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]