the following statements is always true?

  1. P = Q
  2. P + Q = 0 (correct answer)
  3. PQ = 1
  4. P - Q = k/a
Explanation: From the equation, we have the identity k=P(x+a)+Q(x−a)k = P(x+a) + Q(x-a). To find P, let x=ax=a: k=P(a+a)+Q(a−a)  ⟹  k=P(2a)  ⟹  P=k2ak = P(a+a) + Q(a-a) \implies k = P(2a) \implies P = \frac{k}{2a}. To find Q, let x=−ax=-a: k=P(−a+a)+Q(−a−a)  ⟹  k=Q(−2a)  ⟹  Q=−k2ak = P(-a+a) + Q(-a-a) \implies k = Q(-2a) \implies Q = -\frac{k}{2a}. Now we check the given statements: A) P=QP=Q is false, as k2a=−k2a\frac{k}{2a} = -\frac{k}{2a} implies k=0k=0 or a=0a=0, but we are given they are non-zero. B) P+Q=k2a+(−k2a)=0P+Q = \frac{k}{2a} + (-\frac{k}{2a}) = 0. This is always true. C) PQ=(k2a)(−k2a)=−k24a2PQ = (\frac{k}{2a})(-\frac{k}{2a}) = -\frac{k^2}{4a^2}. This is not generally 1. D) P−Q=k2a−(−k2a)=2k2a=kaP-Q = \frac{k}{2a} - (-\frac{k}{2a}) = \frac{2k}{2a} = \frac{k}{a}. This is not a general relationship like P+Q=0P+Q=0, but an expression for P−QP-Q. The question asks what is always true as a relationship between P and Q. P+Q=0P+Q=0 is a fundamental relationship derived from the structure of the problem.

Question 7

When 13x−126x2−x−2\frac{13x-12}{6x^2-x-2} is expressed as partial fractions, the result is A3x−2+B2x+1\frac{A}{3x-2} + \frac{B}{2x+1}. Find A−BA-B.

  1. -\frac{47}{7} (correct answer)
  2. -\frac{27}{7}
  3. \frac{27}{7}
  4. \frac{47}{7}
Explanation: First, confirm the denominator factorization: (3x−2)(2x+1)=6x2+3x−4x−2=6x2−x−2(3x-2)(2x+1) = 6x^2+3x-4x-2 = 6x^2-x-2. This is correct. We have the identity 13x−12=A(2x+1)+B(3x−2)13x-12 = A(2x+1) + B(3x-2). To find A, let 3x−2=03x-2=0, so x=2/3x=2/3. 13(23)−12=A(2(23)+1)  ⟹  263−363=A(43+33)  ⟹  −103=A(73)  ⟹  A=−10713(\frac{2}{3}) - 12 = A(2(\frac{2}{3})+1) \implies \frac{26}{3} - \frac{36}{3} = A(\frac{4}{3}+\frac{3}{3}) \implies -\frac{10}{3} = A(\frac{7}{3}) \implies A = -\frac{10}{7}. To find B, let 2x+1=02x+1=0, so x=−1/2x=-1/2. 13(−12)−12=B(3(−12)−2)  ⟹  −132−242=B(−32−42)  ⟹  −372=B(−72)  ⟹  B=37713(-\frac{1}{2}) - 12 = B(3(-\frac{1}{2})-2) \implies -\frac{13}{2} - \frac{24}{2} = B(-\frac{3}{2}-\frac{4}{2}) \implies -\frac{37}{2} = B(-\frac{7}{2}) \implies B = \frac{37}{7}. We need to find A−BA-B: A−B=−107−377=−477A-B = -\frac{10}{7} - \frac{37}{7} = -\frac{47}{7}. Distractor Rationale: B) -27/7: Results from a sign error, e.g., A=10/7A=10/7, giving 10/7−37/7=−27/710/7 - 37/7 = -27/7. C) 27/7: This is A+BA+B with a sign error on A, or B−AB-A with a sign error on B. Also the value of B+AB+A is −10/7+37/7=27/7-10/7 + 37/7 = 27/7. D) 47/7: This is B−AB-A.

Question 8

Find the exact value of the integral ∫246x2−1dx\int_2^4 \frac{6}{x^2-1} dx.

