(i)     Find \({f^{ - 1}}(x)\).

(ii)     State the domain of \({f^{ - 1}}\).

The function \(g\) is defined as \(g(x) = \ln x,{\text{ }}x \in {\mathbb{R}^ + }\).

The graph of \(y = g(x)\) and the graph of \(y = {f^{ - 1}}(x)\) intersect at the point \(P\).

Find the coordinates of \(P\).

The graph of \(y = g(x)\) intersects the \(x\)-axis at the point \(Q\).

Show that the equation of the tangent \(T\) to the graph of \(y = g(x)\) at the point \(Q\) is \(y = x - 1\).

A region \(R\) is bounded by the graphs of \(y = g(x)\), the tangent \(T\) and the line \(x = {\text{e}}\).

Find the area of the region \(R\).

A region \(R\) is bounded by the graphs of \(y = g(x)\), the tangent \(T\) and the line \(x = {\text{e}}\).

(i)     Show that \(g(x) \le x - 1,{\text{ }}x \in {\mathbb{R}^ + }\).

(ii)     By replacing \(x\) with \(\frac{1}{x}\) in part (e)(i), show that \(\frac{{x - 1}}{x} \le g(x),{\text{ }}x \in {\mathbb{R}^ + }\).

(i)     \(x = {{\text{e}}^{3y + 1}}\)     M1

Note:     The M1 is for switching variables and can be awarded at any stage.

Further marks do not rely on this mark being awarded.

taking the natural logarithm of both sides and attempting to transpose     M1

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

(ii)     \(x \in {\mathbb{R}^ + }\) or equivalent, for example \(x > 0\).     A1

[4 marks]

\(\ln x = \frac{1}{3}(\ln x - 1) \Rightarrow \ln x - \frac{1}{3}\ln x =  - \frac{1}{3}\) (or equivalent)     M1A1

\(\ln x =  - \frac{1}{2}\) (or equivalent)     A1

\(x = {{\text{e}}^{ - \frac{1}{2}}}\)     A1

coordinates of \(P\) are \(\left( {{{\text{e}}^{ - \frac{1}{2}}},{\text{ }} - \frac{1}{2}} \right)\)     A1

[5 marks]

coordinates of \(Q\) are (\(1,{\rm{ }}0\)) seen anywhere     A1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{x}\)     M1

at \({\text{Q, }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1\)     A1

\(y = x - 1\)     AG

[3 marks]

let the required area be \(A\)

\(A = \int_1^e {x - 1{\text{d}}x - \int_1^e {\ln x{\text{d}}x} } \)     M1

Note:     The M1 is for a difference of integrals. Condone absence of limits here.

attempting to use integration by parts to find \(\int {\ln x{\text{d}}x} \)     (M1)

\( = \left[ {\frac{{{x^2}}}{2} - x} \right]_1^{\text{e}} - [x\ln x - x]_1^{\text{e}}\)     A1A1

Note:     Award A1 for \(\frac{{{x^2}}}{2} - x\) and A1 for \(x\ln x - x\).

Note:     The second M1 and second A1 are independent of the first M1 and the first A1.

\( = \frac{{{{\text{e}}^2}}}{2} - {\text{e}} - \frac{1}{2}\left( { = \frac{{{{\text{e}}^2} - 2{\text{e}} - 1}}{2}} \right)\)     A1

[5 marks]

(i)     METHOD 1

consider for example \(h(x) = x - 1 - \ln x\)

\(h(1) = 0\;\;\;{\text{and}}\;\;\;h'(x) = 1 - \frac{1}{x}\)     (A1)

as \(h'(x) \ge 0\;\;\;{\text{for}}\;\;\;x \ge 1,\;\;\;{\text{then}}\;\;\;h(x) \ge 0\;\;\;{\text{for}}\;\;\;x \ge 1\)     R1

as \(h'(x) \le 0\;\;\;{\text{for}}\;\;\;0 < x \le 1,\;\;\;{\text{then}}\;\;\;h(x) \ge 0\;\;\;{\text{for}}\;\;\;0 < x \le 1\)     R1

so \(g(x) \le x - 1,{\text{ }}x \in {\mathbb{R}^ + }\)     AG

METHOD 2

\(g''(x) =  - \frac{1}{{{x^2}}}\)     A1

\(g''(x) < 0\;\;\;\)(concave down) for\(\;\;\;x \in {\mathbb{R}^ + }\)     R1

the graph of \(y = g(x)\) is below its tangent \((y = x - 1\;\;\;{\text{at}}\;\;\;x = 1)\)     R1

so \(g(x) \le x - 1,{\text{ }}x \in {\mathbb{R}^ + }\)     AG

Note:     The reasoning may be supported by drawn graphical arguments.

METHOD 3

clear correct graphs of \(y = x - 1\;\;\;{\text{and}}\;\;\;\ln x\;\;\;{\text{for}}\;\;\;x > 0\)     A1A1

statement to the effect that the graph of \(\ln x\) is below the graph of its tangent at \(x = 1\)     R1AG

(ii)     replacing \(x\) by \(\frac{1}{x}\) to obtain \(\ln \left( {\frac{1}{x}} \right) \le \frac{1}{x} - 1\left( { = \frac{{1 - x}}{x}} \right)\)     M1

\( - \ln x \le \frac{1}{x} - 1\left( { = \frac{{1 - x}}{x}} \right)\)     (A1)

\(\ln x \ge 1 - \frac{1}{x}\left( { = \frac{{x - 1}}{x}} \right)\)     A1

so \(\frac{{x - 1}}{x} \le g(x),{\text{ }}x \in {\mathbb{R}^ + }\)     AG

[6 marks]

Total [23 marks]

Generally very well done, even by candidates who had shown considerable weaknesses elsewhere on the paper.

Generally very well done, even by candidates who had shown considerable weaknesses elsewhere on the paper.

Generally very well done, even by candidates who had shown considerable weaknesses elsewhere on the paper.

A productive question for many candidates, but some didn’t realise that a difference of areas/integrals was required.

(i)     Many candidates adopted a graphical approach, but sometimes with unconvincing reasoning.

(ii)     Poorly answered. Many candidates applied the suggested substitution only to one side of the inequality, and then had to fudge the answer.