Find an expression for \((f \circ f)(x)\) .

Let \({F_n}(x) = \frac{x}{{{2^n} - ({2^n} - 1)x}}\), where \(0 \leqslant x \leqslant 1\).

Use mathematical induction to show that for any \(n \in {\mathbb{Z}^ + }\)

\[\underbrace {(f \circ f \circ ... \circ f)}_{n{\text{ times}}}(x) = {F_n}(x)\] .

Show that \({F_{ - n}}(x)\) is an expression for the inverse of \({F_n}\) .

(i)     State \({F_n}(0){\text{ and }}{F_n}(1)\) .

(ii)     Show that \({F_n}(x) < x\) , given 0 < x < 1, \(n \in {\mathbb{Z}^ + }\) .

(iii)     For \(n \in {\mathbb{Z}^ + }\) , let \({A_n}\) be the area of the region enclosed by the graph of \(F_n^{ - 1}\) , the x-axis and the line x = 1. Find the area \({B_n}\) of the region enclosed by \({F_n}\) and \(F_n^{ - 1}\) in terms of \({A_n}\) .

\((f \circ f)(x) = f\left( {\frac{x}{{2 - x}}} \right) = \frac{{\frac{x}{{2 - x}}}}{{2 - \frac{x}{{2 - x}}}}\)     M1A1

\((f \circ f)(x) = \frac{x}{{4 - 3x}}\)     A1

[3 marks]

\(P(n):\underbrace {(f \circ f \circ ... \circ f)}_{n{\text{ times}}}(x) = {F_n}(x)\)

\(P(1):{\text{ }}f(x) = {F_1}(x)\)

\(LHS = f(x) = \frac{x}{{2 - x}}{\text{ and }}RHS = {F_1}(x) = \frac{x}{{{2^1} - ({2^1} - 1)x}} = \frac{x}{{2 - x}}\)     A1A1

\(\therefore P(1){\text{ true}}\)

assume that P(k) is true, i.e., \(\underbrace {(f \circ f \circ ... \circ f)}_{{\text{k times}}}(x) = {F_k}(x)\)     M1

consider \(P(k + 1)\)

EITHER

\(\underbrace {(f \circ f \circ ... \circ f)}_{{\text{k}} + 1{\text{ times}}}(x) = \left( {f \circ \underbrace {f \circ f \circ ... \circ f}_{{\text{k times}}}} \right)(x) = f\left( {{F_k}(x)} \right)\)     (M1)

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

\( = \frac{x}{{2\left( {{2^k} - ({2^k} - 1)x} \right) - x}} = \frac{x}{{{2^{k + 1}} - ({2^{k + 1}} - 2)x - x}}\)     A1

OR

\(\underbrace {(f \circ f \circ ... \circ f)}_{{\text{k}} + 1{\text{ times}}}(x) = \left( {f \circ \underbrace {f \circ f \circ ... \circ f}_{{\text{k times}}}} \right)(x) = {F_k}(f(x))\)     (M1)

\( = {F_k}\left( {\frac{x}{{2 - x}}} \right) = \frac{{\frac{x}{{2 - x}}}}{{{2^k} - ({2^k} - 1)\frac{x}{{2 - x}}}}\)     A1

\( = \frac{x}{{{2^{k + 1}} - {2^k}x - {2^k}x + x}}\)     A1

THEN

\( = \frac{x}{{{2^{k + 1}} - ({2^{k + 1}} - 1)x}} = {F_{k + 1}}(x)\)     A1

P(k) true implies P(k + 1) true, P(1) true so P(n) true for all \(n \in {\mathbb{Z}^ + }\)     R1

[8 marks]

METHOD 1

\(x = \frac{y}{{{2^n} - ({2^n} - 1)y}} \Rightarrow {2^n}x - ({2^n} - 1)xy = y\)     M1A1

\( \Rightarrow {2^n}x = \left( {({2^n} - 1)x + 1} \right)y \Rightarrow y = \frac{{{2^n}x}}{{({2^n} - 1)x + 1}}\)     A1