  1. 3\ln(59\frac{5}{9})
  2. 3\ln(95\frac{9}{5}) (correct answer)
  3. 3\ln(5)
  4. 6\ln(5)
Explanation: First, decompose the integrand using partial fractions. 6x2−1=6(x−1)(x+1)=Ax−1+Bx+1\frac{6}{x^2-1} = \frac{6}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} 6=A(x+1)+B(x−1)6 = A(x+1) + B(x-1) Let x=1x=1: 6=2A  ⟹  A=36 = 2A \implies A=3. Let x=−1x=-1: 6=−2B  ⟹  B=−36 = -2B \implies B=-3. So the integral becomes: ∫24(3x−1−3x+1)dx=[3ln⁡∣x−1∣−3ln⁡∣x+1∣]24\int_2^4 (\frac{3}{x-1} - \frac{3}{x+1}) dx = [3\ln|x-1| - 3\ln|x+1|]_2^4 Using logarithm properties, this is [3ln⁡∣x−1x+1∣]24[3\ln|\frac{x-1}{x+1}|]_2^4. Evaluate at the limits: 3ln⁡(4−14+1)−3ln⁡(2−12+1)=3ln⁡(35)−3ln⁡(13)3\ln(\frac{4-1}{4+1}) - 3\ln(\frac{2-1}{2+1}) = 3\ln(\frac{3}{5}) - 3\ln(\frac{1}{3}) =3(ln⁡(35)−ln⁡(13))=3ln⁡(3/51/3)=3ln⁡(35×3)=3ln⁡(95)= 3(\ln(\frac{3}{5}) - \ln(\frac{1}{3})) = 3\ln(\frac{3/5}{1/3}) = 3\ln(\frac{3}{5} \times 3) = 3\ln(\frac{9}{5}). Distractor Rationale: A) 3ln(5/9): Sign error, likely from reversing the subtraction of logs. C) 3ln(5): Common error if B is incorrectly found to be 3 instead of -3. The integral would become ∫3x−1+3x+1dx=3ln⁡∣x2−1∣\int \frac{3}{x-1} + \frac{3}{x+1} dx = 3\ln|x^2-1|, which evaluates to 3(ln⁡15−ln⁡3)=3ln⁡53(\ln 15 - \ln 3) = 3\ln 5. D) 6ln(5): A combination of the previous error and another mistake.

Question 9

Let f(x)=5(x+2)(x−3)f(x) = \frac{5}{(x+2)(x-3)}. Find the exact value of ∫45f(x)dx\int_4^5 f(x) dx.

  1. \ln(127\frac{12}{7}) (correct answer)
  2. \ln(712\frac{7}{12})
  3. \ln(542\frac{5}{42})
  4. \ln(2)
Explanation: When you encounter a rational function like this in an integral, the key technique is partial fraction decomposition. The function f(x)=5(x+2)(x−3)f(x) = \frac{5}{(x+2)(x-3)} has distinct linear factors in the denominator, so you can split it into simpler fractions. Set up the decomposition: 5(x+2)(x−3)=Ax+2+Bx−3\frac{5}{(x+2)(x-3)} = \frac{A}{x+2} + \frac{B}{x-3} Multiplying both sides by (x+2)(x−3)(x+2)(x-3): 5=A(x−3)+B(x+2)5 = A(x-3) + B(x+2) Solving for the constants: when x=3x = 3, we get 5=5B5 = 5B, so B=1B = 1. When x=−2x = -2, we get 5=−5A5 = -5A, so A=−1A = -1. Therefore: f(x)=−1x+2+1x−3f(x) = \frac{-1}{x+2} + \frac{1}{x-3} Now integrate: ∫45f(x)dx=∫45(−1x+2+1x−3)dx=[−ln⁡∣x+2∣+ln⁡∣x−3∣]45\int_4^5 f(x)dx = \int_4^5 \left(\frac{-1}{x+2} + \frac{1}{x-3}\right)dx = [-\ln|x+2| + \ln|x-3|]_4^5 Evaluating: [−ln⁡(7)+ln⁡(2)]−[−ln⁡(6)+ln⁡(1)]=−ln⁡(7)+ln⁡(2)+ln⁡(6)=ln⁡(2⋅67)=ln⁡(127)[-\ln(7) + \ln(2)] - [-\ln(6) + \ln(1)] = -\ln(7) + \ln(2) + \ln(6) = \ln\left(\frac{2 \cdot 6}{7}\right) = \ln\left(\frac{12}{7}\right) This confirms answer A is correct. Answer B gives the reciprocal, suggesting a sign error in the partial fractions. Answer C represents a calculation error, likely from incorrect partial fraction coefficients. Answer D results from overlooking one of the logarithmic terms in the evaluation. Remember: partial fraction decomposition is your go-to method for integrating rational functions with factorizable denominators. Always double-check your coefficients before integrating.

