PARTIAL FRACTIONS
This question bank contains 11 questions covering partial fraction decomposition techniques, including linear factors, repeated factors, and quadratic factors, distributed across different paper types according to IB AAHL curriculum standards.
📌 Multiple Choice Questions (2 Questions)
MCQ 1. The partial fraction decomposition of \(\frac{3x+2}{(x-1)(x+3)}\) is:
A) \(\frac{5/4}{x-1} + \frac{7/4}{x+3}\) B) \(\frac{7/4}{x-1} + \frac{5/4}{x+3}\) C) \(\frac{5/4}{x-1} – \frac{7/4}{x+3}\) D) \(\frac{-1/4}{x-1} + \frac{11/4}{x+3}\)
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Solution:
Set up: \(\frac{3x+2}{(x-1)(x+3)} = \frac{A}{x-1} + \frac{B}{x+3}\)
Clear denominators: \(3x+2 = A(x+3) + B(x-1)\)
x = 1: 3(1)+2 = A(4) → 5 = 4A → A = 5/4
x = -3: 3(-3)+2 = B(-4) → -7 = -4B → B = 7/4
✅ Answer: A) \(\frac{5/4}{x-1} + \frac{7/4}{x+3}\)
MCQ 2. Which setup is correct for \(\frac{2x^2+x+1}{x(x-1)^2}\)?
A) \(\frac{A}{x} + \frac{B}{(x-1)^2}\) B) \(\frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}\) C) \(\frac{Ax+B}{x(x-1)^2}\) D) \(\frac{A}{x} + \frac{Bx+C}{(x-1)^2}\)
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Solution:
For simple linear factor x, we need \(\frac{A}{x}\)
For repeated linear factor (x-1)², we need both \(\frac{B}{x-1}\) and \(\frac{C}{(x-1)^2}\)
Complete setup: \(\frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}\)
✅ Answer: B) \(\frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}\)
📌 Paper 1 Questions (No Calculator) – 4 Questions
Paper 1 – Q1. Express \(\frac{5x-7}{x^2-x-2}\) in partial fractions.
[5 marks]
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Solution:
Step 1: Factor the denominator
\(x^2-x-2 = (x-2)(x+1)\)
Step 2: Set up partial fractions
\(\frac{5x-7}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}\)
Step 3: Clear denominators
\(5x-7 = A(x+1) + B(x-2)\)
Step 4: Find coefficients using cover-up method
x = 2: 5(2)-7 = A(3) → 3 = 3A → A = 1
x = -1: 5(-1)-7 = B(-3) → -12 = -3B → B = 4
✅ Answer: \(\frac{1}{x-2} + \frac{4}{x+1}\)
Paper 1 – Q2. Find the partial fraction decomposition of \(\frac{3x+5}{(x+1)^2}\).
[4 marks]
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Solution:
Step 1: Set up for repeated factor
\(\frac{3x+5}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}\)
Step 2: Clear denominators
\(3x+5 = A(x+1) + B\)
Step 3: Find B using substitution
x = -1: 3(-1)+5 = B → B = 2
Step 4: Find A by equating coefficients
\(3x+5 = Ax+A+B = Ax+A+2\)
Coefficient of x: 3 = A
✅ Answer: \(\frac{3}{x+1} + \frac{2}{(x+1)^2}\)
Paper 1 – Q3. Express \(\frac{x+1}{x^2+1}\) in partial fractions.
[3 marks]
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Solution:
Step 1: Check if x²+1 is irreducible
Discriminant: b²-4ac = 0²-4(1)(1) = -4 < 0, so irreducible
Step 2: Since x²+1 cannot be factored further
The expression \(\frac{x+1}{x^2+1}\) is already in its simplest partial fraction form
Alternative representation:
\(\frac{x+1}{x^2+1} = \frac{x}{x^2+1} + \frac{1}{x^2+1}\)
✅ Answer: \(\frac{x+1}{x^2+1}\) (already in simplest form)
Paper 1 – Q4. Decompose \(\frac{2x^2+3x+1}{(x+1)(x^2+1)}\) into partial fractions.
[6 marks]
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Solution:
Step 1: Set up partial fractions
\(\frac{2x^2+3x+1}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1}\)
Step 2: Clear denominators
\(2x^2+3x+1 = A(x^2+1) + (Bx+C)(x+1)\)
Step 3: Find A using substitution
x = -1: 2(-1)²+3(-1)+1 = A(2) → 0 = 2A → A = 0
Step 4: Simplify and find B, C
\(2x^2+3x+1 = (Bx+C)(x+1) = Bx^2+Bx+Cx+C\)
Coefficient of x²: 2 = B
Coefficient of x: 3 = B + C = 2 + C → C = 1
✅ Answer: \(\frac{2x+1}{x^2+1}\)
📌 Paper 2 Questions (Calculator Allowed) – 5 Questions
Paper 2 – Q1. Express \(\frac{x^3+2x^2+3x+2}{x^2+x-2}\) in partial fractions.
