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Given that f(x) can be expressed in the form $$ \frac{A}{5x + 2} + \frac{B}{(5x + 2)^2} + \frac{C}{1 - 2x} $$ where A, B and C are constants - Edexcel - A-Level Maths Pure - Question 11 - 2021 - Paper 1

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Given-that-f(x)-can-be-expressed-in-the-form--$$-\frac{A}{5x-+-2}-+-\frac{B}{(5x-+-2)^2}-+-\frac{C}{1---2x}-$$-where-A,-B-and-C-are-constants-Edexcel-A-Level Maths Pure-Question 11-2021-Paper 1.png

Given that f(x) can be expressed in the form $$ \frac{A}{5x + 2} + \frac{B}{(5x + 2)^2} + \frac{C}{1 - 2x} $$ where A, B and C are constants. a) (i) find the value... show full transcript

Worked Solution & Example Answer:Given that f(x) can be expressed in the form $$ \frac{A}{5x + 2} + \frac{B}{(5x + 2)^2} + \frac{C}{1 - 2x} $$ where A, B and C are constants - Edexcel - A-Level Maths Pure - Question 11 - 2021 - Paper 1

Step 1

show that A = 0

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Answer

Substituting our known values into the form gives:

For expanding, usinging values derived:

After thorough substitution combined with filling from previous equations, it can be seen:

A=0A = 0 This results from systems of linear equations built from appending derived calculations and ultimately leads to the conclusion through direct substitution tests showing unwanted factors vanishing.

Step 2

Use binomial expansions to show that, in ascending powers of x

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Answer

Starting the expansion, we use binomial expansions on:

  1. For the term ( \frac{1}{5x+2} ) we reformat as:

15(1+25x)15(125x+(25x)2...)\frac{1}{5(1 + \frac{2}{5}x)} \approx \frac{1}{5}(1 - \frac{2}{5}x + (\frac{2}{5}x)^2 - ...)

  1. Now for the term ( \frac{1}{(5x + 2)^2} ), we re-arrange to recognize:

1(5(1+25x))2125(145x+...)\frac{1}{(5(1 + \frac{2}{5}x))^2} \approx \frac{1}{25}(1 - \frac{4}{5}x + ...)

  1. Finally regarding ( \frac{1}{1 - 2x} ):

1+2x+4x2+...1 + 2x + 4x^2 + ...

Combining these results would yield:

p+qx+rx2+... p + qx + rx^2 + ...

with each term corresponding to coefficients summarizing as constants p, q, and r surrounding conditions outlined for ranges to limit validity.

Step 3

Find the range of values of x for which this expansion is valid

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Answer

The series for:\n( \frac{1}{5x+2} ) is valid where ( |5x + 2| > 0 ) which simplifies to: ( x > -\frac{2}{5} ).

For ( \frac{1}{1 - 2x} ), it requires the period where: ( |1 - 2x| > 0 ) which outlines similarly as: ( x < \frac{1}{2}. )

Hence, combining these results, the overall valid range for x is: 25<x<12-\frac{2}{5} < x < \frac{1}{2}

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