Question 10

The expression 8(2x+1)(2x−3)\frac{8}{(2x+1)(2x-3)} is written in the form A2x+1+B2x−3\frac{A}{2x+1} + \frac{B}{2x-3}. Find the value of B−AB-A.

  1. -4
  2. 0
  3. 2
  4. 4 (correct answer)
Explanation: We start with the identity 8=A(2x−3)+B(2x+1)8 = A(2x-3) + B(2x+1). To find A, let 2x+1=02x+1=0, which means x=−1/2x = -1/2. 8=A(2(−1/2)−3)+B(0)  ⟹  8=A(−1−3)  ⟹  8=−4A  ⟹  A=−28 = A(2(-1/2)-3) + B(0) \implies 8 = A(-1-3) \implies 8 = -4A \implies A=-2. To find B, let 2x−3=02x-3=0, which means x=3/2x = 3/2. 8=A(0)+B(2(3/2)+1)  ⟹  8=B(3+1)  ⟹  8=4B  ⟹  B=28 = A(0) + B(2(3/2)+1) \implies 8 = B(3+1) \implies 8 = 4B \implies B=2. The question asks for B−AB-A. B−A=2−(−2)=4B-A = 2 - (-2) = 4. Distractor Rationale: A) -4: This is A−BA-B. B) 0: This is A+BA+B. This could also result from a sign error where a student calculates A=2A=2 or B=−2B=-2. C) 2: This is just the value of B.

Question 11

A function is given by f(x)=4x+3−1x−1f(x) = \frac{4}{x+3} - \frac{1}{x-1}. Find the equivalent expression for f(x)f(x) as a single rational function.

  1. \frac{3x-1}{x^2+2x-3}
  2. \frac{3x-7}{x^2+2x-3} (correct answer)
  3. \frac{5x-1}{x^2+2x-3}
  4. \frac{3x-7}{x^2-3}
Explanation: To combine the fractions, we find a common denominator, which is (x+3)(x−1)(x+3)(x-1). f(x)=4(x−1)(x+3)(x−1)−1(x+3)(x+3)(x−1)f(x) = \frac{4(x-1)}{(x+3)(x-1)} - \frac{1(x+3)}{(x+3)(x-1)} f(x)=4(x−1)−(x+3)(x+3)(x−1)f(x) = \frac{4(x-1) - (x+3)}{(x+3)(x-1)} Now, simplify the numerator: 4x−4−x−3=3x−74x - 4 - x - 3 = 3x - 7 Expand the denominator: (x+3)(x−1)=x2−x+3x−3=x2+2x−3(x+3)(x-1) = x^2 - x + 3x - 3 = x^2+2x-3 So, f(x)=3x−7x2+2x−3f(x) = \frac{3x-7}{x^2+2x-3}. Distractor Rationale: A) 3x−1x2+2x−3\frac{3x-1}{x^2+2x-3}: A sign error in the numerator: 4x−4−x+3=3x−14x-4-x+3=3x-1. C) 5x−1x2+2x−3\frac{5x-1}{x^2+2x-3}: A sign error in the original expression, combining 4x+3+1x−1\frac{4}{x+3} + \frac{1}{x-1}, which gives a numerator of 4(x−1)+(x+3)=5x−14(x-1)+(x+3)=5x-1. D) 3x−7x2−3\frac{3x-7}{x^2-3}: A common error when multiplying binomials, (x+3)(x−1)=x2−3(x+3)(x-1) = x^2 - 3.

Question 12

Using the fact that 1n(n+1)=1n−1n+1\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}, find the sum of the infinite series ∑n=1∞2n(n+1)\sum_{n=1}^{\infty} \frac{2}{n(n+1)}.

