Find an expression for \({f^{ - 1}}(x)\).

Given that \(f(x)\) can be written in the form \(f(x) = A + \frac{B}{{2x - 1}}\), find the values of the constants \(A\) and \(B\).

Hence, write down \(\int {\frac{{3x - 2}}{{2x - 1}}} {\text{d}}x\).

\(f:x \to y = \frac{{3x - 2}}{{2x - 1}}\;\;\;{f^{ - 1}}:y \to x\)

\(y = \frac{{3x - 2}}{{2x - 1}} \Rightarrow 3x - 2 = 2xy - y\)     M1

\( \Rightarrow 3x - 2xy =  - y + 2\)     M1

\(x(3 - 2y) = 2 - y\)

\(x = \frac{{2 - y}}{{3 - 2y}}\)     A1

\(\left( {{f^{ - 1}}(y) = \frac{{2 - y}}{{3 - 2y}}} \right)\)

\({f^{ - 1}}(x) = \frac{{2 - x}}{{3 - 2x}}\;\;\;\left( {x \ne \frac{3}{2}} \right)\)     A1

Note:     \(x\) and \(y\) might be interchanged earlier.

Note:     First M1 is for interchange of variables second M1 for manipulation

Note:     Final answer must be a function of \(x\)

[4 marks]

\(\frac{{3x - 2}}{{2x - 1}} = A + \frac{B}{{2x - 1}} \Rightarrow 3x - 2 = A(2x - 1) + B\)

equating coefficients \(3 = 2A\) and \( - 2 =  - A + B\)     (M1)

\(A = \frac{3}{2}\) and \(B =  - \frac{1}{2}\)     A1

Note:     Could also be done by division or substitution of values.

[2 marks]

\(\int {f(x){\text{d}}x = \frac{3}{2}x - \frac{1}{4}\ln \left| {2x - 1} \right| + c} \)     A1

Note:     accept equivalent e.g. \(\ln \left| {4x - 2} \right|\)

[1 mark]

Total [7 marks]

Well done. Only a few candidates confused inverse with derivative or reciprocal.

Not enough had the method of polynomial division.

Reasonable if they had an answer to (b) (follow through was given) usual mistakes with not allowing for the derivative of the bracket.