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Given the equation: $$\frac{2(4x^2 + 1)}{(2x + 1)(2x - 1)} = \frac{A}{(2x + 1)} + \frac{B}{(2x - 1)} + \frac{C}{(2x - 1)}.$$ (a) Find the values of the constants A, B and C - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 7
Question 6
Given the equation: $$\frac{2(4x^2 + 1)}{(2x + 1)(2x - 1)} = \frac{A}{(2x + 1)} + \frac{B}{(2x - 1)} + \frac{C}{(2x - 1)}.$$ (a) Find the values of the constant... show full transcript
Worked Solution & Example Answer:Given the equation: $$\frac{2(4x^2 + 1)}{(2x + 1)(2x - 1)} = \frac{A}{(2x + 1)} + \frac{B}{(2x - 1)} + \frac{C}{(2x - 1)}.$$ (a) Find the values of the constants A, B and C - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 7
Step 1
Find the values of the constants A, B and C.
Answer
To find the constants A, B, and C, we will use partial fraction decomposition. We start by performing polynomial long division on the left-hand side.
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Long Division:
Divide (2(4x^2 + 1)) by ((2x + 1)(2x - 1)), which gives: [ A = 2 ]
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Setting Up the Fractions:
Now, we can equate: [ 2(4x^2 + 1) = A(2x + 1)(2x - 1) + B(2x - 1) + C(2x + 1) ]
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Finding B and C:
Next, by substituting convenient values for (x):
- Let (x = \frac{-1}{2} \Rightarrow B = -2)
- Let (x = \frac{1}{2} \Rightarrow C = 2)
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Final Values:
Thus, the final values obtained are:
- ( A = 2 )
- ( B = -2 )
- ( C = 2 )
Step 2
Hence show that the exact value of \( \int_0^2 \frac{2(4x^2 + 1)}{(2x + 1)(2x - 1)} \, dx \) is \( 2 + \ln k \), giving the value of the constant k.
Answer
Now we will evaluate the integral to confirm the given condition and find the constant k.
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Separate the Integral:
Using the values found for A, B, and C, we rewrite the integral: [ \int_0^2 \left( \frac{2}{(2x + 1)} + \frac{-2}{(2x - 1)} + \frac{2}{(2x - 1)} \right) , dx ]
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Integration:
This separates into: [ \int_0^2 \frac{2}{(2x + 1)} , dx - \int_0^2 \frac{2}{(2x - 1)} , dx ]
Evaluating these yields: [ 2\ln(2x + 1) |{0}^{2} + 2\ln(2 - 1) |{0}^{2} ]
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Calculating Limits:
Substituting the limits gives: [ 2\ln(5) - (0) + 2\ln(1) - (0) ]
Simplifying, it results in: [ 2\ln(5) - 0 = 2 + \ln(5) ]
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Final Result:
Therefore, comparing with the format ( 2 + \ln k ), we see that ( k = 5 ).