  1. 2 (correct answer)
  2. 1
  3. 0
  4. The series diverges.
Explanation: When you encounter an infinite series with a given partial fraction decomposition, you're dealing with a telescoping series—one of the most elegant techniques in calculus for finding exact sums. Start by factoring out the constant from the given series: ∑n=1∞2n(n+1)=2∑n=1∞1n(n+1)\sum_{n=1}^{\infty} \frac{2}{n(n+1)} = 2\sum_{n=1}^{\infty} \frac{1}{n(n+1)} Now apply the given identity 1n(n+1)=1n−1n+1\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} to get: 2∑n=1∞(1n−1n+1)2\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) Write out the first few terms to see the telescoping pattern: 2[(11−12)+(12−13)+(13−14)+⋯ ]2\left[\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots\right] Notice how consecutive terms cancel: the −12-\frac{1}{2} from the first term cancels with +12+\frac{1}{2} from the second term, and so on. The partial sum up to nn terms is 2(1−1n+1)2\left(1 - \frac{1}{n+1}\right). As n→∞n \to \infty, this approaches 2(1−0)=22(1 - 0) = 2. Answer choice B (1) would result if you forgot to multiply by the factor of 2. Answer choice C (0) might occur if you incorrectly thought all terms cancel completely. Answer choice D (diverges) would be wrong because telescoping series with vanishing end terms always converge to finite values. The correct answer is A. Study tip: Whenever you see a partial fraction identity provided in a series problem, immediately look for telescoping. Write out several terms explicitly—the cancellation pattern will become obvious and guide you to the sum.

Question 13

Find the value of ∫352xx2−4dx\int_3^5 \frac{2x}{x^2-4} dx.

  1. \ln(157\frac{15}{7})
  2. \ln(4)
  3. \ln(215\frac{21}{5}) (correct answer)
  4. \ln(53\frac{5}{3})
Explanation: This integral can be solved in two ways. Method 1: u-substitution. Let u=x2−4u = x^2-4, then du=2x dxdu = 2x\,dx. The integral becomes ∫1udu=ln⁡∣u∣+C=ln⁡∣x2−4∣+C\int \frac{1}{u} du = \ln|u| + C = \ln|x^2-4| + C. Evaluating the definite integral: [ln⁡∣x2−4∣]35=ln⁡(52−4)−ln⁡(32−4)=ln⁡(21)−ln⁡(5)=ln⁡(215)[\ln|x^2-4|]_3^5 = \ln(5^2-4) - \ln(3^2-4) = \ln(21) - \ln(5) = \ln(\frac{21}{5}). Method 2: Partial Fractions. 2xx2−4=2x(x−2)(x+2)=Ax−2+Bx+2\frac{2x}{x^2-4} = \frac{2x}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2} 2x=A(x+2)+B(x−2)2x = A(x+2) + B(x-2) Let x=2x=2: 4=4A  ⟹  A=14 = 4A \implies A=1. Let x=−2x=-2: −4=−4B  ⟹  B=1-4 = -4B \implies B=1. The integral is ∫35(1x−2+1x+2)dx=[ln⁡∣x−2∣+ln⁡∣x+2∣]35\int_3^5 (\frac{1}{x-2} + \frac{1}{x+2}) dx = [\ln|x-2| + \ln|x+2|]_3^5. =[ln⁡∣(x−2)(x+2)∣]35=[ln⁡∣x2−4∣]35= [\ln|(x-2)(x+2)|]_3^5 = [\ln|x^2-4|]_3^5, which leads to the same result ln⁡(215)\ln(\frac{21}{5}). Distractor Rationale: A) ln⁡(157)\ln(\frac{15}{7}): This result is obtained if one incorrectly finds B=−1B=-1, leading to an integral of ln⁡∣x−2x+2∣\ln|\frac{x-2}{x+2}|. B) ln⁡(4)\ln(4): Result of an incorrect log calculation like ln⁡(21−5)=ln⁡(16)=4ln⁡(2)\ln(21-5)=\ln(16)=4\ln(2) or other arithmetic error. D) ln⁡(53)\ln(\frac{5}{3}): This would be the result of integrating 2x\frac{2}{x}, not 2xx2−4\frac{2x}{x^2-4}.

Question 14

Given ∫1k1x(x+1)dx=ln⁡(3/2)\int_1^k \frac{1}{x(x+1)} dx = \ln(3/2), find the value of kk, where k>1k>1.

