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]