Given the function: $$ f(x) = \frac{1}{x(3x - 1)^2} = \frac{A}{x} + \frac{B}{(3x - 1)} + \frac{C}{(3x - 1)^2} $$ (a) Find the values of the constants A, B and C. (b) Hence find $ \int f(x) \, dx $ dx. (i) Find $ \int^2 f(x) \, dx $, leaving your answer in the form a + ln b, where a and b are constants.

A-Level Edexcel Maths Pure Binomial Expansion: Given the function: $$ f(x) = \

Given the function: $$ f(x) = \frac{1}{x(3x - 1)^2} = \frac{A}{x} + \frac{B}{(3x - 1)} + \frac{C}{(3x - 1)^2} $$ (a) Find the values of the constants A, B and C - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 7

Question 3

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Given the function: $$ f(x) = \frac{1}{x(3x - 1)^2} = \frac{A}{x} + \frac{B}{(3x - 1)} + \frac{C}{(3x - 1)^2} $$ (a) Find the values of the constants A, B and C. ... show full transcript

Worked Solution & Example Answer:Given the function: $$ f(x) = \frac{1}{x(3x - 1)^2} = \frac{A}{x} + \frac{B}{(3x - 1)} + \frac{C}{(3x - 1)^2} $$ (a) Find the values of the constants A, B and C - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 7

Step 1

Find the values of the constants A, B and C

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Answer

To find the constants A, B, and C, we first combine the fractions on the right-hand side and set them equal to the left-hand side:

$f(x) = \frac{A(3x - 1)^2 + Bx(3x - 1) + Cx}{x(3x - 1)^2}$

Then, expanding the numerators and combining like terms gives:

Set x = 0: $1 = A(3(0) - 1)^2$
Hence, $A = 4$
Set x approaching ( \frac{1}{3} ): $1 = C \rightarrow C = 3 \text{ (considering the limit)}$
Compare coefficients: From the coefficient of ( x^2 ): $0 = 9A + 3B$
$0 = 9(4) + 3B \Rightarrow B = -3$

Thus, the constants are:

A = 4, B = -3, C = 3.

Step 2

Hence find $ \int f(x) \, dx $

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Answer

Now, substituting the values of A, B, and C back into the integral:

$f(x) = \frac{4}{x} + \frac{-3}{(3x - 1)} + \frac{3}{(3x - 1)^2}$

We can integrate term by term:

( \int \frac{4}{x} , dx = 4 \ln |x| + C_1 )
( \int \frac{-3}{3x - 1} , dx = - \ln |3x - 1| + C_2 )
For ( \int \frac{3}{(3x - 1)^2} , dx ):
- Use substitution ( u = 3x - 1 \Rightarrow du = 3 , dx \Rightarrow dx = \frac{du}{3} )
- This results in: $-\frac{1}{3(3x-1)} + C_3$

Combining all parts gives:

$\int f(x) \, dx = 4 \ln |x| - \ln |3x - 1| - \frac{1}{3(3x - 1)} + C$ .

Step 3

Find $ \int^2 f(x) \, dx $

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Answer

Now we evaluate:

$\int^2 f(x) \, dx = \left[ 4 \ln |x| - \ln |3x - 1| - \frac{1}{3(3x - 1)} \right]_{0}^{2}$ .

Substituting x = 2:

Calculate: ( 4 \ln 2 - \ln(5) - \frac{1}{3 imes 5} )

Since we want the answer in the form a + ln b:

Identify constants a and b: $a = 4 \ln 2 - \frac{1}{15}, \ b = 5$ .

Given the function: $$ f(x) = \frac{1}{x(3x - 1)^2} = \frac{A}{x} + \frac{B}{(3x - 1)} + \frac{C}{(3x - 1)^2} $$ (a) Find the values of the constants A, B and C - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 7

Worked Solution & Example Answer:Given the function: $$ f(x) = \frac{1}{x(3x - 1)^2} = \frac{A}{x} + \frac{B}{(3x - 1)} + \frac{C}{(3x - 1)^2} $$ (a) Find the values of the constants A, B and C - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 7

Find the values of the constants A, B and C

Hence find \( \int f(x) \, dx \)

Find \( \int^2 f(x) \, dx \)

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