[6 marks]
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Solution:
Step 1: Check if proper fraction
Degree of numerator (3) > degree of denominator (2), so improper
Step 2: Perform polynomial long division
\(\frac{x^3+2x^2+3x+2}{x^2+x-2} = x + 1 + \frac{2x+4}{x^2+x-2}\)
Step 3: Factor denominator of remainder
\(x^2+x-2 = (x+2)(x-1)\)
Step 4: Decompose remainder
\(\frac{2x+4}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}\)
x = -2: 2(-2)+4 = A(-3) → 0 = -3A → A = 0
x = 1: 2(1)+4 = B(3) → 6 = 3B → B = 2
✅ Answer: \(x + 1 + \frac{2}{x-1}\)
Paper 2 – Q2. Find the partial fraction decomposition of \(\frac{x^2-3x+2}{x(x-1)(x-2)}\).
[5 marks]
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Solution:
Step 1: Set up partial fractions
\(\frac{x^2-3x+2}{x(x-1)(x-2)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-2}\)
Step 2: Clear denominators
\(x^2-3x+2 = A(x-1)(x-2) + Bx(x-2) + Cx(x-1)\)
Step 3: Find coefficients using cover-up method
x = 0: 0-0+2 = A(-1)(-2) → 2 = 2A → A = 1
x = 1: 1-3+2 = B(1)(-1) → 0 = -B → B = 0
x = 2: 4-6+2 = C(2)(1) → 0 = 2C → C = 0
✅ Answer: \(\frac{1}{x}\)
Paper 2 – Q3. Express \(\frac{3x^2+x+4}{(x+1)^2(x-2)}\) in partial fractions.
[7 marks]
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Solution:
Step 1: Set up with repeated factor
\(\frac{3x^2+x+4}{(x+1)^2(x-2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-2}\)
Step 2: Clear denominators
\(3x^2+x+4 = A(x+1)(x-2) + B(x-2) + C(x+1)^2\)
Step 3: Find coefficients using substitution
x = -1: 3(-1)²+(-1)+4 = B(-3) → 6 = -3B → B = -2
x = 2: 3(4)+2+4 = C(9) → 18 = 9C → C = 2
Step 4: Find A using coefficient comparison
Expanding and equating x² coefficient: 3 = A + C = A + 2 → A = 1
✅ Answer: \(\frac{1}{x+1} + \frac{-2}{(x+1)^2} + \frac{2}{x-2}\)
Paper 2 – Q4. A rational function has the form \(\frac{ax+b}{(x-1)(x+2)}\). If the partial fraction decomposition is \(\frac{2}{x-1} + \frac{3}{x+2}\), find the values of a and b.
[4 marks]
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Solution:
Step 1: Combine the given partial fractions
\(\frac{2}{x-1} + \frac{3}{x+2} = \frac{2(x+2) + 3(x-1)}{(x-1)(x+2)}\)
Step 2: Expand the numerator
Numerator = 2(x+2) + 3(x-1) = 2x+4+3x-3 = 5x+1
Step 3: Compare with given form
\(\frac{ax+b}{(x-1)(x+2)} = \frac{5x+1}{(x-1)(x+2)}\)
Step 4: Identify coefficients
Comparing ax+b = 5x+1: a = 5, b = 1
✅ Answer: a = 5, b = 1
Paper 2 – Q5. Find the partial fraction decomposition of \(\frac{2x^3-x^2+3x-1}{(x^2+1)^2}\).
[8 marks]
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Solution:
Step 1: Check if proper fraction
Degree of numerator (3) < degree of denominator (4), so proper
Step 2: Set up for repeated quadratic factor
\(\frac{2x^3-x^2+3x-1}{(x^2+1)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}\)
Step 3: Clear denominators
\(2x^3-x^2+3x-1 = (Ax+B)(x^2+1) + (Cx+D)\)
Step 4: Expand and equate coefficients
\(2x^3-x^2+3x-1 = Ax^3+Ax+Bx^2+B+Cx+D\)
x³: 2 = A
x²: -1 = B
x¹: 3 = A + C = 2 + C → C = 1
x⁰: -1 = B + D = -1 + D → D = 0
✅ Answer: \(\frac{2x-1}{x^2+1} + \frac{x}{(x^2+1)^2}\)