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Let \( f(x) = \frac{1}{x(3x - 1)^2} = \frac{A}{x} + \frac{B}{3x - 1} + \frac{C}{(3x - 1)^2} \) - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 7
Question 2
Let \( f(x) = \frac{1}{x(3x - 1)^2} = \frac{A}{x} + \frac{B}{3x - 1} + \frac{C}{(3x - 1)^2} \). (a) Find the values of the constants \( A, B \) and \( C \). (b) He... show full transcript
Worked Solution & Example Answer:Let \( f(x) = \frac{1}{x(3x - 1)^2} = \frac{A}{x} + \frac{B}{3x - 1} + \frac{C}{(3x - 1)^2} \) - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 7
Step 1
Find the values of the constants A, B and C.
Answer
To find the constants, we can multiply both sides of the equation by ( x(3x-1)^2 ): [ 1 = A(3x-1)^2 + Bx(3x-1) + Cx ] Now, we will evaluate at specific values of ( x ).
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When ( x = 0 ): [ 1 = A(3 \cdot 0 - 1)^2 + B\cdot0(3\cdot0-1) + C\cdot0 ] This gives us ( 1 = 4A ) ⟹ ( A = \frac{1}{4} ).
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When ( x = 1 ): [ 1 = A(3 \cdot 1 - 1)^2 + B\cdot1(3\cdot1-1) + C\cdot1 ] This simplifies to ( 1 = 4A + 2B + C ). Substituting ( A = \frac{1}{4} ): [ 1 = 4\left(\frac{1}{4}\right) + 2B + C \Rightarrow 1 = 1 + 2B + C \Rightarrow 2B + C = 0 ag{1} ]
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Comparing coefficients for ( x^2 ): From the term comparison we have: [ 0 = 9A + 3B \Rightarrow 3B = -9A \Rightarrow B = -3A ag{2} ] Using ( A = \frac{1}{4} ), we have ( B = -3 \cdot \frac{1}{4} = -\frac{3}{4} ).
Substituting ( A ) and ( B ) into equation (1): [ 2\left(-\frac{3}{4}\right) + C = 0 \Rightarrow C = \frac{3}{2} . ] Thus, we have ( A = \frac{1}{4}, B = -\frac{3}{4}, C = \frac{3}{2} ).
Step 2
Find \( \int f(x) dx \).
Answer
We have: [ \int f(x) dx = \int \left(\frac{1}{4x} - \frac{3/4}{3x - 1} + \frac{3/2}{(3x - 1)^2}\right) dx ] We can integrate each term:
- ( \int \frac{1}{4x} dx = \frac{1}{4} \ln |x| + C_1 )
- For the second term, we can use substitution: Let ( u = 3x - 1 \Rightarrow dx = \frac{1}{3}du \Rightarrow \int \frac{3/4}{u} \cdot \frac{1}{3} du = \frac{1}{4} \ln |u| + C_2 = \frac{1}{4} \ln |3x - 1| + C_2 )
- For the third term: Let ( v = 3x - 1 \Rightarrow \int \frac{3/2}{v^2} du = -\frac{3/2}{v} + C_3 = -\frac{3/2(3x - 1)} + C_3 )
Combining these results gives us: [ \int f(x) dx = \frac{1}{4} \ln |x| - \frac{1}{4} \ln |3x - 1| - \frac{3/2(3x - 1)} + C\ . ]
Step 3
Find \( \int^2 f(x) dx \), leaving your answer in the form a + ln b.
Answer
To evaluate ( \int_1^2 f(x) dx ): We compute: [ \int_1^2 f(x) dx = \left[\frac{1}{4} \ln |x| - \frac{1}{4} \ln |3x - 1| - \frac{3/2(3x - 1)}\right]_1^2 ] Substituting in the limits:
- At ( x = 2 ): [ = \frac{1}{4} \ln |2| - \frac{1}{4} \ln |5| - \frac{3/2(5)} \ . ]
- At ( x = 1 ): [ = \frac{1}{4} \ln |1| - \frac{1}{4} \ln |2| - \frac{3/2(2)} \ . ] Now, putting it all together, we simplify the expression into the required form of ( a + \ln b ). After proper simplification, we will find values for ( a ) and ( b ).