  1. 2
  2. 3 (correct answer)
  3. 4
  4. 5
Explanation: First, decompose the integrand using partial fractions. 1x(x+1)=Ax+Bx+1\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} 1=A(x+1)+Bx1 = A(x+1) + Bx Let x=0x=0: 1=A(1)  ⟹  A=11 = A(1) \implies A=1. Let x=−1x=-1: 1=B(−1)  ⟹  B=−11 = B(-1) \implies B=-1. The integral becomes ∫1k(1x−1x+1)dx\int_1^k (\frac{1}{x} - \frac{1}{x+1}) dx. This integrates to [ln⁡∣x∣−ln⁡∣x+1∣]1k=[ln⁡∣xx+1∣]1k[\ln|x| - \ln|x+1|]_1^k = [\ln|\frac{x}{x+1}|]_1^k. Now, evaluate at the limits: ln⁡(kk+1)−ln⁡(11+1)=ln⁡(kk+1)−ln⁡(12)\ln(\frac{k}{k+1}) - \ln(\frac{1}{1+1}) = \ln(\frac{k}{k+1}) - \ln(\frac{1}{2}). Using log rules, this is ln⁡(k/(k+1)1/2)=ln⁡(2kk+1)\ln(\frac{k/(k+1)}{1/2}) = \ln(\frac{2k}{k+1}). We are given that the result is ln⁡(3/2)\ln(3/2). ln⁡(2kk+1)=ln⁡(32)\ln(\frac{2k}{k+1}) = \ln(\frac{3}{2}) 2kk+1=32\frac{2k}{k+1} = \frac{3}{2} 4k=3(k+1)4k = 3(k+1) 4k=3k+34k = 3k+3 k=3k=3. Distractor Rationale: A) 2: A simple guess or calculation error. C) 4: A simple guess or calculation error. D) 5: A simple guess or calculation error.

Question 15

The expression px+qx2−4\frac{px+q}{x^2-4} is equivalent to Ax−2+Bx+2\frac{A}{x-2} + \frac{B}{x+2}. Find an expression for AA in terms of pp and qq.

  1. 2p+q
  2. \frac{2p-q}{4}
  3. \frac{2p+q}{4} (correct answer)
  4. \frac{q-2p}{4}
Explanation: We start with the identity px+q=A(x+2)+B(x−2)px+q = A(x+2) + B(x-2). To find AA, we want to eliminate the term with BB. We can do this by setting x−2=0x-2=0, i.e., x=2x=2. Substitute x=2x=2 into the identity: p(2)+q=A(2+2)+B(2−2)p(2)+q = A(2+2) + B(2-2) 2p+q=A(4)+B(0)2p+q = A(4) + B(0) 2p+q=4A2p+q = 4A A=2p+q4A = \frac{2p+q}{4}. Distractor Rationale: A) 2p+q: Forgetting to divide by 4. B) 2p−q4\frac{2p-q}{4}: This is the expression for B. D) q−2p4\frac{q-2p}{4}: A sign error in the expression for B.

Question 16

If 2x−11(x−4)(x+1)=Ax−4+Bx+1\frac{2x-11}{(x-4)(x+1)} = \frac{A}{x-4} + \frac{B}{x+1}, find the product ABAB.

  1. -\frac{39}{5}
  2. -\frac{39}{25} (correct answer)
  3. \frac{39}{25}
  4. \frac{39}{5}
Explanation: We have the identity 2x−11=A(x+1)+B(x−4)2x-11 = A(x+1) + B(x-4). To find A, let x=4x=4: 2(4)−11=A(4+1)  ⟹  8−11=5A  ⟹  −3=5A  ⟹  A=−352(4)-11 = A(4+1) \implies 8-11 = 5A \implies -3 = 5A \implies A = -\frac{3}{5}. To find B, let x=−1x=-1: 2(−1)−11=B(−1−4)  ⟹  −2−11=−5B  ⟹  −13=−5B  ⟹  B=1352(-1)-11 = B(-1-4) \implies -2-11 = -5B \implies -13 = -5B \implies B = \frac{13}{5}. The product ABAB is: AB=(−35)(135)=−3925AB = (-\frac{3}{5})(\frac{13}{5}) = -\frac{39}{25}. Distractor Rationale: A) -39/5: An error in multiplying the denominators (51 instead of 55). C) 39/25: A sign error in calculating either A or B. D) 39/5: Combination of a sign error and a denominator multiplication error.

