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The function $f$ is defined by $$f(x) = \frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25}, \; x > 4.$$ (a) Show that $f(x) = \frac{A}{Bx + C}$ where $A, B$ and $C$ are constants to be found - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 5
Question 4
The function $f$ is defined by $$f(x) = \frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25}, \; x > 4.$$ (a) Show that $f(x) = \frac{A}{Bx + C}$ where $A, B$ and... show full transcript
Worked Solution & Example Answer:The function $f$ is defined by $$f(x) = \frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25}, \; x > 4.$$ (a) Show that $f(x) = \frac{A}{Bx + C}$ where $A, B$ and $C$ are constants to be found - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 5
Step 1
Show that $f(x) = \frac{A}{Bx + C}$ where $A, B$ and $C$ are constants to be found.
Answer
To show that the function can be expressed in the desired form, we first find a common denominator for the terms in the function:
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The common denominator for the terms is ( (2x + 5)(2x - 5)(4x^2 - 25) ). Notice that ( 4x^2 - 25 ) factors to ( (2x + 5)(2x - 5) ).
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We rewrite each fraction:
- ( \frac{6}{2x + 5} = \frac{6(2x - 5)(4x^2 - 25)}{(2x + 5)(2x - 5)(4x^2 - 25)} )
- ( \frac{2}{2x - 5} = \frac{2(2x + 5)(4x^2 - 25)}{(2x + 5)(2x - 5)(4x^2 - 25)} )
- ( \frac{60}{4x^2 - 25} = \frac{60}{(2x + 5)(2x - 5)} )
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Combine all terms: Hence, ( f(x) = \frac{16x + 40}{(2x + 5)(2x - 5)} ) which simplifies further to ( \frac{8}{(2x + 5)(x - 5/2)} ). Therefore, we can represent ( f(x) ) as ( \frac{8}{2x - 5} ) in the form ( \frac{A}{Bx + C} ). Here, we identify A = 8, B = 1, C = -5/2.
Step 2
Find $f^{-1}(x)$ and state its domain.
Answer
To find the inverse, we start with:
- Set ( y = f(x) = \frac{8}{2x - 5} )
- Swap ( x ) and ( y ):
- Rearranging this gives: Thus, the inverse is:
- To find the domain of ( f^{-1}(x) ), we need to determine where ( x \neq 0 ) since it's in the denominator. Therefore, the domain of ( f^{-1}(x) ) is all real numbers except for zero: .