Edexcel IAL January 2023 P4 (WMA14/01) Q1 Partial Fractions, Binomial Expansion
\( f(x) = \frac{5x+10}{(1-x)(2+3x)} \)
a. Write \( f(x) \) in partial fraction form.
(3)
(3)
b. Hence find, in ascending powers of x up to and including the terms in \( x^2 \), the binomial series expansion of \( f(x) \). Give each coefficient as a simplified fraction.
(5)
(5)
c. Find the range of values of x for which this expansion is valid.
(1)
(1)
Solution to Part a: Expression of f(x) in partial fractions
Key Concepts Used:
- Partial Fractions – Distinct Linear Factors
Step-by-Step Working:
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Set \( \frac{5x+10}{(1-x)(2+3x)} \) identical to \( \frac{A}{1-x} + \frac{B}{2+3x} \).
\( \frac{5x+10}{(1-x)(2+3x)} = \frac{A}{1-x} + \frac{B}{2+3x} \)
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Let’s find the value of A & B.
Add the two fractions.\( \frac{5x+10}{(1-x)(2+3x)} = \frac{A(2+3x) + B(1-x)}{(1-x)(2+3x)} \)
\( 5x+10 = A(2+3x) + B(1-x) \)
\( 5x + 10 = A(2 + 3x) + B(1 – x) \)
To find A, substitute \( x = 1 \).
\( 5(1) + 10 = A(2 + 3 \times 1) + B(1 – 1) \)
\( 15 = 5A \)
\( A = 3 \)
To find B, substitute \( x = -\frac{2}{3} \).
\( 5\left(-\frac{2}{3}\right) + 10 = A\left(2 + 3 \times -\frac{2}{3}\right) + B\left(1 – \left(-\frac{2}{3}\right)\right) \)
\( \frac{20}{3} = \frac{5}{3}B \)
\( B = 3 \)
3. Hence,
\( \frac{5x+10}{(1-x)(2+3x)} = \frac{3}{1-x} + \frac{4}{2+3x} \)
\( \frac{5x+10}{(1-x)(2+3x)} = \frac{3}{1-x} + \frac{4}{2+3x} \)
Final Answer:
\( \frac{5x+10}{(1-x)(2+3x)} = \frac{3}{1-x} + \frac{4}{2+3x} \)
Solution to Part b: Binomial Expansion
Key Concepts Used:
- Binomial Expansion
Step-by-Step Working:
Key Points
- Expansion Formula:
\( (1+y)^n \approx 1 + ny + \frac{n(n-1)}{2}y^2 + \ldots \) for \( |y| < 1 \). - Rewrite Fractions:
\( \frac{3}{1-x} = 3(1-x)^{-1} \)
\( \frac{4}{2+3x} = \frac{4}{2(1+\frac{3}{2}x)} = 2(1+\frac{3}{2}x)^{-1} \)
Step-by-Step Solution
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First, write in the index form.
\( f(x) = \frac{3}{1-x} + \frac{4}{2+3x} \)
\( f(x) = 3(1-x)^{-1} + 4(2+3x)^{-1} \)
\( (1-x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots + \frac{n(n-1)\ldots(n-r+1)}{r!}x^r \) -
Let us find out the expansion of \( (1-x)^{-1} \) first separately using binomial expansion.
Replace n by -1. And since n is not positive, so no term of the expansion would ever be positive.\( (1-x)^{-1} = 1 + (-1)(-x) + \frac{-1(-2)(-x)^2}{2} + \ldots \)
\( (1-x)^{-1} = 1 + x + x^2 + \ldots \)Now, find out the expansion of \( (2+3x)^{-1} \) separately using binomial expansion.
\( (2+3x)^{-1} = [2(1+\frac{3}{2}x)]^{-1} \) -
Take out a factor of \( 2^{-1} \).
\( (2+3x)^{-1} = 2^{-1}(1+\frac{3}{2}x)^{-1} \)
\( = \frac{1}{2}(1 + \frac{3}{2}x)^{-1} \)
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Now, finding the expansion.
\( \frac{1}{2}(1 + \frac{3}{2}x)^{-1} = \frac{1}{2}\left[1 – \frac{3}{2}x + \frac{-1(-2)}{2}\left(\frac{3}{2}x\right)^2 + \ldots \right] \)
So,
\( (2 + 3x)^{-1} = \frac{1}{2}(1 – \frac{3}{2}x + \frac{9}{4}x^2) \)Hence, combining the expansions.
\( 3(1-x)^{-1} + 4(2+3x)^{-1} = 3(1 + x + x^2 + \ldots) + 4 \cdot \frac{1}{2}(1 – \frac{3}{2}x + \frac{9}{4}x^2) \)
\( 3(1-x)^{-1} + 4(2+3x)^{-1} = 3 + 3x + 3x^2 + 2 – 3x + \frac{9}{2}x^2 \)
\( = 5 + \frac{15}{2}x^2 \)
Final Answer:
\( 5 + \frac{15}{2}x^2 \)
Solution to Part c: Validity of Expansion
Key Concepts Used:
- Convergence Condition: \( |y| < 1 \) for \( (1+y)^n \).
- Apply to Each Term:
- \( -x \) : Valid for \( |x| < 1 \).
- \( \frac{3}{2}x \) : Valid for \( |\frac{3}{2}x| < 1 \implies |x| < \frac{2}{3} \).
Step-by-Step Solution
\( 3(1-x)^{-1} + 2(1 + \frac{3}{2}x)^{-1} = 5 + \frac{15}{2}x^2 \)
The expansion of \( (1-x)^{-1} \) is valid as long as \( |x| < 1 \) or in other words \( -1 < x < 1 \).
Whereas for \( (1 + \frac{3}{2}x)^{-1} \) is valid as long as \( \left| \frac{3x}{2} \right| < 1 \) or in other words \( -\frac{2}{3} < x < \frac{2}{3} \).
The overlapping interval is
\( -\frac{2}{3} < x < \frac{2}{3} \) Hence, the expansion is valid for the above interval only.