Show that $f’(0) = p$.

Show that ${f^{(n + 2)}}(0) = ({n^2} - {p^2}){f^{(n)}}(0)$.

For $p \in \mathbb{R}\backslash \{  \pm 1,{\text{ }} \pm 3\}$, show that the Maclaurin series for $f(x)$, up to and including the ${x^5}$ term, is

$px + \frac{{p(1 - {p^2})}}{{3!}}{x^3} + \frac{{p(9 - {p^2})(1 - {p^2})}}{{5!}}{x^5}$.

Hence or otherwise, find $\mathop {\lim }\limits_{x \to 0} \frac{{\sin (p\arcsin x)}}{x}$.

If $p$ is an odd integer, prove that the Maclaurin series for $f(x)$ is a polynomial of degree $p$.

$f’(x) = \frac{{p\cos (p\arcsin x)}}{{\sqrt {1 - {x^2}} }}$     (M1)A1

Note: Award M1 for attempting to use the chain rule.

$f’(0) = p$     AG

[2 marks]

EITHER

${f^{(n + 2)}}(0) + ({p^2} - {n^2}){f^{(n)}}(0) = 0$     A1

OR

for eg, $(1 - {x^2}){f^{(n + 2)}}(x) = (2n + 1)x{f^{(n + 1)}}(x) - ({p^2} - {n^2}){f^{(n)}}(x)$     A1

Note: Award A1 for eg, $(1 - {x^2}){f^{(n + 2)}}(x) - (2n + 1)x{f^{(n + 1)}}(x) =  - ({p^2} - {n^2}){f^{(n)}}(x)$.

THEN

${f^{(n + 2)}}(0) = ({n^2} - {p^2}){f^{(n)}}(0)$     AG

[1 mark]

considering $f$ and its derivatives at $x = 0$     (M1)

$f(0) = 0$ and $f’(0) = p$ from (a)     A1

$f’’(0) = 0,{\text{ }}{f^{(4)}}(0) = 0$     A1

${f^{(3)}}(0) = (1 - {p^2}){f^{(1)}}(0) = (1 - {p^2})p$,

${f^{(5)}}(0) = (9 - {p^2}){f^{(3)}}(0) = (9 - {p^2})(1 - {p^2})p$     A1

Note:     Only award the last A1 if either ${f^{(3)}}(0) = (1 - {p^2}){f^{(1)}}(0)$ and ${f^{(5)}}(0) = (9 - {p^2}){f^{(3)}}(0)$ have been stated or the general Maclaurin series has been stated and used.

$px + \frac{{p(1 - {p^2})}}{{3!}}{x^3} + \frac{{p(9 - {p^2})(1 - {p^2})}}{{5!}}{x^5}$     AG

[4 marks]

METHOD 1

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (p\arcsin x)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{px + \frac{{p(1 - {p^2})}}{{3!}}{x^3} +  \ldots }}{3}$     M1

$= p$     A1

METHOD 2

by l’Hôpital’s rule $\mathop {\lim }\limits_{x \to 0} \frac{{\sin (p\arcsin x)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{p\cos (p\arcsin x)}}{{\sqrt {1 - {x^2}} }}$     M1

$= p$     A1

[2 marks]

the coefficients of all even powers of $x$ are zero     A1

the coefficient of ${x^p}$ for ($p$ odd) is non-zero (or equivalent eg,

the coefficients of all odd powers of $x$ up to $p$ are non-zero)     A1

${f^{(p + 2)}}(0) = ({p^2} - {p^2}){f^{(p)}}(0) = 0$ and so the coefficient of ${x^{p + 2}}$ is zero     A1

the coefficients of all odd powers of $x$ greater than $p + 2$ are zero (or equivalent)     A1

so the Maclaurin series for $f(x)$ is a polynomial of degree $p$     AG

[4 marks]