f(x) = \frac{6}{\sqrt{9 - 4x}}; \quad |x| < \frac{9}{4} (a) Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x^3. Give each coefficient in its simplest form. Use your answer to part (a) to find the binomial expansion in ascending powers of x, up to and including the term in x^3, of (b) g(x) = \frac{6}{\sqrt{9 + 4x}}; \quad |x| < \frac{9}{4} (c) h(x) = \frac{6}{\sqrt{9 - 8x}}; \quad |x| < \frac{9}{8}

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f(x) = \frac{6}{\sqrt{9 - 4x}}; \quad |x| < \frac{9}{4} (a) Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x^3 - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 7

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f(x) = \frac{6}{\sqrt{9 - 4x}}; \quad |x| < \frac{9}{4} (a) Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x^3. Give... show full transcript

Worked Solution & Example Answer:f(x) = \frac{6}{\sqrt{9 - 4x}}; \quad |x| < \frac{9}{4} (a) Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x^3 - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 7

Step 1

Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x^3.

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Answer

To find the binomial expansion of the function $f(x) = \frac{6}{\sqrt{9 - 4x}}$ , we can rewrite it as:

$f(x) = 6(9 - 4x)^{-\frac{1}{2}}$

Using the binomial series expansion, we have:

$(1 + u)^n = 1 + nu + \frac{n(n - 1)}{2}u^2 + \frac{n(n - 1)(n - 2)}{6}u^3 + \ldots$

where $u = -\frac{4x}{9}$ . Thus:

$f(x) = 6 \left[ 1 + \left(-\frac{1}{2}\right)\left(-\frac{4x}{9}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\frac{4x}{9}\right)^{2} + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}\left(-\frac{4x}{9}\right)^{3} \right] + \ldots$

Calculating each term:

The first term is $6 \cdot 1 = 6$ .
The second term is: $6 \cdot \left(-\frac{1}{2}\right)\left(-\frac{4x}{9}\right) = \frac{12x}{9} = \frac{4x}{3}.$
The third term is: $6 \cdot \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\frac{4x}{9}\right)^{2} = 6 \cdot \frac{3}{4} \cdot \frac{16x^{2}}{81} = \frac{24x^{2}}{81} = \frac{8x^{2}}{27}.$
The fourth term is: $6 \cdot \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}\left(-\frac{4x}{9}\right)^{3} = 6 \cdot \frac{5}{8} \cdot \frac{-64x^{3}}{729} = -\frac{5 \cdot 64x^{3}}{729 \cdot 8} = -\frac{40x^{3}}{729}.$

Combining everything, we get:

$f(x) = 6 + \frac{4x}{3} + \frac{8x^{2}}{27} - \frac{40x^{3}}{729} + \ldots$

Thus, the binomial expansion of $f(x)$ up to $x^3$ is: $f(x) = 6 + \frac{4}{3}x + \frac{8}{27}x^2 - \frac{40}{729}x^3 + \ldots$

Step 2

Use your answer to part (a) to find the binomial expansion in ascending powers of x, up to and including the term in x^3, of g(x) = \frac{6}{\sqrt{9 + 4x}}; \quad |x| < \frac{9}{4}.

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Answer

Using the result from part (a), we can express $g(x)$ as:

$g(x) = \frac{6}{\sqrt{9 + 4x}} = \frac{6}{\sqrt{9(1 + \frac{4x}{9})}} = \frac{2}{3} \cdot \frac{6}{\sqrt{1 + \frac{4x}{9}}}.$

We know the expansion for $(1 + u)^{-\frac{1}{2}}$ gives: $$ 1 - \frac{1}{2}u + \frac{3}{8}u^2 - \frac{5}{16}u^3 + \ldots $,

where $u = \frac{4x}{9}$ :

$= 1 - \frac{2x}{9} + \frac{6x^2}{243} - \frac{20x^3}{1296} + \ldots$

Thus: $g(x) = \frac{2}{3} \left( 1 - \frac{2x}{9} + \frac{6x^2}{243} - \frac{20x^3}{1296} + \ldots \right) = \frac{2}{3} - \frac{4x}{27} + \frac{12x^2}{729} - \frac{40x^3}{3888} + \ldots$

Step 3

Use your answer to part (a) to find the binomial expansion in ascending powers of x, up to and including the term in x^3, of h(x) = \frac{6}{\sqrt{9 - 8x}}; \quad |x| < \frac{9}{8}.

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Answer

Similarly, for $h(x)$ , we can rewrite it as:

$h(x) = \frac{6}{\sqrt{9 - 8x}} = \frac{2}{3} \cdot \frac{6}{\sqrt{1 - \frac{8x}{9}}}.$

Using the binomial expansion again, we have: $1 + \frac{4x}{9} + \frac{12x^2}{81} + \frac{40x^3}{729} + \ldots$

Therefore, we have: $h(x) = \frac{2}{3} \left( 1 + \frac{4x}{9} + \frac{12x^2}{81} + \frac{40x^3}{729} + \ldots \right) = \frac{2}{3} + \frac{8x}{27} + \frac{24x^2}{243} + \frac{80x^3}{2187} + \ldots$

f(x) = \frac{6}{\sqrt{9 - 4x}}; \quad |x| < \frac{9}{4} (a) Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x^3 - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 7

Worked Solution & Example Answer:f(x) = \frac{6}{\sqrt{9 - 4x}}; \quad |x| < \frac{9}{4} (a) Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x^3 - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 7

Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x^3.

Use your answer to part (a) to find the binomial expansion in ascending powers of x, up to and including the term in x^3, of g(x) = \frac{6}{\sqrt{9 + 4x}}; \quad |x| < \frac{9}{4}.

Use your answer to part (a) to find the binomial expansion in ascending powers of x, up to and including the term in x^3, of h(x) = \frac{6}{\sqrt{9 - 8x}}; \quad |x| < \frac{9}{8}.

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