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4. (a) Find the values of the constants A, B and C - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 8
Question 6
4. (a) Find the values of the constants A, B and C. (b) Hence show that the exact value of \( \int_{2}^{4} \frac{2(4x^{2}+1)}{(2x+1)(2x-1)} dx\) is \(2 + \ln k\),... show full transcript
Worked Solution & Example Answer:4. (a) Find the values of the constants A, B and C - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 8
Step 1
Find the values of the constants A, B and C.
Answer
To start, we perform polynomial long division on the expression ( \frac{2(4x^{2}+1)}{(2x+1)(2x-1)} ).
Using the long division method:
- Divide the leading term: (2(4x^2) / (2x^2) = 2), thus (A = 2).
- Multiply and subtract to find the remainder:
- (2(2x^2-1) = 4x^2 - 2)
- So, subtracting: (2(4x^2 + 1) - (4x^2 - 2) = 4)
- Now we have the simplified form: ( \frac{4}{(2x+1)(2x-1)} = \frac{B}{2x+1} + \frac{C}{2x-1}).
- We must equate coefficients to find B and C.
- Let (x=-\frac{1}{2}) to find C:
- From the equation, we can conclude that (C = 2).
- Substitute C back into the equation and find B:
- Equating with another suitable x can yield (B = -2).
- Let (x=-\frac{1}{2}) to find C:
Thus, the values are: (A = 2), (B = -2), and (C = 2).
Step 2
Hence show that the exact value of \( \int_{2}^{4} \frac{2(4x^{2}+1)}{(2x+1)(2x-1)} dx \) is \(2 + \ln k\), giving the value of the constant k.
Answer
To compute the integral:
[ \int \frac{2(4x^{2}+1)}{(2x+1)(2x-1)} dx = \int \left( 2 + \frac{-2}{2x+1} + \frac{2}{2x-1} \right) dx ]
- Breaking this down, we get: [ \int 2 dx + \int \left( -\frac{2}{2x+1} \right) dx + \int \left( \frac{2}{2x-1} \right) dx ]
- Evaluating each integral:
- The first integral: (2x)
- The second integral: (-2 \ln |2x+1|)
- The third: (2 \ln |2x-1|)
- Combine the results: [ 2x - 2 \ln |2x + 1| + 2 \ln |2x - 1| \Bigg|_{2}^{4} ]
- Substitute the limits from 2 to 4 and simplify to find the constants:
- After substitution: (2(4) - 2 \ln 9 + 2 \ln 7 - (4 - 2 \ln 5 + 2 \ln 3))
- Simplifying gives:
(2 + \ln k) where you will isolate (k) such that:
- Final expression gives (k = \frac{9}{5}).
Thus, we conclude that: (k = \frac{9}{5}) and the exact value of the integral is (2 + \ln k).