Photo AI
3. f(x) = \frac{6}{\sqrt{9 - 4x}} , \ |x| < \frac{9}{4} (a) Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x³ - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 7
Question 5
3. f(x) = \frac{6}{\sqrt{9 - 4x}} , \ |x| < \frac{9}{4} (a) Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x³. Give... show full transcript
Worked Solution & Example Answer:3. f(x) = \frac{6}{\sqrt{9 - 4x}} , \ |x| < \frac{9}{4} (a) Find the binomial expansion of f(x) in ascending powers of x, up to and including the term in x³ - 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 x³.
Answer
To find the binomial expansion of ( f(x) = \frac{6}{\sqrt{9 - 4x}} ), we first rewrite it in a suitable form:
[ f(x) = 6(9 - 4x)^{-\frac{1}{2}} ]
Using the binomial expansion formula ( (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots ), we set:
[ u = -\frac{4x}{9} \ \text{and thus} \ n = -\frac{1}{2} ]
Now substituting:
[ f(x) = 6 \left( 1 - \frac{4x}{9} \right)^{-\frac{1}{2}} = 6 \left( 1 + \frac{1}{2} \left(-\frac{4x}{9}\right) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2}\left(-\frac{4x}{9}\right)^2 + \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3!}\left(-\frac{4x}{9}\right)^3 + \ldots\right) ]
Calculating further,
[= 6 \left( 1 + \left(-\frac{2x}{9}\right) + \frac{6}{27}x^2 + \frac{40}{729}x^3 + \ldots \right) ]
Thus,
[ f(x) = 6 \left( 1 - \frac{2x}{9} + \frac{2}{9}x^2 + \frac{40}{729}x^3 + \ldots \right) ]
Finally, we factor:
[ f(x) = 6 \left( 1 + \frac{4}{9}x^2 + \frac{40}{729}x^3\right) ]
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 x³, of g(x) = \( \frac{6}{\sqrt{9 + 4x}} \).
Answer
To find the binomial expansion for ( g(x) = \frac{6}{\sqrt{9 + 4x}} ), we rewrite it:
[ g(x) = 6(9 + 4x)^{-\frac{1}{2}} ]
Using the same binomial expansion method, set ( u = \frac{4x}{9} ) and apply the formula:
[ g(x) = 6 \left(1 + \frac{4x}{9}\right)^{-\frac{1}{2}} \approx 6 \left( 1 - \frac{2x}{9} + \frac{2}{9}x^2 - \frac{40}{729}x^3\right) ]
Thus,
[ g(x) \approx 6 \left(1 - \frac{2}{9}x + \frac{8}{729}x^2 - \frac{320}{729}x^3\right) ]
Step 3
Find the binomial expansion in ascending powers of x, up to and including the term in x³, of h(x) = \( \frac{6}{\sqrt{9 - 8x}} \).
Answer
For ( h(x) = \frac{6}{\sqrt{9 - 8x}} ), we write:
[ h(x) = 6(9 - 8x)^{-\frac{1}{2}} ]
Using the binomial expansion formula as before, let ( u = -\frac{8x}{9} ):
[ h(x) = 6 \left(1 + \frac{8x}{9}\right)^{-\frac{1}{2}} \approx 6 \left( 1 + \frac{4}{9} x + \frac{16}{729} x^2 + \frac{320}{729} x^3\right) ]
Thus,
[ h(x) \approx 6 \left(1 + \frac{4}{9}x + \frac{16}{729}x^2 + \frac{320}{729}x^3\right) ]