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2. (a) Express \[ \frac{4x-1}{2(x-1)} - \frac{3}{2(x-1)(2x-1)} \] as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 4
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2. (a) Express \[ \frac{4x-1}{2(x-1)} - \frac{3}{2(x-1)(2x-1)} \] as a single fraction in its simplest form. Given that \[ f(x) = \frac{4x-1}{2(x-1)} - \frac{3}{2... show full transcript
Worked Solution & Example Answer:2. (a) Express \[ \frac{4x-1}{2(x-1)} - \frac{3}{2(x-1)(2x-1)} \] as a single fraction in its simplest form - Edexcel - A-Level Maths Pure - Question 4 - 2011 - Paper 4
Step 1
Express \( \frac{4x-1}{2(x-1)} - \frac{3}{2(x-1)(2x-1)} \) as a single fraction
Answer
To begin, we need to have a common denominator for the two fractions. The common denominator is ( 2(x-1)(2x-1) ).
The first fraction can be rewritten as:
[ \frac{4x-1}{2(x-1)} = \frac{(4x-1)(2x-1)}{2(x-1)(2x-1)} = \frac{(8x^2 - 4x - 2x + 1)}{2(x-1)(2x-1)} = \frac{8x^2 - 6x + 1}{2(x-1)(2x-1)} ]
Now, substituting this back into the expression we have:
[ \frac{8x^2 - 6x + 1}{2(x-1)(2x-1)} - \frac{3}{2(x-1)(2x-1)} = \frac{8x^2 - 6x + 1 - 3}{2(x-1)(2x-1)} = \frac{8x^2 - 6x - 2}{2(x-1)(2x-1)} ]
This simplifies to:
[ \frac{4x^2 - 3x - 1}{(x-1)(2x-1)} ]
Thus, the final answer is:\n [\frac{4x^2 - 3x - 1}{(x-1)(2x-1)}]
Step 2
show that \( f(x) = \frac{3}{2x-1} \)
Answer
From the definition of ( f(x) ):
[ f(x) = \frac{4x-1}{2(x-1)} - \frac{3}{2(x-1)(2x-1)} - 2 ]
Substituting the expression we found from part (a):
[ f(x) = \frac{4x-1}{2(x-1)} - \frac{3}{2(x-1)(2x-1)} ]
We have:
[ f(x) = \frac{4x - 1 - 3}{2(x-1)(2x-1)} = \frac{3}{2x-1} ]
Thus proving the equation.
Step 3
Hence differentiate \( f(x) \) and find \( f'(2) \)
Answer
To differentiate ( f(x) = \frac{3}{2x-1} ), we use the quotient rule:
[ f'(x) = \frac{3(0) - 3(2)}{(2x-1)^2} = \frac{-6}{(2x-1)^2} ]
Now, substituting ( x = 2 ) into the derivative:
[ f'(2) = \frac{-6}{(2(2)-1)^2} = \frac{-6}{(4-1)^2} = \frac{-6}{9} = -\frac{2}{3} ]
Thus, ( f'(2) = -\frac{2}{3} ).