f(x) = \frac{27x^3 + 32x^2 + 16}{(3x + 2)(1-x)} \\ |x| < \frac{2}{3}\\ Given that f(x) can be expressed in the form\\ f(x) = \frac{A}{(3x+2)} + \frac{B}{(3x+2)} + \frac{C}{(1-x)}.\\ (a) find the values of B and C and show that A = 0. \\ \\ (b) Hence, or otherwise, find the series expansion of f(x), in ascending powers of x, up to and including the term in x^2. Simplify each term. \\ \\ (c) Find the percentage error made in using the series expansion in part (b) to estimate the value of f(0.2). Give your answer to 2 significant figures.

A-Level Edexcel Maths Pure Equation of a Straight Line: f(x) = \frac{27x^3 + 32x^2 + 16}

f(x) = \frac{27x^3 + 32x^2 + 16}{(3x + 2)(1-x)} \\ |x| < \frac{2}{3}\\ Given that f(x) can be expressed in the form\\ f(x) = \frac{A}{(3x+2)} + \frac{B}{(3x+2)} + \frac{C}{(1-x)}.\\ (a) find the values of B and C and show that A = 0 - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 3

Question 4

$f(x)-=-\frac{27x^3-+-32x^2-+-16}{(3x-+-2)(1-x)}-\\-|x|-<-\frac{2}{3}\\--Given-that-f(x)-can-be-expressed-in-the-form\\-f(x)-=-\frac{A}{(3x+2)}-+-\frac{B}{(3x+2)}-+-\frac{C}{(1-x)}.\\--(a)-find-the-values-of-B-and-C-and-show-that-A-=-0-Edexcel-A-Level Maths Pure-Question 4-2009-Paper 3.png$

f(x) = \frac{27x^3 + 32x^2 + 16}{(3x + 2)(1-x)} \\ |x| < \frac{2}{3}\\ Given that f(x) can be expressed in the form\\ f(x) = \frac{A}{(3x+2)} + \frac{B}{(3x+2)} + \... show full transcript

Worked Solution & Example Answer:f(x) = \frac{27x^3 + 32x^2 + 16}{(3x + 2)(1-x)} \\ |x| < \frac{2}{3}\\ Given that f(x) can be expressed in the form\\ f(x) = \frac{A}{(3x+2)} + \frac{B}{(3x+2)} + \frac{C}{(1-x)}.\\ (a) find the values of B and C and show that A = 0 - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 3

Step 1

Find the percentage error made in using the series expansion in part (b) to estimate the value of f(0.2). Give your answer to 2 significant figures.

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Answer

To estimate f(0.2) using the series expansion:

\text{Estimate} = 2 + 3(0.2) + \frac{9}{4}(0.2)^2 = 2 + 0.6 + 0.09 = 2.69.

The actual value of f(0.2) is:

\text{Actual } f(0.2) = \frac{27(0.2)^3 + 32(0.2)^2 + 16}{(3(0.2) + 2)(1 - 0.2)} = \frac{1.08 + 6.4 + 16}{(0.6)(0.8)} = \frac{23.48}{0.48} = 48.87.

Calculating the percentage error:

\text{Percentage error} = \frac{|\text{Actual} - \text{Estimate}|}{\text{Actual}} \times 100 = \frac{|48.87 - 2.69|}{48.87} \times 100 \approx 94.5\%.

f(x) = \frac{27x^3 + 32x^2 + 16}{(3x + 2)(1-x)} \\ |x| < \frac{2}{3}\\ Given that f(x) can be expressed in the form\\ f(x) = \frac{A}{(3x+2)} + \frac{B}{(3x+2)} + \frac{C}{(1-x)}.\\ (a) find the values of B and C and show that A = 0 - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 3

Find the percentage error made in using the series expansion in part (b) to estimate the value of f(0.2). Give your answer to 2 significant figures.

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