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4. (a) Express \( \frac{25}{x^2(2x+1)} \) in partial fractions - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 8

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4. (a) Express \( \frac{25}{x^2(2x+1)} \) in partial fractions. (b) Use calculus to find the exact volume of the solid of revolution generated, giving your answer... 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 6 - 2014 - Paper 8

Step 1

Express \( \frac{25}{x^2(2x+1)} \) in partial fractions

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Answer

To express ( \frac{25}{x^2(2x+1)} ) in partial fractions, we assume it can be decomposed as follows:

25x2(2x+1)=Ax+Bx2+C2x+1\frac{25}{x^2(2x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x+1}

Multiplying both sides by ( x^2(2x+1) ) gives:

25=Ax(2x+1)+B(2x+1)+Cx225 = A x(2x+1) + B(2x+1) + C x^2

Setting up equations by substituting suitable values for ( x ):

  • Let ( x = 0 ):
    25=B(1)B=2525 = B(1) \Rightarrow B = 25

  • Let ( x = 1 ):
    25=A(1)(3)+25(3)+C(1)25=3A+75+CC=503A25 = A(1)(3) + 25(3) + C(1) \Rightarrow 25 = 3A + 75 + C \Rightarrow C = -50 - 3A

  • For the coefficient of ( x^2 ) in ( 25 = (A + C)x^2 + (2B + 2C)x + (B) ):
    0=A+C ext(coefficientofx2)0=A503A2A=50A=250 = A + C\ ext{(coefficient of } x^2\text{)} \Rightarrow 0 = A - 50 - 3A \Rightarrow 2A = 50 \Rightarrow A = 25

Substituting ( A = 25 ) back gives ( C = -50 - 75 \Rightarrow C = -125 ).

Thus, we get:

25x2(2x+1)=25x+25x21252x+1\frac{25}{x^2(2x+1)} = \frac{25}{x} + \frac{25}{x^2} - \frac{125}{2x+1}

Step 2

Use calculus to find the exact volume of the solid of revolution generated

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Answer

To find the volume ( V ) of the solid of revolution generated when region ( R ) is rotated around the x-axis, we use the formula:

V=14π[f(x)]2dxV = \int_{1}^{4} \pi [f(x)]^2 \, dx

In this case, ( f(x) = \frac{5}{\sqrt{2x+1}} ) so:

V=π14(52x+1)2dx=π14252x+1dxV = \pi \int_{1}^{4} \left(\frac{5}{\sqrt{2x+1}}\right)^2 \, dx = \pi \int_{1}^{4} \frac{25}{2x+1} \, dx

Evaluating the integral:

V=π[25ln(2x+1)]14V = \pi \cdot \left[ 25 \ln(2x+1) \right]_{1}^{4}

This simplifies to:

V=π(25ln(9)25ln(3))=25πln(3)V = \pi \left( 25 \ln(9) - 25 \ln(3) \right) = 25\pi \ln(3)

Finally, we represent this in the required form:

V=25π(ln(3)ln(1))+25πV = 25\pi(\ln(3) - \ln(1)) + 25\pi

So, letting ( a = 25\pi, b = 25\pi, c = 3 ) gives us:

V=a+bln(c)V = a + b\ln(c)

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