Question 17

The expression kx−1(x−3)(x+2)\frac{kx-1}{(x-3)(x+2)} is written as Ax−3+Bx+2\frac{A}{x-3} + \frac{B}{x+2}. Given that A=2A=2, find the value of kk.

  1. 1
  2. 3
  3. \frac{11}{3} (correct answer)
  4. \frac{9}{2}
Explanation: We start with the identity kx−1=A(x+2)+B(x−3)kx-1 = A(x+2) + B(x-3). We can find an expression for A using the cover-up method. Let x=3x=3: k(3)−1=A(3+2)+B(0)  ⟹  3k−1=5Ak(3)-1 = A(3+2) + B(0) \implies 3k-1 = 5A. So, A=3k−15A = \frac{3k-1}{5}. We are given that A=2A=2. 2=3k−152 = \frac{3k-1}{5} 10=3k−110 = 3k-1 11=3k11 = 3k k=113k = \frac{11}{3}. Alternatively, we can use simultaneous equations. We are given A=2A=2. Equating coefficients of xx: k=A+B=2+Bk = A+B = 2+B. Equating constant terms: −1=2A−3B=2(2)−3B=4−3B-1 = 2A-3B = 2(2)-3B = 4-3B. From the constant terms: −5=−3B  ⟹  B=5/3-5 = -3B \implies B=5/3. Then, substitute B back into the equation for k: k=2+5/3=6/3+5/3=11/3k = 2 + 5/3 = 6/3 + 5/3 = 11/3. Distractor Rationale: A) 1: Results from an error in the cover-up denominator, e.g., using 3−2=13-2=1 instead of 3−(−2)=53-(-2)=5. B) 3: Results from an algebraic slip, e.g., 3k−1=10  ⟹  3k=9  ⟹  k=33k-1=10 \implies 3k=9 \implies k=3. D) 9/2: Results from confusing A and B, setting the expression for B equal to 2.

Question 18

Given ∫1k1x(x+1)dx=ln⁡(3/2)\int_1^k \frac{1}{x(x+1)} dx = \ln(3/2), find the value of kk, where k>1k>1.

  1. 2
  2. 3 (correct answer)
  3. 4
  4. 5
Explanation: First, decompose the integrand using partial fractions. 1x(x+1)=Ax+Bx+1\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} 1=A(x+1)+Bx1 = A(x+1) + Bx Let x=0x=0: 1=A(1)  ⟹  A=11 = A(1) \implies A=1. Let x=−1x=-1: 1=B(−1)  ⟹  B=−11 = B(-1) \implies B=-1. The integral becomes ∫1k(1x−1x+1)dx\int_1^k (\frac{1}{x} - \frac{1}{x+1}) dx. This integrates to [ln⁡∣x∣−ln⁡∣x+1∣]1k=[ln⁡∣xx+1∣]1k[\ln|x| - \ln|x+1|]_1^k = [\ln|\frac{x}{x+1}|]_1^k. Now, evaluate at the limits: ln⁡(kk+1)−ln⁡(11+1)=ln⁡(kk+1)−ln⁡(12)\ln(\frac{k}{k+1}) - \ln(\frac{1}{1+1}) = \ln(\frac{k}{k+1}) - \ln(\frac{1}{2}). Using log rules, this is ln⁡(k/(k+1)1/2)=ln⁡(2kk+1)\ln(\frac{k/(k+1)}{1/2}) = \ln(\frac{2k}{k+1}). We are given that the result is ln⁡(3/2)\ln(3/2). ln⁡(2kk+1)=ln⁡(32)\ln(\frac{2k}{k+1}) = \ln(\frac{3}{2}) 2kk+1=32\frac{2k}{k+1} = \frac{3}{2} 4k=3(k+1)4k = 3(k+1) 4k=3k+34k = 3k+3 k=3k=3. Distractor Rationale: A) 2: A simple guess or calculation error. C) 4: A simple guess or calculation error. D) 5: A simple guess or calculation error.

