Determine whether \({f_n}\) is an odd or even function, justifying your answer.

By using mathematical induction, prove that

\({f_n}(x) = \frac{{\sin {2^{n + 1}}x}}{{{2^n}\sin 2x}},{\text{ }}x \ne \frac{{m\pi }}{2}\) where \(m \in \mathbb{Z}\).

Hence or otherwise, find an expression for the derivative of \({f_n}(x)\) with respect to \(x\).

Show that, for \(n > 1\), the equation of the tangent to the curve \(y = {f_n}(x)\) at \(x = \frac{\pi }{4}\) is \(4x - 2y - \pi  = 0\).

even function     A1

since \(\cos kx = \cos ( - kx)\) and \({f_n}(x)\) is a product of even functions     R1

OR

even function     A1

since \((\cos 2x)(\cos 4x) \ldots  = \left( {\cos ( - 2x)} \right)\left( {\cos ( - 4x)} \right) \ldots \)     R1

Note:     Do not award A0R1.

[2 marks]

consider the case \(n = 1\)

\(\frac{{\sin 4x}}{{2\sin 2x}} = \frac{{2\sin 2x\cos 2x}}{{2\sin 2x}} = \cos 2x\)     M1

hence true for \(n = 1\)     R1

assume true for \(n = k\), ie, \((\cos 2x)(\cos 4x) \ldots (\cos {2^k}x) = \frac{{\sin {2^{k + 1}}x}}{{{2^k}\sin 2x}}\)     M1

Note:     Do not award M1 for “let \(n = k\)” or “assume \(n = k\)” or equivalent.

consider \(n = k + 1\):

\({f_{k + 1}}(x) = {f_k}(x)(\cos {2^{k + 1}}x)\)     (M1)

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

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

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

so \(n = 1\) true and \(n = k\) true \( \Rightarrow n = k + 1\) true. Hence true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note:     To obtain the final R1, all the previous M marks must have been awarded.

[8 marks]

attempt to use \(f’ = \frac{{vu' - uv'}}{{{v^2}}}\) (or correct product rule)     M1

\({f’_n}(x) = \frac{{({2^n}\sin 2x)({2^{n + 1}}\cos {2^{n + 1}}x) - (\sin {2^{n + 1}}x)({2^{n + 1}}\cos 2x)}}{{{{({2^n}\sin 2x)}^2}}}\)     A1A1

Note:     Award A1 for correct numerator and A1 for correct denominator.

[3 marks]

\({f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{\left( {{2^n}\sin \frac{\pi }{2}} \right)\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right) - \left( {\sin {2^{n + 1}}\frac{\pi }{4}} \right)\left( {{2^{n + 1}}\cos \frac{\pi }{2}} \right)}}{{{{\left( {{2^n}\sin \frac{\pi }{2}} \right)}^2}}}\)     (M1)(A1)

\({f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{({2^n})\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right)}}{{{{({2^n})}^2}}}\)     (A1)

\( = 2\cos {2^{n + 1}}\frac{\pi }{4}{\text{ }}( = 2\cos {2^{n - 1}}\pi )\)     A1

\({f’_n}\left( {\frac{\pi }{4}} \right) = 2\)     A1

\({f_n}\left( {\frac{\pi }{4}} \right) = 0\)     A1

Note:     This A mark is independent from the previous marks.

\(y = 2\left( {x - \frac{\pi }{4}} \right)\)     M1A1

\(4x - 2y - \pi  = 0\)     AG

[8 marks]