IB Mathematics AA Partial fractions Study Notes
IB Mathematics AA Partial fractions Study Notes
IB Mathematics AA Partial fractions Study Notes Offer a clear explanation of Partial fractions , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Partial fractions.
Partial Fractions
Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions. This technique is particularly useful for integration, algebraic manipulation, and solving differential equations.
Key Concepts
- Rational Functions:
- A rational function is of the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials.
- The degree of \( P(x) \) must be less than the degree of \( Q(x) \) for partial fraction decomposition to apply.
- Partial Fraction Form:
- For a denominator with distinct linear factors:
\( \frac{P(x)}{(x – a)(x – b)} = \frac{A}{(x – a)} + \frac{B}{(x – b)} \).
- For a denominator with distinct linear factors:
Guidance & Clarifications:
- Maximum of Two Linear Factors:
- The denominator may have at most two distinct linear terms.
- Example: \( \frac{2x + 1}{x^2 + x – 2} = \frac{1}{(x – 1)} + \frac{1}{(x + 2)} \).
- Integration Applications:
- Partial fractions are linked to rearranging integrands for easier evaluation (AHL 5.15).
- This includes breaking down rational functions into simpler forms suitable for standard integration techniques.
Steps for Partial Fraction Decomposition
- Factorize the Denominator: Write \( Q(x) \) as a product of distinct linear factors.
- Set Up the Decomposition:
\( \frac{P(x)}{Q(x)} = \frac{A}{(x – a)} + \frac{B}{(x – b)} + \dots \).
- Determine Coefficients:
- Multiply through by \( Q(x) \) to eliminate the denominators.
- Equate coefficients of like powers of \( x \) to solve for \( A, B, \dots \).
- Rewrite the Expression: Substitute the coefficients back into the partial fraction form.
Examples
Example : Simple Linear Factors
Decompose \( \frac{2x + 1}{x^2 + x – 2} \):
Factorize: \( x^2 + x – 2 = (x – 1)(x + 2) \).
Set up: \( \frac{2x + 1}{x^2 + x – 2} = \frac{A}{(x – 1)} + \frac{B}{(x + 2)} \).
Eliminate denominators: \( 2x + 1 = A(x + 2) + B(x – 1) \).
Expand and solve: Equate coefficients to find \( A = 1 \) and \( B = 1 \).
Result: \( \frac{2x + 1}{x^2 + x – 2} = \frac{1}{(x – 1)} + \frac{1}{(x + 2)} \).
IB Mathematics AA SL Partial fractions Exam Style Worked Out Questions
Question
Write the partial fraction decomposition of the following expression.
(20x + 35)/(x + 4)2
▶️Answer/Explanation
Solution:
(20x + 35)/(x + 4)2
(20x + 35)/(x + 4)2 = [A/(x + 4)] + [B/(x + 4)2]
(20x + 35)/(x + 4)2 = [A(x + 4) + B]/ (x + 4)2
Now, equating the numerators,
20x + 35 = A(x + 4) + B
20x + 35 = Ax + 4A + B
20x + 35 = Ax + (4A + B)
By equating the coefficients,
A = 20
4A + B = 35
4(20) + B = 35
B = 35 – 80 = -45
Therefore, (20x + 35)/(x + 4)2 = [20/(x + 4)] – [45/(x + 4)2]
Question
Decompose the given expression into partial fractions.
(x2 + 1)/ (x3 + 3x2 + 3x + 2)
▶️Answer/Explanation
Solution:
(x2 + 1)/ (x3 + 3x2 + 3x + 2)
Using the factor theorem, x + 2 is a factor of x3 + 3x2 + 3x + 2.
Thus, x3 + 3x2 + 3x + 2 = (x + 2)(x2 + x + 1)
Now, the given expression can be written as:
(x2 + 1)/ (x3 + 3x2 + 3x + 2) = (x2 + 1)/ [(x + 2)(x2 + x + 1)]
By the method of decomposition,
(x2 + 1)/(x + 2)(x2 + x + 1) = [A/(x + 2)] + [(Bx + C)/(x2 + x + 1)]
(x2 + 1)/(x + 2)(x2 + x + 1) = [A(x2 + x + 1) + (Bx + C)(x + 2)]/ [(x + 2)(x2 + x + 1)]
= [(A + B)x2 + (A + 2B + C)x + A + 2C]/ [(x + 2)(x2 + x + 1)]
Equating the coefficients in the numerators of both LHS and RHS,
A + B = 1
A + 2B + C = 0
A + 2C = 1
Solving these equations,
A = 5/3, B = -2/3 and C = -1/3
(x2 + 1)/(x + 2)(x2 + x + 1) = [5/3(x + 2)] – [(2x + 1)/3(x2 + x + 1)]