Consider the function: $$f(x) = \frac{3x^2 + 16}{(1-3x)(2+x)^2}$$ (a) Find the values of A and C and show that B = 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.

A-Level Edexcel Maths Pure Differentiation: Consider the function: $$f(x)

Consider the function: $$f(x) = \frac{3x^2 + 16}{(1-3x)(2+x)^2}$$ (a) Find the values of A and C and show that B = 0 - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 7

Question 7

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Consider the function: $$f(x) = \frac{3x^2 + 16}{(1-3x)(2+x)^2}$$ (a) Find the values of A and C and show that B = 0. (b) Hence, or otherwise, find the series ... show full transcript

Worked Solution & Example Answer:Consider the function: $$f(x) = \frac{3x^2 + 16}{(1-3x)(2+x)^2}$$ (a) Find the values of A and C and show that B = 0 - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 7

Step 1

Find the values of A and C and show that B = 0.

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Answer

To solve for the constants A, B, and C, we start by equating the expression:
$3x^2 + 16 = A(2+x)^2 + B(1-3x) + C(1-3x)(2+x)^2$
Substituting specific values for x helps isolate the constants:

Letting $x = 0$ gives: $16 = 4A + B + 4C$$$
Next, let $x = 1/3$ : $3(\frac{1}{9}) + 16 = A(\frac{7}{3})^2 + 0 + C(0)\ 16 + \frac{1}{3} = \frac{49}{9}A\ C = 4\ A = 3\ B = 0$$$

Thus, we find that A = 3, C = 4, and show that B = 0.

Step 2

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.

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Answer

The function can be written as:
$f(x) = \frac{(3x^2 + 16)}{(1-3x)(2+x)^2}$
We can expand the denominator terms using Taylor series expansions. Let's first consider:

For $(1-3x)^{-1}$ , the series expansion is: $1 + 3x + 9x^2 + \ldots$
For $(2+x)^{-2}$ , we expand using the derivative: $\frac{1}{4}(1 - \frac{x}{2})^{-2} = \frac{1}{4}(1 + \frac{x}{2} + \frac{3x^2}{4} + \ldots)$

By multiplying these series together and collecting terms, we obtain the series expansion of f(x).

The simplified result for $f(x)$ up to the $x^2$ term is: $4 + 8x + 2.5x^2$

Consider the function: $$f(x) = \frac{3x^2 + 16}{(1-3x)(2+x)^2}$$ (a) Find the values of A and C and show that B = 0 - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 7

Worked Solution & Example Answer:Consider the function: $$f(x) = \frac{3x^2 + 16}{(1-3x)(2+x)^2}$$ (a) Find the values of A and C and show that B = 0 - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 7

Find the values of A and C and show that B = 0.

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.

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