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4. (a) Express \[ \frac{25}{x^2(2x + 1)} \] in partial fractions - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 8
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4. (a) Express \[ \frac{25}{x^2(2x + 1)} \] in partial fractions. (4) (b) Using calculus to find the exact volume of the solid of revolution generated, giving you... show full transcript
Worked Solution & Example Answer:4. (a) Express \[ \frac{25}{x^2(2x + 1)} \] in partial fractions - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 8
Step 1
Express \( \frac{25}{x^2(2x + 1)} \) in partial fractions
Answer
To express ( \frac{25}{x^2(2x + 1)} ) in partial fractions, we start by assuming a form for the decomposition:
[ \frac{25}{x^2(2x + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x + 1} ]
Multiplying through by the common denominator ( x^2(2x + 1) ) gives:
[ 25 = A x (2x + 1) + B (2x + 1) + C x^2 ]
Next, we can substitute suitable values of ( x ) to simplify finding ( A ), ( B ), ) and ( C ).
- Let ( x = 0 ): [ 25 = B \cdot 1 \Rightarrow B = 25 ]
- Let ( x = -\frac{1}{2} ): [ 25 = A(-\frac{1}{2})(0) + B(0) + C(-\frac{1}{2})^2 ] [ 25 = C \cdot \frac{1}{4} \Rightarrow C = 100 ]
- To find ( A ), we can substitute back into our equation and determine from the coefficients of like terms.
the equation reduces to [ 25 = A(2x^2 + x) + 25(2x + 1) + 100x^2 = (2A + 100)x^2 + (A + 50)x + 25 ]
Equating coefficients gives:
- Comparing coefficients: ( 2A + 100 = 0 \rightarrow A = -50 )
Thus, we can rewrite the partial fractions as:
[ \frac{25}{x^2(2x + 1)} = \frac{-50}{x} + \frac{25}{x^2} + \frac{100}{2x + 1} ]
Step 2
Using calculus to find the volume of the solid of revolution
Answer
To find the volume of the solid of revolution generated by rotating region ( R ) about the x-axis, we use the formula:
[ V = \pi \int_{a}^{b} f(x)^2 , dx ]
Where ( f(x) = \frac{5}{\sqrt{2x + 1}} ) and the limits are from ( x = 1 ) to ( x = 4 ).
Thus, we calculate:
[ V = \pi \int_{1}^{4} \left( \frac{5}{\sqrt{2x + 1}} \right)^2 , dx = 25\pi \int_{1}^{4} \frac{1}{2x + 1} , dx ]
Now we apply integration:
[ = 25\pi \left( \frac{1}{2} \ln(2x + 1) \right) \biggr|_{1}^{4} = \frac{25\pi}{2} \left( \ln(9) - \ln(3) \right) = \frac{25\pi}{2} \ln(3) ]
The final volume is thus expressed in the required form:
[ V = 12.5\pi \ln(3) ]