Show that \(f'\left( 0 \right) = 0\).

By differentiating the above equation twice, show that

\[\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 5x{f^{\left( 3 \right)}}\left( x \right) - 4f''\left( x \right) = 0\]

where \({f^{\left( 3 \right)}}\left( x \right)\) and \({f^{\left( 4 \right)}}\left( x \right)\) denote the 3rd and 4th derivative of \(f\left( x \right)\) respectively.

Hence show that the Maclaurin series for \(f\left( x \right)\) up to and including the term in \({x^4}\) is \({x^2} + \frac{1}{3}{x^4}\).

Use this series approximation for \(f\left( x \right)\) with \(x = \frac{1}{2}\) to find an approximate value for \({\pi ^2}\).

\(f'\left( x \right) = \frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 - {x^2}} }}\)     M1A1

Note: Award M1 for an attempt at chain rule differentiation.
Award M0A0 for \(f'\left( x \right) = 2\,{\text{arcsin}}\,\left( x \right)\).

\(f'\left( 0 \right) = 0\)     AG

[2 marks]

differentiating gives \(\left( {1 - {x^2}} \right){f^{\left( 3 \right)}}\left( x \right) - 2xf''\left( x \right) - f'\left( x \right) - xf''\left( x \right)\left( { = 0} \right)\)      M1A1

differentiating again gives \(\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 2x{f^{\left( 3 \right)}}\left( x \right) - 3f''\left( x \right) - 3x{f^{\left( 3 \right)}}\left( x \right) - f''\left( x \right)\left( { = 0} \right)\)     M1A1

Note: Award M1 for an attempt at product rule differentiation of at least one product in each of the above two lines.
Do not penalise candidates who use poor notation.

\(\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 5x{f^{\left( 3 \right)}}\left( x \right) - 4f''\left( x \right) = 0\)      AG

[4 marks]

attempting to find one of \(f''\left( 0 \right)\), \({f^{\left( 3 \right)}}\left( 0 \right)\) or \({f^{\left( 4 \right)}}\left( 0 \right)\) by substituting \(x = 0\) into relevant differential equation(s)       (M1)

Note: Condone \(f''\left( 0 \right)\) found by calculating \(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 - {x^2}} }}} \right)\) at \(x = 0\).

\(\left( {f\left( 0 \right) = 0,\,f'\left( 0 \right) = 0} \right)\)

\(f''\left( 0 \right) = 2\) and \({f^{\left( 4 \right)}}\left( 0 \right) - 4f''\left( 0 \right) = 0 \Rightarrow {f^{\left( 4 \right)}}\left( 0 \right) = 8\)      A1

\({f^{\left( 3 \right)}}\left( 0 \right) = 0\) and so \(\frac{2}{{2{\text{!}}}}{x^2} + \frac{8}{{4{\text{!}}}}{x^4}\)     A1

Note: Only award the above A1, for correct first differentiation in part (b) leading to \({f^{\left( 3 \right)}}\left( 0 \right) = 0\) stated or \({f^{\left( 3 \right)}}\left( 0 \right) = 0\) seen from use of the general Maclaurin series.
Special Case: Award (M1)A0A1 if \({f^{\left( 4 \right)}}\left( 0 \right) = 8\) is stated without justification or found by working backwards from the general Maclaurin series.

so the Maclaurin series for \(f\left( x \right)\) up to and including the term in \({x^4}\) is \({x^2} + \frac{1}{3}{x^4}\)     AG

[3 marks]

substituting \(x = \frac{1}{2}\) into \({x^2} + \frac{1}{3}{x^4}\)      M1

the series approximation gives a value of \(\frac{{13}}{{48}}\)

so \({\pi ^2} \simeq \frac{{13}}{{48}} \times 36\)

\( \simeq 9.75\,\,\left( { \simeq \frac{{39}}{4}} \right)\)     A1

Note: Accept 9.76.

[2 marks]