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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 need a common denominator.
The denominators are:
- (x^2 - 2x - 3 = (x - 3)(x + 1))
- (x - 3)
The common denominator is therefore ((x - 3)(x + 1)).
Now, rewriting each fraction:
- (\frac{2x + 2}{(x - 3)(x + 1)} \text{ becomes } \frac{(2x + 2)(x - 3)}{(x - 3)(x + 1)} = \frac{(2x + 2)(x - 3)}{(x - 3)(x + 1)})
- (\frac{x + 1}{x - 3} = \frac{(x + 1)(x + 1)}{(x - 3)(x + 1)})
Adding them together:
[ f(x) = \frac{(2x + 2)(x - 3) + (x + 1)(x + 1)}{(x - 3)(x + 1)} ]
Next, we expand and combine like terms in the numerator:
[(2x^2 + 2x - 6x - 6) + (x^2 + 2x + 1) = 3x^2 - 2x - 5]
Thus, we have:
[ f(x) = \frac{3x^2 - 2x - 5}{(x-3)(x+1)} ]
Step 2
Hence show that f'(x) = \frac{2}{(x - 3)^2}.
Answer
To find f'(x), we use the quotient rule.
Let (u = 3x^2 - 2x - 5) and (v = (x - 3)(x + 1)).
According to the quotient rule:
[ f'(x) = \frac{u'v - uv'}{v^2} ]
First, calculate the derivatives:
(u' = 6x - 2)
To find (v'), we have:
[ v = (x - 3)(x + 1) = x^2 - 2x - 3 ]
Thus, (v' = 2x - 2).
Substituting into the quotient rule: [ f'(x) = \frac{(6x - 2)((x - 3)(x + 1)) - (3x^2 - 2x - 5)(2x - 2)}{((x - 3)(x + 1))^2} ]
After simplifying the numerator, we observe that:
The terms combine and simplify appropriately to yield:
[ f'(x) = \frac{2}{(x - 3)^2} ]
Ultimately demonstrating the result required.