\(F_n^{ - 1}(x) = \frac{{{2^n}x}}{{({2^n} - 1)x + 1}}\)     A1

\(F_n^{ - 1}(x) = \frac{x}{{\frac{{{2^n} - 1}}{{{2^n}}}x + \frac{1}{{{2^n}}}}}\)     M1

\(F_n^{ - 1}(x) = \frac{x}{{(1 - {2^{ - n}})x + {2^{ - n}}}}\)     A1

\(F_n^{ - 1}(x) = \frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}\)     AG

METHOD 2

attempt \({F_{ - n}}\left( {{F_n}(x)} \right)\)     M1

\( = {F_{ - n}}\left( {\frac{x}{{{2^n} - ({2^n} - 1)x}}} \right) = \frac{{\frac{x}{{{2^n} - ({2^n} - 1)x}}}}{{{2^{ - n}} - ({2^{ - n}} - 1)\frac{x}{{{2^n} - ({2^n} - 1)x}}}}\)     A1A1

\( = \frac{x}{{{2^{ - n}}({2^n} - ({2^n} - 1)x) - ({2^{ - n}} - 1)x}}\)     A1A1

Note: Award A1 marks for numerators and denominators.

\( = \frac{x}{1} = x\)     A1AG

METHOD 3

attempt \({F_n}\left( {{F_{ - n}}(x)} \right)\)     M1

\( = {F_n}\left( {\frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}} \right) = \frac{{\frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}}}{{{2^n} - ({2^n} - 1)\frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}}}\)     A1A1

\( = \frac{x}{{{2^n}({2^{ - n}} - ({2^{ - n}} - 1)x) - ({2^n} - 1)x}}\)     A1A1

Note: Award A1 marks for numerators and denominators.

\( = \frac{x}{1} = x\)     A1AG

[6 marks]

(i)     \({F_n}(0) = 0,{\text{ }}{F_n}(1) = 1\)     A1

(ii)     METHOD 1

\({2^n} - ({2^n} - 1)x - 1 = ({2^n} - 1)(1 - x)\)     (M1)

\( > 0{\text{ if }}0 < x < 1{\text{ and }}n \in {\mathbb{Z}^ + }\)     A1

so \({2^n} - ({2^n} - 1)x > 1{\text{ and }}{F_n}(x) = \frac{x}{{{2^n} - ({2^n} - 1)x}} < \frac{x}{1}( < x)\)     R1

\({F_n}(x) = \frac{x}{{{2^n} - ({2^n} - 1)x}} < x{\text{ for }}0 < x < 1{\text{ and }}n \in {\mathbb{Z}^ + }\)     AG

METHOD 2

\(\frac{x}{{{2^n} - ({2^n} - 1)x}} < x \Leftrightarrow {2^n} - ({2^n} - 1)x > 1\)     (M1)

\( \Leftrightarrow ({2^n} - 1)x < {2^n} - 1\)     A1

\( \Leftrightarrow x < \frac{{{2^n} - 1}}{{{2^n} - 1}} = 1\) true in the interval \(\left] {0,{\text{ }}1} \right[\)     R1

(iii)     \({B_n} = 2\left( {{A_n} - \frac{1}{2}} \right){\text{ }}( = 2{A_n} - 1)\)     (M1)A1

[6 marks]

Part b) was often answered well, and candidates were well prepared in this session for this type of question. Candidates still need to take care when showing explicitly that P(1) is true, and some are still writing ‘Let n = k’ which gains no marks. The inductive step was often well argued, and given in clear detail, though the final inductive reasoning step was incorrect, or appeared rushed, even from the better candidates. ‘True for n =1, n = k and n = k + 1’ is still disappointingly seen, as were some even more unconvincing variations.

Part c) was again very well answered by the majority. A few weaker candidates attempted to find an inverse for the individual case n = 1 , but gained no credit for this.

Part d) was not at all well understood, with virtually no candidates able to tie together the hints given by connecting the different parts of the question. Rash, and often thoughtless attempts were made at each part, though by this stage some seemed to be struggling through lack of time. The inequality part of the question tended to be ‘fudged’, with arguments seen by examiners being largely unconvincing and lacking clarity. A tiny number of candidates provided the correct answer to the final part, though a surprising number persisted with what should have been recognised as fruitless working – usually in the form of long-winded integration attempts.