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4. rac{2(4x^2 + 1)}{(2x + 1)(2x - 1)} = A + rac{B}{(2x + 1)} + rac{C}{(2x - 1)} - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 8
Question 6
4. rac{2(4x^2 + 1)}{(2x + 1)(2x - 1)} = A + rac{B}{(2x + 1)} + rac{C}{(2x - 1)}. (a) Find the values of the constants A, B and C. (b) Hence show th... show full transcript
Worked Solution & Example Answer:4. rac{2(4x^2 + 1)}{(2x + 1)(2x - 1)} = A + rac{B}{(2x + 1)} + rac{C}{(2x - 1)} - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 8
Step 1
Find the values of the constants A, B and C.
Answer
To find the values of the constants A, B, and C, we start by performing polynomial long division on the expression ( \frac{2(4x^2 + 1)}{(2x + 1)(2x - 1)} ).
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Calculate A:
We first identify the degree of the numerator and the denominator. The term with the highest degree in the numerator is ( 8x^2 ). For the denominator, its highest degree term is also ( 8x^2 ). Therefore, we can conclude that ( A = 2 ). -
Determine B and C:
After dividing, we can express the remainder as follows:
[ 2(4x^2 + 1) = 2(2x + 1)(2x - 1) + B(2x - 1) + C(2x + 1) ]
From this expression, we can find the coefficients.- Setting up the equation:
[ 4 = B(2x - 1) + C(2x + 1) ] - Choosing appropriate values for ( x ):
If we set ( x = 1 ), we can isolate terms to solve for ( B ) and ( C ). Thus:
[ B = 2 \text{ and } C = -2. ]
Finally, we conclude with the values:
[ A = 2, \quad B = 2, \quad C = -2. ]
- Setting up the equation:
Step 2
Hence show that the exact value of \int_{1}^{2} \frac{2(4x^2 + 1)}{(2x + 1)(2x - 1)} \, dx is 2 + \ln k, giving the value of the constant k.
Answer
To evaluate the integral ( \int_{1}^{2} \frac{2(4x^2 + 1)}{(2x + 1)(2x - 1)} , dx ):
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Rewriting the expression:
Substitute the values of A, B, and C into the integral:
[ \int_{1}^{2} \left( 2 + \frac{2}{2x + 1} - \frac{2}{2x - 1} \right) dx. ] -
Evaluate the integral:
The integral can be separated into three simpler integrals:- [ \int_{1}^{2} 2 : dx = 2[x]_{1}^{2} = 2(2 - 1) = 2. ]
- For the second integral, [ \int \frac{2}{2x + 1} , dx = \ln(2x + 1) + C. ]
- The third integral yields, [ \int -\frac{2}{2x - 1} , dx = -\ln(2x - 1) + C. ]
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Combine results:
Substituting the limits from 1 to 2 into the computed integrals yields:
[ 2 + \left[ \ln(5) - \ln(1) \right] - \left[ \ln(3) - \ln(1) \right] = 2 + \ln(\frac{5}{3}). ] -
Identify k:
To match the result to ( 2 + \ln k ), we have:
[ k = \frac{5}{3}. ]