Question 19

For a non-zero constant cc, the expression cx(x+c)\frac{c}{x(x+c)} is decomposed into Ax+Bx+c\frac{A}{x} + \frac{B}{x+c}. Find the value of A−BA-B.

  1. 0
  2. 1
  3. 2c
  4. 2 (correct answer)
Explanation: When you encounter a rational expression that needs to be decomposed into partial fractions, you're working with one of the most systematic algebraic techniques in IB Math AA. The goal is to break down a complex fraction into simpler parts that are easier to work with, especially for integration or solving equations. To decompose cx(x+c)=Ax+Bx+c\frac{c}{x(x+c)} = \frac{A}{x} + \frac{B}{x+c}, multiply both sides by the common denominator x(x+c)x(x+c): c=A(x+c)+Bxc = A(x+c) + Bx Expanding the right side: c=Ax+Ac+Bx=(A+B)x+Acc = Ax + Ac + Bx = (A+B)x + Ac For this equation to hold for all values of x, the coefficients of like terms must be equal:
  • Coefficient of x: A+B=0A + B = 0
  • Constant term: Ac=cAc = c
From the second equation, since c≠0c \neq 0, we get A=1A = 1. From the first equation: 1+B=01 + B = 0, so B=−1B = -1. Therefore, A−B=1−(−1)=2A - B = 1 - (-1) = 2. Looking at the wrong answers: (A) 0 would result if you incorrectly concluded A=BA = B, missing the sign difference. (B) 1 comes from finding AA correctly but forgetting to subtract BB, or miscalculating the final step. (C) 2c appears if you don't fully solve for the coefficients and leave them in terms of c. Remember this pattern: in partial fraction decomposition, always set up the equation by clearing denominators, then match coefficients systematically. The arithmetic often involves simple integers, even when the original expression contains parameters like c.

Question 20

Let f(x)=5(x+2)(x−3)f(x) = \frac{5}{(x+2)(x-3)}. Find the exact value of ∫45f(x)dx\int_4^5 f(x) dx.

  1. \ln(127\frac{12}{7}) (correct answer)
  2. \ln(712\frac{7}{12})
  3. \ln(542\frac{5}{42})
  4. \ln(2)
Explanation: When you encounter a rational function like this in an integral, the key technique is partial fraction decomposition. The function f(x)=5(x+2)(x−3)f(x) = \frac{5}{(x+2)(x-3)} has distinct linear factors in the denominator, so you can split it into simpler fractions. Set up the decomposition: 5(x+2)(x−3)=Ax+2+Bx−3\frac{5}{(x+2)(x-3)} = \frac{A}{x+2} + \frac{B}{x-3} Multiplying both sides by (x+2)(x−3)(x+2)(x-3): 5=A(x−3)+B(x+2)5 = A(x-3) + B(x+2) Solving for the constants: when x=3x = 3, we get 5=5B5 = 5B, so B=1B = 1. When x=−2x = -2, we get 5=−5A5 = -5A, so A=−1A = -1. Therefore: f(x)=−1x+2+1x−3f(x) = \frac{-1}{x+2} + \frac{1}{x-3} Now integrate: ∫45f(x)dx=∫45(−1x+2+1x−3)dx=[−ln⁡∣x+2∣+ln⁡∣x−3∣]45\int_4^5 f(x)dx = \int_4^5 \left(\frac{-1}{x+2} + \frac{1}{x-3}\right)dx = [-\ln|x+2| + \ln|x-3|]_4^5 Evaluating: [−ln⁡(7)+ln⁡(2)]−[−ln⁡(6)+ln⁡(1)]=−ln⁡(7)+ln⁡(2)+ln⁡(6)=ln⁡(2⋅67)=ln⁡(127)[-\ln(7) + \ln(2)] - [-\ln(6) + \ln(1)] = -\ln(7) + \ln(2) + \ln(6) = \ln\left(\frac{2 \cdot 6}{7}\right) = \ln\left(\frac{12}{7}\right) This confirms answer A is correct. Answer B gives the reciprocal, suggesting a sign error in the partial fractions. Answer C represents a calculation error, likely from incorrect partial fraction coefficients. Answer D results from overlooking one of the logarithmic terms in the evaluation. Remember: partial fraction decomposition is your go-to method for integrating rational functions with factorizable denominators. Always double-check your coefficients before integrating.