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f(x) = \frac{2x + 2}{x^2 - 2x - 3} + \frac{x + 1}{x - 3} (a) Express f(x) as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 2
Question 4
f(x) = \frac{2x + 2}{x^2 - 2x - 3} + \frac{x + 1}{x - 3} (a) Express f(x) as a single fraction in its simplest form. (b) Hence show that f'(x) = \frac{2}{(x - 3)^2... show full transcript
Worked Solution & Example Answer:f(x) = \frac{2x + 2}{x^2 - 2x - 3} + \frac{x + 1}{x - 3} (a) Express f(x) as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2009 - Paper 2
Step 1
Express f(x) as a single fraction in its simplest form.
Answer
To express f(x) as a single fraction, we first need a common denominator, which will be ( (x^2 - 2x - 3)(x - 3) ).
- Factorize the quadratic: ( x^2 - 2x - 3 = (x - 3)(x + 1) ).
- The common denominator is therefore ( (x - 3)(x + 1) ).
- Rewrite each fraction with the common denominator: [ f(x) = \frac{(2x + 2)(x - 3) + (x + 1)(x^2 - 2x - 3)}{(x^2 - 2x - 3)(x - 3)} ]
- Expand the numerators: [ (2x + 2)(x - 3) = 2x^2 - 6x + 2x - 6 = 2x^2 - 4x - 6 ] [ (x + 1)(x^2 - 2x - 3) = x^3 - 2x^2 - 3x + x^2 - 2x - 3 = x^3 - x^2 - 5x - 3 ]
- Combine the results together: [ f(x) = \frac{2x^2 - 4x - 6 + x^3 - x^2 - 5x - 3}{(x - 3)(x + 1)} ]
- Simplifying: [ f(x) = \frac{x^3 + x^2 - 9x - 9}{(x - 3)(x + 1)} ]
Step 2
Hence show that f'(x) = \frac{2}{(x - 3)^2}
Answer
To find f'(x), we first differentiate f(x) using the quotient rule:
Given ( f(x) = \frac{g(x)}{h(x)} ), where ( g(x) = 2x + 2 + x + 1 ) and ( h(x) = (x - 3)(x + 1) ), we have:
- Differentiate using the quotient rule: [ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} ]
- Calculate ( g'(x) ) and ( h'(x) ): ( g'(x) = 2 ) ( h'(x) = (1)(x + 1) + (x - 3)(1) = 2x - 2 )
- Substitute back into the quotient rule: [ f'(x) = \frac{2[(x - 3)(x + 1)] - (2x + 3)(2x - 2)}{[(x - 3)(x + 1)]^2} ]
- Simplify the numerator to show: [ f'(x) = \frac{2}{(x - 3)^2} ]